Question

我有一个列表如下所示：

a = [['he', 'goes'],
     ['he does'],
     ['one time'],
     [('he','is'), ('she', 'went'), ('they', 'are')],
     ['he', 'must'],
     ['they use']]

我试图将列表展平，以便它只是一个没有元组的列表列表。例如：

a = [['he', 'goes'],
     ['he does'],
     ['one time'],
     ['he','is'], 
     ['she', 'went'],
     ['they', 'are'],
     ['he', 'must'],
     ['they use']]

我尝试过使用itertools.chain.from_iterable()，但会将所有嵌套列表展平。

Answer 1

b = []
for x in a:
    if isinstance(x[0], tuple):
        b.extend([list(y) for y in x])
    else:
        b.append(x)

Answer 2

使用yield from和python3：

from collections import Iterable
def conv(l):
    for ele in l:
        if isinstance(ele[0], Iterable) and not isinstance(ele[0],str):
            yield from map(list,ele)
        else:
            yield ele

print(list(conv(a)))
[['he', 'goes'], ['he does'], ['one time'], ['he', 'is'], ['she', 'went'], ['they', 'are'], ['he', 'must'], ['they use']]

对于python2，您可以遍历itertools.imap对象：

from collections import Iterable
from itertools import imap

def conv(l):
    for ele in l:
        if isinstance(ele[0], Iterable) and not isinstance(ele[0],basestring):
            for sub in imap(list, ele):
                yield sub
        else:
            yield ele

print(list(conv(a)))

如果您有嵌套元组，则需要添加更多逻辑。

Answer 3

这解决了你的例子：

a = [list(strings) for sublist in a for strings in
     ([sublist] if isinstance(sublist[0], str) else sublist)]

对于已经是字符串列表的每个子列表，只需使用该子列表。否则遍历该子列表。

这还不够，或者您的实际数据是否更复杂？

具有元组列表的列表列表

3 个答案: