Cycle a list from alternating sides

问题:

Given a list

a = [0,1,2,3,4,5,6,7,8,9]

how can I get

b = [0,9,1,8,2,7,3,6,4,5]

That is, produce a new list in which each successive element is alternately taken from the two sides of the original list?

回答1:

>>> [a[-i//2] if i % 2 else a[i//2] for i in range(len(a))]
[0, 9, 1, 8, 2, 7, 3, 6, 4, 5]

Explanation:
This code picks numbers from the beginning (a[i//2]) and from the end (a[-i//2]) of a, alternatingly (if i%2 else). A total of len(a) numbers are picked, so this produces no ill effects even if len(a) is odd.
[-i//2 for i in range(len(a))] yields 0, -1, -1, -2, -2, -3, -3, -4, -4, -5,
[ i//2 for i in range(len(a))] yields 0, 0, 1, 1, 2, 2, 3, 3, 4, 4,
and i%2 alternates between False and True,
so the indices we extract from a are: 0, -1, 1, -2, 2, -3, 3, -4, 4, -5.

My assessment of pythonicness:
The nice thing about this one-liner is that it's short and shows symmetry (+i//2 and -i//2).
The bad thing, though, is that this symmetry is deceptive:
One might think that -i//2 were the same as i//2 with the sign flipped. But in Python, integer division returns the floor of the result instead of truncating towards zero. So -1//2 == -1.
Also, I find accessing list elements by index less pythonic than iteration.

回答2:

cycle between getting items from the forward iter and the reversed one. Just make sure you stop at len(a) with islice.

from itertools import islice, cycle

iters = cycle((iter(a), reversed(a)))
b = [next(it) for it in islice(iters, len(a))]

>>> b
[0, 9, 1, 8, 2, 7, 3, 6, 4, 5]

This can easily be put into a single line but then it becomes much more difficult to read:

[next(it) for it in islice(cycle((iter(a),reversed(a))),len(a))]

Putting it in one line would also prevent you from using the other half of the iterators if you wanted to:

>>> iters = cycle((iter(a), reversed(a)))
>>> [next(it) for it in islice(iters, len(a))]
[0, 9, 1, 8, 2, 7, 3, 6, 4, 5]
>>> [next(it) for it in islice(iters, len(a))]
[5, 4, 6, 3, 7, 2, 8, 1, 9, 0]

回答3:

A very nice one-liner in Python 2.7:

results = list(sum(zip(a, reversed(a))[:len(a)/2], ()))
>>>> [0, 9, 1, 8, 2, 7, 3, 6, 4, 5]

First you zip the list with its reverse, take half that list, sum the tuples to form one tuple, and then convert to list.

In Python 3, zip returns a generator, so you have have to use islice from itertools:

from itertools import islice
results = list(sum(islice(zip(a, reversed(a)),0,int(len(a)/2)),()))

Edit: It appears this only works perfectly for even-list lengths - odd-list lengths will omit the middle element :( A small correction for int(len(a)/2) to int(len(a)/2) + 1 will give you a duplicate middle value, so be warned.

回答4:

You can just pop back and forth:

b = [a.pop(-1 if i%2 else 0) for i in range(len(a))]

Note: This destroys the original list, a.

回答5:

Use the right toolz.

from toolz import interleave, take

b = list(take(len(a), interleave((a, reversed(a)))))

First, I tried something similar to Raymond Hettinger's solution with itertools (Python 3).

from itertools import chain, islice

interleaved = chain.from_iterable(zip(a, reversed(a)))
b = list(islice(interleaved, len(a)))

回答6:

Not terribly different from some of the other answers, but it avoids a conditional expression for determining the sign of the index.

a = range(10)
b = [a[i // (2*(-1)**(i&1))] for i in a]

i & 1 alternates between 0 and 1. This causes the exponent to alternate between 1 and -1. This causes the index divisor to alternate between 2 and -2, which causes the index to alternate from end to end as i increases. The sequence is a[0], a[-1], a[1], a[-2], a[2], a[-3], etc.

(I iterate i over a since in this case each value of a is equal to its index. In general, iterate over range(len(a)).)

回答7:

The basic principle behind your question is a so-called roundrobin algorithm. The itertools-documentation-page contains a possible implementation of it:

from itertools import cycle, islice

def roundrobin(*iterables):
    """This function is taken from the python documentation!
    roundrobin('ABC', 'D', 'EF') --> A D E B F C
    Recipe credited to George Sakkis"""
    pending = len(iterables)
    nexts = cycle(iter(it).__next__ for it in iterables) # next instead of __next__ for py2
    while pending:
        try:
            for next in nexts:
                yield next()
        except StopIteration:
            pending -= 1
            nexts = cycle(islice(nexts, pending))

so all you have to do is split your list into two sublists one starting from the left end and one from the right end:

import math
mid = math.ceil(len(a)/2) # Just so that the next line doesn't need to calculate it twice

list(roundrobin(a[:mid], a[:mid-1:-1]))
# Gives you the desired result: [0, 9, 1, 8, 2, 7, 3, 6, 4, 5]

alternatively you could create a longer list (containing alternating items from sequence going from left to right and the items of the complete sequence going right to left) and only take the relevant elements:

list(roundrobin(a, reversed(a)))[:len(a)]

or using it as explicit generator with next:

rr = roundrobin(a, reversed(a))
[next(rr) for _ in range(len(a))]

or the speedy variant suggested by @Tadhg McDonald-Jensen (thank you!):

list(islice(roundrobin(a,reversed(a)),len(a)))

回答8:

Not sure, whether this can be written more compactly, but it is efficient as it only uses iterators / generators

a = [0,1,2,3,4,5,6,7,8,9]

iter1 = iter(a)
iter2 = reversed(a)
b = [item for n, item in enumerate(
        next(iter) for _ in a for iter in (iter1, iter2)
    ) if n < len(a)]

回答9:

For fun, here is an itertools variant:

>>> a = [0,1,2,3,4,5,6,7,8,9]
>>> list(chain.from_iterable(izip(islice(a, len(a)//2), reversed(a))))
[0, 9, 1, 8, 2, 7, 3, 6, 4, 5]

This works where len(a) is even. It would need a special code for odd-lengthened input.

Enjoy!

回答10:

Not at all elegant, but it is a clumsy one-liner:

a = range(10)
[val for pair in zip(a[:len(a)//2],a[-1:(len(a)//2-1):-1]) for val in pair]

Note that it assumes you are doing this for a list of even length. If that breaks, then this breaks (it drops the middle term). Note that I got some of the idea from here.

回答11:

Two versions not seen yet:

b = list(sum(zip(a, a[::-1]), ())[:len(a)])

and

import itertools as it

b = [a[j] for j in it.accumulate(i*(-1)**i for i in range(len(a)))]

回答12:

mylist = [0,1,2,3,4,5,6,7,8,9]
result = []

for i in mylist:
    result += [i, mylist.pop()]

Note:

Beware: Just like @Tadhg McDonald-Jensen has said (see the comment below) it'll destroy half of original list object.

回答13:

One way to do this for even-sized lists (inspired by this post):

a = range(10)

b = [val for pair in zip(a[:5], a[5:][::-1]) for val in pair]

回答14:

I would do something like this

a = [0,1,2,3,4,5,6,7,8,9]
b = []
i = 0
j = len(a) - 1
mid = (i + j) / 2
while i <= j:
    if i == mid and len(a) % 2 == 1:
        b.append(a[i])
        break
    b.extend([a[i], a[j]])
    i = i + 1
    j = j - 1

print b

回答15:

You can partition the list into two parts about the middle, reverse the second half and zip the two partitions, like so:

a = [0,1,2,3,4,5,6,7,8,9]
mid = len(a)//2
l = []
for x, y in zip(a[:mid], a[:mid-1:-1]):
    l.append(x)
    l.append(y)
# if the length is odd
if len(a) % 2 == 1:
    l.append(a[mid])
print(l)

Output:

[0, 9, 1, 8, 2, 7, 3, 6, 4, 5]