交替附加来自两个列表的元素

问题描述

我有三个包含元素的列表:

a = [[0,1],[2,3],...]b = [[5,6],[7,8],...]c = []

我想将 a 和 b 中的元素附加到 c 中以获得:

c = [ [0,1],[5,6],[2,3],[7,8],.... ]

解决方案

另一种使用字符串切片(以及最高效)的非常简单的方法:

>>>a = [[0,1],[2,3]]>>>b = [[5,6],[7,8]]>>>c = a + b # 创建一个大小为 len(a) + len(b) 的列表>>>c[::2], c[1::2] = a, b # 交替插入值>>>C[[0, 1], [5, 6], [2, 3], [7, 8]]

以下是此处提到的答案与 timeit 的结果比较(Python 版本:2.7):

1. 使用字符串切片:每个循环 0.586 usec

   moin@moin-pc:~$ python -m "timeit" -s "a = [[0,1],[2,3]]; b = [[5,6],[7,8]];""c = a + b; c[::2], c[1::2] = a, b"1000000 个循环，最好的 3 个:每个循环 0.586 微秒

2. 使用 itertools.chain():每个循环 1.89 usec

   moin@moin-pc:~$ python -m "timeit" -s "from itertools import chain; a = [[0,1],[2,3]]; b = [[5,6],[7,8]];""c = list(chain(*zip(a, b)))"1000000 个循环，最好的 3 个:每个循环 1.89 微秒

3. 使用 reduce():每个循环 0.829 usec

   moin@moin-pc:~$ python -m "timeit" -s "import operator; a = [[0,1],[2,3]]; b = [[5,6],[7,8]];""c = reduce(operator.concat, zip(a, b))"1000000 个循环，3 个最佳:每个循环 0.829 微秒

4. 使用 list.extend():每个循环 0.824 usec

    moin@moin-pc:~$ python -m "timeit" -s "a = [[0,1],[2,3]]; b = [[5,6],[7,8]]; c=[]" "对于 zip(a,b) 中的对:c.extend(pair)"1000000 个循环，最好的 3 个:每个循环 0.824 微秒

5. 使用 list.append() 两次:每个循环 1.04 usec

   moin@moin-pc:~$ python -m "timeit" -s "a = [[0,1],[2,3]]; b = [[5,6],[7,8]]; c=[]" "对于 zip(a, b) 中的 a_element, b_element: c.append(a_element); c.append(b_element)"1000000 个循环，最好的 3 个:每个循环 1.04 微秒

I have three lists with elements :

a = [[0,1],[2,3],...]
b = [[5,6],[7,8],...]

c = []

I want to append elements from a and b into c to get:

c = [ [0,1],[5,6],[2,3],[7,8],.... ]

解决方案

Another very simple approach using string slicing (and most performance efficient) as:

>>> a = [[0,1],[2,3]]
>>> b = [[5,6],[7,8]]
>>> c = a + b # create a list with size = len(a) + len(b)
>>> c[::2], c[1::2] = a, b  # alternately insert the value
>>> c
[[0, 1], [5, 6], [2, 3], [7, 8]]

Below is the comparison of results with timeit for the answers mentioned here (Python version: 2.7):

1. Using string slicing: 0.586 usec per loop

   moin@moin-pc:~$ python -m "timeit" -s "a = [[0,1],[2,3]]; b = [[5,6],[7,8]];" "c = a + b; c[::2], c[1::2] = a, b"
   1000000 loops, best of 3: 0.586 usec per loop
   

2. Using itertools.chain(): 1.89 usec per loop

   moin@moin-pc:~$ python -m "timeit" -s "from itertools import chain; a = [[0,1],[2,3]]; b = [[5,6],[7,8]];" "c = list(chain(*zip(a, b)))"
   1000000 loops, best of 3: 1.89 usec per loop
   

3. Using reduce(): 0.829 usec per loop

   moin@moin-pc:~$ python -m "timeit" -s "import operator; a = [[0,1],[2,3]]; b = [[5,6],[7,8]];" "c = reduce(operator.concat, zip(a, b))"
   1000000 loops, best of 3: 0.829 usec per loop
   

4. Using list.extend(): 0.824 usec per loop

    moin@moin-pc:~$ python -m "timeit" -s "a = [[0,1],[2,3]]; b = [[5,6],[7,8]]; c=[]" "for pair in zip(a,b): c.extend(pair)"
    1000000 loops, best of 3: 0.824 usec per loop
   

5. Using list.append() twice: 1.04 usec per loop

   moin@moin-pc:~$ python -m "timeit" -s "a = [[0,1],[2,3]]; b = [[5,6],[7,8]]; c=[]" "for a_element, b_element in zip(a, b): c.append(a_element); c.append(b_element)"
   1000000 loops, best of 3: 1.04 usec per loop
   

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