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nzbToMedia/libs/common/more_itertools/more.py
Labrys of Knossos 684cca8c9b Updated more-itertools to 5.0.0
2022-11-29 01:26:47 -05:00

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from __future__ import print_function
from collections import Counter, defaultdict, deque
from functools import partial, wraps
from heapq import merge
from itertools import (
chain,
compress,
count,
cycle,
dropwhile,
groupby,
islice,
repeat,
starmap,
takewhile,
tee
)
from operator import itemgetter, lt, gt, sub
from sys import maxsize, version_info
try:
from collections.abc import Sequence
except ImportError:
from collections import Sequence
from six import binary_type, string_types, text_type
from six.moves import filter, map, range, zip, zip_longest
from .recipes import consume, flatten, take
__all__ = [
'adjacent',
'always_iterable',
'always_reversible',
'bucket',
'chunked',
'circular_shifts',
'collapse',
'collate',
'consecutive_groups',
'consumer',
'count_cycle',
'difference',
'distinct_permutations',
'distribute',
'divide',
'exactly_n',
'first',
'groupby_transform',
'ilen',
'interleave_longest',
'interleave',
'intersperse',
'islice_extended',
'iterate',
'last',
'locate',
'lstrip',
'make_decorator',
'map_reduce',
'numeric_range',
'one',
'padded',
'peekable',
'replace',
'rlocate',
'rstrip',
'run_length',
'seekable',
'SequenceView',
'side_effect',
'sliced',
'sort_together',
'split_at',
'split_after',
'split_before',
'split_into',
'spy',
'stagger',
'strip',
'substrings',
'unique_to_each',
'unzip',
'windowed',
'with_iter',
'zip_offset',
]
_marker = object()
def chunked(iterable, n):
"""Break *iterable* into lists of length *n*:
>>> list(chunked([1, 2, 3, 4, 5, 6], 3))
[[1, 2, 3], [4, 5, 6]]
If the length of *iterable* is not evenly divisible by *n*, the last
returned list will be shorter:
>>> list(chunked([1, 2, 3, 4, 5, 6, 7, 8], 3))
[[1, 2, 3], [4, 5, 6], [7, 8]]
To use a fill-in value instead, see the :func:`grouper` recipe.
:func:`chunked` is useful for splitting up a computation on a large number
of keys into batches, to be pickled and sent off to worker processes. One
example is operations on rows in MySQL, which does not implement
server-side cursors properly and would otherwise load the entire dataset
into RAM on the client.
"""
return iter(partial(take, n, iter(iterable)), [])
def first(iterable, default=_marker):
"""Return the first item of *iterable*, or *default* if *iterable* is
empty.
>>> first([0, 1, 2, 3])
0
>>> first([], 'some default')
'some default'
If *default* is not provided and there are no items in the iterable,
raise ``ValueError``.
:func:`first` is useful when you have a generator of expensive-to-retrieve
values and want any arbitrary one. It is marginally shorter than
``next(iter(iterable), default)``.
"""
try:
return next(iter(iterable))
except StopIteration:
# I'm on the edge about raising ValueError instead of StopIteration. At
# the moment, ValueError wins, because the caller could conceivably
# want to do something different with flow control when I raise the
# exception, and it's weird to explicitly catch StopIteration.
if default is _marker:
raise ValueError('first() was called on an empty iterable, and no '
'default value was provided.')
return default
def last(iterable, default=_marker):
"""Return the last item of *iterable*, or *default* if *iterable* is
empty.
>>> last([0, 1, 2, 3])
3
>>> last([], 'some default')
'some default'
If *default* is not provided and there are no items in the iterable,
raise ``ValueError``.
"""
try:
try:
# Try to access the last item directly
return iterable[-1]
except (TypeError, AttributeError, KeyError):
# If not slice-able, iterate entirely using length-1 deque
return deque(iterable, maxlen=1)[0]
except IndexError: # If the iterable was empty
if default is _marker:
raise ValueError('last() was called on an empty iterable, and no '
'default value was provided.')
return default
class peekable(object):
"""Wrap an iterator to allow lookahead and prepending elements.
Call :meth:`peek` on the result to get the value that will be returned
by :func:`next`. This won't advance the iterator:
>>> p = peekable(['a', 'b'])
>>> p.peek()
'a'
>>> next(p)
'a'
Pass :meth:`peek` a default value to return that instead of raising
``StopIteration`` when the iterator is exhausted.
>>> p = peekable([])
>>> p.peek('hi')
'hi'
peekables also offer a :meth:`prepend` method, which "inserts" items
at the head of the iterable:
>>> p = peekable([1, 2, 3])
>>> p.prepend(10, 11, 12)
>>> next(p)
10
>>> p.peek()
11
>>> list(p)
[11, 12, 1, 2, 3]
peekables can be indexed. Index 0 is the item that will be returned by
:func:`next`, index 1 is the item after that, and so on:
The values up to the given index will be cached.
>>> p = peekable(['a', 'b', 'c', 'd'])
>>> p[0]
'a'
>>> p[1]
'b'
>>> next(p)
'a'
Negative indexes are supported, but be aware that they will cache the
remaining items in the source iterator, which may require significant
storage.
To check whether a peekable is exhausted, check its truth value:
>>> p = peekable(['a', 'b'])
>>> if p: # peekable has items
... list(p)
['a', 'b']
>>> if not p: # peekable is exhaused
... list(p)
[]
"""
def __init__(self, iterable):
self._it = iter(iterable)
self._cache = deque()
def __iter__(self):
return self
def __bool__(self):
try:
self.peek()
except StopIteration:
return False
return True
def __nonzero__(self):
# For Python 2 compatibility
return self.__bool__()
def peek(self, default=_marker):
"""Return the item that will be next returned from ``next()``.
Return ``default`` if there are no items left. If ``default`` is not
provided, raise ``StopIteration``.
"""
if not self._cache:
try:
self._cache.append(next(self._it))
except StopIteration:
if default is _marker:
raise
return default
return self._cache[0]
def prepend(self, *items):
"""Stack up items to be the next ones returned from ``next()`` or
``self.peek()``. The items will be returned in
first in, first out order::
>>> p = peekable([1, 2, 3])
>>> p.prepend(10, 11, 12)
>>> next(p)
10
>>> list(p)
[11, 12, 1, 2, 3]
It is possible, by prepending items, to "resurrect" a peekable that
previously raised ``StopIteration``.
>>> p = peekable([])
>>> next(p)
Traceback (most recent call last):
...
StopIteration
>>> p.prepend(1)
>>> next(p)
1
>>> next(p)
Traceback (most recent call last):
...
StopIteration
"""
self._cache.extendleft(reversed(items))
def __next__(self):
if self._cache:
return self._cache.popleft()
return next(self._it)
next = __next__ # For Python 2 compatibility
def _get_slice(self, index):
# Normalize the slice's arguments
step = 1 if (index.step is None) else index.step
if step > 0:
start = 0 if (index.start is None) else index.start
stop = maxsize if (index.stop is None) else index.stop
elif step < 0:
start = -1 if (index.start is None) else index.start
stop = (-maxsize - 1) if (index.stop is None) else index.stop
else:
raise ValueError('slice step cannot be zero')
# If either the start or stop index is negative, we'll need to cache
# the rest of the iterable in order to slice from the right side.
if (start < 0) or (stop < 0):
self._cache.extend(self._it)
# Otherwise we'll need to find the rightmost index and cache to that
# point.
else:
n = min(max(start, stop) + 1, maxsize)
cache_len = len(self._cache)
if n >= cache_len:
self._cache.extend(islice(self._it, n - cache_len))
return list(self._cache)[index]
def __getitem__(self, index):
if isinstance(index, slice):
return self._get_slice(index)
cache_len = len(self._cache)
if index < 0:
self._cache.extend(self._it)
elif index >= cache_len:
self._cache.extend(islice(self._it, index + 1 - cache_len))
return self._cache[index]
def _collate(*iterables, **kwargs):
"""Helper for ``collate()``, called when the user is using the ``reverse``
or ``key`` keyword arguments on Python versions below 3.5.
"""
key = kwargs.pop('key', lambda a: a)
reverse = kwargs.pop('reverse', False)
min_or_max = partial(max if reverse else min, key=itemgetter(0))
peekables = [peekable(it) for it in iterables]
peekables = [p for p in peekables if p] # Kill empties.
while peekables:
_, p = min_or_max((key(p.peek()), p) for p in peekables)
yield next(p)
peekables = [x for x in peekables if x]
def collate(*iterables, **kwargs):
"""Return a sorted merge of the items from each of several already-sorted
*iterables*.
>>> list(collate('ACDZ', 'AZ', 'JKL'))
['A', 'A', 'C', 'D', 'J', 'K', 'L', 'Z', 'Z']
Works lazily, keeping only the next value from each iterable in memory. Use
:func:`collate` to, for example, perform a n-way mergesort of items that
don't fit in memory.
If a *key* function is specified, the iterables will be sorted according
to its result:
>>> key = lambda s: int(s) # Sort by numeric value, not by string
>>> list(collate(['1', '10'], ['2', '11'], key=key))
['1', '2', '10', '11']
If the *iterables* are sorted in descending order, set *reverse* to
``True``:
>>> list(collate([5, 3, 1], [4, 2, 0], reverse=True))
[5, 4, 3, 2, 1, 0]
If the elements of the passed-in iterables are out of order, you might get
unexpected results.
On Python 2.7, this function delegates to :func:`heapq.merge` if neither
of the keyword arguments are specified. On Python 3.5+, this function
is an alias for :func:`heapq.merge`.
"""
if not kwargs:
return merge(*iterables)
return _collate(*iterables, **kwargs)
# If using Python version 3.5 or greater, heapq.merge() will be faster than
# collate - use that instead.
if version_info >= (3, 5, 0):
_collate_docstring = collate.__doc__
collate = partial(merge)
collate.__doc__ = _collate_docstring
def consumer(func):
"""Decorator that automatically advances a PEP-342-style "reverse iterator"
to its first yield point so you don't have to call ``next()`` on it
manually.
>>> @consumer
... def tally():
... i = 0
... while True:
... print('Thing number %s is %s.' % (i, (yield)))
... i += 1
...
>>> t = tally()
>>> t.send('red')
Thing number 0 is red.
>>> t.send('fish')
Thing number 1 is fish.
Without the decorator, you would have to call ``next(t)`` before
``t.send()`` could be used.
"""
@wraps(func)
def wrapper(*args, **kwargs):
gen = func(*args, **kwargs)
next(gen)
return gen
return wrapper
def ilen(iterable):
"""Return the number of items in *iterable*.
>>> ilen(x for x in range(1000000) if x % 3 == 0)
333334
This consumes the iterable, so handle with care.
"""
# This approach was selected because benchmarks showed it's likely the
# fastest of the known implementations at the time of writing.
# See GitHub tracker: #236, #230.
counter = count()
deque(zip(iterable, counter), maxlen=0)
return next(counter)
def iterate(func, start):
"""Return ``start``, ``func(start)``, ``func(func(start))``, ...
>>> from itertools import islice
>>> list(islice(iterate(lambda x: 2*x, 1), 10))
[1, 2, 4, 8, 16, 32, 64, 128, 256, 512]
"""
while True:
yield start
start = func(start)
def with_iter(context_manager):
"""Wrap an iterable in a ``with`` statement, so it closes once exhausted.
For example, this will close the file when the iterator is exhausted::
upper_lines = (line.upper() for line in with_iter(open('foo')))
Any context manager which returns an iterable is a candidate for
``with_iter``.
"""
with context_manager as iterable:
for item in iterable:
yield item
def one(iterable, too_short=None, too_long=None):
"""Return the first item from *iterable*, which is expected to contain only
that item. Raise an exception if *iterable* is empty or has more than one
item.
:func:`one` is useful for ensuring that an iterable contains only one item.
For example, it can be used to retrieve the result of a database query
that is expected to return a single row.
If *iterable* is empty, ``ValueError`` will be raised. You may specify a
different exception with the *too_short* keyword:
>>> it = []
>>> one(it) # doctest: +IGNORE_EXCEPTION_DETAIL
Traceback (most recent call last):
...
ValueError: too many items in iterable (expected 1)'
>>> too_short = IndexError('too few items')
>>> one(it, too_short=too_short) # doctest: +IGNORE_EXCEPTION_DETAIL
Traceback (most recent call last):
...
IndexError: too few items
Similarly, if *iterable* contains more than one item, ``ValueError`` will
be raised. You may specify a different exception with the *too_long*
keyword:
>>> it = ['too', 'many']
>>> one(it) # doctest: +IGNORE_EXCEPTION_DETAIL
Traceback (most recent call last):
...
ValueError: too many items in iterable (expected 1)'
>>> too_long = RuntimeError
>>> one(it, too_long=too_long) # doctest: +IGNORE_EXCEPTION_DETAIL
Traceback (most recent call last):
...
RuntimeError
Note that :func:`one` attempts to advance *iterable* twice to ensure there
is only one item. If there is more than one, both items will be discarded.
See :func:`spy` or :func:`peekable` to check iterable contents less
destructively.
"""
it = iter(iterable)
try:
value = next(it)
except StopIteration:
raise too_short or ValueError('too few items in iterable (expected 1)')
try:
next(it)
except StopIteration:
pass
else:
raise too_long or ValueError('too many items in iterable (expected 1)')
return value
def distinct_permutations(iterable):
"""Yield successive distinct permutations of the elements in *iterable*.
>>> sorted(distinct_permutations([1, 0, 1]))
[(0, 1, 1), (1, 0, 1), (1, 1, 0)]
Equivalent to ``set(permutations(iterable))``, except duplicates are not
generated and thrown away. For larger input sequences this is much more
efficient.
Duplicate permutations arise when there are duplicated elements in the
input iterable. The number of items returned is
`n! / (x_1! * x_2! * ... * x_n!)`, where `n` is the total number of
items input, and each `x_i` is the count of a distinct item in the input
sequence.
"""
def perm_unique_helper(item_counts, perm, i):
"""Internal helper function
:arg item_counts: Stores the unique items in ``iterable`` and how many
times they are repeated
:arg perm: The permutation that is being built for output
:arg i: The index of the permutation being modified
The output permutations are built up recursively; the distinct items
are placed until their repetitions are exhausted.
"""
if i < 0:
yield tuple(perm)
else:
for item in item_counts:
if item_counts[item] <= 0:
continue
perm[i] = item
item_counts[item] -= 1
for x in perm_unique_helper(item_counts, perm, i - 1):
yield x
item_counts[item] += 1
item_counts = Counter(iterable)
length = sum(item_counts.values())
return perm_unique_helper(item_counts, [None] * length, length - 1)
def intersperse(e, iterable, n=1):
"""Intersperse filler element *e* among the items in *iterable*, leaving
*n* items between each filler element.
>>> list(intersperse('!', [1, 2, 3, 4, 5]))
[1, '!', 2, '!', 3, '!', 4, '!', 5]
>>> list(intersperse(None, [1, 2, 3, 4, 5], n=2))
[1, 2, None, 3, 4, None, 5]
"""
if n == 0:
raise ValueError('n must be > 0')
elif n == 1:
# interleave(repeat(e), iterable) -> e, x_0, e, e, x_1, e, x_2...
# islice(..., 1, None) -> x_0, e, e, x_1, e, x_2...
return islice(interleave(repeat(e), iterable), 1, None)
else:
# interleave(filler, chunks) -> [e], [x_0, x_1], [e], [x_2, x_3]...
# islice(..., 1, None) -> [x_0, x_1], [e], [x_2, x_3]...
# flatten(...) -> x_0, x_1, e, x_2, x_3...
filler = repeat([e])
chunks = chunked(iterable, n)
return flatten(islice(interleave(filler, chunks), 1, None))
def unique_to_each(*iterables):
"""Return the elements from each of the input iterables that aren't in the
other input iterables.
For example, suppose you have a set of packages, each with a set of
dependencies::
{'pkg_1': {'A', 'B'}, 'pkg_2': {'B', 'C'}, 'pkg_3': {'B', 'D'}}
If you remove one package, which dependencies can also be removed?
If ``pkg_1`` is removed, then ``A`` is no longer necessary - it is not
associated with ``pkg_2`` or ``pkg_3``. Similarly, ``C`` is only needed for
``pkg_2``, and ``D`` is only needed for ``pkg_3``::
>>> unique_to_each({'A', 'B'}, {'B', 'C'}, {'B', 'D'})
[['A'], ['C'], ['D']]
If there are duplicates in one input iterable that aren't in the others
they will be duplicated in the output. Input order is preserved::
>>> unique_to_each("mississippi", "missouri")
[['p', 'p'], ['o', 'u', 'r']]
It is assumed that the elements of each iterable are hashable.
"""
pool = [list(it) for it in iterables]
counts = Counter(chain.from_iterable(map(set, pool)))
uniques = {element for element in counts if counts[element] == 1}
return [list(filter(uniques.__contains__, it)) for it in pool]
def windowed(seq, n, fillvalue=None, step=1):
"""Return a sliding window of width *n* over the given iterable.
>>> all_windows = windowed([1, 2, 3, 4, 5], 3)
>>> list(all_windows)
[(1, 2, 3), (2, 3, 4), (3, 4, 5)]
When the window is larger than the iterable, *fillvalue* is used in place
of missing values::
>>> list(windowed([1, 2, 3], 4))
[(1, 2, 3, None)]
Each window will advance in increments of *step*:
>>> list(windowed([1, 2, 3, 4, 5, 6], 3, fillvalue='!', step=2))
[(1, 2, 3), (3, 4, 5), (5, 6, '!')]
"""
if n < 0:
raise ValueError('n must be >= 0')
if n == 0:
yield tuple()
return
if step < 1:
raise ValueError('step must be >= 1')
it = iter(seq)
window = deque([], n)
append = window.append
# Initial deque fill
for _ in range(n):
append(next(it, fillvalue))
yield tuple(window)
# Appending new items to the right causes old items to fall off the left
i = 0
for item in it:
append(item)
i = (i + 1) % step
if i % step == 0:
yield tuple(window)
# If there are items from the iterable in the window, pad with the given
# value and emit them.
if (i % step) and (step - i < n):
for _ in range(step - i):
append(fillvalue)
yield tuple(window)
def substrings(iterable, join_func=None):
"""Yield all of the substrings of *iterable*.
>>> [''.join(s) for s in substrings('more')]
['m', 'o', 'r', 'e', 'mo', 'or', 're', 'mor', 'ore', 'more']
Note that non-string iterables can also be subdivided.
>>> list(substrings([0, 1, 2]))
[(0,), (1,), (2,), (0, 1), (1, 2), (0, 1, 2)]
"""
# The length-1 substrings
seq = []
for item in iter(iterable):
seq.append(item)
yield (item,)
seq = tuple(seq)
item_count = len(seq)
# And the rest
for n in range(2, item_count + 1):
for i in range(item_count - n + 1):
yield seq[i:i + n]
class bucket(object):
"""Wrap *iterable* and return an object that buckets it iterable into
child iterables based on a *key* function.
>>> iterable = ['a1', 'b1', 'c1', 'a2', 'b2', 'c2', 'b3']
>>> s = bucket(iterable, key=lambda x: x[0])
>>> a_iterable = s['a']
>>> next(a_iterable)
'a1'
>>> next(a_iterable)
'a2'
>>> list(s['b'])
['b1', 'b2', 'b3']
The original iterable will be advanced and its items will be cached until
they are used by the child iterables. This may require significant storage.
By default, attempting to select a bucket to which no items belong will
exhaust the iterable and cache all values.
If you specify a *validator* function, selected buckets will instead be
checked against it.
>>> from itertools import count
>>> it = count(1, 2) # Infinite sequence of odd numbers
>>> key = lambda x: x % 10 # Bucket by last digit
>>> validator = lambda x: x in {1, 3, 5, 7, 9} # Odd digits only
>>> s = bucket(it, key=key, validator=validator)
>>> 2 in s
False
>>> list(s[2])
[]
"""
def __init__(self, iterable, key, validator=None):
self._it = iter(iterable)
self._key = key
self._cache = defaultdict(deque)
self._validator = validator or (lambda x: True)
def __contains__(self, value):
if not self._validator(value):
return False
try:
item = next(self[value])
except StopIteration:
return False
else:
self._cache[value].appendleft(item)
return True
def _get_values(self, value):
"""
Helper to yield items from the parent iterator that match *value*.
Items that don't match are stored in the local cache as they
are encountered.
"""
while True:
# If we've cached some items that match the target value, emit
# the first one and evict it from the cache.
if self._cache[value]:
yield self._cache[value].popleft()
# Otherwise we need to advance the parent iterator to search for
# a matching item, caching the rest.
else:
while True:
try:
item = next(self._it)
except StopIteration:
return
item_value = self._key(item)
if item_value == value:
yield item
break
elif self._validator(item_value):
self._cache[item_value].append(item)
def __getitem__(self, value):
if not self._validator(value):
return iter(())
return self._get_values(value)
def spy(iterable, n=1):
"""Return a 2-tuple with a list containing the first *n* elements of
*iterable*, and an iterator with the same items as *iterable*.
This allows you to "look ahead" at the items in the iterable without
advancing it.
There is one item in the list by default:
>>> iterable = 'abcdefg'
>>> head, iterable = spy(iterable)
>>> head
['a']
>>> list(iterable)
['a', 'b', 'c', 'd', 'e', 'f', 'g']
You may use unpacking to retrieve items instead of lists:
>>> (head,), iterable = spy('abcdefg')
>>> head
'a'
>>> (first, second), iterable = spy('abcdefg', 2)
>>> first
'a'
>>> second
'b'
The number of items requested can be larger than the number of items in
the iterable:
>>> iterable = [1, 2, 3, 4, 5]
>>> head, iterable = spy(iterable, 10)
>>> head
[1, 2, 3, 4, 5]
>>> list(iterable)
[1, 2, 3, 4, 5]
"""
it = iter(iterable)
head = take(n, it)
return head, chain(head, it)
def interleave(*iterables):
"""Return a new iterable yielding from each iterable in turn,
until the shortest is exhausted.
>>> list(interleave([1, 2, 3], [4, 5], [6, 7, 8]))
[1, 4, 6, 2, 5, 7]
For a version that doesn't terminate after the shortest iterable is
exhausted, see :func:`interleave_longest`.
"""
return chain.from_iterable(zip(*iterables))
def interleave_longest(*iterables):
"""Return a new iterable yielding from each iterable in turn,
skipping any that are exhausted.
>>> list(interleave_longest([1, 2, 3], [4, 5], [6, 7, 8]))
[1, 4, 6, 2, 5, 7, 3, 8]
This function produces the same output as :func:`roundrobin`, but may
perform better for some inputs (in particular when the number of iterables
is large).
"""
i = chain.from_iterable(zip_longest(*iterables, fillvalue=_marker))
return (x for x in i if x is not _marker)
def collapse(iterable, base_type=None, levels=None):
"""Flatten an iterable with multiple levels of nesting (e.g., a list of
lists of tuples) into non-iterable types.
>>> iterable = [(1, 2), ([3, 4], [[5], [6]])]
>>> list(collapse(iterable))
[1, 2, 3, 4, 5, 6]
String types are not considered iterable and will not be collapsed.
To avoid collapsing other types, specify *base_type*:
>>> iterable = ['ab', ('cd', 'ef'), ['gh', 'ij']]
>>> list(collapse(iterable, base_type=tuple))
['ab', ('cd', 'ef'), 'gh', 'ij']
Specify *levels* to stop flattening after a certain level:
>>> iterable = [('a', ['b']), ('c', ['d'])]
>>> list(collapse(iterable)) # Fully flattened
['a', 'b', 'c', 'd']
>>> list(collapse(iterable, levels=1)) # Only one level flattened
['a', ['b'], 'c', ['d']]
"""
def walk(node, level):
if (
((levels is not None) and (level > levels)) or
isinstance(node, string_types) or
((base_type is not None) and isinstance(node, base_type))
):
yield node
return
try:
tree = iter(node)
except TypeError:
yield node
return
else:
for child in tree:
for x in walk(child, level + 1):
yield x
for x in walk(iterable, 0):
yield x
def side_effect(func, iterable, chunk_size=None, before=None, after=None):
"""Invoke *func* on each item in *iterable* (or on each *chunk_size* group
of items) before yielding the item.
`func` must be a function that takes a single argument. Its return value
will be discarded.
*before* and *after* are optional functions that take no arguments. They
will be executed before iteration starts and after it ends, respectively.
`side_effect` can be used for logging, updating progress bars, or anything
that is not functionally "pure."
Emitting a status message:
>>> from more_itertools import consume
>>> func = lambda item: print('Received {}'.format(item))
>>> consume(side_effect(func, range(2)))
Received 0
Received 1
Operating on chunks of items:
>>> pair_sums = []
>>> func = lambda chunk: pair_sums.append(sum(chunk))
>>> list(side_effect(func, [0, 1, 2, 3, 4, 5], 2))
[0, 1, 2, 3, 4, 5]
>>> list(pair_sums)
[1, 5, 9]
Writing to a file-like object:
>>> from io import StringIO
>>> from more_itertools import consume
>>> f = StringIO()
>>> func = lambda x: print(x, file=f)
>>> before = lambda: print(u'HEADER', file=f)
>>> after = f.close
>>> it = [u'a', u'b', u'c']
>>> consume(side_effect(func, it, before=before, after=after))
>>> f.closed
True
"""
try:
if before is not None:
before()
if chunk_size is None:
for item in iterable:
func(item)
yield item
else:
for chunk in chunked(iterable, chunk_size):
func(chunk)
for item in chunk:
yield item
finally:
if after is not None:
after()
def sliced(seq, n):
"""Yield slices of length *n* from the sequence *seq*.
>>> list(sliced((1, 2, 3, 4, 5, 6), 3))
[(1, 2, 3), (4, 5, 6)]
If the length of the sequence is not divisible by the requested slice
length, the last slice will be shorter.
>>> list(sliced((1, 2, 3, 4, 5, 6, 7, 8), 3))
[(1, 2, 3), (4, 5, 6), (7, 8)]
This function will only work for iterables that support slicing.
For non-sliceable iterables, see :func:`chunked`.
"""
return takewhile(bool, (seq[i: i + n] for i in count(0, n)))
def split_at(iterable, pred):
"""Yield lists of items from *iterable*, where each list is delimited by
an item where callable *pred* returns ``True``. The lists do not include
the delimiting items.
>>> list(split_at('abcdcba', lambda x: x == 'b'))
[['a'], ['c', 'd', 'c'], ['a']]
>>> list(split_at(range(10), lambda n: n % 2 == 1))
[[0], [2], [4], [6], [8], []]
"""
buf = []
for item in iterable:
if pred(item):
yield buf
buf = []
else:
buf.append(item)
yield buf
def split_before(iterable, pred):
"""Yield lists of items from *iterable*, where each list starts with an
item where callable *pred* returns ``True``:
>>> list(split_before('OneTwo', lambda s: s.isupper()))
[['O', 'n', 'e'], ['T', 'w', 'o']]
>>> list(split_before(range(10), lambda n: n % 3 == 0))
[[0, 1, 2], [3, 4, 5], [6, 7, 8], [9]]
"""
buf = []
for item in iterable:
if pred(item) and buf:
yield buf
buf = []
buf.append(item)
yield buf
def split_after(iterable, pred):
"""Yield lists of items from *iterable*, where each list ends with an
item where callable *pred* returns ``True``:
>>> list(split_after('one1two2', lambda s: s.isdigit()))
[['o', 'n', 'e', '1'], ['t', 'w', 'o', '2']]
>>> list(split_after(range(10), lambda n: n % 3 == 0))
[[0], [1, 2, 3], [4, 5, 6], [7, 8, 9]]
"""
buf = []
for item in iterable:
buf.append(item)
if pred(item) and buf:
yield buf
buf = []
if buf:
yield buf
def split_into(iterable, sizes):
"""Yield a list of sequential items from *iterable* of length 'n' for each
integer 'n' in *sizes*.
>>> list(split_into([1,2,3,4,5,6], [1,2,3]))
[[1], [2, 3], [4, 5, 6]]
If the sum of *sizes* is smaller than the length of *iterable*, then the
remaining items of *iterable* will not be returned.
>>> list(split_into([1,2,3,4,5,6], [2,3]))
[[1, 2], [3, 4, 5]]
If the sum of *sizes* is larger than the length of *iterable*, fewer items
will be returned in the iteration that overruns *iterable* and further
lists will be empty:
>>> list(split_into([1,2,3,4], [1,2,3,4]))
[[1], [2, 3], [4], []]
When a ``None`` object is encountered in *sizes*, the returned list will
contain items up to the end of *iterable* the same way that itertools.slice
does:
>>> list(split_into([1,2,3,4,5,6,7,8,9,0], [2,3,None]))
[[1, 2], [3, 4, 5], [6, 7, 8, 9, 0]]
:func:`split_into` can be useful for grouping a series of items where the
sizes of the groups are not uniform. An example would be where in a row
from a table, multiple columns represent elements of the same feature
(e.g. a point represented by x,y,z) but, the format is not the same for
all columns.
"""
# convert the iterable argument into an iterator so its contents can
# be consumed by islice in case it is a generator
it = iter(iterable)
for size in sizes:
if size is None:
yield list(it)
return
else:
yield list(islice(it, size))
def padded(iterable, fillvalue=None, n=None, next_multiple=False):
"""Yield the elements from *iterable*, followed by *fillvalue*, such that
at least *n* items are emitted.
>>> list(padded([1, 2, 3], '?', 5))
[1, 2, 3, '?', '?']
If *next_multiple* is ``True``, *fillvalue* will be emitted until the
number of items emitted is a multiple of *n*::
>>> list(padded([1, 2, 3, 4], n=3, next_multiple=True))
[1, 2, 3, 4, None, None]
If *n* is ``None``, *fillvalue* will be emitted indefinitely.
"""
it = iter(iterable)
if n is None:
for item in chain(it, repeat(fillvalue)):
yield item
elif n < 1:
raise ValueError('n must be at least 1')
else:
item_count = 0
for item in it:
yield item
item_count += 1
remaining = (n - item_count) % n if next_multiple else n - item_count
for _ in range(remaining):
yield fillvalue
def distribute(n, iterable):
"""Distribute the items from *iterable* among *n* smaller iterables.
>>> group_1, group_2 = distribute(2, [1, 2, 3, 4, 5, 6])
>>> list(group_1)
[1, 3, 5]
>>> list(group_2)
[2, 4, 6]
If the length of *iterable* is not evenly divisible by *n*, then the
length of the returned iterables will not be identical:
>>> children = distribute(3, [1, 2, 3, 4, 5, 6, 7])
>>> [list(c) for c in children]
[[1, 4, 7], [2, 5], [3, 6]]
If the length of *iterable* is smaller than *n*, then the last returned
iterables will be empty:
>>> children = distribute(5, [1, 2, 3])
>>> [list(c) for c in children]
[[1], [2], [3], [], []]
This function uses :func:`itertools.tee` and may require significant
storage. If you need the order items in the smaller iterables to match the
original iterable, see :func:`divide`.
"""
if n < 1:
raise ValueError('n must be at least 1')
children = tee(iterable, n)
return [islice(it, index, None, n) for index, it in enumerate(children)]
def stagger(iterable, offsets=(-1, 0, 1), longest=False, fillvalue=None):
"""Yield tuples whose elements are offset from *iterable*.
The amount by which the `i`-th item in each tuple is offset is given by
the `i`-th item in *offsets*.
>>> list(stagger([0, 1, 2, 3]))
[(None, 0, 1), (0, 1, 2), (1, 2, 3)]
>>> list(stagger(range(8), offsets=(0, 2, 4)))
[(0, 2, 4), (1, 3, 5), (2, 4, 6), (3, 5, 7)]
By default, the sequence will end when the final element of a tuple is the
last item in the iterable. To continue until the first element of a tuple
is the last item in the iterable, set *longest* to ``True``::
>>> list(stagger([0, 1, 2, 3], longest=True))
[(None, 0, 1), (0, 1, 2), (1, 2, 3), (2, 3, None), (3, None, None)]
By default, ``None`` will be used to replace offsets beyond the end of the
sequence. Specify *fillvalue* to use some other value.
"""
children = tee(iterable, len(offsets))
return zip_offset(
*children, offsets=offsets, longest=longest, fillvalue=fillvalue
)
def zip_offset(*iterables, **kwargs):
"""``zip`` the input *iterables* together, but offset the `i`-th iterable
by the `i`-th item in *offsets*.
>>> list(zip_offset('0123', 'abcdef', offsets=(0, 1)))
[('0', 'b'), ('1', 'c'), ('2', 'd'), ('3', 'e')]
This can be used as a lightweight alternative to SciPy or pandas to analyze
data sets in which some series have a lead or lag relationship.
By default, the sequence will end when the shortest iterable is exhausted.
To continue until the longest iterable is exhausted, set *longest* to
``True``.
>>> list(zip_offset('0123', 'abcdef', offsets=(0, 1), longest=True))
[('0', 'b'), ('1', 'c'), ('2', 'd'), ('3', 'e'), (None, 'f')]
By default, ``None`` will be used to replace offsets beyond the end of the
sequence. Specify *fillvalue* to use some other value.
"""
offsets = kwargs['offsets']
longest = kwargs.get('longest', False)
fillvalue = kwargs.get('fillvalue', None)
if len(iterables) != len(offsets):
raise ValueError("Number of iterables and offsets didn't match")
staggered = []
for it, n in zip(iterables, offsets):
if n < 0:
staggered.append(chain(repeat(fillvalue, -n), it))
elif n > 0:
staggered.append(islice(it, n, None))
else:
staggered.append(it)
if longest:
return zip_longest(*staggered, fillvalue=fillvalue)
return zip(*staggered)
def sort_together(iterables, key_list=(0,), reverse=False):
"""Return the input iterables sorted together, with *key_list* as the
priority for sorting. All iterables are trimmed to the length of the
shortest one.
This can be used like the sorting function in a spreadsheet. If each
iterable represents a column of data, the key list determines which
columns are used for sorting.
By default, all iterables are sorted using the ``0``-th iterable::
>>> iterables = [(4, 3, 2, 1), ('a', 'b', 'c', 'd')]
>>> sort_together(iterables)
[(1, 2, 3, 4), ('d', 'c', 'b', 'a')]
Set a different key list to sort according to another iterable.
Specifying multiple keys dictates how ties are broken::
>>> iterables = [(3, 1, 2), (0, 1, 0), ('c', 'b', 'a')]
>>> sort_together(iterables, key_list=(1, 2))
[(2, 3, 1), (0, 0, 1), ('a', 'c', 'b')]
Set *reverse* to ``True`` to sort in descending order.
>>> sort_together([(1, 2, 3), ('c', 'b', 'a')], reverse=True)
[(3, 2, 1), ('a', 'b', 'c')]
"""
return list(zip(*sorted(zip(*iterables),
key=itemgetter(*key_list),
reverse=reverse)))
def unzip(iterable):
"""The inverse of :func:`zip`, this function disaggregates the elements
of the zipped *iterable*.
The ``i``-th iterable contains the ``i``-th element from each element
of the zipped iterable. The first element is used to to determine the
length of the remaining elements.
>>> iterable = [('a', 1), ('b', 2), ('c', 3), ('d', 4)]
>>> letters, numbers = unzip(iterable)
>>> list(letters)
['a', 'b', 'c', 'd']
>>> list(numbers)
[1, 2, 3, 4]
This is similar to using ``zip(*iterable)``, but it avoids reading
*iterable* into memory. Note, however, that this function uses
:func:`itertools.tee` and thus may require significant storage.
"""
head, iterable = spy(iter(iterable))
if not head:
# empty iterable, e.g. zip([], [], [])
return ()
# spy returns a one-length iterable as head
head = head[0]
iterables = tee(iterable, len(head))
def itemgetter(i):
def getter(obj):
try:
return obj[i]
except IndexError:
# basically if we have an iterable like
# iter([(1, 2, 3), (4, 5), (6,)])
# the second unzipped iterable would fail at the third tuple
# since it would try to access tup[1]
# same with the third unzipped iterable and the second tuple
# to support these "improperly zipped" iterables,
# we create a custom itemgetter
# which just stops the unzipped iterables
# at first length mismatch
raise StopIteration
return getter
return tuple(map(itemgetter(i), it) for i, it in enumerate(iterables))
def divide(n, iterable):
"""Divide the elements from *iterable* into *n* parts, maintaining
order.
>>> group_1, group_2 = divide(2, [1, 2, 3, 4, 5, 6])
>>> list(group_1)
[1, 2, 3]
>>> list(group_2)
[4, 5, 6]
If the length of *iterable* is not evenly divisible by *n*, then the
length of the returned iterables will not be identical:
>>> children = divide(3, [1, 2, 3, 4, 5, 6, 7])
>>> [list(c) for c in children]
[[1, 2, 3], [4, 5], [6, 7]]
If the length of the iterable is smaller than n, then the last returned
iterables will be empty:
>>> children = divide(5, [1, 2, 3])
>>> [list(c) for c in children]
[[1], [2], [3], [], []]
This function will exhaust the iterable before returning and may require
significant storage. If order is not important, see :func:`distribute`,
which does not first pull the iterable into memory.
"""
if n < 1:
raise ValueError('n must be at least 1')
seq = tuple(iterable)
q, r = divmod(len(seq), n)
ret = []
for i in range(n):
start = (i * q) + (i if i < r else r)
stop = ((i + 1) * q) + (i + 1 if i + 1 < r else r)
ret.append(iter(seq[start:stop]))
return ret
def always_iterable(obj, base_type=(text_type, binary_type)):
"""If *obj* is iterable, return an iterator over its items::
>>> obj = (1, 2, 3)
>>> list(always_iterable(obj))
[1, 2, 3]
If *obj* is not iterable, return a one-item iterable containing *obj*::
>>> obj = 1
>>> list(always_iterable(obj))
[1]
If *obj* is ``None``, return an empty iterable:
>>> obj = None
>>> list(always_iterable(None))
[]
By default, binary and text strings are not considered iterable::
>>> obj = 'foo'
>>> list(always_iterable(obj))
['foo']
If *base_type* is set, objects for which ``isinstance(obj, base_type)``
returns ``True`` won't be considered iterable.
>>> obj = {'a': 1}
>>> list(always_iterable(obj)) # Iterate over the dict's keys
['a']
>>> list(always_iterable(obj, base_type=dict)) # Treat dicts as a unit
[{'a': 1}]
Set *base_type* to ``None`` to avoid any special handling and treat objects
Python considers iterable as iterable:
>>> obj = 'foo'
>>> list(always_iterable(obj, base_type=None))
['f', 'o', 'o']
"""
if obj is None:
return iter(())
if (base_type is not None) and isinstance(obj, base_type):
return iter((obj,))
try:
return iter(obj)
except TypeError:
return iter((obj,))
def adjacent(predicate, iterable, distance=1):
"""Return an iterable over `(bool, item)` tuples where the `item` is
drawn from *iterable* and the `bool` indicates whether
that item satisfies the *predicate* or is adjacent to an item that does.
For example, to find whether items are adjacent to a ``3``::
>>> list(adjacent(lambda x: x == 3, range(6)))
[(False, 0), (False, 1), (True, 2), (True, 3), (True, 4), (False, 5)]
Set *distance* to change what counts as adjacent. For example, to find
whether items are two places away from a ``3``:
>>> list(adjacent(lambda x: x == 3, range(6), distance=2))
[(False, 0), (True, 1), (True, 2), (True, 3), (True, 4), (True, 5)]
This is useful for contextualizing the results of a search function.
For example, a code comparison tool might want to identify lines that
have changed, but also surrounding lines to give the viewer of the diff
context.
The predicate function will only be called once for each item in the
iterable.
See also :func:`groupby_transform`, which can be used with this function
to group ranges of items with the same `bool` value.
"""
# Allow distance=0 mainly for testing that it reproduces results with map()
if distance < 0:
raise ValueError('distance must be at least 0')
i1, i2 = tee(iterable)
padding = [False] * distance
selected = chain(padding, map(predicate, i1), padding)
adjacent_to_selected = map(any, windowed(selected, 2 * distance + 1))
return zip(adjacent_to_selected, i2)
def groupby_transform(iterable, keyfunc=None, valuefunc=None):
"""An extension of :func:`itertools.groupby` that transforms the values of
*iterable* after grouping them.
*keyfunc* is a function used to compute a grouping key for each item.
*valuefunc* is a function for transforming the items after grouping.
>>> iterable = 'AaaABbBCcA'
>>> keyfunc = lambda x: x.upper()
>>> valuefunc = lambda x: x.lower()
>>> grouper = groupby_transform(iterable, keyfunc, valuefunc)
>>> [(k, ''.join(g)) for k, g in grouper]
[('A', 'aaaa'), ('B', 'bbb'), ('C', 'cc'), ('A', 'a')]
*keyfunc* and *valuefunc* default to identity functions if they are not
specified.
:func:`groupby_transform` is useful when grouping elements of an iterable
using a separate iterable as the key. To do this, :func:`zip` the iterables
and pass a *keyfunc* that extracts the first element and a *valuefunc*
that extracts the second element::
>>> from operator import itemgetter
>>> keys = [0, 0, 1, 1, 1, 2, 2, 2, 3]
>>> values = 'abcdefghi'
>>> iterable = zip(keys, values)
>>> grouper = groupby_transform(iterable, itemgetter(0), itemgetter(1))
>>> [(k, ''.join(g)) for k, g in grouper]
[(0, 'ab'), (1, 'cde'), (2, 'fgh'), (3, 'i')]
Note that the order of items in the iterable is significant.
Only adjacent items are grouped together, so if you don't want any
duplicate groups, you should sort the iterable by the key function.
"""
valuefunc = (lambda x: x) if valuefunc is None else valuefunc
return ((k, map(valuefunc, g)) for k, g in groupby(iterable, keyfunc))
def numeric_range(*args):
"""An extension of the built-in ``range()`` function whose arguments can
be any orderable numeric type.
With only *stop* specified, *start* defaults to ``0`` and *step*
defaults to ``1``. The output items will match the type of *stop*:
>>> list(numeric_range(3.5))
[0.0, 1.0, 2.0, 3.0]
With only *start* and *stop* specified, *step* defaults to ``1``. The
output items will match the type of *start*:
>>> from decimal import Decimal
>>> start = Decimal('2.1')
>>> stop = Decimal('5.1')
>>> list(numeric_range(start, stop))
[Decimal('2.1'), Decimal('3.1'), Decimal('4.1')]
With *start*, *stop*, and *step* specified the output items will match
the type of ``start + step``:
>>> from fractions import Fraction
>>> start = Fraction(1, 2) # Start at 1/2
>>> stop = Fraction(5, 2) # End at 5/2
>>> step = Fraction(1, 2) # Count by 1/2
>>> list(numeric_range(start, stop, step))
[Fraction(1, 2), Fraction(1, 1), Fraction(3, 2), Fraction(2, 1)]
If *step* is zero, ``ValueError`` is raised. Negative steps are supported:
>>> list(numeric_range(3, -1, -1.0))
[3.0, 2.0, 1.0, 0.0]
Be aware of the limitations of floating point numbers; the representation
of the yielded numbers may be surprising.
"""
argc = len(args)
if argc == 1:
stop, = args
start = type(stop)(0)
step = 1
elif argc == 2:
start, stop = args
step = 1
elif argc == 3:
start, stop, step = args
else:
err_msg = 'numeric_range takes at most 3 arguments, got {}'
raise TypeError(err_msg.format(argc))
values = (start + (step * n) for n in count())
if step > 0:
return takewhile(partial(gt, stop), values)
elif step < 0:
return takewhile(partial(lt, stop), values)
else:
raise ValueError('numeric_range arg 3 must not be zero')
def count_cycle(iterable, n=None):
"""Cycle through the items from *iterable* up to *n* times, yielding
the number of completed cycles along with each item. If *n* is omitted the
process repeats indefinitely.
>>> list(count_cycle('AB', 3))
[(0, 'A'), (0, 'B'), (1, 'A'), (1, 'B'), (2, 'A'), (2, 'B')]
"""
iterable = tuple(iterable)
if not iterable:
return iter(())
counter = count() if n is None else range(n)
return ((i, item) for i in counter for item in iterable)
def locate(iterable, pred=bool, window_size=None):
"""Yield the index of each item in *iterable* for which *pred* returns
``True``.
*pred* defaults to :func:`bool`, which will select truthy items:
>>> list(locate([0, 1, 1, 0, 1, 0, 0]))
[1, 2, 4]
Set *pred* to a custom function to, e.g., find the indexes for a particular
item.
>>> list(locate(['a', 'b', 'c', 'b'], lambda x: x == 'b'))
[1, 3]
If *window_size* is given, then the *pred* function will be called with
that many items. This enables searching for sub-sequences:
>>> iterable = [0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3]
>>> pred = lambda *args: args == (1, 2, 3)
>>> list(locate(iterable, pred=pred, window_size=3))
[1, 5, 9]
Use with :func:`seekable` to find indexes and then retrieve the associated
items:
>>> from itertools import count
>>> from more_itertools import seekable
>>> source = (3 * n + 1 if (n % 2) else n // 2 for n in count())
>>> it = seekable(source)
>>> pred = lambda x: x > 100
>>> indexes = locate(it, pred=pred)
>>> i = next(indexes)
>>> it.seek(i)
>>> next(it)
106
"""
if window_size is None:
return compress(count(), map(pred, iterable))
if window_size < 1:
raise ValueError('window size must be at least 1')
it = windowed(iterable, window_size, fillvalue=_marker)
return compress(count(), starmap(pred, it))
def lstrip(iterable, pred):
"""Yield the items from *iterable*, but strip any from the beginning
for which *pred* returns ``True``.
For example, to remove a set of items from the start of an iterable:
>>> iterable = (None, False, None, 1, 2, None, 3, False, None)
>>> pred = lambda x: x in {None, False, ''}
>>> list(lstrip(iterable, pred))
[1, 2, None, 3, False, None]
This function is analogous to to :func:`str.lstrip`, and is essentially
an wrapper for :func:`itertools.dropwhile`.
"""
return dropwhile(pred, iterable)
def rstrip(iterable, pred):
"""Yield the items from *iterable*, but strip any from the end
for which *pred* returns ``True``.
For example, to remove a set of items from the end of an iterable:
>>> iterable = (None, False, None, 1, 2, None, 3, False, None)
>>> pred = lambda x: x in {None, False, ''}
>>> list(rstrip(iterable, pred))
[None, False, None, 1, 2, None, 3]
This function is analogous to :func:`str.rstrip`.
"""
cache = []
cache_append = cache.append
for x in iterable:
if pred(x):
cache_append(x)
else:
for y in cache:
yield y
del cache[:]
yield x
def strip(iterable, pred):
"""Yield the items from *iterable*, but strip any from the
beginning and end for which *pred* returns ``True``.
For example, to remove a set of items from both ends of an iterable:
>>> iterable = (None, False, None, 1, 2, None, 3, False, None)
>>> pred = lambda x: x in {None, False, ''}
>>> list(strip(iterable, pred))
[1, 2, None, 3]
This function is analogous to :func:`str.strip`.
"""
return rstrip(lstrip(iterable, pred), pred)
def islice_extended(iterable, *args):
"""An extension of :func:`itertools.islice` that supports negative values
for *stop*, *start*, and *step*.
>>> iterable = iter('abcdefgh')
>>> list(islice_extended(iterable, -4, -1))
['e', 'f', 'g']
Slices with negative values require some caching of *iterable*, but this
function takes care to minimize the amount of memory required.
For example, you can use a negative step with an infinite iterator:
>>> from itertools import count
>>> list(islice_extended(count(), 110, 99, -2))
[110, 108, 106, 104, 102, 100]
"""
s = slice(*args)
start = s.start
stop = s.stop
if s.step == 0:
raise ValueError('step argument must be a non-zero integer or None.')
step = s.step or 1
it = iter(iterable)
if step > 0:
start = 0 if (start is None) else start
if (start < 0):
# Consume all but the last -start items
cache = deque(enumerate(it, 1), maxlen=-start)
len_iter = cache[-1][0] if cache else 0
# Adjust start to be positive
i = max(len_iter + start, 0)
# Adjust stop to be positive
if stop is None:
j = len_iter
elif stop >= 0:
j = min(stop, len_iter)
else:
j = max(len_iter + stop, 0)
# Slice the cache
n = j - i
if n <= 0:
return
for index, item in islice(cache, 0, n, step):
yield item
elif (stop is not None) and (stop < 0):
# Advance to the start position
next(islice(it, start, start), None)
# When stop is negative, we have to carry -stop items while
# iterating
cache = deque(islice(it, -stop), maxlen=-stop)
for index, item in enumerate(it):
cached_item = cache.popleft()
if index % step == 0:
yield cached_item
cache.append(item)
else:
# When both start and stop are positive we have the normal case
for item in islice(it, start, stop, step):
yield item
else:
start = -1 if (start is None) else start
if (stop is not None) and (stop < 0):
# Consume all but the last items
n = -stop - 1
cache = deque(enumerate(it, 1), maxlen=n)
len_iter = cache[-1][0] if cache else 0
# If start and stop are both negative they are comparable and
# we can just slice. Otherwise we can adjust start to be negative
# and then slice.
if start < 0:
i, j = start, stop
else:
i, j = min(start - len_iter, -1), None
for index, item in list(cache)[i:j:step]:
yield item
else:
# Advance to the stop position
if stop is not None:
m = stop + 1
next(islice(it, m, m), None)
# stop is positive, so if start is negative they are not comparable
# and we need the rest of the items.
if start < 0:
i = start
n = None
# stop is None and start is positive, so we just need items up to
# the start index.
elif stop is None:
i = None
n = start + 1
# Both stop and start are positive, so they are comparable.
else:
i = None
n = start - stop
if n <= 0:
return
cache = list(islice(it, n))
for item in cache[i::step]:
yield item
def always_reversible(iterable):
"""An extension of :func:`reversed` that supports all iterables, not
just those which implement the ``Reversible`` or ``Sequence`` protocols.
>>> print(*always_reversible(x for x in range(3)))
2 1 0
If the iterable is already reversible, this function returns the
result of :func:`reversed()`. If the iterable is not reversible,
this function will cache the remaining items in the iterable and
yield them in reverse order, which may require significant storage.
"""
try:
return reversed(iterable)
except TypeError:
return reversed(list(iterable))
def consecutive_groups(iterable, ordering=lambda x: x):
"""Yield groups of consecutive items using :func:`itertools.groupby`.
The *ordering* function determines whether two items are adjacent by
returning their position.
By default, the ordering function is the identity function. This is
suitable for finding runs of numbers:
>>> iterable = [1, 10, 11, 12, 20, 30, 31, 32, 33, 40]
>>> for group in consecutive_groups(iterable):
... print(list(group))
[1]
[10, 11, 12]
[20]
[30, 31, 32, 33]
[40]
For finding runs of adjacent letters, try using the :meth:`index` method
of a string of letters:
>>> from string import ascii_lowercase
>>> iterable = 'abcdfgilmnop'
>>> ordering = ascii_lowercase.index
>>> for group in consecutive_groups(iterable, ordering):
... print(list(group))
['a', 'b', 'c', 'd']
['f', 'g']
['i']
['l', 'm', 'n', 'o', 'p']
"""
for k, g in groupby(
enumerate(iterable), key=lambda x: x[0] - ordering(x[1])
):
yield map(itemgetter(1), g)
def difference(iterable, func=sub):
"""By default, compute the first difference of *iterable* using
:func:`operator.sub`.
>>> iterable = [0, 1, 3, 6, 10]
>>> list(difference(iterable))
[0, 1, 2, 3, 4]
This is the opposite of :func:`accumulate`'s default behavior:
>>> from more_itertools import accumulate
>>> iterable = [0, 1, 2, 3, 4]
>>> list(accumulate(iterable))
[0, 1, 3, 6, 10]
>>> list(difference(accumulate(iterable)))
[0, 1, 2, 3, 4]
By default *func* is :func:`operator.sub`, but other functions can be
specified. They will be applied as follows::
A, B, C, D, ... --> A, func(B, A), func(C, B), func(D, C), ...
For example, to do progressive division:
>>> iterable = [1, 2, 6, 24, 120] # Factorial sequence
>>> func = lambda x, y: x // y
>>> list(difference(iterable, func))
[1, 2, 3, 4, 5]
"""
a, b = tee(iterable)
try:
item = next(b)
except StopIteration:
return iter([])
return chain([item], map(lambda x: func(x[1], x[0]), zip(a, b)))
class SequenceView(Sequence):
"""Return a read-only view of the sequence object *target*.
:class:`SequenceView` objects are analogous to Python's built-in
"dictionary view" types. They provide a dynamic view of a sequence's items,
meaning that when the sequence updates, so does the view.
>>> seq = ['0', '1', '2']
>>> view = SequenceView(seq)
>>> view
SequenceView(['0', '1', '2'])
>>> seq.append('3')
>>> view
SequenceView(['0', '1', '2', '3'])
Sequence views support indexing, slicing, and length queries. They act
like the underlying sequence, except they don't allow assignment:
>>> view[1]
'1'
>>> view[1:-1]
['1', '2']
>>> len(view)
4
Sequence views are useful as an alternative to copying, as they don't
require (much) extra storage.
"""
def __init__(self, target):
if not isinstance(target, Sequence):
raise TypeError
self._target = target
def __getitem__(self, index):
return self._target[index]
def __len__(self):
return len(self._target)
def __repr__(self):
return '{}({})'.format(self.__class__.__name__, repr(self._target))
class seekable(object):
"""Wrap an iterator to allow for seeking backward and forward. This
progressively caches the items in the source iterable so they can be
re-visited.
Call :meth:`seek` with an index to seek to that position in the source
iterable.
To "reset" an iterator, seek to ``0``:
>>> from itertools import count
>>> it = seekable((str(n) for n in count()))
>>> next(it), next(it), next(it)
('0', '1', '2')
>>> it.seek(0)
>>> next(it), next(it), next(it)
('0', '1', '2')
>>> next(it)
'3'
You can also seek forward:
>>> it = seekable((str(n) for n in range(20)))
>>> it.seek(10)
>>> next(it)
'10'
>>> it.seek(20) # Seeking past the end of the source isn't a problem
>>> list(it)
[]
>>> it.seek(0) # Resetting works even after hitting the end
>>> next(it), next(it), next(it)
('0', '1', '2')
The cache grows as the source iterable progresses, so beware of wrapping
very large or infinite iterables.
You may view the contents of the cache with the :meth:`elements` method.
That returns a :class:`SequenceView`, a view that updates automatically:
>>> it = seekable((str(n) for n in range(10)))
>>> next(it), next(it), next(it)
('0', '1', '2')
>>> elements = it.elements()
>>> elements
SequenceView(['0', '1', '2'])
>>> next(it)
'3'
>>> elements
SequenceView(['0', '1', '2', '3'])
"""
def __init__(self, iterable):
self._source = iter(iterable)
self._cache = []
self._index = None
def __iter__(self):
return self
def __next__(self):
if self._index is not None:
try:
item = self._cache[self._index]
except IndexError:
self._index = None
else:
self._index += 1
return item
item = next(self._source)
self._cache.append(item)
return item
next = __next__
def elements(self):
return SequenceView(self._cache)
def seek(self, index):
self._index = index
remainder = index - len(self._cache)
if remainder > 0:
consume(self, remainder)
class run_length(object):
"""
:func:`run_length.encode` compresses an iterable with run-length encoding.
It yields groups of repeated items with the count of how many times they
were repeated:
>>> uncompressed = 'abbcccdddd'
>>> list(run_length.encode(uncompressed))
[('a', 1), ('b', 2), ('c', 3), ('d', 4)]
:func:`run_length.decode` decompresses an iterable that was previously
compressed with run-length encoding. It yields the items of the
decompressed iterable:
>>> compressed = [('a', 1), ('b', 2), ('c', 3), ('d', 4)]
>>> list(run_length.decode(compressed))
['a', 'b', 'b', 'c', 'c', 'c', 'd', 'd', 'd', 'd']
"""
@staticmethod
def encode(iterable):
return ((k, ilen(g)) for k, g in groupby(iterable))
@staticmethod
def decode(iterable):
return chain.from_iterable(repeat(k, n) for k, n in iterable)
def exactly_n(iterable, n, predicate=bool):
"""Return ``True`` if exactly ``n`` items in the iterable are ``True``
according to the *predicate* function.
>>> exactly_n([True, True, False], 2)
True
>>> exactly_n([True, True, False], 1)
False
>>> exactly_n([0, 1, 2, 3, 4, 5], 3, lambda x: x < 3)
True
The iterable will be advanced until ``n + 1`` truthy items are encountered,
so avoid calling it on infinite iterables.
"""
return len(take(n + 1, filter(predicate, iterable))) == n
def circular_shifts(iterable):
"""Return a list of circular shifts of *iterable*.
>>> circular_shifts(range(4))
[(0, 1, 2, 3), (1, 2, 3, 0), (2, 3, 0, 1), (3, 0, 1, 2)]
"""
lst = list(iterable)
return take(len(lst), windowed(cycle(lst), len(lst)))
def make_decorator(wrapping_func, result_index=0):
"""Return a decorator version of *wrapping_func*, which is a function that
modifies an iterable. *result_index* is the position in that function's
signature where the iterable goes.
This lets you use itertools on the "production end," i.e. at function
definition. This can augment what the function returns without changing the
function's code.
For example, to produce a decorator version of :func:`chunked`:
>>> from more_itertools import chunked
>>> chunker = make_decorator(chunked, result_index=0)
>>> @chunker(3)
... def iter_range(n):
... return iter(range(n))
...
>>> list(iter_range(9))
[[0, 1, 2], [3, 4, 5], [6, 7, 8]]
To only allow truthy items to be returned:
>>> truth_serum = make_decorator(filter, result_index=1)
>>> @truth_serum(bool)
... def boolean_test():
... return [0, 1, '', ' ', False, True]
...
>>> list(boolean_test())
[1, ' ', True]
The :func:`peekable` and :func:`seekable` wrappers make for practical
decorators:
>>> from more_itertools import peekable
>>> peekable_function = make_decorator(peekable)
>>> @peekable_function()
... def str_range(*args):
... return (str(x) for x in range(*args))
...
>>> it = str_range(1, 20, 2)
>>> next(it), next(it), next(it)
('1', '3', '5')
>>> it.peek()
'7'
>>> next(it)
'7'
"""
# See https://sites.google.com/site/bbayles/index/decorator_factory for
# notes on how this works.
def decorator(*wrapping_args, **wrapping_kwargs):
def outer_wrapper(f):
def inner_wrapper(*args, **kwargs):
result = f(*args, **kwargs)
wrapping_args_ = list(wrapping_args)
wrapping_args_.insert(result_index, result)
return wrapping_func(*wrapping_args_, **wrapping_kwargs)
return inner_wrapper
return outer_wrapper
return decorator
def map_reduce(iterable, keyfunc, valuefunc=None, reducefunc=None):
"""Return a dictionary that maps the items in *iterable* to categories
defined by *keyfunc*, transforms them with *valuefunc*, and
then summarizes them by category with *reducefunc*.
*valuefunc* defaults to the identity function if it is unspecified.
If *reducefunc* is unspecified, no summarization takes place:
>>> keyfunc = lambda x: x.upper()
>>> result = map_reduce('abbccc', keyfunc)
>>> sorted(result.items())
[('A', ['a']), ('B', ['b', 'b']), ('C', ['c', 'c', 'c'])]
Specifying *valuefunc* transforms the categorized items:
>>> keyfunc = lambda x: x.upper()
>>> valuefunc = lambda x: 1
>>> result = map_reduce('abbccc', keyfunc, valuefunc)
>>> sorted(result.items())
[('A', [1]), ('B', [1, 1]), ('C', [1, 1, 1])]
Specifying *reducefunc* summarizes the categorized items:
>>> keyfunc = lambda x: x.upper()
>>> valuefunc = lambda x: 1
>>> reducefunc = sum
>>> result = map_reduce('abbccc', keyfunc, valuefunc, reducefunc)
>>> sorted(result.items())
[('A', 1), ('B', 2), ('C', 3)]
You may want to filter the input iterable before applying the map/reduce
procedure:
>>> all_items = range(30)
>>> items = [x for x in all_items if 10 <= x <= 20] # Filter
>>> keyfunc = lambda x: x % 2 # Evens map to 0; odds to 1
>>> categories = map_reduce(items, keyfunc=keyfunc)
>>> sorted(categories.items())
[(0, [10, 12, 14, 16, 18, 20]), (1, [11, 13, 15, 17, 19])]
>>> summaries = map_reduce(items, keyfunc=keyfunc, reducefunc=sum)
>>> sorted(summaries.items())
[(0, 90), (1, 75)]
Note that all items in the iterable are gathered into a list before the
summarization step, which may require significant storage.
The returned object is a :obj:`collections.defaultdict` with the
``default_factory`` set to ``None``, such that it behaves like a normal
dictionary.
"""
valuefunc = (lambda x: x) if (valuefunc is None) else valuefunc
ret = defaultdict(list)
for item in iterable:
key = keyfunc(item)
value = valuefunc(item)
ret[key].append(value)
if reducefunc is not None:
for key, value_list in ret.items():
ret[key] = reducefunc(value_list)
ret.default_factory = None
return ret
def rlocate(iterable, pred=bool, window_size=None):
"""Yield the index of each item in *iterable* for which *pred* returns
``True``, starting from the right and moving left.
*pred* defaults to :func:`bool`, which will select truthy items:
>>> list(rlocate([0, 1, 1, 0, 1, 0, 0])) # Truthy at 1, 2, and 4
[4, 2, 1]
Set *pred* to a custom function to, e.g., find the indexes for a particular
item:
>>> iterable = iter('abcb')
>>> pred = lambda x: x == 'b'
>>> list(rlocate(iterable, pred))
[3, 1]
If *window_size* is given, then the *pred* function will be called with
that many items. This enables searching for sub-sequences:
>>> iterable = [0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3]
>>> pred = lambda *args: args == (1, 2, 3)
>>> list(rlocate(iterable, pred=pred, window_size=3))
[9, 5, 1]
Beware, this function won't return anything for infinite iterables.
If *iterable* is reversible, ``rlocate`` will reverse it and search from
the right. Otherwise, it will search from the left and return the results
in reverse order.
See :func:`locate` to for other example applications.
"""
if window_size is None:
try:
len_iter = len(iterable)
return (
len_iter - i - 1 for i in locate(reversed(iterable), pred)
)
except TypeError:
pass
return reversed(list(locate(iterable, pred, window_size)))
def replace(iterable, pred, substitutes, count=None, window_size=1):
"""Yield the items from *iterable*, replacing the items for which *pred*
returns ``True`` with the items from the iterable *substitutes*.
>>> iterable = [1, 1, 0, 1, 1, 0, 1, 1]
>>> pred = lambda x: x == 0
>>> substitutes = (2, 3)
>>> list(replace(iterable, pred, substitutes))
[1, 1, 2, 3, 1, 1, 2, 3, 1, 1]
If *count* is given, the number of replacements will be limited:
>>> iterable = [1, 1, 0, 1, 1, 0, 1, 1, 0]
>>> pred = lambda x: x == 0
>>> substitutes = [None]
>>> list(replace(iterable, pred, substitutes, count=2))
[1, 1, None, 1, 1, None, 1, 1, 0]
Use *window_size* to control the number of items passed as arguments to
*pred*. This allows for locating and replacing subsequences.
>>> iterable = [0, 1, 2, 5, 0, 1, 2, 5]
>>> window_size = 3
>>> pred = lambda *args: args == (0, 1, 2) # 3 items passed to pred
>>> substitutes = [3, 4] # Splice in these items
>>> list(replace(iterable, pred, substitutes, window_size=window_size))
[3, 4, 5, 3, 4, 5]
"""
if window_size < 1:
raise ValueError('window_size must be at least 1')
# Save the substitutes iterable, since it's used more than once
substitutes = tuple(substitutes)
# Add padding such that the number of windows matches the length of the
# iterable
it = chain(iterable, [_marker] * (window_size - 1))
windows = windowed(it, window_size)
n = 0
for w in windows:
# If the current window matches our predicate (and we haven't hit
# our maximum number of replacements), splice in the substitutes
# and then consume the following windows that overlap with this one.
# For example, if the iterable is (0, 1, 2, 3, 4...)
# and the window size is 2, we have (0, 1), (1, 2), (2, 3)...
# If the predicate matches on (0, 1), we need to zap (0, 1) and (1, 2)
if pred(*w):
if (count is None) or (n < count):
n += 1
for s in substitutes:
yield s
consume(windows, window_size - 1)
continue
# If there was no match (or we've reached the replacement limit),
# yield the first item from the window.
if w and (w[0] is not _marker):
yield w[0]
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