`!itertools` --- Functions creating iterators for efficient looping

accumulate(iterable[, function, *, initial=None])

Make an iterator that returns accumulated sums or accumulated results from other binary functions.

The function defaults to addition. The function should accept two arguments, an accumulated total and a value from the iterable.

If an initial value is provided, the accumulation will start with that value and the output will have one more element than the input iterable.

Roughly equivalent to:

def accumulate(iterable, function=operator.add, *, initial=None):
    'Return running totals'
    # accumulate([1,2,3,4,5]) → 1 3 6 10 15
    # accumulate([1,2,3,4,5], initial=100) → 100 101 103 106 110 115
    # accumulate([1,2,3,4,5], operator.mul) → 1 2 6 24 120

    iterator = iter(iterable)
    total = initial
    if initial is None:
        try:
            total = next(iterator)
        except StopIteration:
            return

    yield total
    for element in iterator:
        total = function(total, element)
        yield total

To compute a running minimum, set function to min. For a running maximum, set function to max. Or for a running product, set function to operator.mul. To build an amortization table, accumulate the interest and apply payments:

See functools.reduce for a similar function that returns only the final accumulated value.

batched(iterable, n, *, strict=False)

Batch data from the iterable into tuples of length n. The last batch may be shorter than n.

If strict is true, will raise a ValueError if the final batch is shorter than n.

Loops over the input iterable and accumulates data into tuples up to size n. The input is consumed lazily, just enough to fill a batch. The result is yielded as soon as the batch is full or when the input iterable is exhausted:

Roughly equivalent to:

def batched(iterable, n, *, strict=False):
    # batched('ABCDEFG', 3) → ABC DEF G
    if n < 1:
        raise ValueError('n must be at least one')
    iterator = iter(iterable)
    while batch := tuple(islice(iterator, n)):
        if strict and len(batch) != n:
            raise ValueError('batched(): incomplete batch')
        yield batch

chain(*iterables)

Make an iterator that returns elements from the first iterable until it is exhausted, then proceeds to the next iterable, until all of the iterables are exhausted. This combines multiple data sources into a single iterator. Roughly equivalent to:

def chain(*iterables):
    # chain('ABC', 'DEF') → A B C D E F
    for iterable in iterables:
        yield from iterable

Note that unpacking in comprehensions <unpacking-comprehensions> provides similar functionality so that list(chain(p, q)) could be written as [*s for s in (p, q)].

combinations(iterable, r)

Return r length subsequences of elements from the input iterable.

The output is a subsequence of product keeping only entries that are subsequences of the iterable. The length of the output is given by math.comb which computes n! / r! / (n - r)! when 0 ≤ r ≤ n or zero when r > n.

The combination tuples are emitted in lexicographic order according to the order of the input iterable. If the input iterable is sorted, the output tuples will be produced in sorted order.

Elements are treated as unique based on their position, not on their value. If the input elements are unique, there will be no repeated values within each combination.

Roughly equivalent to:

def combinations(iterable, r):
    # combinations('ABCD', 2) → AB AC AD BC BD CD
    # combinations(range(4), 3) → 012 013 023 123

    pool = tuple(iterable)
    n = len(pool)
    if r > n:
        return
    indices = list(range(r))

    yield tuple(pool[i] for i in indices)
    while True:
        for i in reversed(range(r)):
            if indices[i] != i + n - r:
                break
        else:
            return
        indices[i] += 1
        for j in range(i+1, r):
            indices[j] = indices[j-1] + 1
        yield tuple(pool[i] for i in indices)

combinations_with_replacement(iterable, r)

Return r length subsequences of elements from the input iterable allowing individual elements to be repeated more than once.

The output is a subsequence of product that keeps only entries that are subsequences (with possible repeated elements) of the iterable. The number of subsequence returned is (n + r - 1)! / r! / (n - 1)! when n > 0.

The combination tuples are emitted in lexicographic order according to the order of the input iterable. if the input iterable is sorted, the output tuples will be produced in sorted order.

Elements are treated as unique based on their position, not on their value. If the input elements are unique, the generated combinations will also be unique.

Roughly equivalent to:

def combinations_with_replacement(iterable, r):
    # combinations_with_replacement('ABC', 2) → AA AB AC BB BC CC

    pool = tuple(iterable)
    n = len(pool)
    if not n and r:
        return
    indices = [0] * r

    yield tuple(pool[i] for i in indices)
    while True:
        for i in reversed(range(r)):
            if indices[i] != n - 1:
                break
        else:
            return
        indices[i:] = [indices[i] + 1] * (r - i)
        yield tuple(pool[i] for i in indices)

compress(data, selectors)

Make an iterator that returns elements from data where the corresponding element in selectors is true. Stops when either the data or selectors iterables have been exhausted. Roughly equivalent to:

def compress(data, selectors):
    # compress('ABCDEF', [1,0,1,0,1,1]) → A C E F
    return (datum for datum, selector in zip(data, selectors) if selector)

count(start=0, step=1)

Make an iterator that returns evenly spaced values beginning with start. Can be used with map to generate consecutive data points or with zip to add sequence numbers. Roughly equivalent to:

def count(start=0, step=1):
    # count(10) → 10 11 12 13 14 ...
    # count(2.5, 0.5) → 2.5 3.0 3.5 ...
    n = start
    while True:
        yield n
        n += step

When counting with floating-point numbers, better accuracy can sometimes be achieved by substituting multiplicative code such as: (start + step * i for i in count()).

cycle(iterable)

Make an iterator returning elements from the iterable and saving a copy of each. When the iterable is exhausted, return elements from the saved copy. Repeats indefinitely. Roughly equivalent to:

def cycle(iterable):
    # cycle('ABCD') → A B C D A B C D A B C D ...

    saved = []
    for element in iterable:
        yield element
        saved.append(element)

    while saved:
        for element in saved:
            yield element

This itertool may require significant auxiliary storage (depending on the length of the iterable).

dropwhile(predicate, iterable)

Make an iterator that drops elements from the iterable while the predicate is true and afterwards returns every element. Roughly equivalent to:

def dropwhile(predicate, iterable):
    # dropwhile(lambda x: x<5, [1,4,6,3,8]) → 6 3 8

    iterator = iter(iterable)
    for x in iterator:
        if not predicate(x):
            yield x
            break

    for x in iterator:
        yield x

Note this does not produce any output until the predicate first becomes false, so this itertool may have a lengthy start-up time.

filterfalse(predicate, iterable)

Make an iterator that filters elements from the iterable returning only those for which the predicate returns a false value. If predicate is None, returns the items that are false. Roughly equivalent to:

def filterfalse(predicate, iterable):
    # filterfalse(lambda x: x<5, [1,4,6,3,8]) → 6 8

    if predicate is None:
        predicate = bool

    for x in iterable:
        if not predicate(x):
            yield x

groupby(iterable, key=None)

Make an iterator that returns consecutive keys and groups from the iterable. The key is a function computing a key value for each element. If not specified or is None, key defaults to an identity function and returns the element unchanged. Generally, the iterable needs to already be sorted on the same key function.

The operation of groupby is similar to the uniq filter in Unix. It generates a break or new group every time the value of the key function changes (which is why it is usually necessary to have sorted the data using the same key function). That behavior differs from SQL's GROUP BY which aggregates common elements regardless of their input order.

The returned group is itself an iterator that shares the underlying iterable with groupby. Because the source is shared, when the groupby object is advanced, the previous group is no longer visible. So, if that data is needed later, it should be stored as a list:

groups = []
uniquekeys = []
data = sorted(data, key=keyfunc)
for k, g in groupby(data, keyfunc):
    groups.append(list(g))      # Store group iterator as a list
    uniquekeys.append(k)

groupby is roughly equivalent to:

def groupby(iterable, key=None):
    # [k for k, g in groupby('AAAABBBCCDAABBB')] → A B C D A B
    # [list(g) for k, g in groupby('AAAABBBCCD')] → AAAA BBB CC D

    keyfunc = (lambda x: x) if key is None else key
    iterator = iter(iterable)
    exhausted = False

    def _grouper(target_key):
        nonlocal curr_value, curr_key, exhausted
        yield curr_value
        for curr_value in iterator:
            curr_key = keyfunc(curr_value)
            if curr_key != target_key:
                return
            yield curr_value
        exhausted = True

    try:
        curr_value = next(iterator)
    except StopIteration:
        return
    curr_key = keyfunc(curr_value)

    while not exhausted:
        target_key = curr_key
        curr_group = _grouper(target_key)
        yield curr_key, curr_group
        if curr_key == target_key:
            for _ in curr_group:
                pass

islice(iterable, stop) islice(iterable, start, stop[, step])

Make an iterator that returns selected elements from the iterable. Works like sequence slicing but does not support negative values for start, stop, or step.

If start is zero or None, iteration starts at zero. Otherwise, elements from the iterable are skipped until start is reached.

If stop is None, iteration continues until the input is exhausted, if at all. Otherwise, it stops at the specified position.

If step is None, the step defaults to one. Elements are returned consecutively unless step is set higher than one which results in items being skipped.

Roughly equivalent to:

def islice(iterable, *args):
    # islice('ABCDEFG', 2) → A B
    # islice('ABCDEFG', 2, 4) → C D
    # islice('ABCDEFG', 2, None) → C D E F G
    # islice('ABCDEFG', 0, None, 2) → A C E G

    s = slice(*args)
    start = 0 if s.start is None else s.start
    stop = s.stop
    step = 1 if s.step is None else s.step
    if start < 0 or (stop is not None and stop < 0) or step <= 0:
        raise ValueError

    indices = count() if stop is None else range(max(start, stop))
    next_i = start
    for i, element in zip(indices, iterable):
        if i == next_i:
            yield element
            next_i += step

If the input is an iterator, then fully consuming the islice advances the input iterator by max(start, stop) steps regardless of the step value.

pairwise(iterable)

Return successive overlapping pairs taken from the input iterable.

The number of 2-tuples in the output iterator will be one fewer than the number of inputs. It will be empty if the input iterable has fewer than two values.

Roughly equivalent to:

def pairwise(iterable):
    # pairwise('ABCDEFG') → AB BC CD DE EF FG

    iterator = iter(iterable)
    a = next(iterator, None)

    for b in iterator:
        yield a, b
        a = b

permutations(iterable, r=None)

Return successive r length permutations of elements from the iterable.

If r is not specified or is None, then r defaults to the length of the iterable and all possible full-length permutations are generated.

The output is a subsequence of product where entries with repeated elements have been filtered out. The length of the output is given by math.perm which computes n! / (n - r)! when 0 ≤ r ≤ n or zero when r > n.

The permutation tuples are emitted in lexicographic order according to the order of the input iterable. If the input iterable is sorted, the output tuples will be produced in sorted order.

Elements are treated as unique based on their position, not on their value. If the input elements are unique, there will be no repeated values within a permutation.

Roughly equivalent to:

def permutations(iterable, r=None):
    # permutations('ABCD', 2) → AB AC AD BA BC BD CA CB CD DA DB DC
    # permutations(range(3)) → 012 021 102 120 201 210

    pool = tuple(iterable)
    n = len(pool)
    r = n if r is None else r
    if r > n:
        return

    indices = list(range(n))
    cycles = list(range(n, n-r, -1))
    yield tuple(pool[i] for i in indices[:r])

    while n:
        for i in reversed(range(r)):
            cycles[i] -= 1
            if cycles[i] == 0:
                indices[i:] = indices[i+1:] + indices[i:i+1]
                cycles[i] = n - i
            else:
                j = cycles[i]
                indices[i], indices[-j] = indices[-j], indices[i]
                yield tuple(pool[i] for i in indices[:r])
                break
        else:
            return

product(*iterables, repeat=1)

Cartesian product of the input iterables.

Roughly equivalent to nested for-loops in a generator expression. For example, product(A, B) returns the same as ((x,y) for x in A for y in B).

The nested loops cycle like an odometer with the rightmost element advancing on every iteration. This pattern creates a lexicographic ordering so that if the input's iterables are sorted, the product tuples are emitted in sorted order.

To compute the product of an iterable with itself, specify the number of repetitions with the optional repeat keyword argument. For example, product(A, repeat=4) means the same as product(A, A, A, A).

This function is roughly equivalent to the following code, except that the actual implementation does not build up intermediate results in memory:

def product(*iterables, repeat=1):
    # product('ABCD', 'xy') → Ax Ay Bx By Cx Cy Dx Dy
    # product(range(2), repeat=3) → 000 001 010 011 100 101 110 111

    if repeat < 0:
        raise ValueError('repeat argument cannot be negative')
    pools = [tuple(pool) for pool in iterables] * repeat

    result = [[]]
    for pool in pools:
        result = [x+[y] for x in result for y in pool]

    for prod in result:
        yield tuple(prod)

Before product runs, it completely consumes the input iterables, keeping pools of values in memory to generate the products. Accordingly, it is only useful with finite inputs.

repeat(object[, times])

Make an iterator that returns object over and over again. Runs indefinitely unless the times argument is specified.

Roughly equivalent to:

def repeat(object, times=None):
    # repeat(10, 3) → 10 10 10
    if times is None:
        while True:
            yield object
    else:
        for i in range(times):
            yield object

A common use for repeat is to supply a stream of constant values to map or zip:

starmap(function, iterable)

Make an iterator that computes the function using arguments obtained from the iterable. Used instead of map when argument parameters have already been "pre-zipped" into tuples.

The difference between map and starmap parallels the distinction between function(a,b) and function(*c). Roughly equivalent to:

def starmap(function, iterable):
    # starmap(pow, [(2,5), (3,2), (10,3)]) → 32 9 1000
    for args in iterable:
        yield function(*args)

takewhile(predicate, iterable)

Make an iterator that returns elements from the iterable as long as the predicate is true. Roughly equivalent to:

def takewhile(predicate, iterable):
    # takewhile(lambda x: x<5, [1,4,6,3,8]) → 1 4
    for x in iterable:
        if not predicate(x):
            break
        yield x

Note, the element that first fails the predicate condition is consumed from the input iterator and there is no way to access it. This could be an issue if an application wants to further consume the input iterator after takewhile has been run to exhaustion <exhausted>. To work around this problem, consider using more-itertools before_and_after() instead.

tee(iterable, n=2)

Return n independent iterators from a single iterable.

Roughly equivalent to:

def tee(iterable, n=2):
    if n < 0:
        raise ValueError
    if n == 0:
        return ()
    iterator = _tee(iterable)
    result = [iterator]
    for _ in range(n - 1):
        result.append(_tee(iterator))
    return tuple(result)

class _tee:

    def __init__(self, iterable):
        it = iter(iterable)
        if isinstance(it, _tee):
            self.iterator = it.iterator
            self.link = it.link
        else:
            self.iterator = it
            self.link = [None, None]

    def __iter__(self):
        return self

    def __next__(self):
        link = self.link
        if link[1] is None:
            link[0] = next(self.iterator)
            link[1] = [None, None]
        value, self.link = link
        return value

When the input iterable is already a tee iterator object, all members of the return tuple are constructed as if they had been produced by the upstream tee call. This "flattening step" allows nested tee calls to share the same underlying data chain and to have a single update step rather than a chain of calls.

The flattening property makes tee iterators efficiently peekable:

tee iterators are not threadsafe. A RuntimeError may be raised when simultaneously using iterators returned by the same tee call, even if the original iterable is threadsafe.

This itertool may require significant auxiliary storage (depending on how much temporary data needs to be stored). In general, if one iterator uses most or all of the data before another iterator starts, it is faster to use list instead of tee.

zip_longest(*iterables, fillvalue=None)

Make an iterator that aggregates elements from each of the iterables.

If the iterables are of uneven length, missing values are filled-in with fillvalue. If not specified, fillvalue defaults to None.

Iteration continues until the longest iterable is exhausted.

Roughly equivalent to:

def zip_longest(*iterables, fillvalue=None):
    # zip_longest('ABCD', 'xy', fillvalue='-') → Ax By C- D-

    iterators = list(map(iter, iterables))
    num_active = len(iterators)
    if not num_active:
        return

    while True:
        values = []
        for i, iterator in enumerate(iterators):
            try:
                value = next(iterator)
            except StopIteration:
                num_active -= 1
                if not num_active:
                    return
                iterators[i] = repeat(fillvalue)
                value = fillvalue
            values.append(value)
        yield tuple(values)

If one of the iterables is potentially infinite, then the zip_longest function should be wrapped with something that limits the number of calls (for example islice or takewhile).