API Reference

Grouping

These tools yield groups of items from a source iterable.


New itertools

more_itertools.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 grouper() recipe.

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.

more_itertools.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 chunked().

more_itertools.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 itertools.tee() and may require significant storage. If you need the order items in the smaller iterables to match the original iterable, see divide().

more_itertools.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 distribute(), which does not first pull the iterable into memory.

more_itertools.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]]
more_itertools.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]]
more_itertools.bucket(iterable, key)

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 s: s[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.

Be aware that attempting to select a bucket that no items correspond to will exhaust the iterable and cache all values.


Itertools recipes

more_itertools.grouper(n, iterable, fillvalue=None)

Collect data into fixed-length chunks or blocks.

>>> list(grouper(3, 'ABCDEFG', 'x'))
[('A', 'B', 'C'), ('D', 'E', 'F'), ('G', 'x', 'x')]
more_itertools.partition(pred, iterable)

Returns a 2-tuple of iterables derived from the input iterable. The first yields the items that have pred(item) == False. The first yields the items that have pred(item) == False.

>>> is_odd = lambda x: x % 2 != 0
>>> iterable = range(10)
>>> even_items, odd_items = partition(is_odd, iterable)
>>> list(even_items), list(odd_items)
([0, 2, 4, 6, 8], [1, 3, 5, 7, 9])

Lookahead

These tools peek at an iterable’s values without advancing it.


New itertools

more_itertools.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]
class more_itertools.peekable(iterable)

Wrap an iterator to allow lookahead and prepending elements.

Call peek() on the result to get the value that will be returned by next(). This won’t advance the iterator:

>>> p = peekable(['a', 'b'])
>>> p.peek()
'a'
>>> next(p)
'a'

Pass 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 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 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)
[]

Windowing

These tools yield windows of items from an iterable.


New itertools

more_itertools.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, '!')]
more_itertools.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.


Itertools recipes

more_itertools.pairwise(iterable)

Returns an iterator of paired items, overlapping, from the original

>>> take(4, pairwise(count()))
[(0, 1), (1, 2), (2, 3), (3, 4)]

Augmenting

These tools yield items from an iterable, plus additional data.


New itertools

more_itertools.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')]
more_itertools.intersperse(e, iterable)

Intersperse object e between the items of iterable.

>>> list(intersperse('x', 'ABCD'))
['A', 'x', 'B', 'x', 'C', 'x', 'D']
>>> list(intersperse(None, [1, 2, 3]))
[1, None, 2, None, 3]
more_itertools.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.

more_itertools.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 groupby_transform(), which can be used with this function to group ranges of items with the same bool value.

more_itertools.groupby_transform(iterable, keyfunc=None, valuefunc=None)

An extension of 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.

groupby_transform() is useful when grouping elements of an iterable using a separate iterable as the key. To do this, 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')]

Itertools recipes

more_itertools.padnone(iterable)

Returns the sequence of elements and then returns None indefinitely.

>>> take(5, padnone(range(3)))
[0, 1, 2, None, None]

Useful for emulating the behavior of the built-in map() function.

See also padded().

more_itertools.ncycles(iterable, n)

Returns the sequence elements n times

>>> list(ncycles(["a", "b"], 3))
['a', 'b', 'a', 'b', 'a', 'b']

Combining

These tools combine multiple iterables.


New itertools

more_itertools.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']]
more_itertools.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 mutliple 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')]
more_itertools.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 interleave_longest().

more_itertools.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]
more_itertools.collate(*iterables, key=lambda a: a, reverse=False)

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 collate() to, for example, perform a n-way mergesort of items that don’t fit in memory.

Parameters:
  • key – A function that returns a comparison value for an item. Defaults to the identity function.
  • reverse – If reverse=True, yield results in descending order rather than ascending. iterables must also yield their elements in descending order.

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

If neither of the keyword arguments are specified, this function delegates to heapq.merge().

more_itertools.zip_offset(*iterables, offsets, longest=False, fillvalue=None)

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 somes 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.


Itertools recipes

more_itertools.dotproduct(vec1, vec2)

Returns the dot product of the two iterables.

>>> dotproduct([10, 10], [20, 20])
400
more_itertools.flatten(listOfLists)

Return an iterator flattening one level of nesting in a list of lists.

>>> list(flatten([[0, 1], [2, 3]]))
[0, 1, 2, 3]

See also collapse(), which can flatten multiple levels of nesting.

more_itertools.roundrobin(*iterables)

Yields an item from each iterable, alternating between them.

>>> list(roundrobin('ABC', 'D', 'EF'))
['A', 'D', 'E', 'B', 'F', 'C']

See interleave_longest() for a slightly faster implementation.

Summarizing

These tools return summarized or aggregated data from an iterable.


New itertools

more_itertools.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.

more_itertools.first(iterable[, default])

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.

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).

more_itertools.one(iterable)

Return the only element from the iterable.

Raise ValueError if the iterable is empty or longer than 1 element. For example, assert that a DB query returns a single, unique result.

>>> one(['val'])
'val'
>>> one(['val', 'other'])  
Traceback (most recent call last):
...
ValueError: too many values to unpack (expected 1)
>>> one([])  
Traceback (most recent call last):
...
ValueError: not enough values to unpack (expected 1, got 0)

one() attempts to advance the iterable twice in order to ensure there aren’t further items. Because this discards any second item, one() is not suitable in situations where you want to catch its exception and then try an alternative treatment of the iterable. It should be used only when a iterable longer than 1 item is, in fact, an error.

more_itertools.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.

more_itertools.locate(iterable, pred=<type 'bool'>)

Yield the index of each item in iterable for which pred returns True.

pred defaults to 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]

Itertools recipes

more_itertools.all_equal(iterable)

Returns True if all the elements are equal to each other.

>>> all_equal('aaaa')
True
>>> all_equal('aaab')
False
more_itertools.first_true(iterable, default=False, pred=None)

Returns the first true value in the iterable.

If no true value is found, returns default

If pred is not None, returns the first item for which pred(item) == True .

>>> first_true(range(10))
1
>>> first_true(range(10), pred=lambda x: x > 5)
6
>>> first_true(range(10), default='missing', pred=lambda x: x > 9)
'missing'
more_itertools.nth(iterable, n, default=None)

Returns the nth item or a default value.

>>> l = range(10)
>>> nth(l, 3)
3
>>> nth(l, 20, "zebra")
'zebra'
more_itertools.quantify(iterable, pred=<type 'bool'>)

Return the how many times the predicate is true.

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

Selecting

These yools yield certain items from an iterable.


Itertools recipes

more_itertools.take(n, iterable)

Return first n items of the iterable as a list.

>>> take(3, range(10))
[0, 1, 2]
>>> take(5, range(3))
[0, 1, 2]

Effectively a short replacement for next based iterator consumption when you want more than one item, but less than the whole iterator.

more_itertools.tail(n, iterable)

Return an iterator over the last n items of iterable.

>>> t = tail(3, 'ABCDEFG')
>>> list(t)
['E', 'F', 'G']
more_itertools.unique_everseen(iterable, key=None)

Yield unique elements, preserving order.

>>> list(unique_everseen('AAAABBBCCDAABBB'))
['A', 'B', 'C', 'D']
>>> list(unique_everseen('ABBCcAD', str.lower))
['A', 'B', 'C', 'D']

Sequences with a mix of hashable and unhashable items can be used. The function will be slower (i.e., O(n^2)) for unhashable items.

more_itertools.unique_justseen(iterable, key=None)

Yields elements in order, ignoring serial duplicates

>>> list(unique_justseen('AAAABBBCCDAABBB'))
['A', 'B', 'C', 'D', 'A', 'B']
>>> list(unique_justseen('ABBCcAD', str.lower))
['A', 'B', 'C', 'A', 'D']

Combinatorics

These tools yield combinatorial arrangements of items from iterables.


New itertools

more_itertools.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.


Itertools recipes

more_itertools.powerset(iterable)

Yields all possible subsets of the iterable.

>>> list(powerset([1,2,3]))
[(), (1,), (2,), (3,), (1, 2), (1, 3), (2, 3), (1, 2, 3)]
more_itertools.random_product(*args, **kwds)

Draw an item at random from each of the input iterables.

>>> random_product('abc', range(4), 'XYZ')  
('c', 3, 'Z')

If repeat is provided as a keyword argument, that many items will be drawn from each iterable.

>>> random_product('abcd', range(4), repeat=2)  
('a', 2, 'd', 3)

This equivalent to taking a random selection from itertools.product(*args, **kwarg).

more_itertools.random_permutation(iterable, r=None)

Return a random r length permutation of the elements in iterable.

If r is not specified or is None, then r defaults to the length of iterable.

>>> random_permutation(range(5))  
(3, 4, 0, 1, 2)

This equivalent to taking a random selection from itertools.permutations(iterable, r).

more_itertools.random_combination(iterable, r)

Return a random r length subsequence of the elements in iterable.

>>> random_combination(range(5), 3)  
(2, 3, 4)

This equivalent to taking a random selection from itertools.combinations(iterable, r).

more_itertools.random_combination_with_replacement(iterable, r)

Return a random r length subsequence of elements in iterable, allowing individual elements to be repeated.

>>> random_combination_with_replacement(range(3), 5) 
(0, 0, 1, 2, 2)

This equivalent to taking a random selection from itertools.combinations_with_replacement(iterable, r).

Wrapping

These tools provide wrappers to smooth working with objects that produce or consume iterables.


New itertools

more_itertools.always_iterable(obj)

Given an object, always return an iterable.

If the object is not already iterable, return a tuple containing containing the object:

>>> always_iterable(1)
(1,)

If the object is None, return an empty iterable:

>>> always_iterable(None)
()

Otherwise, return the object itself:

>>> always_iterable([1, 2, 3])
[1, 2, 3]

Strings (binary or unicode) are not considered to be iterable:

>>> always_iterable('foo')
('foo',)

This function is useful in applications where a passed parameter may be either a single item or a collection of items:

>>> def item_sum(param):
...     total = 0
...     for item in always_iterable(param):
...         total += item
...
...     return total
>>> item_sum(10)
10
>>> item_sum([10, 20])
30
more_itertools.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.

more_itertools.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.


Itertools recipes

more_itertools.iter_except(func, exception, first=None)

Yields results from a function repeatedly until an exception is raised.

Converts a call-until-exception interface to an iterator interface. Like iter(func, sentinel), but uses an exception instead of a sentinel to end the loop.

>>> l = [0, 1, 2]
>>> list(iter_except(l.pop, IndexError))
[2, 1, 0]

Others

New itertools

more_itertools.numeric_range(start, stop, step)

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.

more_itertools.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
more_itertools.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]

Itertools recipes

more_itertools.consume(iterator, n=None)

Advance iterable by n steps. If n is None, consume it entirely.

Efficiently exhausts an iterator without returning values. Defaults to consuming the whole iterator, but an optional second argument may be provided to limit consumption.

>>> i = (x for x in range(10))
>>> next(i)
0
>>> consume(i, 3)
>>> next(i)
4
>>> consume(i)
>>> next(i)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
StopIteration

If the iterator has fewer items remaining than the provided limit, the whole iterator will be consumed.

>>> i = (x for x in range(3))
>>> consume(i, 5)
>>> next(i)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
StopIteration
more_itertools.accumulate(iterable, func=<built-in function add>)

Return an iterator whose items are the accumulated results of a function (specified by the optional func argument) that takes two arguments. By default, returns accumulated sums with operator.add().

>>> list(accumulate([1, 2, 3, 4, 5]))  # Running sum
[1, 3, 6, 10, 15]
>>> list(accumulate([1, 2, 3], func=operator.mul))  # Running product
[1, 2, 6]
>>> list(accumulate([0, 1, -1, 2, 3, 2], func=max))  # Running maximum
[0, 1, 1, 2, 3, 3]

This function is available in the itertools module for Python 3.2 and greater.

more_itertools.tabulate(function, start=0)

Return an iterator over the results of func(start), func(start + 1), func(start + 2)...

func should be a function that accepts one integer argument.

If start is not specified it defaults to 0. It will be incremented each time the iterator is advanced.

>>> square = lambda x: x ** 2
>>> iterator = tabulate(square, -3)
>>> take(4, iterator)
[9, 4, 1, 0]
more_itertools.repeatfunc(func, times=None, *args)

Call func with args repeatedly, returning an iterable over the results.

If times is specified, the iterable will terminate after that many repetitions:

>>> from operator import add
>>> times = 4
>>> args = 3, 5
>>> list(repeatfunc(add, times, *args))
[8, 8, 8, 8]

If times is None the iterable will not terminate:

>>> from random import randrange
>>> times = None
>>> args = 1, 11
>>> take(6, repeatfunc(randrange, times, *args))  
[2, 4, 8, 1, 8, 4]