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from __future__ import print_function 

 

from collections import Counter, defaultdict, deque 

from functools import partial, wraps 

from heapq import merge 

from itertools import ( 

chain, 

compress, 

count, 

cycle, 

dropwhile, 

groupby, 

islice, 

repeat, 

starmap, 

takewhile, 

tee 

) 

from operator import itemgetter, lt, gt, sub 

from sys import maxsize, version_info 

try: 

from collections.abc import Sequence 

except ImportError: 

from collections import Sequence 

 

from six import binary_type, string_types, text_type 

from six.moves import filter, map, range, zip, zip_longest 

 

from .recipes import consume, flatten, take 

 

__all__ = [ 

'adjacent', 

'always_iterable', 

'always_reversible', 

'bucket', 

'chunked', 

'circular_shifts', 

'collapse', 

'collate', 

'consecutive_groups', 

'consumer', 

'count_cycle', 

'difference', 

'distinct_permutations', 

'distribute', 

'divide', 

'exactly_n', 

'first', 

'groupby_transform', 

'ilen', 

'interleave_longest', 

'interleave', 

'intersperse', 

'islice_extended', 

'iterate', 

'last', 

'locate', 

'lstrip', 

'make_decorator', 

'map_reduce', 

'numeric_range', 

'one', 

'padded', 

'peekable', 

'replace', 

'rlocate', 

'rstrip', 

'run_length', 

'seekable', 

'SequenceView', 

'side_effect', 

'sliced', 

'sort_together', 

'split_at', 

'split_after', 

'split_before', 

'split_into', 

'spy', 

'stagger', 

'strip', 

'substrings', 

'unique_to_each', 

'unzip', 

'windowed', 

'with_iter', 

'zip_offset', 

] 

 

_marker = object() 

 

 

def chunked(iterable, n): 

"""Break *iterable* into lists of length *n*: 

 

>>> list(chunked([1, 2, 3, 4, 5, 6], 3)) 

[[1, 2, 3], [4, 5, 6]] 

 

If the length of *iterable* is not evenly divisible by *n*, the last 

returned list will be shorter: 

 

>>> list(chunked([1, 2, 3, 4, 5, 6, 7, 8], 3)) 

[[1, 2, 3], [4, 5, 6], [7, 8]] 

 

To use a fill-in value instead, see the :func:`grouper` recipe. 

 

:func:`chunked` is useful for splitting up a computation on a large number 

of keys into batches, to be pickled and sent off to worker processes. One 

example is operations on rows in MySQL, which does not implement 

server-side cursors properly and would otherwise load the entire dataset 

into RAM on the client. 

 

""" 

return iter(partial(take, n, iter(iterable)), []) 

 

 

def first(iterable, default=_marker): 

"""Return the first item of *iterable*, or *default* if *iterable* is 

empty. 

 

>>> first([0, 1, 2, 3]) 

0 

>>> first([], 'some default') 

'some default' 

 

If *default* is not provided and there are no items in the iterable, 

raise ``ValueError``. 

 

:func:`first` is useful when you have a generator of expensive-to-retrieve 

values and want any arbitrary one. It is marginally shorter than 

``next(iter(iterable), default)``. 

 

""" 

try: 

return next(iter(iterable)) 

except StopIteration: 

# I'm on the edge about raising ValueError instead of StopIteration. At 

# the moment, ValueError wins, because the caller could conceivably 

# want to do something different with flow control when I raise the 

# exception, and it's weird to explicitly catch StopIteration. 

if default is _marker: 

raise ValueError('first() was called on an empty iterable, and no ' 

'default value was provided.') 

return default 

 

 

def last(iterable, default=_marker): 

"""Return the last item of *iterable*, or *default* if *iterable* is 

empty. 

 

>>> last([0, 1, 2, 3]) 

3 

>>> last([], 'some default') 

'some default' 

 

If *default* is not provided and there are no items in the iterable, 

raise ``ValueError``. 

""" 

try: 

try: 

# Try to access the last item directly 

return iterable[-1] 

except (TypeError, AttributeError, KeyError): 

# If not slice-able, iterate entirely using length-1 deque 

return deque(iterable, maxlen=1)[0] 

except IndexError: # If the iterable was empty 

if default is _marker: 

raise ValueError('last() was called on an empty iterable, and no ' 

'default value was provided.') 

return default 

 

 

class peekable(object): 

"""Wrap an iterator to allow lookahead and prepending elements. 

 

Call :meth:`peek` on the result to get the value that will be returned 

by :func:`next`. This won't advance the iterator: 

 

>>> p = peekable(['a', 'b']) 

>>> p.peek() 

'a' 

>>> next(p) 

'a' 

 

Pass :meth:`peek` a default value to return that instead of raising 

``StopIteration`` when the iterator is exhausted. 

 

>>> p = peekable([]) 

>>> p.peek('hi') 

'hi' 

 

peekables also offer a :meth:`prepend` method, which "inserts" items 

at the head of the iterable: 

 

>>> p = peekable([1, 2, 3]) 

>>> p.prepend(10, 11, 12) 

>>> next(p) 

10 

>>> p.peek() 

11 

>>> list(p) 

[11, 12, 1, 2, 3] 

 

peekables can be indexed. Index 0 is the item that will be returned by 

:func:`next`, index 1 is the item after that, and so on: 

The values up to the given index will be cached. 

 

>>> p = peekable(['a', 'b', 'c', 'd']) 

>>> p[0] 

'a' 

>>> p[1] 

'b' 

>>> next(p) 

'a' 

 

Negative indexes are supported, but be aware that they will cache the 

remaining items in the source iterator, which may require significant 

storage. 

 

To check whether a peekable is exhausted, check its truth value: 

 

>>> p = peekable(['a', 'b']) 

>>> if p: # peekable has items 

... list(p) 

['a', 'b'] 

>>> if not p: # peekable is exhaused 

... list(p) 

[] 

 

""" 

def __init__(self, iterable): 

self._it = iter(iterable) 

self._cache = deque() 

 

def __iter__(self): 

return self 

 

def __bool__(self): 

try: 

self.peek() 

except StopIteration: 

return False 

return True 

 

def __nonzero__(self): 

# For Python 2 compatibility 

return self.__bool__() 

 

def peek(self, default=_marker): 

"""Return the item that will be next returned from ``next()``. 

 

Return ``default`` if there are no items left. If ``default`` is not 

provided, raise ``StopIteration``. 

 

""" 

if not self._cache: 

try: 

self._cache.append(next(self._it)) 

except StopIteration: 

if default is _marker: 

raise 

return default 

return self._cache[0] 

 

def prepend(self, *items): 

"""Stack up items to be the next ones returned from ``next()`` or 

``self.peek()``. The items will be returned in 

first in, first out order:: 

 

>>> p = peekable([1, 2, 3]) 

>>> p.prepend(10, 11, 12) 

>>> next(p) 

10 

>>> list(p) 

[11, 12, 1, 2, 3] 

 

It is possible, by prepending items, to "resurrect" a peekable that 

previously raised ``StopIteration``. 

 

>>> p = peekable([]) 

>>> next(p) 

Traceback (most recent call last): 

... 

StopIteration 

>>> p.prepend(1) 

>>> next(p) 

1 

>>> next(p) 

Traceback (most recent call last): 

... 

StopIteration 

 

""" 

self._cache.extendleft(reversed(items)) 

 

def __next__(self): 

if self._cache: 

return self._cache.popleft() 

 

return next(self._it) 

 

next = __next__ # For Python 2 compatibility 

 

def _get_slice(self, index): 

# Normalize the slice's arguments 

step = 1 if (index.step is None) else index.step 

if step > 0: 

start = 0 if (index.start is None) else index.start 

stop = maxsize if (index.stop is None) else index.stop 

elif step < 0: 

start = -1 if (index.start is None) else index.start 

stop = (-maxsize - 1) if (index.stop is None) else index.stop 

else: 

raise ValueError('slice step cannot be zero') 

 

# If either the start or stop index is negative, we'll need to cache 

# the rest of the iterable in order to slice from the right side. 

if (start < 0) or (stop < 0): 

self._cache.extend(self._it) 

# Otherwise we'll need to find the rightmost index and cache to that 

# point. 

else: 

n = min(max(start, stop) + 1, maxsize) 

cache_len = len(self._cache) 

if n >= cache_len: 

self._cache.extend(islice(self._it, n - cache_len)) 

 

return list(self._cache)[index] 

 

def __getitem__(self, index): 

if isinstance(index, slice): 

return self._get_slice(index) 

 

cache_len = len(self._cache) 

if index < 0: 

self._cache.extend(self._it) 

elif index >= cache_len: 

self._cache.extend(islice(self._it, index + 1 - cache_len)) 

 

return self._cache[index] 

 

 

def _collate(*iterables, **kwargs): 

"""Helper for ``collate()``, called when the user is using the ``reverse`` 

or ``key`` keyword arguments on Python versions below 3.5. 

 

""" 

key = kwargs.pop('key', lambda a: a) 

reverse = kwargs.pop('reverse', False) 

 

min_or_max = partial(max if reverse else min, key=itemgetter(0)) 

peekables = [peekable(it) for it in iterables] 

peekables = [p for p in peekables if p] # Kill empties. 

while peekables: 

_, p = min_or_max((key(p.peek()), p) for p in peekables) 

yield next(p) 

peekables = [x for x in peekables if x] 

 

 

def collate(*iterables, **kwargs): 

"""Return a sorted merge of the items from each of several already-sorted 

*iterables*. 

 

>>> list(collate('ACDZ', 'AZ', 'JKL')) 

['A', 'A', 'C', 'D', 'J', 'K', 'L', 'Z', 'Z'] 

 

Works lazily, keeping only the next value from each iterable in memory. Use 

:func:`collate` to, for example, perform a n-way mergesort of items that 

don't fit in memory. 

 

If a *key* function is specified, the iterables will be sorted according 

to its result: 

 

>>> key = lambda s: int(s) # Sort by numeric value, not by string 

>>> list(collate(['1', '10'], ['2', '11'], key=key)) 

['1', '2', '10', '11'] 

 

 

If the *iterables* are sorted in descending order, set *reverse* to 

``True``: 

 

>>> list(collate([5, 3, 1], [4, 2, 0], reverse=True)) 

[5, 4, 3, 2, 1, 0] 

 

If the elements of the passed-in iterables are out of order, you might get 

unexpected results. 

 

On Python 2.7, this function delegates to :func:`heapq.merge` if neither 

of the keyword arguments are specified. On Python 3.5+, this function 

is an alias for :func:`heapq.merge`. 

 

""" 

if not kwargs: 

return merge(*iterables) 

 

return _collate(*iterables, **kwargs) 

 

 

# If using Python version 3.5 or greater, heapq.merge() will be faster than 

# collate - use that instead. 

if version_info >= (3, 5, 0): 

_collate_docstring = collate.__doc__ 

collate = partial(merge) 

collate.__doc__ = _collate_docstring 

 

 

def consumer(func): 

"""Decorator that automatically advances a PEP-342-style "reverse iterator" 

to its first yield point so you don't have to call ``next()`` on it 

manually. 

 

>>> @consumer 

... def tally(): 

... i = 0 

... while True: 

... print('Thing number %s is %s.' % (i, (yield))) 

... i += 1 

... 

>>> t = tally() 

>>> t.send('red') 

Thing number 0 is red. 

>>> t.send('fish') 

Thing number 1 is fish. 

 

Without the decorator, you would have to call ``next(t)`` before 

``t.send()`` could be used. 

 

""" 

@wraps(func) 

def wrapper(*args, **kwargs): 

gen = func(*args, **kwargs) 

next(gen) 

return gen 

return wrapper 

 

 

def ilen(iterable): 

"""Return the number of items in *iterable*. 

 

>>> ilen(x for x in range(1000000) if x % 3 == 0) 

333334 

 

This consumes the iterable, so handle with care. 

 

""" 

# This approach was selected because benchmarks showed it's likely the 

# fastest of the known implementations at the time of writing. 

# See GitHub tracker: #236, #230. 

counter = count() 

deque(zip(iterable, counter), maxlen=0) 

return next(counter) 

 

 

def iterate(func, start): 

"""Return ``start``, ``func(start)``, ``func(func(start))``, ... 

 

>>> from itertools import islice 

>>> list(islice(iterate(lambda x: 2*x, 1), 10)) 

[1, 2, 4, 8, 16, 32, 64, 128, 256, 512] 

 

""" 

while True: 

yield start 

start = func(start) 

 

 

def with_iter(context_manager): 

"""Wrap an iterable in a ``with`` statement, so it closes once exhausted. 

 

For example, this will close the file when the iterator is exhausted:: 

 

upper_lines = (line.upper() for line in with_iter(open('foo'))) 

 

Any context manager which returns an iterable is a candidate for 

``with_iter``. 

 

""" 

with context_manager as iterable: 

for item in iterable: 

yield item 

 

 

def one(iterable, too_short=None, too_long=None): 

"""Return the first item from *iterable*, which is expected to contain only 

that item. Raise an exception if *iterable* is empty or has more than one 

item. 

 

:func:`one` is useful for ensuring that an iterable contains only one item. 

For example, it can be used to retrieve the result of a database query 

that is expected to return a single row. 

 

If *iterable* is empty, ``ValueError`` will be raised. You may specify a 

different exception with the *too_short* keyword: 

 

>>> it = [] 

>>> one(it) # doctest: +IGNORE_EXCEPTION_DETAIL 

Traceback (most recent call last): 

... 

ValueError: too many items in iterable (expected 1)' 

>>> too_short = IndexError('too few items') 

>>> one(it, too_short=too_short) # doctest: +IGNORE_EXCEPTION_DETAIL 

Traceback (most recent call last): 

... 

IndexError: too few items 

 

Similarly, if *iterable* contains more than one item, ``ValueError`` will 

be raised. You may specify a different exception with the *too_long* 

keyword: 

 

>>> it = ['too', 'many'] 

>>> one(it) # doctest: +IGNORE_EXCEPTION_DETAIL 

Traceback (most recent call last): 

... 

ValueError: too many items in iterable (expected 1)' 

>>> too_long = RuntimeError 

>>> one(it, too_long=too_long) # doctest: +IGNORE_EXCEPTION_DETAIL 

Traceback (most recent call last): 

... 

RuntimeError 

 

Note that :func:`one` attempts to advance *iterable* twice to ensure there 

is only one item. If there is more than one, both items will be discarded. 

See :func:`spy` or :func:`peekable` to check iterable contents less 

destructively. 

 

""" 

it = iter(iterable) 

 

try: 

value = next(it) 

except StopIteration: 

raise too_short or ValueError('too few items in iterable (expected 1)') 

 

try: 

next(it) 

except StopIteration: 

pass 

else: 

raise too_long or ValueError('too many items in iterable (expected 1)') 

 

return value 

 

 

def distinct_permutations(iterable): 

"""Yield successive distinct permutations of the elements in *iterable*. 

 

>>> sorted(distinct_permutations([1, 0, 1])) 

[(0, 1, 1), (1, 0, 1), (1, 1, 0)] 

 

Equivalent to ``set(permutations(iterable))``, except duplicates are not 

generated and thrown away. For larger input sequences this is much more 

efficient. 

 

Duplicate permutations arise when there are duplicated elements in the 

input iterable. The number of items returned is 

`n! / (x_1! * x_2! * ... * x_n!)`, where `n` is the total number of 

items input, and each `x_i` is the count of a distinct item in the input 

sequence. 

 

""" 

def perm_unique_helper(item_counts, perm, i): 

"""Internal helper function 

 

:arg item_counts: Stores the unique items in ``iterable`` and how many 

times they are repeated 

:arg perm: The permutation that is being built for output 

:arg i: The index of the permutation being modified 

 

The output permutations are built up recursively; the distinct items 

are placed until their repetitions are exhausted. 

""" 

if i < 0: 

yield tuple(perm) 

else: 

for item in item_counts: 

if item_counts[item] <= 0: 

continue 

perm[i] = item 

item_counts[item] -= 1 

for x in perm_unique_helper(item_counts, perm, i - 1): 

yield x 

item_counts[item] += 1 

 

item_counts = Counter(iterable) 

length = sum(item_counts.values()) 

 

return perm_unique_helper(item_counts, [None] * length, length - 1) 

 

 

def intersperse(e, iterable, n=1): 

"""Intersperse filler element *e* among the items in *iterable*, leaving 

*n* items between each filler element. 

 

>>> list(intersperse('!', [1, 2, 3, 4, 5])) 

[1, '!', 2, '!', 3, '!', 4, '!', 5] 

 

>>> list(intersperse(None, [1, 2, 3, 4, 5], n=2)) 

[1, 2, None, 3, 4, None, 5] 

 

""" 

if n == 0: 

raise ValueError('n must be > 0') 

elif n == 1: 

# interleave(repeat(e), iterable) -> e, x_0, e, e, x_1, e, x_2... 

# islice(..., 1, None) -> x_0, e, e, x_1, e, x_2... 

return islice(interleave(repeat(e), iterable), 1, None) 

else: 

# interleave(filler, chunks) -> [e], [x_0, x_1], [e], [x_2, x_3]... 

# islice(..., 1, None) -> [x_0, x_1], [e], [x_2, x_3]... 

# flatten(...) -> x_0, x_1, e, x_2, x_3... 

filler = repeat([e]) 

chunks = chunked(iterable, n) 

return flatten(islice(interleave(filler, chunks), 1, None)) 

 

 

def unique_to_each(*iterables): 

"""Return the elements from each of the input iterables that aren't in the 

other input iterables. 

 

For example, suppose you have a set of packages, each with a set of 

dependencies:: 

 

{'pkg_1': {'A', 'B'}, 'pkg_2': {'B', 'C'}, 'pkg_3': {'B', 'D'}} 

 

If you remove one package, which dependencies can also be removed? 

 

If ``pkg_1`` is removed, then ``A`` is no longer necessary - it is not 

associated with ``pkg_2`` or ``pkg_3``. Similarly, ``C`` is only needed for 

``pkg_2``, and ``D`` is only needed for ``pkg_3``:: 

 

>>> unique_to_each({'A', 'B'}, {'B', 'C'}, {'B', 'D'}) 

[['A'], ['C'], ['D']] 

 

If there are duplicates in one input iterable that aren't in the others 

they will be duplicated in the output. Input order is preserved:: 

 

>>> unique_to_each("mississippi", "missouri") 

[['p', 'p'], ['o', 'u', 'r']] 

 

It is assumed that the elements of each iterable are hashable. 

 

""" 

pool = [list(it) for it in iterables] 

counts = Counter(chain.from_iterable(map(set, pool))) 

uniques = {element for element in counts if counts[element] == 1} 

return [list(filter(uniques.__contains__, it)) for it in pool] 

 

 

def windowed(seq, n, fillvalue=None, step=1): 

"""Return a sliding window of width *n* over the given iterable. 

 

>>> all_windows = windowed([1, 2, 3, 4, 5], 3) 

>>> list(all_windows) 

[(1, 2, 3), (2, 3, 4), (3, 4, 5)] 

 

When the window is larger than the iterable, *fillvalue* is used in place 

of missing values:: 

 

>>> list(windowed([1, 2, 3], 4)) 

[(1, 2, 3, None)] 

 

Each window will advance in increments of *step*: 

 

>>> list(windowed([1, 2, 3, 4, 5, 6], 3, fillvalue='!', step=2)) 

[(1, 2, 3), (3, 4, 5), (5, 6, '!')] 

 

""" 

if n < 0: 

raise ValueError('n must be >= 0') 

if n == 0: 

yield tuple() 

return 

if step < 1: 

raise ValueError('step must be >= 1') 

 

it = iter(seq) 

window = deque([], n) 

append = window.append 

 

# Initial deque fill 

for _ in range(n): 

append(next(it, fillvalue)) 

yield tuple(window) 

 

# Appending new items to the right causes old items to fall off the left 

i = 0 

for item in it: 

append(item) 

i = (i + 1) % step 

if i % step == 0: 

yield tuple(window) 

 

# If there are items from the iterable in the window, pad with the given 

# value and emit them. 

if (i % step) and (step - i < n): 

for _ in range(step - i): 

append(fillvalue) 

yield tuple(window) 

 

 

def substrings(iterable, join_func=None): 

"""Yield all of the substrings of *iterable*. 

 

>>> [''.join(s) for s in substrings('more')] 

['m', 'o', 'r', 'e', 'mo', 'or', 're', 'mor', 'ore', 'more'] 

 

Note that non-string iterables can also be subdivided. 

 

>>> list(substrings([0, 1, 2])) 

[(0,), (1,), (2,), (0, 1), (1, 2), (0, 1, 2)] 

 

""" 

# The length-1 substrings 

seq = [] 

for item in iter(iterable): 

seq.append(item) 

yield (item,) 

seq = tuple(seq) 

item_count = len(seq) 

 

# And the rest 

for n in range(2, item_count + 1): 

for i in range(item_count - n + 1): 

yield seq[i:i + n] 

 

 

class bucket(object): 

"""Wrap *iterable* and return an object that buckets it iterable into 

child iterables based on a *key* function. 

 

>>> iterable = ['a1', 'b1', 'c1', 'a2', 'b2', 'c2', 'b3'] 

>>> s = bucket(iterable, key=lambda x: x[0]) 

>>> a_iterable = s['a'] 

>>> next(a_iterable) 

'a1' 

>>> next(a_iterable) 

'a2' 

>>> list(s['b']) 

['b1', 'b2', 'b3'] 

 

The original iterable will be advanced and its items will be cached until 

they are used by the child iterables. This may require significant storage. 

 

By default, attempting to select a bucket to which no items belong will 

exhaust the iterable and cache all values. 

If you specify a *validator* function, selected buckets will instead be 

checked against it. 

 

>>> from itertools import count 

>>> it = count(1, 2) # Infinite sequence of odd numbers 

>>> key = lambda x: x % 10 # Bucket by last digit 

>>> validator = lambda x: x in {1, 3, 5, 7, 9} # Odd digits only 

>>> s = bucket(it, key=key, validator=validator) 

>>> 2 in s 

False 

>>> list(s[2]) 

[] 

 

""" 

def __init__(self, iterable, key, validator=None): 

self._it = iter(iterable) 

self._key = key 

self._cache = defaultdict(deque) 

self._validator = validator or (lambda x: True) 

 

def __contains__(self, value): 

if not self._validator(value): 

return False 

 

try: 

item = next(self[value]) 

except StopIteration: 

return False 

else: 

self._cache[value].appendleft(item) 

 

return True 

 

def _get_values(self, value): 

""" 

Helper to yield items from the parent iterator that match *value*. 

Items that don't match are stored in the local cache as they 

are encountered. 

""" 

while True: 

# If we've cached some items that match the target value, emit 

# the first one and evict it from the cache. 

if self._cache[value]: 

yield self._cache[value].popleft() 

# Otherwise we need to advance the parent iterator to search for 

# a matching item, caching the rest. 

else: 

while True: 

try: 

item = next(self._it) 

except StopIteration: 

return 

item_value = self._key(item) 

if item_value == value: 

yield item 

break 

elif self._validator(item_value): 

self._cache[item_value].append(item) 

 

def __getitem__(self, value): 

if not self._validator(value): 

return iter(()) 

 

return self._get_values(value) 

 

 

def spy(iterable, n=1): 

"""Return a 2-tuple with a list containing the first *n* elements of 

*iterable*, and an iterator with the same items as *iterable*. 

This allows you to "look ahead" at the items in the iterable without 

advancing it. 

 

There is one item in the list by default: 

 

>>> iterable = 'abcdefg' 

>>> head, iterable = spy(iterable) 

>>> head 

['a'] 

>>> list(iterable) 

['a', 'b', 'c', 'd', 'e', 'f', 'g'] 

 

You may use unpacking to retrieve items instead of lists: 

 

>>> (head,), iterable = spy('abcdefg') 

>>> head 

'a' 

>>> (first, second), iterable = spy('abcdefg', 2) 

>>> first 

'a' 

>>> second 

'b' 

 

The number of items requested can be larger than the number of items in 

the iterable: 

 

>>> iterable = [1, 2, 3, 4, 5] 

>>> head, iterable = spy(iterable, 10) 

>>> head 

[1, 2, 3, 4, 5] 

>>> list(iterable) 

[1, 2, 3, 4, 5] 

 

""" 

it = iter(iterable) 

head = take(n, it) 

 

return head, chain(head, it) 

 

 

def interleave(*iterables): 

"""Return a new iterable yielding from each iterable in turn, 

until the shortest is exhausted. 

 

>>> list(interleave([1, 2, 3], [4, 5], [6, 7, 8])) 

[1, 4, 6, 2, 5, 7] 

 

For a version that doesn't terminate after the shortest iterable is 

exhausted, see :func:`interleave_longest`. 

 

""" 

return chain.from_iterable(zip(*iterables)) 

 

 

def interleave_longest(*iterables): 

"""Return a new iterable yielding from each iterable in turn, 

skipping any that are exhausted. 

 

>>> list(interleave_longest([1, 2, 3], [4, 5], [6, 7, 8])) 

[1, 4, 6, 2, 5, 7, 3, 8] 

 

This function produces the same output as :func:`roundrobin`, but may 

perform better for some inputs (in particular when the number of iterables 

is large). 

 

""" 

i = chain.from_iterable(zip_longest(*iterables, fillvalue=_marker)) 

return (x for x in i if x is not _marker) 

 

 

def collapse(iterable, base_type=None, levels=None): 

"""Flatten an iterable with multiple levels of nesting (e.g., a list of 

lists of tuples) into non-iterable types. 

 

>>> iterable = [(1, 2), ([3, 4], [[5], [6]])] 

>>> list(collapse(iterable)) 

[1, 2, 3, 4, 5, 6] 

 

String types are not considered iterable and will not be collapsed. 

To avoid collapsing other types, specify *base_type*: 

 

>>> iterable = ['ab', ('cd', 'ef'), ['gh', 'ij']] 

>>> list(collapse(iterable, base_type=tuple)) 

['ab', ('cd', 'ef'), 'gh', 'ij'] 

 

Specify *levels* to stop flattening after a certain level: 

 

>>> iterable = [('a', ['b']), ('c', ['d'])] 

>>> list(collapse(iterable)) # Fully flattened 

['a', 'b', 'c', 'd'] 

>>> list(collapse(iterable, levels=1)) # Only one level flattened 

['a', ['b'], 'c', ['d']] 

 

""" 

def walk(node, level): 

if ( 

((levels is not None) and (level > levels)) or 

isinstance(node, string_types) or 

((base_type is not None) and isinstance(node, base_type)) 

): 

yield node 

return 

 

try: 

tree = iter(node) 

except TypeError: 

yield node 

return 

else: 

for child in tree: 

for x in walk(child, level + 1): 

yield x 

 

for x in walk(iterable, 0): 

yield x 

 

 

def side_effect(func, iterable, chunk_size=None, before=None, after=None): 

"""Invoke *func* on each item in *iterable* (or on each *chunk_size* group 

of items) before yielding the item. 

 

`func` must be a function that takes a single argument. Its return value 

will be discarded. 

 

*before* and *after* are optional functions that take no arguments. They 

will be executed before iteration starts and after it ends, respectively. 

 

`side_effect` can be used for logging, updating progress bars, or anything 

that is not functionally "pure." 

 

Emitting a status message: 

 

>>> from more_itertools import consume 

>>> func = lambda item: print('Received {}'.format(item)) 

>>> consume(side_effect(func, range(2))) 

Received 0 

Received 1 

 

Operating on chunks of items: 

 

>>> pair_sums = [] 

>>> func = lambda chunk: pair_sums.append(sum(chunk)) 

>>> list(side_effect(func, [0, 1, 2, 3, 4, 5], 2)) 

[0, 1, 2, 3, 4, 5] 

>>> list(pair_sums) 

[1, 5, 9] 

 

Writing to a file-like object: 

 

>>> from io import StringIO 

>>> from more_itertools import consume 

>>> f = StringIO() 

>>> func = lambda x: print(x, file=f) 

>>> before = lambda: print(u'HEADER', file=f) 

>>> after = f.close 

>>> it = [u'a', u'b', u'c'] 

>>> consume(side_effect(func, it, before=before, after=after)) 

>>> f.closed 

True 

 

""" 

try: 

if before is not None: 

before() 

 

if chunk_size is None: 

for item in iterable: 

func(item) 

yield item 

else: 

for chunk in chunked(iterable, chunk_size): 

func(chunk) 

for item in chunk: 

yield item 

finally: 

if after is not None: 

after() 

 

 

def sliced(seq, n): 

"""Yield slices of length *n* from the sequence *seq*. 

 

>>> list(sliced((1, 2, 3, 4, 5, 6), 3)) 

[(1, 2, 3), (4, 5, 6)] 

 

If the length of the sequence is not divisible by the requested slice 

length, the last slice will be shorter. 

 

>>> list(sliced((1, 2, 3, 4, 5, 6, 7, 8), 3)) 

[(1, 2, 3), (4, 5, 6), (7, 8)] 

 

This function will only work for iterables that support slicing. 

For non-sliceable iterables, see :func:`chunked`. 

 

""" 

return takewhile(bool, (seq[i: i + n] for i in count(0, n))) 

 

 

def split_at(iterable, pred): 

"""Yield lists of items from *iterable*, where each list is delimited by 

an item where callable *pred* returns ``True``. The lists do not include 

the delimiting items. 

 

>>> list(split_at('abcdcba', lambda x: x == 'b')) 

[['a'], ['c', 'd', 'c'], ['a']] 

 

>>> list(split_at(range(10), lambda n: n % 2 == 1)) 

[[0], [2], [4], [6], [8], []] 

""" 

buf = [] 

for item in iterable: 

if pred(item): 

yield buf 

buf = [] 

else: 

buf.append(item) 

yield buf 

 

 

def split_before(iterable, pred): 

"""Yield lists of items from *iterable*, where each list starts with an 

item where callable *pred* returns ``True``: 

 

>>> list(split_before('OneTwo', lambda s: s.isupper())) 

[['O', 'n', 'e'], ['T', 'w', 'o']] 

 

>>> list(split_before(range(10), lambda n: n % 3 == 0)) 

[[0, 1, 2], [3, 4, 5], [6, 7, 8], [9]] 

 

""" 

buf = [] 

for item in iterable: 

if pred(item) and buf: 

yield buf 

buf = [] 

buf.append(item) 

yield buf 

 

 

def split_after(iterable, pred): 

"""Yield lists of items from *iterable*, where each list ends with an 

item where callable *pred* returns ``True``: 

 

>>> list(split_after('one1two2', lambda s: s.isdigit())) 

[['o', 'n', 'e', '1'], ['t', 'w', 'o', '2']] 

 

>>> list(split_after(range(10), lambda n: n % 3 == 0)) 

[[0], [1, 2, 3], [4, 5, 6], [7, 8, 9]] 

 

""" 

buf = [] 

for item in iterable: 

buf.append(item) 

if pred(item) and buf: 

yield buf 

buf = [] 

if buf: 

yield buf 

 

 

def split_into(iterable, sizes): 

"""Yield a list of sequential items from *iterable* of length 'n' for each 

integer 'n' in *sizes*. 

 

>>> list(split_into([1,2,3,4,5,6], [1,2,3])) 

[[1], [2, 3], [4, 5, 6]] 

 

If the sum of *sizes* is smaller than the length of *iterable*, then the 

remaining items of *iterable* will not be returned. 

 

>>> list(split_into([1,2,3,4,5,6], [2,3])) 

[[1, 2], [3, 4, 5]] 

 

If the sum of *sizes* is larger than the length of *iterable*, fewer items 

will be returned in the iteration that overruns *iterable* and further 

lists will be empty: 

 

>>> list(split_into([1,2,3,4], [1,2,3,4])) 

[[1], [2, 3], [4], []] 

 

When a ``None`` object is encountered in *sizes*, the returned list will 

contain items up to the end of *iterable* the same way that itertools.slice 

does: 

 

>>> list(split_into([1,2,3,4,5,6,7,8,9,0], [2,3,None])) 

[[1, 2], [3, 4, 5], [6, 7, 8, 9, 0]] 

 

:func:`split_into` can be useful for grouping a series of items where the 

sizes of the groups are not uniform. An example would be where in a row 

from a table, multiple columns represent elements of the same feature 

(e.g. a point represented by x,y,z) but, the format is not the same for 

all columns. 

""" 

# convert the iterable argument into an iterator so its contents can 

# be consumed by islice in case it is a generator 

it = iter(iterable) 

 

for size in sizes: 

if size is None: 

yield list(it) 

return 

else: 

yield list(islice(it, size)) 

 

 

def padded(iterable, fillvalue=None, n=None, next_multiple=False): 

"""Yield the elements from *iterable*, followed by *fillvalue*, such that 

at least *n* items are emitted. 

 

>>> list(padded([1, 2, 3], '?', 5)) 

[1, 2, 3, '?', '?'] 

 

If *next_multiple* is ``True``, *fillvalue* will be emitted until the 

number of items emitted is a multiple of *n*:: 

 

>>> list(padded([1, 2, 3, 4], n=3, next_multiple=True)) 

[1, 2, 3, 4, None, None] 

 

If *n* is ``None``, *fillvalue* will be emitted indefinitely. 

 

""" 

it = iter(iterable) 

if n is None: 

for item in chain(it, repeat(fillvalue)): 

yield item 

elif n < 1: 

raise ValueError('n must be at least 1') 

else: 

item_count = 0 

for item in it: 

yield item 

item_count += 1 

 

remaining = (n - item_count) % n if next_multiple else n - item_count 

for _ in range(remaining): 

yield fillvalue 

 

 

def distribute(n, iterable): 

"""Distribute the items from *iterable* among *n* smaller iterables. 

 

>>> group_1, group_2 = distribute(2, [1, 2, 3, 4, 5, 6]) 

>>> list(group_1) 

[1, 3, 5] 

>>> list(group_2) 

[2, 4, 6] 

 

If the length of *iterable* is not evenly divisible by *n*, then the 

length of the returned iterables will not be identical: 

 

>>> children = distribute(3, [1, 2, 3, 4, 5, 6, 7]) 

>>> [list(c) for c in children] 

[[1, 4, 7], [2, 5], [3, 6]] 

 

If the length of *iterable* is smaller than *n*, then the last returned 

iterables will be empty: 

 

>>> children = distribute(5, [1, 2, 3]) 

>>> [list(c) for c in children] 

[[1], [2], [3], [], []] 

 

This function uses :func:`itertools.tee` and may require significant 

storage. If you need the order items in the smaller iterables to match the 

original iterable, see :func:`divide`. 

 

""" 

if n < 1: 

raise ValueError('n must be at least 1') 

 

children = tee(iterable, n) 

return [islice(it, index, None, n) for index, it in enumerate(children)] 

 

 

def stagger(iterable, offsets=(-1, 0, 1), longest=False, fillvalue=None): 

"""Yield tuples whose elements are offset from *iterable*. 

The amount by which the `i`-th item in each tuple is offset is given by 

the `i`-th item in *offsets*. 

 

>>> list(stagger([0, 1, 2, 3])) 

[(None, 0, 1), (0, 1, 2), (1, 2, 3)] 

>>> list(stagger(range(8), offsets=(0, 2, 4))) 

[(0, 2, 4), (1, 3, 5), (2, 4, 6), (3, 5, 7)] 

 

By default, the sequence will end when the final element of a tuple is the 

last item in the iterable. To continue until the first element of a tuple 

is the last item in the iterable, set *longest* to ``True``:: 

 

>>> list(stagger([0, 1, 2, 3], longest=True)) 

[(None, 0, 1), (0, 1, 2), (1, 2, 3), (2, 3, None), (3, None, None)] 

 

By default, ``None`` will be used to replace offsets beyond the end of the 

sequence. Specify *fillvalue* to use some other value. 

 

""" 

children = tee(iterable, len(offsets)) 

 

return zip_offset( 

*children, offsets=offsets, longest=longest, fillvalue=fillvalue 

) 

 

 

def zip_offset(*iterables, **kwargs): 

"""``zip`` the input *iterables* together, but offset the `i`-th iterable 

by the `i`-th item in *offsets*. 

 

>>> list(zip_offset('0123', 'abcdef', offsets=(0, 1))) 

[('0', 'b'), ('1', 'c'), ('2', 'd'), ('3', 'e')] 

 

This can be used as a lightweight alternative to SciPy or pandas to analyze 

data sets in which some series have a lead or lag relationship. 

 

By default, the sequence will end when the shortest iterable is exhausted. 

To continue until the longest iterable is exhausted, set *longest* to 

``True``. 

 

>>> list(zip_offset('0123', 'abcdef', offsets=(0, 1), longest=True)) 

[('0', 'b'), ('1', 'c'), ('2', 'd'), ('3', 'e'), (None, 'f')] 

 

By default, ``None`` will be used to replace offsets beyond the end of the 

sequence. Specify *fillvalue* to use some other value. 

 

""" 

offsets = kwargs['offsets'] 

longest = kwargs.get('longest', False) 

fillvalue = kwargs.get('fillvalue', None) 

 

if len(iterables) != len(offsets): 

raise ValueError("Number of iterables and offsets didn't match") 

 

staggered = [] 

for it, n in zip(iterables, offsets): 

if n < 0: 

staggered.append(chain(repeat(fillvalue, -n), it)) 

elif n > 0: 

staggered.append(islice(it, n, None)) 

else: 

staggered.append(it) 

 

if longest: 

return zip_longest(*staggered, fillvalue=fillvalue) 

 

return zip(*staggered) 

 

 

def sort_together(iterables, key_list=(0,), reverse=False): 

"""Return the input iterables sorted together, with *key_list* as the 

priority for sorting. All iterables are trimmed to the length of the 

shortest one. 

 

This can be used like the sorting function in a spreadsheet. If each 

iterable represents a column of data, the key list determines which 

columns are used for sorting. 

 

By default, all iterables are sorted using the ``0``-th iterable:: 

 

>>> iterables = [(4, 3, 2, 1), ('a', 'b', 'c', 'd')] 

>>> sort_together(iterables) 

[(1, 2, 3, 4), ('d', 'c', 'b', 'a')] 

 

Set a different key list to sort according to another iterable. 

Specifying multiple keys dictates how ties are broken:: 

 

>>> iterables = [(3, 1, 2), (0, 1, 0), ('c', 'b', 'a')] 

>>> sort_together(iterables, key_list=(1, 2)) 

[(2, 3, 1), (0, 0, 1), ('a', 'c', 'b')] 

 

Set *reverse* to ``True`` to sort in descending order. 

 

>>> sort_together([(1, 2, 3), ('c', 'b', 'a')], reverse=True) 

[(3, 2, 1), ('a', 'b', 'c')] 

 

""" 

return list(zip(*sorted(zip(*iterables), 

key=itemgetter(*key_list), 

reverse=reverse))) 

 

 

def unzip(iterable): 

"""The inverse of :func:`zip`, this function disaggregates the elements 

of the zipped *iterable*. 

 

The ``i``-th iterable contains the ``i``-th element from each element 

of the zipped iterable. The first element is used to to determine the 

length of the remaining elements. 

 

>>> iterable = [('a', 1), ('b', 2), ('c', 3), ('d', 4)] 

>>> letters, numbers = unzip(iterable) 

>>> list(letters) 

['a', 'b', 'c', 'd'] 

>>> list(numbers) 

[1, 2, 3, 4] 

 

This is similar to using ``zip(*iterable)``, but it avoids reading 

*iterable* into memory. Note, however, that this function uses 

:func:`itertools.tee` and thus may require significant storage. 

 

""" 

head, iterable = spy(iter(iterable)) 

if not head: 

# empty iterable, e.g. zip([], [], []) 

return () 

# spy returns a one-length iterable as head 

head = head[0] 

iterables = tee(iterable, len(head)) 

 

def itemgetter(i): 

def getter(obj): 

try: 

return obj[i] 

except IndexError: 

# basically if we have an iterable like 

# iter([(1, 2, 3), (4, 5), (6,)]) 

# the second unzipped iterable would fail at the third tuple 

# since it would try to access tup[1] 

# same with the third unzipped iterable and the second tuple 

# to support these "improperly zipped" iterables, 

# we create a custom itemgetter 

# which just stops the unzipped iterables 

# at first length mismatch 

raise StopIteration 

return getter 

 

return tuple(map(itemgetter(i), it) for i, it in enumerate(iterables)) 

 

 

def divide(n, iterable): 

"""Divide the elements from *iterable* into *n* parts, maintaining 

order. 

 

>>> group_1, group_2 = divide(2, [1, 2, 3, 4, 5, 6]) 

>>> list(group_1) 

[1, 2, 3] 

>>> list(group_2) 

[4, 5, 6] 

 

If the length of *iterable* is not evenly divisible by *n*, then the 

length of the returned iterables will not be identical: 

 

>>> children = divide(3, [1, 2, 3, 4, 5, 6, 7]) 

>>> [list(c) for c in children] 

[[1, 2, 3], [4, 5], [6, 7]] 

 

If the length of the iterable is smaller than n, then the last returned 

iterables will be empty: 

 

>>> children = divide(5, [1, 2, 3]) 

>>> [list(c) for c in children] 

[[1], [2], [3], [], []] 

 

This function will exhaust the iterable before returning and may require 

significant storage. If order is not important, see :func:`distribute`, 

which does not first pull the iterable into memory. 

 

""" 

if n < 1: 

raise ValueError('n must be at least 1') 

 

seq = tuple(iterable) 

q, r = divmod(len(seq), n) 

 

ret = [] 

for i in range(n): 

start = (i * q) + (i if i < r else r) 

stop = ((i + 1) * q) + (i + 1 if i + 1 < r else r) 

ret.append(iter(seq[start:stop])) 

 

return ret 

 

 

def always_iterable(obj, base_type=(text_type, binary_type)): 

"""If *obj* is iterable, return an iterator over its items:: 

 

>>> obj = (1, 2, 3) 

>>> list(always_iterable(obj)) 

[1, 2, 3] 

 

If *obj* is not iterable, return a one-item iterable containing *obj*:: 

 

>>> obj = 1 

>>> list(always_iterable(obj)) 

[1] 

 

If *obj* is ``None``, return an empty iterable: 

 

>>> obj = None 

>>> list(always_iterable(None)) 

[] 

 

By default, binary and text strings are not considered iterable:: 

 

>>> obj = 'foo' 

>>> list(always_iterable(obj)) 

['foo'] 

 

If *base_type* is set, objects for which ``isinstance(obj, base_type)`` 

returns ``True`` won't be considered iterable. 

 

>>> obj = {'a': 1} 

>>> list(always_iterable(obj)) # Iterate over the dict's keys 

['a'] 

>>> list(always_iterable(obj, base_type=dict)) # Treat dicts as a unit 

[{'a': 1}] 

 

Set *base_type* to ``None`` to avoid any special handling and treat objects 

Python considers iterable as iterable: 

 

>>> obj = 'foo' 

>>> list(always_iterable(obj, base_type=None)) 

['f', 'o', 'o'] 

""" 

if obj is None: 

return iter(()) 

 

if (base_type is not None) and isinstance(obj, base_type): 

return iter((obj,)) 

 

try: 

return iter(obj) 

except TypeError: 

return iter((obj,)) 

 

 

def adjacent(predicate, iterable, distance=1): 

"""Return an iterable over `(bool, item)` tuples where the `item` is 

drawn from *iterable* and the `bool` indicates whether 

that item satisfies the *predicate* or is adjacent to an item that does. 

 

For example, to find whether items are adjacent to a ``3``:: 

 

>>> list(adjacent(lambda x: x == 3, range(6))) 

[(False, 0), (False, 1), (True, 2), (True, 3), (True, 4), (False, 5)] 

 

Set *distance* to change what counts as adjacent. For example, to find 

whether items are two places away from a ``3``: 

 

>>> list(adjacent(lambda x: x == 3, range(6), distance=2)) 

[(False, 0), (True, 1), (True, 2), (True, 3), (True, 4), (True, 5)] 

 

This is useful for contextualizing the results of a search function. 

For example, a code comparison tool might want to identify lines that 

have changed, but also surrounding lines to give the viewer of the diff 

context. 

 

The predicate function will only be called once for each item in the 

iterable. 

 

See also :func:`groupby_transform`, which can be used with this function 

to group ranges of items with the same `bool` value. 

 

""" 

# Allow distance=0 mainly for testing that it reproduces results with map() 

if distance < 0: 

raise ValueError('distance must be at least 0') 

 

i1, i2 = tee(iterable) 

padding = [False] * distance 

selected = chain(padding, map(predicate, i1), padding) 

adjacent_to_selected = map(any, windowed(selected, 2 * distance + 1)) 

return zip(adjacent_to_selected, i2) 

 

 

def groupby_transform(iterable, keyfunc=None, valuefunc=None): 

"""An extension of :func:`itertools.groupby` that transforms the values of 

*iterable* after grouping them. 

*keyfunc* is a function used to compute a grouping key for each item. 

*valuefunc* is a function for transforming the items after grouping. 

 

>>> iterable = 'AaaABbBCcA' 

>>> keyfunc = lambda x: x.upper() 

>>> valuefunc = lambda x: x.lower() 

>>> grouper = groupby_transform(iterable, keyfunc, valuefunc) 

>>> [(k, ''.join(g)) for k, g in grouper] 

[('A', 'aaaa'), ('B', 'bbb'), ('C', 'cc'), ('A', 'a')] 

 

*keyfunc* and *valuefunc* default to identity functions if they are not 

specified. 

 

:func:`groupby_transform` is useful when grouping elements of an iterable 

using a separate iterable as the key. To do this, :func:`zip` the iterables 

and pass a *keyfunc* that extracts the first element and a *valuefunc* 

that extracts the second element:: 

 

>>> from operator import itemgetter 

>>> keys = [0, 0, 1, 1, 1, 2, 2, 2, 3] 

>>> values = 'abcdefghi' 

>>> iterable = zip(keys, values) 

>>> grouper = groupby_transform(iterable, itemgetter(0), itemgetter(1)) 

>>> [(k, ''.join(g)) for k, g in grouper] 

[(0, 'ab'), (1, 'cde'), (2, 'fgh'), (3, 'i')] 

 

Note that the order of items in the iterable is significant. 

Only adjacent items are grouped together, so if you don't want any 

duplicate groups, you should sort the iterable by the key function. 

 

""" 

valuefunc = (lambda x: x) if valuefunc is None else valuefunc 

return ((k, map(valuefunc, g)) for k, g in groupby(iterable, keyfunc)) 

 

 

def numeric_range(*args): 

"""An extension of the built-in ``range()`` function whose arguments can 

be any orderable numeric type. 

 

With only *stop* specified, *start* defaults to ``0`` and *step* 

defaults to ``1``. The output items will match the type of *stop*: 

 

>>> list(numeric_range(3.5)) 

[0.0, 1.0, 2.0, 3.0] 

 

With only *start* and *stop* specified, *step* defaults to ``1``. The 

output items will match the type of *start*: 

 

>>> from decimal import Decimal 

>>> start = Decimal('2.1') 

>>> stop = Decimal('5.1') 

>>> list(numeric_range(start, stop)) 

[Decimal('2.1'), Decimal('3.1'), Decimal('4.1')] 

 

With *start*, *stop*, and *step* specified the output items will match 

the type of ``start + step``: 

 

>>> from fractions import Fraction 

>>> start = Fraction(1, 2) # Start at 1/2 

>>> stop = Fraction(5, 2) # End at 5/2 

>>> step = Fraction(1, 2) # Count by 1/2 

>>> list(numeric_range(start, stop, step)) 

[Fraction(1, 2), Fraction(1, 1), Fraction(3, 2), Fraction(2, 1)] 

 

If *step* is zero, ``ValueError`` is raised. Negative steps are supported: 

 

>>> list(numeric_range(3, -1, -1.0)) 

[3.0, 2.0, 1.0, 0.0] 

 

Be aware of the limitations of floating point numbers; the representation 

of the yielded numbers may be surprising. 

 

""" 

argc = len(args) 

if argc == 1: 

stop, = args 

start = type(stop)(0) 

step = 1 

elif argc == 2: 

start, stop = args 

step = 1 

elif argc == 3: 

start, stop, step = args 

else: 

err_msg = 'numeric_range takes at most 3 arguments, got {}' 

raise TypeError(err_msg.format(argc)) 

 

values = (start + (step * n) for n in count()) 

if step > 0: 

return takewhile(partial(gt, stop), values) 

elif step < 0: 

return takewhile(partial(lt, stop), values) 

else: 

raise ValueError('numeric_range arg 3 must not be zero') 

 

 

def count_cycle(iterable, n=None): 

"""Cycle through the items from *iterable* up to *n* times, yielding 

the number of completed cycles along with each item. If *n* is omitted the 

process repeats indefinitely. 

 

>>> list(count_cycle('AB', 3)) 

[(0, 'A'), (0, 'B'), (1, 'A'), (1, 'B'), (2, 'A'), (2, 'B')] 

 

""" 

iterable = tuple(iterable) 

if not iterable: 

return iter(()) 

counter = count() if n is None else range(n) 

return ((i, item) for i in counter for item in iterable) 

 

 

def locate(iterable, pred=bool, window_size=None): 

"""Yield the index of each item in *iterable* for which *pred* returns 

``True``. 

 

*pred* defaults to :func:`bool`, which will select truthy items: 

 

>>> list(locate([0, 1, 1, 0, 1, 0, 0])) 

[1, 2, 4] 

 

Set *pred* to a custom function to, e.g., find the indexes for a particular 

item. 

 

>>> list(locate(['a', 'b', 'c', 'b'], lambda x: x == 'b')) 

[1, 3] 

 

If *window_size* is given, then the *pred* function will be called with 

that many items. This enables searching for sub-sequences: 

 

>>> iterable = [0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3] 

>>> pred = lambda *args: args == (1, 2, 3) 

>>> list(locate(iterable, pred=pred, window_size=3)) 

[1, 5, 9] 

 

Use with :func:`seekable` to find indexes and then retrieve the associated 

items: 

 

>>> from itertools import count 

>>> from more_itertools import seekable 

>>> source = (3 * n + 1 if (n % 2) else n // 2 for n in count()) 

>>> it = seekable(source) 

>>> pred = lambda x: x > 100 

>>> indexes = locate(it, pred=pred) 

>>> i = next(indexes) 

>>> it.seek(i) 

>>> next(it) 

106 

 

""" 

if window_size is None: 

return compress(count(), map(pred, iterable)) 

 

if window_size < 1: 

raise ValueError('window size must be at least 1') 

 

it = windowed(iterable, window_size, fillvalue=_marker) 

return compress(count(), starmap(pred, it)) 

 

 

def lstrip(iterable, pred): 

"""Yield the items from *iterable*, but strip any from the beginning 

for which *pred* returns ``True``. 

 

For example, to remove a set of items from the start of an iterable: 

 

>>> iterable = (None, False, None, 1, 2, None, 3, False, None) 

>>> pred = lambda x: x in {None, False, ''} 

>>> list(lstrip(iterable, pred)) 

[1, 2, None, 3, False, None] 

 

This function is analogous to to :func:`str.lstrip`, and is essentially 

an wrapper for :func:`itertools.dropwhile`. 

 

""" 

return dropwhile(pred, iterable) 

 

 

def rstrip(iterable, pred): 

"""Yield the items from *iterable*, but strip any from the end 

for which *pred* returns ``True``. 

 

For example, to remove a set of items from the end of an iterable: 

 

>>> iterable = (None, False, None, 1, 2, None, 3, False, None) 

>>> pred = lambda x: x in {None, False, ''} 

>>> list(rstrip(iterable, pred)) 

[None, False, None, 1, 2, None, 3] 

 

This function is analogous to :func:`str.rstrip`. 

 

""" 

cache = [] 

cache_append = cache.append 

for x in iterable: 

if pred(x): 

cache_append(x) 

else: 

for y in cache: 

yield y 

del cache[:] 

yield x 

 

 

def strip(iterable, pred): 

"""Yield the items from *iterable*, but strip any from the 

beginning and end for which *pred* returns ``True``. 

 

For example, to remove a set of items from both ends of an iterable: 

 

>>> iterable = (None, False, None, 1, 2, None, 3, False, None) 

>>> pred = lambda x: x in {None, False, ''} 

>>> list(strip(iterable, pred)) 

[1, 2, None, 3] 

 

This function is analogous to :func:`str.strip`. 

 

""" 

return rstrip(lstrip(iterable, pred), pred) 

 

 

def islice_extended(iterable, *args): 

"""An extension of :func:`itertools.islice` that supports negative values 

for *stop*, *start*, and *step*. 

 

>>> iterable = iter('abcdefgh') 

>>> list(islice_extended(iterable, -4, -1)) 

['e', 'f', 'g'] 

 

Slices with negative values require some caching of *iterable*, but this 

function takes care to minimize the amount of memory required. 

 

For example, you can use a negative step with an infinite iterator: 

 

>>> from itertools import count 

>>> list(islice_extended(count(), 110, 99, -2)) 

[110, 108, 106, 104, 102, 100] 

 

""" 

s = slice(*args) 

start = s.start 

stop = s.stop 

if s.step == 0: 

raise ValueError('step argument must be a non-zero integer or None.') 

step = s.step or 1 

 

it = iter(iterable) 

 

if step > 0: 

start = 0 if (start is None) else start 

 

if (start < 0): 

# Consume all but the last -start items 

cache = deque(enumerate(it, 1), maxlen=-start) 

len_iter = cache[-1][0] if cache else 0 

 

# Adjust start to be positive 

i = max(len_iter + start, 0) 

 

# Adjust stop to be positive 

if stop is None: 

j = len_iter 

elif stop >= 0: 

j = min(stop, len_iter) 

else: 

j = max(len_iter + stop, 0) 

 

# Slice the cache 

n = j - i 

if n <= 0: 

return 

 

for index, item in islice(cache, 0, n, step): 

yield item 

elif (stop is not None) and (stop < 0): 

# Advance to the start position 

next(islice(it, start, start), None) 

 

# When stop is negative, we have to carry -stop items while 

# iterating 

cache = deque(islice(it, -stop), maxlen=-stop) 

 

for index, item in enumerate(it): 

cached_item = cache.popleft() 

if index % step == 0: 

yield cached_item 

cache.append(item) 

else: 

# When both start and stop are positive we have the normal case 

for item in islice(it, start, stop, step): 

yield item 

else: 

start = -1 if (start is None) else start 

 

if (stop is not None) and (stop < 0): 

# Consume all but the last items 

n = -stop - 1 

cache = deque(enumerate(it, 1), maxlen=n) 

len_iter = cache[-1][0] if cache else 0 

 

# If start and stop are both negative they are comparable and 

# we can just slice. Otherwise we can adjust start to be negative 

# and then slice. 

if start < 0: 

i, j = start, stop 

else: 

i, j = min(start - len_iter, -1), None 

 

for index, item in list(cache)[i:j:step]: 

yield item 

else: 

# Advance to the stop position 

if stop is not None: 

m = stop + 1 

next(islice(it, m, m), None) 

 

# stop is positive, so if start is negative they are not comparable 

# and we need the rest of the items. 

if start < 0: 

i = start 

n = None 

# stop is None and start is positive, so we just need items up to 

# the start index. 

elif stop is None: 

i = None 

n = start + 1 

# Both stop and start are positive, so they are comparable. 

else: 

i = None 

n = start - stop 

if n <= 0: 

return 

 

cache = list(islice(it, n)) 

 

for item in cache[i::step]: 

yield item 

 

 

def always_reversible(iterable): 

"""An extension of :func:`reversed` that supports all iterables, not 

just those which implement the ``Reversible`` or ``Sequence`` protocols. 

 

>>> print(*always_reversible(x for x in range(3))) 

2 1 0 

 

If the iterable is already reversible, this function returns the 

result of :func:`reversed()`. If the iterable is not reversible, 

this function will cache the remaining items in the iterable and 

yield them in reverse order, which may require significant storage. 

""" 

try: 

return reversed(iterable) 

except TypeError: 

return reversed(list(iterable)) 

 

 

def consecutive_groups(iterable, ordering=lambda x: x): 

"""Yield groups of consecutive items using :func:`itertools.groupby`. 

The *ordering* function determines whether two items are adjacent by 

returning their position. 

 

By default, the ordering function is the identity function. This is 

suitable for finding runs of numbers: 

 

>>> iterable = [1, 10, 11, 12, 20, 30, 31, 32, 33, 40] 

>>> for group in consecutive_groups(iterable): 

... print(list(group)) 

[1] 

[10, 11, 12] 

[20] 

[30, 31, 32, 33] 

[40] 

 

For finding runs of adjacent letters, try using the :meth:`index` method 

of a string of letters: 

 

>>> from string import ascii_lowercase 

>>> iterable = 'abcdfgilmnop' 

>>> ordering = ascii_lowercase.index 

>>> for group in consecutive_groups(iterable, ordering): 

... print(list(group)) 

['a', 'b', 'c', 'd'] 

['f', 'g'] 

['i'] 

['l', 'm', 'n', 'o', 'p'] 

 

""" 

for k, g in groupby( 

enumerate(iterable), key=lambda x: x[0] - ordering(x[1]) 

): 

yield map(itemgetter(1), g) 

 

 

def difference(iterable, func=sub): 

"""By default, compute the first difference of *iterable* using 

:func:`operator.sub`. 

 

>>> iterable = [0, 1, 3, 6, 10] 

>>> list(difference(iterable)) 

[0, 1, 2, 3, 4] 

 

This is the opposite of :func:`accumulate`'s default behavior: 

 

>>> from more_itertools import accumulate 

>>> iterable = [0, 1, 2, 3, 4] 

>>> list(accumulate(iterable)) 

[0, 1, 3, 6, 10] 

>>> list(difference(accumulate(iterable))) 

[0, 1, 2, 3, 4] 

 

By default *func* is :func:`operator.sub`, but other functions can be 

specified. They will be applied as follows:: 

 

A, B, C, D, ... --> A, func(B, A), func(C, B), func(D, C), ... 

 

For example, to do progressive division: 

 

>>> iterable = [1, 2, 6, 24, 120] # Factorial sequence 

>>> func = lambda x, y: x // y 

>>> list(difference(iterable, func)) 

[1, 2, 3, 4, 5] 

 

""" 

a, b = tee(iterable) 

try: 

item = next(b) 

except StopIteration: 

return iter([]) 

return chain([item], map(lambda x: func(x[1], x[0]), zip(a, b))) 

 

 

class SequenceView(Sequence): 

"""Return a read-only view of the sequence object *target*. 

 

:class:`SequenceView` objects are analogous to Python's built-in 

"dictionary view" types. They provide a dynamic view of a sequence's items, 

meaning that when the sequence updates, so does the view. 

 

>>> seq = ['0', '1', '2'] 

>>> view = SequenceView(seq) 

>>> view 

SequenceView(['0', '1', '2']) 

>>> seq.append('3') 

>>> view 

SequenceView(['0', '1', '2', '3']) 

 

Sequence views support indexing, slicing, and length queries. They act 

like the underlying sequence, except they don't allow assignment: 

 

>>> view[1] 

'1' 

>>> view[1:-1] 

['1', '2'] 

>>> len(view) 

4 

 

Sequence views are useful as an alternative to copying, as they don't 

require (much) extra storage. 

 

""" 

def __init__(self, target): 

if not isinstance(target, Sequence): 

raise TypeError 

self._target = target 

 

def __getitem__(self, index): 

return self._target[index] 

 

def __len__(self): 

return len(self._target) 

 

def __repr__(self): 

return '{}({})'.format(self.__class__.__name__, repr(self._target)) 

 

 

class seekable(object): 

"""Wrap an iterator to allow for seeking backward and forward. This 

progressively caches the items in the source iterable so they can be 

re-visited. 

 

Call :meth:`seek` with an index to seek to that position in the source 

iterable. 

 

To "reset" an iterator, seek to ``0``: 

 

>>> from itertools import count 

>>> it = seekable((str(n) for n in count())) 

>>> next(it), next(it), next(it) 

('0', '1', '2') 

>>> it.seek(0) 

>>> next(it), next(it), next(it) 

('0', '1', '2') 

>>> next(it) 

'3' 

 

You can also seek forward: 

 

>>> it = seekable((str(n) for n in range(20))) 

>>> it.seek(10) 

>>> next(it) 

'10' 

>>> it.seek(20) # Seeking past the end of the source isn't a problem 

>>> list(it) 

[] 

>>> it.seek(0) # Resetting works even after hitting the end 

>>> next(it), next(it), next(it) 

('0', '1', '2') 

 

The cache grows as the source iterable progresses, so beware of wrapping 

very large or infinite iterables. 

 

You may view the contents of the cache with the :meth:`elements` method. 

That returns a :class:`SequenceView`, a view that updates automatically: 

 

>>> it = seekable((str(n) for n in range(10))) 

>>> next(it), next(it), next(it) 

('0', '1', '2') 

>>> elements = it.elements() 

>>> elements 

SequenceView(['0', '1', '2']) 

>>> next(it) 

'3' 

>>> elements 

SequenceView(['0', '1', '2', '3']) 

 

""" 

 

def __init__(self, iterable): 

self._source = iter(iterable) 

self._cache = [] 

self._index = None 

 

def __iter__(self): 

return self 

 

def __next__(self): 

if self._index is not None: 

try: 

item = self._cache[self._index] 

except IndexError: 

self._index = None 

else: 

self._index += 1 

return item 

 

item = next(self._source) 

self._cache.append(item) 

return item 

 

next = __next__ 

 

def elements(self): 

return SequenceView(self._cache) 

 

def seek(self, index): 

self._index = index 

remainder = index - len(self._cache) 

if remainder > 0: 

consume(self, remainder) 

 

 

class run_length(object): 

""" 

:func:`run_length.encode` compresses an iterable with run-length encoding. 

It yields groups of repeated items with the count of how many times they 

were repeated: 

 

>>> uncompressed = 'abbcccdddd' 

>>> list(run_length.encode(uncompressed)) 

[('a', 1), ('b', 2), ('c', 3), ('d', 4)] 

 

:func:`run_length.decode` decompresses an iterable that was previously 

compressed with run-length encoding. It yields the items of the 

decompressed iterable: 

 

>>> compressed = [('a', 1), ('b', 2), ('c', 3), ('d', 4)] 

>>> list(run_length.decode(compressed)) 

['a', 'b', 'b', 'c', 'c', 'c', 'd', 'd', 'd', 'd'] 

 

""" 

 

@staticmethod 

def encode(iterable): 

return ((k, ilen(g)) for k, g in groupby(iterable)) 

 

@staticmethod 

def decode(iterable): 

return chain.from_iterable(repeat(k, n) for k, n in iterable) 

 

 

def exactly_n(iterable, n, predicate=bool): 

"""Return ``True`` if exactly ``n`` items in the iterable are ``True`` 

according to the *predicate* function. 

 

>>> exactly_n([True, True, False], 2) 

True 

>>> exactly_n([True, True, False], 1) 

False 

>>> exactly_n([0, 1, 2, 3, 4, 5], 3, lambda x: x < 3) 

True 

 

The iterable will be advanced until ``n + 1`` truthy items are encountered, 

so avoid calling it on infinite iterables. 

 

""" 

return len(take(n + 1, filter(predicate, iterable))) == n 

 

 

def circular_shifts(iterable): 

"""Return a list of circular shifts of *iterable*. 

 

>>> circular_shifts(range(4)) 

[(0, 1, 2, 3), (1, 2, 3, 0), (2, 3, 0, 1), (3, 0, 1, 2)] 

""" 

lst = list(iterable) 

return take(len(lst), windowed(cycle(lst), len(lst))) 

 

 

def make_decorator(wrapping_func, result_index=0): 

"""Return a decorator version of *wrapping_func*, which is a function that 

modifies an iterable. *result_index* is the position in that function's 

signature where the iterable goes. 

 

This lets you use itertools on the "production end," i.e. at function 

definition. This can augment what the function returns without changing the 

function's code. 

 

For example, to produce a decorator version of :func:`chunked`: 

 

>>> from more_itertools import chunked 

>>> chunker = make_decorator(chunked, result_index=0) 

>>> @chunker(3) 

... def iter_range(n): 

... return iter(range(n)) 

... 

>>> list(iter_range(9)) 

[[0, 1, 2], [3, 4, 5], [6, 7, 8]] 

 

To only allow truthy items to be returned: 

 

>>> truth_serum = make_decorator(filter, result_index=1) 

>>> @truth_serum(bool) 

... def boolean_test(): 

... return [0, 1, '', ' ', False, True] 

... 

>>> list(boolean_test()) 

[1, ' ', True] 

 

The :func:`peekable` and :func:`seekable` wrappers make for practical 

decorators: 

 

>>> from more_itertools import peekable 

>>> peekable_function = make_decorator(peekable) 

>>> @peekable_function() 

... def str_range(*args): 

... return (str(x) for x in range(*args)) 

... 

>>> it = str_range(1, 20, 2) 

>>> next(it), next(it), next(it) 

('1', '3', '5') 

>>> it.peek() 

'7' 

>>> next(it) 

'7' 

 

""" 

# See https://sites.google.com/site/bbayles/index/decorator_factory for 

# notes on how this works. 

def decorator(*wrapping_args, **wrapping_kwargs): 

def outer_wrapper(f): 

def inner_wrapper(*args, **kwargs): 

result = f(*args, **kwargs) 

wrapping_args_ = list(wrapping_args) 

wrapping_args_.insert(result_index, result) 

return wrapping_func(*wrapping_args_, **wrapping_kwargs) 

 

return inner_wrapper 

 

return outer_wrapper 

 

return decorator 

 

 

def map_reduce(iterable, keyfunc, valuefunc=None, reducefunc=None): 

"""Return a dictionary that maps the items in *iterable* to categories 

defined by *keyfunc*, transforms them with *valuefunc*, and 

then summarizes them by category with *reducefunc*. 

 

*valuefunc* defaults to the identity function if it is unspecified. 

If *reducefunc* is unspecified, no summarization takes place: 

 

>>> keyfunc = lambda x: x.upper() 

>>> result = map_reduce('abbccc', keyfunc) 

>>> sorted(result.items()) 

[('A', ['a']), ('B', ['b', 'b']), ('C', ['c', 'c', 'c'])] 

 

Specifying *valuefunc* transforms the categorized items: 

 

>>> keyfunc = lambda x: x.upper() 

>>> valuefunc = lambda x: 1 

>>> result = map_reduce('abbccc', keyfunc, valuefunc) 

>>> sorted(result.items()) 

[('A', [1]), ('B', [1, 1]), ('C', [1, 1, 1])] 

 

Specifying *reducefunc* summarizes the categorized items: 

 

>>> keyfunc = lambda x: x.upper() 

>>> valuefunc = lambda x: 1 

>>> reducefunc = sum 

>>> result = map_reduce('abbccc', keyfunc, valuefunc, reducefunc) 

>>> sorted(result.items()) 

[('A', 1), ('B', 2), ('C', 3)] 

 

You may want to filter the input iterable before applying the map/reduce 

procedure: 

 

>>> all_items = range(30) 

>>> items = [x for x in all_items if 10 <= x <= 20] # Filter 

>>> keyfunc = lambda x: x % 2 # Evens map to 0; odds to 1 

>>> categories = map_reduce(items, keyfunc=keyfunc) 

>>> sorted(categories.items()) 

[(0, [10, 12, 14, 16, 18, 20]), (1, [11, 13, 15, 17, 19])] 

>>> summaries = map_reduce(items, keyfunc=keyfunc, reducefunc=sum) 

>>> sorted(summaries.items()) 

[(0, 90), (1, 75)] 

 

Note that all items in the iterable are gathered into a list before the 

summarization step, which may require significant storage. 

 

The returned object is a :obj:`collections.defaultdict` with the 

``default_factory`` set to ``None``, such that it behaves like a normal 

dictionary. 

 

""" 

valuefunc = (lambda x: x) if (valuefunc is None) else valuefunc 

 

ret = defaultdict(list) 

for item in iterable: 

key = keyfunc(item) 

value = valuefunc(item) 

ret[key].append(value) 

 

if reducefunc is not None: 

for key, value_list in ret.items(): 

ret[key] = reducefunc(value_list) 

 

ret.default_factory = None 

return ret 

 

 

def rlocate(iterable, pred=bool, window_size=None): 

"""Yield the index of each item in *iterable* for which *pred* returns 

``True``, starting from the right and moving left. 

 

*pred* defaults to :func:`bool`, which will select truthy items: 

 

>>> list(rlocate([0, 1, 1, 0, 1, 0, 0])) # Truthy at 1, 2, and 4 

[4, 2, 1] 

 

Set *pred* to a custom function to, e.g., find the indexes for a particular 

item: 

 

>>> iterable = iter('abcb') 

>>> pred = lambda x: x == 'b' 

>>> list(rlocate(iterable, pred)) 

[3, 1] 

 

If *window_size* is given, then the *pred* function will be called with 

that many items. This enables searching for sub-sequences: 

 

>>> iterable = [0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3] 

>>> pred = lambda *args: args == (1, 2, 3) 

>>> list(rlocate(iterable, pred=pred, window_size=3)) 

[9, 5, 1] 

 

Beware, this function won't return anything for infinite iterables. 

If *iterable* is reversible, ``rlocate`` will reverse it and search from 

the right. Otherwise, it will search from the left and return the results 

in reverse order. 

 

See :func:`locate` to for other example applications. 

 

""" 

if window_size is None: 

try: 

len_iter = len(iterable) 

return ( 

len_iter - i - 1 for i in locate(reversed(iterable), pred) 

) 

except TypeError: 

pass 

 

return reversed(list(locate(iterable, pred, window_size))) 

 

 

def replace(iterable, pred, substitutes, count=None, window_size=1): 

"""Yield the items from *iterable*, replacing the items for which *pred* 

returns ``True`` with the items from the iterable *substitutes*. 

 

>>> iterable = [1, 1, 0, 1, 1, 0, 1, 1] 

>>> pred = lambda x: x == 0 

>>> substitutes = (2, 3) 

>>> list(replace(iterable, pred, substitutes)) 

[1, 1, 2, 3, 1, 1, 2, 3, 1, 1] 

 

If *count* is given, the number of replacements will be limited: 

 

>>> iterable = [1, 1, 0, 1, 1, 0, 1, 1, 0] 

>>> pred = lambda x: x == 0 

>>> substitutes = [None] 

>>> list(replace(iterable, pred, substitutes, count=2)) 

[1, 1, None, 1, 1, None, 1, 1, 0] 

 

Use *window_size* to control the number of items passed as arguments to 

*pred*. This allows for locating and replacing subsequences. 

 

>>> iterable = [0, 1, 2, 5, 0, 1, 2, 5] 

>>> window_size = 3 

>>> pred = lambda *args: args == (0, 1, 2) # 3 items passed to pred 

>>> substitutes = [3, 4] # Splice in these items 

>>> list(replace(iterable, pred, substitutes, window_size=window_size)) 

[3, 4, 5, 3, 4, 5] 

 

""" 

if window_size < 1: 

raise ValueError('window_size must be at least 1') 

 

# Save the substitutes iterable, since it's used more than once 

substitutes = tuple(substitutes) 

 

# Add padding such that the number of windows matches the length of the 

# iterable 

it = chain(iterable, [_marker] * (window_size - 1)) 

windows = windowed(it, window_size) 

 

n = 0 

for w in windows: 

# If the current window matches our predicate (and we haven't hit 

# our maximum number of replacements), splice in the substitutes 

# and then consume the following windows that overlap with this one. 

# For example, if the iterable is (0, 1, 2, 3, 4...) 

# and the window size is 2, we have (0, 1), (1, 2), (2, 3)... 

# If the predicate matches on (0, 1), we need to zap (0, 1) and (1, 2) 

if pred(*w): 

if (count is None) or (n < count): 

n += 1 

for s in substitutes: 

yield s 

consume(windows, window_size - 1) 

continue 

 

# If there was no match (or we've reached the replacement limit), 

# yield the first item from the window. 

if w and (w[0] is not _marker): 

yield w[0]