Python Equivalent to Ruby's #each_cons?

Question

Is there a Pythonic equivalent to Ruby’s #each_cons?

In Ruby you can do this:

array = [1,2,3,4]
array.each_cons(2).to_a
=> [[1,2],[2,3],[3,4]]

Asked By: maxhawkins

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Source

Answer 1

For such things, itertools is the module you should be looking at:

from itertools import tee, izip

def pairwise(iterable):
    "s -> (s0,s1), (s1,s2), (s2, s3), ..."
    a, b = tee(iterable)
    next(b, None)
    return izip(a, b)

Then:

>>> list(pairwise([1, 2, 3, 4]))
[(1, 2), (2, 3), (3, 4)]

For an even more general solution, consider this:

def split_subsequences(iterable, length=2, overlap=0):
    it = iter(iterable)
    results = list(itertools.islice(it, length))
    while len(results) == length:
        yield results
        results = results[length - overlap:]
        results.extend(itertools.islice(it, length - overlap))
    if results:
        yield results

This allows arbitrary lengths of subsequences and arbitrary overlapping. Usage:

>> list(split_subsequences([1, 2, 3, 4], length=2))
[[1, 2], [3, 4]]
>> list(split_subsequences([1, 2, 3, 4], length=2, overlap=1))
[[1, 2], [2, 3], [3, 4], [4]]

Answered By: Eli Bendersky

Answer 2

A quick one-liner:

a = [1, 2, 3, 4]

out = [a[i:i + 2] for i in range(len(a) - 1)]

Answered By: Blender

Answer 3

I don’t think there is one, I looked through the built-in module itertools, which is where I would expect it to be. You can simply create one though:

def each_cons(xs, n):
    return [xs[i:i+n] for i in range(len(xs)-n+1)]

Answered By: Gustav Larsson

Answer 4

Python can surely do this. If you don’t want to do it so eagerly, use itertool’s islice and izip. Also, its important to remember that normal slices will create a copy so if memory usage is important you should also consider the itertool equivalents.

each_cons = lambda l: zip(l[:-1], l[1:])

Answered By: dcolish

Answer 5

My solution for lists (Python2):

import itertools
def each_cons(xs, n):
    return itertools.izip(*(xs[i:] for i in xrange(n)))

Edit: With Python 3 itertools.izip is no longer, so you use plain zip:

def each_cons(xs, n):
    return zip(*(xs[i:] for i in range(n)))

Answered By: snipsnipsnip

Answer 6

UPDATE: Nevermind my answer below, just use toolz.itertoolz.sliding_window() — it will do the right thing.

For a truly lazy implementation that preserves the behavior of Ruby’s each_cons when the sequence/generator has insufficient length:

import itertools
def each_cons(sequence, n):
    return itertools.izip(*(itertools.islice(g, i, None)
                          for i, g in
                          enumerate(itertools.tee(sequence, n))))

Examples:

>>> print(list(each_cons(xrange(5), 2)))
[(0, 1), (1, 2), (2, 3), (3, 4)]
>>> print(list(each_cons(xrange(5), 5)))
[(0, 1, 2, 3, 4)]
>>> print(list(each_cons(xrange(5), 6)))
[]
>>> print(list(each_cons((a for a in xrange(5)), 2)))
[(0, 1), (1, 2), (2, 3), (3, 4)]

Note that the tuple unpacking used on the arguments for izip is applied to a tuple of size n resulting of itertools.tee(xs, n) (that is, the “window size”), and not the sequence we want to iterate.

Answered By: Elias Dorneles

Answer 7

Python 3 version of Elias’s code

from itertools import islice, tee

def each_cons(sequence, n):
    return zip(
        *(
            islice(g, i, None)
            for i, g in
            enumerate(tee(sequence, n))
        )
    )

$ ipython

...    

In [2]: a_list = [1, 2, 3, 4, 5]

In [3]: list(each_cons(a_list, 2))
Out[3]: [(1, 2), (2, 3), (3, 4), (4, 5)]

In [4]: list(each_cons(a_list, 3))
Out[4]: [(1, 2, 3), (2, 3, 4), (3, 4, 5)]

In [5]: list(each_cons(a_list, 5))
Out[5]: [(1, 2, 3, 4, 5)]

In [6]: list(each_cons(a_list, 6))
Out[6]: []

Answered By: bwv549

Answer 8

Close to @Blender’s solution but with a fix:

a = [1, 2, 3, 4]
n = 2
out = [a[i:i + n] for i in range(len(a) - n + 1)]
# => [[1, 2], [2, 3], [3, 4]]

Or

a = [1, 2, 3, 4]
n = 3
out = [a[i:i + n] for i in range(len(a) - n + 1)]
# => [[1, 2, 3], [2, 3, 4]]

Answered By: Mark

Answer 9

Here’s an implementation using collections.deque. This supports arbitrary generators as well

from collections import deque

def each_cons(it, n):
    # convert it to an iterator
    it = iter(it)

    # insert first n items to a list first
    deq = deque()
    for _ in range(n):
        try:
            deq.append(next(it))
        except StopIteration:
            for _ in range(n - len(deq)):
                deq.append(None)
            yield tuple(deq)
            return

    yield tuple(deq)

    # main loop
    while True:
        try:
            val = next(it)
        except StopIteration:
            return
        deq.popleft()
        deq.append(val)
        yield tuple(deq)

Usage:

list(each_cons([1,2,3,4], 2))
# => [(1, 2), (2, 3), (3, 4)]

# This supports generators
list(each_cons(range(5), 2))
# => [(0, 1), (1, 2), (2, 3), (3, 4)]

list(each_cons([1,2,3,4], 10))
# => [(1, 2, 3, 4, None, None, None, None, None, None)]

Answered By: James Wong

Python Equivalent to Ruby's #each_cons?

Question:

Answers:

Python 3 version of Elias’s code