setuptools._vendor.more_itertools.more

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.

Returns ``True`` if all the elements of *iterable* are unique (no two elements are equal). >>> all_unique('ABCB') False If a *key* function is specified, it will be used to make comparisons. >>> all_unique('ABCb') True >>> all_unique('ABCb', str.lower) False The function returns as soon as the first non-unique element is encountered. Iterables with a mix of hashable and unhashable items can be used, but the function will be slower for unhashable items.

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']

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.

Break *iterable* into lists of length *n*: >>> list(chunked([1, 2, 3, 4, 5, 6], 3)) [[1, 2, 3], [4, 5, 6]] By the default, the last yielded list will have fewer than *n* elements if the length of *iterable* is not divisible by *n*: >>> 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. If the length of *iterable* is not divisible by *n* and *strict* is ``True``, then ``ValueError`` will be raised before the last list is yielded.

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)]

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] Binary and text strings 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']]

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 3.5+, this function is an alias for :func:`heapq.merge`.

Equivalent to ``list(combinations(iterable, r)).index(element)`` The subsequences of *iterable* that are of length *r* can be ordered lexicographically. :func:`combination_index` computes the index of the first *element*, without computing the previous combinations. >>> combination_index('adf', 'abcdefg') 10 ``ValueError`` will be raised if the given *element* isn't one of the combinations of *iterable*.

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'] Each group of consecutive items is an iterator that shares it source with *iterable*. When an an output group is advanced, the previous group is no longer available unless its elements are copied (e.g., into a ``list``). >>> iterable = [1, 2, 11, 12, 21, 22] >>> saved_groups = [] >>> for group in consecutive_groups(iterable): ... saved_groups.append(list(group)) # Copy group elements >>> saved_groups [[1, 2], [11, 12], [21, 22]]

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.

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')]

This function is the inverse of :func:`itertools.accumulate`. By default it will compute the first difference of *iterable* using :func:`operator.sub`: >>> from itertools import accumulate >>> iterable = accumulate([0, 1, 2, 3, 4]) # produces 0, 1, 3, 6, 10 >>> list(difference(iterable)) [0, 1, 2, 3, 4] *func* defaults to :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] >>> func = lambda x, y: x // y >>> list(difference(iterable, func)) [1, 2, 3, 4, 5] If the *initial* keyword is set, the first element will be skipped when computing successive differences. >>> it = [10, 11, 13, 16] # from accumulate([1, 2, 3], initial=10) >>> list(difference(it, initial=10)) [1, 2, 3]

Yield the distinct combinations of *r* items taken from *iterable*. >>> list(distinct_combinations([0, 0, 1], 2)) [(0, 0), (0, 1)] Equivalent to ``set(combinations(iterable))``, except duplicates are not generated and thrown away. For larger input sequences this is much more efficient.

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. If *r* is given, only the *r*-length permutations are yielded. >>> sorted(distinct_permutations([1, 0, 1], r=2)) [(0, 1), (1, 0), (1, 1)] >>> sorted(distinct_permutations(range(3), r=2)) [(0, 1), (0, 2), (1, 0), (1, 2), (2, 0), (2, 1)]

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`.

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.

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.

Yield the items from *iterable* for which the *validator* function does not raise one of the specified *exceptions*. *validator* is called for each item in *iterable*. It should be a function that accepts one argument and raises an exception if that item is not valid. >>> iterable = ['1', '2', 'three', '4', None] >>> list(filter_except(int, iterable, ValueError, TypeError)) ['1', '2', '4'] If an exception other than one given by *exceptions* is raised by *validator*, it is raised like normal.

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)``.

An extension of :func:`itertools.groupby` that can apply transformations to the grouped data. * *keyfunc* is a function computing a key value for each item in *iterable* * *valuefunc* is a function that transforms the individual items from *iterable* after grouping * *reducefunc* is a function that transforms each group of items >>> iterable = 'aAAbBBcCC' >>> keyfunc = lambda k: k.upper() >>> valuefunc = lambda v: v.lower() >>> reducefunc = lambda g: ''.join(g) >>> list(groupby_transform(iterable, keyfunc, valuefunc, reducefunc)) [('A', 'aaa'), ('B', 'bbb'), ('C', 'ccc')] Each optional argument defaults to an identity function if 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.

Break *iterable* into sub-iterables with *n* elements each. :func:`ichunked` is like :func:`chunked`, but it yields iterables instead of lists. If the sub-iterables are read in order, the elements of *iterable* won't be stored in memory. If they are read out of order, :func:`itertools.tee` is used to cache elements as necessary. >>> from itertools import count >>> all_chunks = ichunked(count(), 4) >>> c_1, c_2, c_3 = next(all_chunks), next(all_chunks), next(all_chunks) >>> list(c_2) # c_1's elements have been cached; c_3's haven't been [4, 5, 6, 7] >>> list(c_1) [0, 1, 2, 3] >>> list(c_3) [8, 9, 10, 11]

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.

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 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).

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]

Returns ``True`` if the items of iterable are in sorted order, and ``False`` otherwise. *key* and *reverse* have the same meaning that they do in the built-in :func:`sorted` function. >>> is_sorted(['1', '2', '3', '4', '5'], key=int) True >>> is_sorted([5, 4, 3, 1, 2], reverse=True) False The function returns ``False`` after encountering the first out-of-order item. If there are no out-of-order items, the iterable is exhausted.

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]

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``.

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

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 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'

Transform each item from *iterable* with *function* and yield the result, unless *function* raises one of the specified *exceptions*. *function* is called to transform each item in *iterable*. It should be a accept one argument. >>> iterable = ['1', '2', 'three', '4', None] >>> list(map_except(int, iterable, ValueError, TypeError)) [1, 2, 4] If an exception other than one given by *exceptions* is raised by *function*, it is raised like normal.

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.

Yield 3-tuples of the form ``(is_first, is_last, item)``. >>> list(mark_ends('ABC')) [(True, False, 'A'), (False, False, 'B'), (False, True, 'C')] Use this when looping over an iterable to take special action on its first and/or last items: >>> iterable = ['Header', 100, 200, 'Footer'] >>> total = 0 >>> for is_first, is_last, item in mark_ends(iterable): ... if is_first: ... continue # Skip the header ... if is_last: ... continue # Skip the footer ... total += item >>> print(total) 300

Return the nth or the last item of *iterable*, or *default* if *iterable* is empty. >>> nth_or_last([0, 1, 2, 3], 2) 2 >>> nth_or_last([0, 1], 2) 1 >>> nth_or_last([], 0, 'some default') 'some default' If *default* is not provided and there are no items in the iterable, raise ``ValueError``.

Equivalent to ``list(permutations(iterable, r))[index]``` The subsequences of *iterable* that are of length *r* where order is important can be ordered lexicographically. :func:`nth_permutation` computes the subsequence at sort position *index* directly, without computing the previous subsequences. >>> nth_permutation('ghijk', 2, 5) ('h', 'i') ``ValueError`` will be raised If *r* is negative or greater than the length of *iterable*. ``IndexError`` will be raised if the given *index* is invalid.

Equivalent to ``list(product(*args))[index]``. The products of *args* can be ordered lexicographically. :func:`nth_product` computes the product at sort position *index* without computing the previous products. >>> nth_product(8, range(2), range(2), range(2), range(2)) (1, 0, 0, 0) ``IndexError`` will be raised if the given *index* is invalid.

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: Expected exactly one item in iterable, but got 'too', 'many', and perhaps more. >>> 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. See :func:`spy` or :func:`peekable` to check iterable contents less destructively.

If *iterable* has only one item, return it. If it has zero items, return *default*. If it has more than one item, raise the exception given by *too_long*, which is ``ValueError`` by default. >>> only([], default='missing') 'missing' >>> only([1]) 1 >>> only([1, 2]) # doctest: +IGNORE_EXCEPTION_DETAIL Traceback (most recent call last): ... ValueError: Expected exactly one item in iterable, but got 1, 2, and perhaps more.' >>> only([1, 2], too_long=TypeError) # doctest: +IGNORE_EXCEPTION_DETAIL Traceback (most recent call last): ... TypeError Note that :func:`only` attempts to advance *iterable* twice to ensure there is only one item. See :func:`spy` or :func:`peekable` to check iterable contents less destructively.

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.

Yield all possible order-preserving partitions of *iterable*. >>> iterable = 'abc' >>> for part in partitions(iterable): ... print([''.join(p) for p in part]) ['abc'] ['a', 'bc'] ['ab', 'c'] ['a', 'b', 'c'] This is unrelated to :func:`partition`.

Equivalent to ``list(permutations(iterable, r)).index(element)``` The subsequences of *iterable* that are of length *r* where order is important can be ordered lexicographically. :func:`permutation_index` computes the index of the first *element* directly, without computing the previous permutations. >>> permutation_index([1, 3, 2], range(5)) 19 ``ValueError`` will be raised if the given *element* isn't one of the permutations of *iterable*.

Equivalent to ``list(product(*args)).index(element)`` The products of *args* can be ordered lexicographically. :func:`product_index` computes the first index of *element* without computing the previous products. >>> product_index([8, 2], range(10), range(5)) 42 ``ValueError`` will be raised if the given *element* isn't in the product of *args*.

After the *iterable* is exhausted, keep yielding its last element. >>> list(islice(repeat_last(range(3)), 5)) [0, 1, 2, 2, 2] If the iterable is empty, yield *default* forever:: >>> list(islice(repeat_last(range(0), 42), 5)) [42, 42, 42, 42, 42]

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]

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.

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`.

Return a *k*-length list of elements chosen (without replacement) from the *iterable*. Like :func:`random.sample`, but works on iterables of unknown length. >>> iterable = range(100) >>> sample(iterable, 5) # doctest: +SKIP [81, 60, 96, 16, 4] An iterable with *weights* may also be given: >>> iterable = range(100) >>> weights = (i * i + 1 for i in range(100)) >>> sampled = sample(iterable, 5, weights=weights) # doctest: +SKIP [79, 67, 74, 66, 78] The algorithm can also be used to generate weighted random permutations. The relative weight of each item determines the probability that it appears late in the permutation. >>> data = "abcdefgh" >>> weights = range(1, len(data) + 1) >>> sample(data, k=len(data), weights=weights) # doctest: +SKIP ['c', 'a', 'b', 'e', 'g', 'd', 'h', 'f']

Yield the set partitions of *iterable* into *k* parts. Set partitions are not order-preserving. >>> iterable = 'abc' >>> for part in set_partitions(iterable, 2): ... print([''.join(p) for p in part]) ['a', 'bc'] ['ab', 'c'] ['b', 'ac'] If *k* is not given, every set partition is generated. >>> iterable = 'abc' >>> for part in set_partitions(iterable): ... print([''.join(p) for p in part]) ['abc'] ['a', 'bc'] ['ab', 'c'] ['b', 'ac'] ['a', 'b', 'c']

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

Yield slices of length *n* from the sequence *seq*. >>> list(sliced((1, 2, 3, 4, 5, 6), 3)) [(1, 2, 3), (4, 5, 6)] By the default, the last yielded slice will have fewer than *n* elements if the length of *seq* is not divisible by *n*: >>> list(sliced((1, 2, 3, 4, 5, 6, 7, 8), 3)) [(1, 2, 3), (4, 5, 6), (7, 8)] If the length of *seq* is not divisible by *n* and *strict* is ``True``, then ``ValueError`` will be raised before the last slice is yielded. This function will only work for iterables that support slicing. For non-sliceable iterables, see :func:`chunked`.

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')] To sort by a function of the elements of the iterable, pass a *key* function. Its arguments are the elements of the iterables corresponding to the key list:: >>> names = ('a', 'b', 'c') >>> lengths = (1, 2, 3) >>> widths = (5, 2, 1) >>> def area(length, width): ... return length * width >>> sort_together([names, lengths, widths], key_list=(1, 2), key=area) [('c', 'b', 'a'), (3, 2, 1), (1, 2, 5)] 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')]

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]] At most *maxsplit* splits are done. If *maxsplit* is not specified or -1, then there is no limit on the number of splits: >>> list(split_after(range(10), lambda n: n % 3 == 0, maxsplit=2)) [[0], [1, 2, 3], [4, 5, 6, 7, 8, 9]]

Yield lists of items from *iterable*, where each list is delimited by an item where callable *pred* returns ``True``. >>> 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], []] At most *maxsplit* splits are done. If *maxsplit* is not specified or -1, then there is no limit on the number of splits: >>> list(split_at(range(10), lambda n: n % 2 == 1, maxsplit=2)) [[0], [2], [4, 5, 6, 7, 8, 9]] By default, the delimiting items are not included in the output. The include them, set *keep_separator* to ``True``. >>> list(split_at('abcdcba', lambda x: x == 'b', keep_separator=True)) [['a'], ['b'], ['c', 'd', 'c'], ['b'], ['a']]

Yield lists of items from *iterable*, where each list ends just before an item for which 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]] At most *maxsplit* splits are done. If *maxsplit* is not specified or -1, then there is no limit on the number of splits: >>> list(split_before(range(10), lambda n: n % 3 == 0, maxsplit=2)) [[0, 1, 2], [3, 4, 5], [6, 7, 8, 9]]

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.

Split *iterable* into pieces based on the output of *pred*. *pred* should be a function that takes successive pairs of items and returns ``True`` if the iterable should be split in between them. For example, to find runs of increasing numbers, split the iterable when element ``i`` is larger than element ``i + 1``: >>> list(split_when([1, 2, 3, 3, 2, 5, 2, 4, 2], lambda x, y: x > y)) [[1, 2, 3, 3], [2, 5], [2, 4], [2]] At most *maxsplit* splits are done. If *maxsplit* is not specified or -1, then there is no limit on the number of splits: >>> list(split_when([1, 2, 3, 3, 2, 5, 2, 4, 2], ... lambda x, y: x > y, maxsplit=2)) [[1, 2, 3, 3], [2, 5], [2, 4, 2]]

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]

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.

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`.

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)]

Yield all substrings and their positions in *seq* The items yielded will be a tuple of the form ``(substr, i, j)``, where ``substr == seq[i:j]``. This function only works for iterables that support slicing, such as ``str`` objects. >>> for item in substrings_indexes('more'): ... print(item) ('m', 0, 1) ('o', 1, 2) ('r', 2, 3) ('e', 3, 4) ('mo', 0, 2) ('or', 1, 3) ('re', 2, 4) ('mor', 0, 3) ('ore', 1, 4) ('more', 0, 4) Set *reverse* to ``True`` to yield the same items in the opposite order.

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.

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.

Yield all arguments passed to the function in the same order in which they were passed. If an argument itself is iterable then iterate over its values. >>> list(value_chain(1, 2, 3, [4, 5, 6])) [1, 2, 3, 4, 5, 6] Binary and text strings are not considered iterable and are emitted as-is: >>> list(value_chain('12', '34', ['56', '78'])) ['12', '34', '56', '78'] Multiple levels of nesting are not flattened.

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, '!')] To slide into the iterable's items, use :func:`chain` to add filler items to the left: >>> iterable = [1, 2, 3, 4] >>> n = 3 >>> padding = [None] * (n - 1) >>> list(windowed(chain(padding, iterable), 3)) [(None, None, 1), (None, 1, 2), (1, 2, 3), (2, 3, 4)]

Yield ``(beginning, middle, end)`` tuples, where: * Each ``middle`` has *n* items from *iterable* * Each ``beginning`` has the items before the ones in ``middle`` * Each ``end`` has the items after the ones in ``middle`` >>> iterable = range(7) >>> n = 3 >>> for beginning, middle, end in windowed_complete(iterable, n): ... print(beginning, middle, end) () (0, 1, 2) (3, 4, 5, 6) (0,) (1, 2, 3) (4, 5, 6) (0, 1) (2, 3, 4) (5, 6) (0, 1, 2) (3, 4, 5) (6,) (0, 1, 2, 3) (4, 5, 6) () Note that *n* must be at least 0 and most equal to the length of *iterable*. This function will exhaust the iterable and may require significant storage.

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``.

``zip`` the input *iterables* together, but raise ``UnequalIterablesError`` if they aren't all the same length. >>> it_1 = range(3) >>> it_2 = iter('abc') >>> list(zip_equal(it_1, it_2)) [(0, 'a'), (1, 'b'), (2, 'c')] >>> it_1 = range(3) >>> it_2 = iter('abcd') >>> list(zip_equal(it_1, it_2)) # doctest: +IGNORE_EXCEPTION_DETAIL Traceback (most recent call last): ... more_itertools.more.UnequalIterablesError: Iterables have different lengths

``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.