如何将可迭代对象拆分为两个具有交替元素的列表

Question

I want to split an iterable into two lists with alternating elements. Here is a working solution. But is there a simpler way to achieve the same?

def zigzag(seq):
    """Return two sequences with alternating elements from `seq`"""
    x, y = [], []
    p, q = x, y
    for e in seq:
        p.append(e)
        p, q = q, p
    return x, y

Sample output:

>>> zigzag('123456')
(['1', '3', '5'], ['2', '4', '6'])

Answer 1

If seq is a sequence, then:

def zigzag(seq):
  return seq[::2], seq[1::2]

If seq is a totally generic iterable, such as possibly a generator:

def zigzag(seq):
  results = [], []
  for i, e in enumerate(seq):
    results[i%2].append(e)
  return results

Answer 2

This takes an iterator and returns two iterators:

import itertools
def zigzag(seq):
    t1,t2 = itertools.tee(seq)
    even = itertools.islice(t1,0,None,2)
    odd = itertools.islice(t2,1,None,2)
    return even,odd

If you prefer lists then you can return list(even),list(odd) .

Answer 3

def zigzag(seq):
    return seq[::2], seq[1::2]

Answer 4

I just wanted to clear something. Say you have a list

list1 = list(range(200))

you can do either :

## OPTION A ##
a = list1[1::2]
b = list1[0::2]

or

## OPTION B ##
a = list1[0:][::2] # even
b = list1[1:][::2] # odd

And get have alternative elements in variable a and b .

But OPTION A is twice as fast