Question

今天下午我很无聊。我该如何转换

['1,9', '3,10', '5,8', '8,11', '2,7', '4,5', '2,6']

进入

[1,9,3,10,5,8,8,11,2,7,4,5,2,6]

Answer 1

>>> L = ['1,9', '3,10', '5,8', '8,11', '2,7', '4,5', '2,6']
>>> [int(y) for x in L for y in x.split(',')]
[1, 9, 3, 10, 5, 8, 8, 11, 2, 7, 4, 5, 2, 6]

这种嵌套列表理解与此等效：

res = []
for x in L:
    for y in x.split(','):
        res.append(int(y))

如您所见，自上而下的结构在列表理解

中从左向右

即

[int(y) 
 for x in L 
 for y in x.split(',')]

现在解开的

看起来与for循环相同。

另一种方式：

>>> [int(x) for x in ','.join(L).split(',')]
[1, 9, 3, 10, 5, 8, 8, 11, 2, 7, 4, 5, 2, 6]

Answer 2

看到有几种不同的方法可以做到这一点我决定运行一些（快速）测试，看看哪个是最快的。

python -m timeit -s "L = ['1,9', '3,10', '5,8', '8,11', '2,7', '4,5', '2,6']" "[int(x) for x in ''.join(L).split(',')]"
>>> 100000 loops, best of 3: 3.2 usec per loop

python -m timeit -s "L = ['1,9', '3,10', '5,8', '8,11', '2,7', '4,5', '2,6']" "[int(y) for x in L for y in x.split(',')]"
>>> 100000 loops, best of 3: 6.38 usec per loop

python -m timeit -s "L = ['1,9', '3,10', '5,8', '8,11', '2,7', '4,5', '2,6'];from itertools import chain" "[int(x) for x in chain.from_iterable(l) if x != ',']"
>>> 100000 loops, best of 3: 6.68 usec per loop

~~似乎[int(x) for x in ''.join(L).split(',')]取得了成功。~~

编辑：根据jamylak的建议，我添加了以下测试：

python -m timeit -s "L = ['1,9', '3,10', '5,8', '8,11', '2,7', '4,5', '2,6']" "map(int, ''.join(L).split(','))"
>>> 100000 loops, best of 3: 2.79 usec per loop

python -m timeit -s "L = ['1,9', '3,10', '5,8', '8,11', '2,7', '4,5', '2,6']" "list(map(int, ''.join(L).split(',')))"
>>> 100000 loops, best of 3: 3.02 usec per loop

因此，python3的map(int, ''.join(L).split(','))或list(map(int, ''.join(L).split(',')))是最好的方法。

将嵌套的字符串列表展平为单个int列表

2 个答案: