more.py

from __future__ import print_function

from collections import Counter, defaultdict, deque
from functools import partial, wraps
from heapq import merge
from itertools import (
    chain,
    compress,
    count,
    dropwhile,
    groupby,
    islice,
    repeat,
    takewhile,
    tee
)
from operator import itemgetter, lt, gt
from sys import maxsize, version_info

from six import binary_type, string_types, text_type
from six.moves import filter, map, range, zip, zip_longest

from .recipes import flatten, take

__all__ = [
    'adjacent',
    'always_iterable',
    'bucket',
    'chunked',
    'collapse',
    'collate',
    'consecutive_groups',
    'consumer',
    'count_cycle',
    'distinct_permutations',
    'distribute',
    'divide',
    'first',
    'groupby_transform',
    'ilen',
    'interleave_longest',
    'interleave',
    'intersperse',
    'islice_extended',
    'iterate',
    'locate',
    'lstrip',
    'numeric_range',
    'one',
    'padded',
    'peekable',
    'rstrip',
    'run_length',
    'side_effect',
    'sliced',
    'sort_together',
    'split_after',
    'split_before',
    'spy',
    'stagger',
    'strip',
    'unique_to_each',
    'windowed',
    'with_iter',
    'zip_offset',
]

_marker = object()


def chunked(iterable, n):
    """Break *iterable* into lists of length *n*:

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

    If the length of *iterable* is not evenly divisible by *n*, the last
    returned list will be shorter:

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

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

    :func:`chunked` is useful for splitting up a computation on a large number
    of keys into batches, to be pickled and sent off to worker processes. One
    example is operations on rows in MySQL, which does not implement
    server-side cursors properly and would otherwise load the entire dataset
    into RAM on the client.

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


def first(iterable, default=_marker):
    """Return the first item of *iterable*, or *default* if *iterable* is
    empty.

        >>> first([0, 1, 2, 3])
        0
        >>> first([], 'some default')
        'some default'

    If *default* is not provided and there are no items in the iterable,
    raise ``ValueError``.

    :func:`first` is useful when you have a generator of expensive-to-retrieve
    values and want any arbitrary one. It is marginally shorter than
    ``next(iter(iterable), default)``.

    """
    try:
        return next(iter(iterable))
    except StopIteration:
        # I'm on the edge about raising ValueError instead of StopIteration. At
        # the moment, ValueError wins, because the caller could conceivably
        # want to do something different with flow control when I raise the
        # exception, and it's weird to explicitly catch StopIteration.
        if default is _marker:
            raise ValueError('first() was called on an empty iterable, and no '
                             'default value was provided.')
        return default


class peekable(object):
    """Wrap an iterator to allow lookahead and prepending elements.

    Call :meth:`peek` on the result to get the value that will be returned
    by :func:`next`. This won't advance the iterator:

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

    Pass :meth:`peek` a default value to return that instead of raising
    ``StopIteration`` when the iterator is exhausted.

        >>> p = peekable([])
        >>> p.peek('hi')
        'hi'

    peekables also offer a :meth:`prepend` method, which "inserts" items
    at the head of the iterable:

        >>> p = peekable([1, 2, 3])
        >>> p.prepend(10, 11, 12)
        >>> next(p)
        10
        >>> p.peek()
        11
        >>> list(p)
        [11, 12, 1, 2, 3]

    peekables can be indexed. Index 0 is the item that will be returned by
    :func:`next`, index 1 is the item after that, and so on:
    The values up to the given index will be cached.

        >>> p = peekable(['a', 'b', 'c', 'd'])
        >>> p[0]
        'a'
        >>> p[1]
        'b'
        >>> next(p)
        'a'

    Negative indexes are supported, but be aware that they will cache the
    remaining items in the source iterator, which may require significant
    storage.

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

        >>> p = peekable(['a', 'b'])
        >>> if p:  # peekable has items
        ...     list(p)
        ['a', 'b']
        >>> if not p:  # peekable is exhaused
        ...     list(p)
        []

    """
    def __init__(self, iterable):
        self._it = iter(iterable)
        self._cache = deque()

    def __iter__(self):
        return self

    def __bool__(self):
        try:
            self.peek()
        except StopIteration:
            return False
        return True

    def __nonzero__(self):
        # For Python 2 compatibility
        return self.__bool__()

    def peek(self, default=_marker):
        """Return the item that will be next returned from ``next()``.

        Return ``default`` if there are no items left. If ``default`` is not
        provided, raise ``StopIteration``.

        """
        if not self._cache:
            try:
                self._cache.append(next(self._it))
            except StopIteration:
                if default is _marker:
                    raise
                return default
        return self._cache[0]

    def prepend(self, *items):
        """Stack up items to be the next ones returned from ``next()`` or
        ``self.peek()``. The items will be returned in
        first in, first out order::

            >>> p = peekable([1, 2, 3])
            >>> p.prepend(10, 11, 12)
            >>> next(p)
            10
            >>> list(p)
            [11, 12, 1, 2, 3]

        It is possible, by prepending items, to "resurrect" a peekable that
        previously raised ``StopIteration``.

            >>> p = peekable([])
            >>> next(p)
            Traceback (most recent call last):
              ...
            StopIteration
            >>> p.prepend(1)
            >>> next(p)
            1
            >>> next(p)
            Traceback (most recent call last):
              ...
            StopIteration

        """
        self._cache.extendleft(reversed(items))

    def __next__(self):
        if self._cache:
            return self._cache.popleft()

        return next(self._it)

    next = __next__  # For Python 2 compatibility

    def _get_slice(self, index):
        # Normalize the slice's arguments
        step = 1 if (index.step is None) else index.step
        if step > 0:
            start = 0 if (index.start is None) else index.start
            stop = maxsize if (index.stop is None) else index.stop
        elif step < 0:
            start = -1 if (index.start is None) else index.start
            stop = (-maxsize - 1) if (index.stop is None) else index.stop
        else:
            raise ValueError('slice step cannot be zero')

        # If either the start or stop index is negative, we'll need to cache
        # the rest of the iterable in order to slice from the right side.
        if (start < 0) or (stop < 0):
            self._cache.extend(self._it)
        # Otherwise we'll need to find the rightmost index and cache to that
        # point.
        else:
            n = min(max(start, stop) + 1, maxsize)
            cache_len = len(self._cache)
            if n >= cache_len:
                self._cache.extend(islice(self._it, n - cache_len))

        return list(self._cache)[index]

    def __getitem__(self, index):
        if isinstance(index, slice):
            return self._get_slice(index)

        cache_len = len(self._cache)
        if index < 0:
            self._cache.extend(self._it)
        elif index >= cache_len:
            self._cache.extend(islice(self._it, index + 1 - cache_len))

        return self._cache[index]


def _collate(*iterables, **kwargs):
    """Helper for ``collate()``, called when the user is using the ``reverse``
    or ``key`` keyword arguments on Python versions below 3.5.

    """
    key = kwargs.pop('key', lambda a: a)
    reverse = kwargs.pop('reverse', False)

    min_or_max = partial(max if reverse else min, key=itemgetter(0))
    peekables = [peekable(it) for it in iterables]
    peekables = [p for p in peekables if p]  # Kill empties.
    while peekables:
        _, p = min_or_max((key(p.peek()), p) for p in peekables)
        yield next(p)
        peekables = [x for x in peekables if x]


def collate(*iterables, **kwargs):
    """Return a sorted merge of the items from each of several already-sorted
    ``iterables``.

        >>> list(collate('ACDZ', 'AZ', 'JKL'))
        ['A', 'A', 'C', 'D', 'J', 'K', 'L', 'Z', 'Z']

    Works lazily, keeping only the next value from each iterable in memory. Use
    :func:`collate` to, for example, perform a n-way mergesort of items that
    don't fit in memory.

    :arg key: A function that returns a comparison value for an item. Defaults
        to the identity function.
    :arg 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 :func:`heapq.merge`.

    """
    if not kwargs:
        return merge(*iterables)

    return _collate(*iterables, **kwargs)


# If using Python version 3.5 or greater, heapq.merge() will be faster than
# collate - use that instead.
if version_info >= (3, 5, 0):
    collate = merge  # noqa


def consumer(func):
    """Decorator that automatically advances a PEP-342-style "reverse iterator"
    to its first yield point so you don't have to call ``next()`` on it
    manually.

        >>> @consumer
        ... def tally():
        ...     i = 0
        ...     while True:
        ...         print('Thing number %s is %s.' % (i, (yield)))
        ...         i += 1
        ...
        >>> t = tally()
        >>> t.send('red')
        Thing number 0 is red.
        >>> t.send('fish')
        Thing number 1 is fish.

    Without the decorator, you would have to call ``next(t)`` before
    ``t.send()`` could be used.

    """
    @wraps(func)
    def wrapper(*args, **kwargs):
        gen = func(*args, **kwargs)
        next(gen)
        return gen
    return wrapper


def ilen(iterable):
    """Return the number of items in *iterable*.

        >>> ilen(x for x in range(1000000) if x % 3 == 0)
        333334

    This consumes the iterable, so handle with care.

    """
    d = deque(enumerate(iterable, 1), maxlen=1)
    return d[0][0] if d else 0


def iterate(func, start):
    """Return ``start``, ``func(start)``, ``func(func(start))``, ...

        >>> from itertools import islice
        >>> list(islice(iterate(lambda x: 2*x, 1), 10))
        [1, 2, 4, 8, 16, 32, 64, 128, 256, 512]

    """
    while True:
        yield start
        start = func(start)


def with_iter(context_manager):
    """Wrap an iterable in a ``with`` statement, so it closes once exhausted.

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

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

    Any context manager which returns an iterable is a candidate for
    ``with_iter``.

    """
    with context_manager as iterable:
        for item in iterable:
            yield item


def one(iterable):
    """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'])  # doctest: +IGNORE_EXCEPTION_DETAIL
        Traceback (most recent call last):
        ...
        ValueError: too many values to unpack (expected 1)

        >>> one([])  # doctest: +IGNORE_EXCEPTION_DETAIL
        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.

    """
    element, = iterable
    return element


def distinct_permutations(iterable):
    """Yield successive distinct permutations of the elements in *iterable*.

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

    Equivalent to ``set(permutations(iterable))``, except duplicates are not
    generated and thrown away. For larger input sequences this is much more
    efficient.

    Duplicate permutations arise when there are duplicated elements in the
    input iterable. The number of items returned is
    `n! / (x_1! * x_2! * ... * x_n!)`, where `n` is the total number of
    items input, and each `x_i` is the count of a distinct item in the input
    sequence.

    """
    def perm_unique_helper(item_counts, perm, i):
        """Internal helper function

        :arg item_counts: Stores the unique items in ``iterable`` and how many
            times they are repeated
        :arg perm: The permutation that is being built for output
        :arg i: The index of the permutation being modified

        The output permutations are built up recursively; the distinct items
        are placed until their repetitions are exhausted.
        """
        if i < 0:
            yield tuple(perm)
        else:
            for item in item_counts:
                if item_counts[item] <= 0:
                    continue
                perm[i] = item
                item_counts[item] -= 1
                for x in perm_unique_helper(item_counts, perm, i - 1):
                    yield x
                item_counts[item] += 1

    item_counts = Counter(iterable)

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


def intersperse(e, iterable, n=1):
    """Intersperse filler element *e* among the items in *iterable*, leaving
    *n* items between each filler element.

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

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

    """
    if n == 0:
        raise ValueError('n must be > 0')
    elif n == 1:
        # interleave(repeat(e), iterable) -> e, x_0, e, e, x_1, e, x_2...
        # islice(..., 1, None) -> x_0, e, e, x_1, e, x_2...
        return islice(interleave(repeat(e), iterable), 1, None)
    else:
        # interleave(filler, chunks) -> [e], [x_0, x_1], [e], [x_2, x_3]...
        # islice(..., 1, None) -> [x_0, x_1], [e], [x_2, x_3]...
        # flatten(...) -> x_0, x_1, e, x_2, x_3...
        filler = repeat([e])
        chunks = chunked(iterable, n)
        return flatten(islice(interleave(filler, chunks), 1, None))


def unique_to_each(*iterables):
    """Return the elements from each of the input iterables that aren't in the
    other input iterables.

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

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

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

    If ``pkg_1`` is removed, then ``A`` is no longer necessary - it is not
    associated with ``pkg_2`` or ``pkg_3``. Similarly, ``C`` is only needed for
    ``pkg_2``, and ``D`` is only needed for ``pkg_3``::

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

    If there are duplicates in one input iterable that aren't in the others
    they will be duplicated in the output. Input order is preserved::

        >>> unique_to_each("mississippi", "missouri")
        [['p', 'p'], ['o', 'u', 'r']]

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

    """
    pool = [list(it) for it in iterables]
    counts = Counter(chain.from_iterable(map(set, pool)))
    uniques = {element for element in counts if counts[element] == 1}
    return [list(filter(uniques.__contains__, it)) for it in pool]


def windowed(seq, n, fillvalue=None, step=1):
    """Return a sliding window of width *n* over the given iterable.

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

    When the window is larger than the iterable, *fillvalue* is used in place
    of missing values::

        >>> list(windowed([1, 2, 3], 4))
        [(1, 2, 3, None)]

    Each window will advance in increments of *step*:

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

    """
    if n < 0:
        raise ValueError('n must be >= 0')
    if n == 0:
        yield tuple()
        return
    if step < 1:
        raise ValueError('step must be >= 1')

    it = iter(seq)
    window = deque([], n)
    append = window.append

    # Initial deque fill
    for _ in range(n):
        append(next(it, fillvalue))
    yield tuple(window)

    # Appending new items to the right causes old items to fall off the left
    i = 0
    for item in it:
        append(item)
        i = (i + 1) % step
        if i % step == 0:
            yield tuple(window)

    # If there are items from the iterable in the window, pad with the given
    # value and emit them.
    if (i % step) and (step - i < n):
        for _ in range(step - i):
            append(fillvalue)
        yield tuple(window)


class bucket(object):
    """Wrap *iterable* and return an object that buckets it iterable into
    child iterables based on a *key* function.

        >>> iterable = ['a1', 'b1', 'c1', 'a2', 'b2', 'c2', 'b3']
        >>> s = bucket(iterable, key=lambda x: x[0])
        >>> a_iterable = s['a']
        >>> next(a_iterable)
        'a1'
        >>> next(a_iterable)
        'a2'
        >>> list(s['b'])
        ['b1', 'b2', 'b3']

    The original iterable will be advanced and its items will be cached until
    they are used by the child iterables. This may require significant storage.

    By default, attempting to select a bucket to which no items belong  will
    exhaust the iterable and cache all values.
    If you specify a *validator* function, selected buckets will instead be
    checked against it.

        >>> from itertools import count
        >>> it = count(1, 2)  # Infinite sequence of odd numbers
        >>> key = lambda x: x % 10  # Bucket by last digit
        >>> validator = lambda x: x in {1, 3, 5, 7, 9}  # Odd digits only
        >>> s = bucket(it, key=key, validator=validator)
        >>> 2 in s
        False
        >>> list(s[2])
        []

    """
    def __init__(self, iterable, key, validator=None):
        self._it = iter(iterable)
        self._key = key
        self._cache = defaultdict(deque)
        self._validator = validator or (lambda x: True)

    def __contains__(self, value):
        if not self._validator(value):
            return False

        try:
            item = next(self[value])
        except StopIteration:
            return False
        else:
            self._cache[value].appendleft(item)

        return True

    def _get_values(self, value):
        """
        Helper to yield items from the parent iterator that match *value*.
        Items that don't match are stored in the local cache as they
        are encountered.
        """
        while True:
            # If we've cached some items that match the target value, emit
            # the first one and evict it from the cache.
            if self._cache[value]:
                yield self._cache[value].popleft()
            # Otherwise we need to advance the parent iterator to search for
            # a matching item, caching the rest.
            else:
                while True:
                    item = next(self._it)
                    item_value = self._key(item)
                    if item_value == value:
                        yield item
                        break
                    elif self._validator(item_value):
                        self._cache[item_value].append(item)

    def __getitem__(self, value):
        if not self._validator(value):
            return iter(())

        return self._get_values(value)


def spy(iterable, n=1):
    """Return a 2-tuple with a list containing the first *n* elements of
    *iterable*, and an iterator with the same items as *iterable*.
    This allows you to "look ahead" at the items in the iterable without
    advancing it.

    There is one item in the list by default:

        >>> iterable = 'abcdefg'
        >>> head, iterable = spy(iterable)
        >>> head
        ['a']
        >>> list(iterable)
        ['a', 'b', 'c', 'd', 'e', 'f', 'g']

    You may use unpacking to retrieve items instead of lists:

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

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

        >>> iterable = [1, 2, 3, 4, 5]
        >>> head, iterable = spy(iterable, 10)
        >>> head
        [1, 2, 3, 4, 5]
        >>> list(iterable)
        [1, 2, 3, 4, 5]

    """
    it = iter(iterable)
    head = take(n, it)

    return head, chain(head, it)


def interleave(*iterables):
    """Return a new iterable yielding from each iterable in turn,
    until the shortest is exhausted.

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

    For a version that doesn't terminate after the shortest iterable is
    exhausted, see :func:`interleave_longest`.

    """
    return chain.from_iterable(zip(*iterables))


def interleave_longest(*iterables):
    """Return a new iterable yielding from each iterable in turn,
    skipping any that are exhausted.

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

    """
    i = chain.from_iterable(zip_longest(*iterables, fillvalue=_marker))
    return filter(lambda x: x is not _marker, i)


def collapse(iterable, base_type=None, levels=None):
    """Flatten an iterable with multiple levels of nesting (e.g., a list of
    lists of tuples) into non-iterable types.

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

    String types are not considered iterable and will not be collapsed.
    To avoid collapsing other types, specify *base_type*:

        >>> iterable = ['ab', ('cd', 'ef'), ['gh', 'ij']]
        >>> list(collapse(iterable, base_type=tuple))
        ['ab', ('cd', 'ef'), 'gh', 'ij']

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

    >>> iterable = [('a', ['b']), ('c', ['d'])]
    >>> list(collapse(iterable))  # Fully flattened
    ['a', 'b', 'c', 'd']
    >>> list(collapse(iterable, levels=1))  # Only one level flattened
    ['a', ['b'], 'c', ['d']]

    """
    def walk(node, level):
        if (
            ((levels is not None) and (level > levels)) or
            isinstance(node, string_types) or
            ((base_type is not None) and isinstance(node, base_type))
        ):
            yield node
            return

        try:
            tree = iter(node)
        except TypeError:
            yield node
            return
        else:
            for child in tree:
                for x in walk(child, level + 1):
                    yield x

    for x in walk(iterable, 0):
        yield x


def side_effect(func, iterable, chunk_size=None, before=None, after=None):
    """Invoke *func* on each item in *iterable* (or on each *chunk_size* group
    of items) before yielding the item.

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

    *before* and *after* are optional functions that take no arguments. They
    will be executed before iteration starts and after it ends, respectively.

    `side_effect` can be used for logging, updating progress bars, or anything
    that is not functionally "pure."

    Emitting a status message:

        >>> from more_itertools import consume
        >>> func = lambda item: print('Received {}'.format(item))
        >>> consume(side_effect(func, range(2)))
        Received 0
        Received 1

    Operating on chunks of items:

        >>> pair_sums = []
        >>> func = lambda chunk: pair_sums.append(sum(chunk))
        >>> list(side_effect(func, [0, 1, 2, 3, 4, 5], 2))
        [0, 1, 2, 3, 4, 5]
        >>> list(pair_sums)
        [1, 5, 9]

    Writing to a file-like object:

        >>> from io import StringIO
        >>> from more_itertools import consume
        >>> f = StringIO()
        >>> func = lambda x: print(x, file=f)
        >>> before = lambda: print(u'HEADER', file=f)
        >>> after = f.close
        >>> it = [u'a', u'b', u'c']
        >>> consume(side_effect(func, it, before=before, after=after))
        >>> f.closed
        True

    """
    try:
        if before is not None:
            before()

        if chunk_size is None:
            for item in iterable:
                func(item)
                yield item
        else:
            for chunk in chunked(iterable, chunk_size):
                func(chunk)
                for item in chunk:
                    yield item
    finally:
        if after is not None:
            after()


def sliced(seq, n):
    """Yield slices of length *n* from the sequence *seq*.

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

    If the length of the sequence is not divisible by the requested slice
    length, the last slice will be shorter.

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

    This function will only work for iterables that support slicing.
    For non-sliceable iterables, see :func:`chunked`.

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


def split_before(iterable, pred):
    """Yield lists of items from *iterable*, where each list starts with an
    item where callable *pred* returns ``True``:

        >>> list(split_before('OneTwo', lambda s: s.isupper()))
        [['O', 'n', 'e'], ['T', 'w', 'o']]

        >>> list(split_before(range(10), lambda n: n % 3 == 0))
        [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9]]

    """
    buf = []
    for item in iterable:
        if pred(item) and buf:
            yield buf
            buf = []
        buf.append(item)
    yield buf


def split_after(iterable, pred):
    """Yield lists of items from *iterable*, where each list ends with an
    item where callable *pred* returns ``True``:

        >>> list(split_after('one1two2', lambda s: s.isdigit()))
        [['o', 'n', 'e', '1'], ['t', 'w', 'o', '2']]

        >>> list(split_after(range(10), lambda n: n % 3 == 0))
        [[0], [1, 2, 3], [4, 5, 6], [7, 8, 9]]

    """
    buf = []
    for item in iterable:
        buf.append(item)
        if pred(item) and buf:
            yield buf
            buf = []
    if buf:
        yield buf


def padded(iterable, fillvalue=None, n=None, next_multiple=False):
    """Yield the elements from *iterable*, followed by *fillvalue*, such that
    at least *n* items are emitted.

        >>> list(padded([1, 2, 3], '?', 5))
        [1, 2, 3, '?', '?']

    If *next_multiple* is ``True``, *fillvalue* will be emitted until the
    number of items emitted is a multiple of *n*::

        >>> list(padded([1, 2, 3, 4], n=3, next_multiple=True))
        [1, 2, 3, 4, None, None]

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

    """
    it = iter(iterable)
    if n is None:
        for item in chain(it, repeat(fillvalue)):
            yield item
    elif n < 1:
        raise ValueError('n must be at least 1')
    else:
        item_count = 0
        for item in it:
            yield item
            item_count += 1

        remaining = (n - item_count) % n if next_multiple else n - item_count
        for _ in range(remaining):
            yield fillvalue


def distribute(n, iterable):
    """Distribute the items from *iterable* among *n* smaller iterables.

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

    If the length of *iterable* is not evenly divisible by *n*, then the
    length of the returned iterables will not be identical:

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

    If the length of *iterable* is smaller than *n*, then the last returned
    iterables will be empty:

        >>> children = distribute(5, [1, 2, 3])
        >>> [list(c) for c in children]
        [[1], [2], [3], [], []]

    This function uses :func:`itertools.tee` and may require significant
    storage. If you need the order items in the smaller iterables to match the
    original iterable, see :func:`divide`.

    """
    if n < 1:
        raise ValueError('n must be at least 1')

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


def stagger(iterable, offsets=(-1, 0, 1), longest=False, fillvalue=None):
    """Yield tuples whose elements are offset from *iterable*.
    The amount by which the `i`-th item in each tuple is offset is given by
    the `i`-th item in *offsets*.

        >>> list(stagger([0, 1, 2, 3]))
        [(None, 0, 1), (0, 1, 2), (1, 2, 3)]
        >>> list(stagger(range(8), offsets=(0, 2, 4)))
        [(0, 2, 4), (1, 3, 5), (2, 4, 6), (3, 5, 7)]

    By default, the sequence will end when the final element of a tuple is the
    last item in the iterable. To continue until the first element of a tuple
    is the last item in the iterable, set *longest* to ``True``::

        >>> list(stagger([0, 1, 2, 3], longest=True))
        [(None, 0, 1), (0, 1, 2), (1, 2, 3), (2, 3, None), (3, None, None)]

    By default, ``None`` will be used to replace offsets beyond the end of the
    sequence. Specify *fillvalue* to use some other value.

    """
    children = tee(iterable, len(offsets))

    return zip_offset(
        *children, offsets=offsets, longest=longest, fillvalue=fillvalue
    )


def zip_offset(*iterables, **kwargs):
    """``zip`` the input *iterables* together, but offset the `i`-th iterable
    by the `i`-th item in *offsets*.

        >>> list(zip_offset('0123', 'abcdef', offsets=(0, 1)))
        [('0', 'b'), ('1', 'c'), ('2', 'd'), ('3', 'e')]

    This can be used as a lightweight alternative to SciPy or pandas to analyze
    data sets in which 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.

    """
    offsets = kwargs['offsets']
    longest = kwargs.get('longest', False)
    fillvalue = kwargs.get('fillvalue', None)

    if len(iterables) != len(offsets):
        raise ValueError("Number of iterables and offsets didn't match")

    staggered = []
    for it, n in zip(iterables, offsets):
        if n < 0:
            staggered.append(chain(repeat(fillvalue, -n), it))
        elif n > 0:
            staggered.append(islice(it, n, None))
        else:
            staggered.append(it)

    if longest:
        return zip_longest(*staggered, fillvalue=fillvalue)

    return zip(*staggered)


def sort_together(iterables, key_list=(0,), reverse=False):
    """Return the input iterables sorted together, with *key_list* as the
    priority for sorting. All iterables are trimmed to the length of the
    shortest one.

    This can be used like the sorting function in a spreadsheet. If each
    iterable represents a column of data, the key list determines which
    columns are used for sorting.

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

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

    Set a different key list to sort according to another iterable.
    Specifying 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')]

    """
    return list(zip(*sorted(zip(*iterables),
                            key=itemgetter(*key_list),
                            reverse=reverse)))


def divide(n, iterable):
    """Divide the elements from *iterable* into *n* parts, maintaining
    order.

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

    If the length of *iterable* is not evenly divisible by *n*, then the
    length of the returned iterables will not be identical:

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

    If the length of the iterable is smaller than n, then the last returned
    iterables will be empty:

        >>> children = divide(5, [1, 2, 3])
        >>> [list(c) for c in children]
        [[1], [2], [3], [], []]

    This function will exhaust the iterable before returning and may require
    significant storage. If order is not important, see :func:`distribute`,
    which does not first pull the iterable into memory.

    """
    if n < 1:
        raise ValueError('n must be at least 1')

    seq = tuple(iterable)
    q, r = divmod(len(seq), n)

    ret = []
    for i in range(n):
        start = (i * q) + (i if i < r else r)
        stop = ((i + 1) * q) + (i + 1 if i + 1 < r else r)
        ret.append(iter(seq[start:stop]))

    return ret


def always_iterable(obj, base_type=(text_type, binary_type)):
    """If *obj* is iterable, return an iterator over its items::

        >>> obj = (1, 2, 3)
        >>> list(always_iterable(obj))
        [1, 2, 3]

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

        >>> obj = 1
        >>> list(always_iterable(obj))
        [1]


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

        >>> obj = None
        >>> list(always_iterable(None))
        []

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

        >>> obj = 'foo'
        >>> list(always_iterable(obj))
        ['foo']

    If *base_type* is set, objects for which ``isinstance(obj, base_type)``
    returns ``True`` won't be considered iterable.

        >>> obj = {'a': 1}
        >>> list(always_iterable(obj))  # Iterate over the dict's keys
        ['a']
        >>> list(always_iterable(obj, base_type=dict))  # Treat dicts as a unit
        [{'a': 1}]

    Set *base_type* to ``None`` to avoid any special handling and treat objects
    Python considers iterable as iterable:

        >>> obj = 'foo'
        >>> list(always_iterable(obj, base_type=None))
        ['f', 'o', 'o']


    """
    if obj is None:
        return iter(())

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

    try:
        return iter(obj)
    except TypeError:
        return iter((obj,))


def adjacent(predicate, iterable, distance=1):
    """Return an iterable over `(bool, item)` tuples where the `item` is
    drawn from *iterable* and the `bool` indicates whether
    that item satisfies the *predicate* or is adjacent to an item that does.

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

        >>> list(adjacent(lambda x: x == 3, range(6)))
        [(False, 0), (False, 1), (True, 2), (True, 3), (True, 4), (False, 5)]

    Set *distance* to change what counts as adjacent. For example, to find
    whether items are two places away from a ``3``:

        >>> list(adjacent(lambda x: x == 3, range(6), distance=2))
        [(False, 0), (True, 1), (True, 2), (True, 3), (True, 4), (True, 5)]

    This is useful for contextualizing the results of a search function.
    For example, a code comparison tool might want to identify lines that
    have changed, but also surrounding lines to give the viewer of the diff
    context.

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

    See also :func:`groupby_transform`, which can be used with this function
    to group ranges of items with the same `bool` value.

    """
    # Allow distance=0 mainly for testing that it reproduces results with map()
    if distance < 0:
        raise ValueError('distance must be at least 0')

    i1, i2 = tee(iterable)
    padding = [False] * distance
    selected = chain(padding, map(predicate, i1), padding)
    adjacent_to_selected = map(any, windowed(selected, 2 * distance + 1))
    return zip(adjacent_to_selected, i2)


def groupby_transform(iterable, keyfunc=None, valuefunc=None):
    """An extension of :func:`itertools.groupby` that transforms the values of
    *iterable* after grouping them.
    *keyfunc* is a function used to compute a grouping key for each item.
    *valuefunc* is a function for transforming the items after grouping.

        >>> iterable = 'AaaABbBCcA'
        >>> keyfunc = lambda x: x.upper()
        >>> valuefunc = lambda x: x.lower()
        >>> grouper = groupby_transform(iterable, keyfunc, valuefunc)
        >>> [(k, ''.join(g)) for k, g in grouper]
        [('A', 'aaaa'), ('B', 'bbb'), ('C', 'cc'), ('A', 'a')]

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

    :func:`groupby_transform` is useful when grouping elements of an iterable
    using a separate iterable as the key. To do this, :func:`zip` the iterables
    and pass a *keyfunc* that extracts the first element and a *valuefunc*
    that extracts the second element::

        >>> from operator import itemgetter
        >>> keys = [0, 0, 1, 1, 1, 2, 2, 2, 3]
        >>> values = 'abcdefghi'
        >>> iterable = zip(keys, values)
        >>> grouper = groupby_transform(iterable, itemgetter(0), itemgetter(1))
        >>> [(k, ''.join(g)) for k, g in grouper]
        [(0, 'ab'), (1, 'cde'), (2, 'fgh'), (3, 'i')]

    """
    valuefunc = (lambda x: x) if valuefunc is None else valuefunc
    return ((k, map(valuefunc, g)) for k, g in groupby(iterable, keyfunc))


def numeric_range(*args):
    """An extension of the built-in ``range()`` function whose arguments can
    be any orderable numeric type.

    With only *stop* specified, *start* defaults to ``0`` and *step*
    defaults to ``1``. The output items will match the type of *stop*:

        >>> list(numeric_range(3.5))
        [0.0, 1.0, 2.0, 3.0]

    With only *start* and *stop* specified, *step* defaults to ``1``. The
    output items will match the type of *start*:

        >>> from decimal import Decimal
        >>> start = Decimal('2.1')
        >>> stop = Decimal('5.1')
        >>> list(numeric_range(start, stop))
        [Decimal('2.1'), Decimal('3.1'), Decimal('4.1')]

    With *start*, *stop*, and *step*  specified the output items will match
    the type of ``start + step``:

        >>> from fractions import Fraction
        >>> start = Fraction(1, 2)  # Start at 1/2
        >>> stop = Fraction(5, 2)  # End at 5/2
        >>> step = Fraction(1, 2)  # Count by 1/2
        >>> list(numeric_range(start, stop, step))
        [Fraction(1, 2), Fraction(1, 1), Fraction(3, 2), Fraction(2, 1)]

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

        >>> list(numeric_range(3, -1, -1.0))
        [3.0, 2.0, 1.0, 0.0]

    Be aware of the limitations of floating point numbers; the representation
    of the yielded numbers may be surprising.

    """
    argc = len(args)
    if argc == 1:
        stop, = args
        start = type(stop)(0)
        step = 1
    elif argc == 2:
        start, stop = args
        step = 1
    elif argc == 3:
        start, stop, step = args
    else:
        err_msg = 'numeric_range takes at most 3 arguments, got {}'
        raise TypeError(err_msg.format(argc))

    values = (start + (step * n) for n in count())
    if step > 0:
        return takewhile(partial(gt, stop), values)
    elif step < 0:
        return takewhile(partial(lt, stop), values)
    else:
        raise ValueError('numeric_range arg 3 must not be zero')


def count_cycle(iterable, n=None):
    """Cycle through the items from *iterable* up to *n* times, yielding
    the number of completed cycles along with each item. If *n* is omitted the
    process repeats indefinitely.

    >>> list(count_cycle('AB', 3))
    [(0, 'A'), (0, 'B'), (1, 'A'), (1, 'B'), (2, 'A'), (2, 'B')]

    """
    iterable = tuple(iterable)
    if not iterable:
        return iter(())
    counter = count() if n is None else range(n)
    return ((i, item) for i in counter for item in iterable)


def locate(iterable, pred=bool):
    """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]

    Use with :func:`windowed` to find the indexes of a sub-sequence:

        >>> from more_itertools import windowed
        >>> iterable = [0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3]
        >>> sub = [1, 2, 3]
        >>> pred = lambda w: w == tuple(sub)  # windowed() returns tuples
        >>> list(locate(windowed(iterable, len(sub)), pred=pred))
        [1, 5, 9]

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


def lstrip(iterable, pred):
    """Yield the items from *iterable*, but strip any from the beginning
    for which *pred* returns ``True``.

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

        >>> iterable = (None, False, None, 1, 2, None, 3, False, None)
        >>> pred = lambda x: x in {None, False, ''}
        >>> list(lstrip(iterable, pred))
        [1, 2, None, 3, False, None]

    This function is analogous to to :func:`str.lstrip`, and is essentially
    an wrapper for :func:`itertools.dropwhile`.

    """
    return dropwhile(pred, iterable)


def rstrip(iterable, pred):
    """Yield the items from *iterable*, but strip any from the end
    for which *pred* returns ``True``.

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

        >>> iterable = (None, False, None, 1, 2, None, 3, False, None)
        >>> pred = lambda x: x in {None, False, ''}
        >>> list(rstrip(iterable, pred))
        [None, False, None, 1, 2, None, 3]

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

    """
    cache = []
    cache_append = cache.append
    for x in iterable:
        if pred(x):
            cache_append(x)
        else:
            for y in cache:
                yield y
            del cache[:]
            yield x


def strip(iterable, pred):
    """Yield the items from *iterable*, but strip any from the
    beginning and end for which *pred* returns ``True``.

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

        >>> iterable = (None, False, None, 1, 2, None, 3, False, None)
        >>> pred = lambda x: x in {None, False, ''}
        >>> list(strip(iterable, pred))
        [1, 2, None, 3]

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

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


def islice_extended(iterable, *args):
    """An extension of :func:`itertools.islice` that supports negative values
    for *stop*, *start*, and *step*.

        >>> iterable = iter('abcdefgh')
        >>> list(islice_extended(iterable, -4, -1))
        ['e', 'f', 'g']

    Slices with negative values require some caching of *iterable*, but this
    function takes care to minimize the amount of memory required.

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

        >>> from itertools import count
        >>> list(islice_extended(count(), 110, 99, -2))
        [110, 108, 106, 104, 102, 100]

    """
    s = slice(*args)
    start = s.start
    stop = s.stop
    if s.step == 0:
        raise ValueError('step argument must be a non-zero integer or None.')
    step = s.step or 1

    it = iter(iterable)

    if step > 0:
        start = 0 if (start is None) else start

        if (start < 0):
            # Consume all but the last -start items
            cache = deque(enumerate(it, 1), maxlen=-start)
            len_iter = cache[-1][0] if cache else 0

            # Adjust start to be positive
            i = max(len_iter + start, 0)

            # Adjust stop to be positive
            if stop is None:
                j = len_iter
            elif stop >= 0:
                j = min(stop, len_iter)
            else:
                j = max(len_iter + stop, 0)

            # Slice the cache
            n = j - i
            if n <= 0:
                return

            for index, item in islice(cache, 0, n, step):
                yield item
        elif (stop is not None) and (stop < 0):
            # Advance to the start position
            next(islice(it, start, start), None)

            # When stop is negative, we have to carry -stop items while
            # iterating
            cache = deque(islice(it, -stop), maxlen=-stop)

            for index, item in enumerate(it):
                cached_item = cache.popleft()
                if index % step == 0:
                    yield cached_item
                cache.append(item)
        else:
            # When both start and stop are positive we have the normal case
            for item in islice(it, start, stop, step):
                yield item
    else:
        start = -1 if (start is None) else start

        if (stop is not None) and (stop < 0):
            # Consume all but the last items
            n = -stop - 1
            cache = deque(enumerate(it, 1), maxlen=n)
            len_iter = cache[-1][0] if cache else 0

            # If start and stop are both negative they are comparable and
            # we can just slice. Otherwise we can adjust start to be negative
            # and then slice.
            if start < 0:
                i, j = start, stop
            else:
                i, j = min(start - len_iter, -1), None

            for index, item in list(cache)[i:j:step]:
                yield item
        else:
            # Advance to the stop position
            if stop is not None:
                m = stop + 1
                next(islice(it, m, m), None)

            # stop is positive, so if start is negative they are not comparable
            # and we need the rest of the items.
            if start < 0:
                i = start
                n = None
            # stop is None and start is positive, so we just need items up to
            # the start index.
            elif stop is None:
                i = None
                n = start + 1
            # Both stop and start are positive, so they are comparable.
            else:
                i = None
                n = start - stop
                if n <= 0:
                    return

            cache = list(islice(it, n))

            for item in cache[i::step]:
                yield item


def consecutive_groups(iterable, ordering=lambda x: x):
    """Yield groups of consecutive items using :func:`itertools.groupby`.
    The *ordering* function determines whether two items are adjacent by
    returning their position.

    By default, the ordering function is the identity function. This is
    suitable for finding runs of numbers:

        >>> iterable = [1, 10, 11, 12, 20, 30, 31, 32, 33, 40]
        >>> for group in consecutive_groups(iterable):
        ...     print(list(group))
        [1]
        [10, 11, 12]
        [20]
        [30, 31, 32, 33]
        [40]

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

        >>> from string import ascii_lowercase
        >>> iterable = 'abcdfgilmnop'
        >>> ordering = ascii_lowercase.index
        >>> for group in consecutive_groups(iterable, ordering):
        ...     print(list(group))
        ['a', 'b', 'c', 'd']
        ['f', 'g']
        ['i']
        ['l', 'm', 'n', 'o', 'p']

    """
    for k, g in groupby(
        enumerate(iterable), key=lambda x: x[0] - ordering(x[1])
    ):
        yield map(itemgetter(1), g)


class run_length(object):
    """
    :func:`run_length.encode` compresses an iterable with run-length encoding.
    It yields groups of repeated items with the count of how many times they
    were repeated:

        >>> uncompressed = 'abbcccdddd'
        >>> list(run_length.encode(uncompressed))
        [('a', 1), ('b', 2), ('c', 3), ('d', 4)]

    :func:`run_length.decode` decompresses an iterable that was previously
    compressed with run-length encoding. It yields the items of the
    decompressed iterable:

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

    """

    @staticmethod
    def encode(iterable):
        return ((k, ilen(g)) for k, g in groupby(iterable))

    @staticmethod
    def decode(iterable):
        return chain.from_iterable(repeat(k, n) for k, n in iterable)