Python：将列表与列表列表组合在一起，同时保留列表元素

Question

是否存在语法上简短的方法？

['a','b','c'] and [[1,2],[1],[1,2,3]] -> [['a', [1,2]], ['b', [1]], ['c',[1,2,3]]]

Answer 1

您正在寻找zip：

>>> a = ['a','b','c'] 
>>> b = [[1,2],[1],[1,2,3]]
>>> list(zip(a, b))
[('a', [1, 2]), ('b', [1]), ('c', [1, 2, 3])]

如果你真的希望元素是列表而不是元组，那么你也可以这样做：

>>> [list(t) for t in zip(a,b)]
[['a', [1, 2]], ['b', [1]], ['c', [1, 2, 3]]]

Answer 2

您正在寻找：     [zip（a，b）中x的列表（x）]

较短版本：map（list，zip（a，b））

这将解释一些事情：

>>> a = ['a','b','c']
>>> b = [[1,2],[1],[1,2,3]]
>>> c = zip(a, b)
>>> c
[('a', [1, 2]), ('b', [1]), ('c', [1, 2, 3])]
>>> d = [list(x) for x in c]
>>> d
[['a', [1, 2]], ['b', [1]], ['c', [1, 2, 3]]]

Answer 3

拉​​链+列表+地图

result = map(list,zip(a,b))

Answer 4

如果列表大小相同，您也可以使用enumerate：

l1  = ['a','b','c']
l2 = [[1,2],[1],[1,2,3]]
[[ele,l2[ind]]for ind, ele in enumerate(l1)]
[['a', [1, 2]], ['b', [1]], ['c', [1, 2, 3]]]

如果您有不同的尺寸列表，则可以使用itertools.izip_longest

l1  = ['a','b','c','d']
l2 = [[1,2],[1],[1,2,3]]

from itertools import izip_longest
print [list(x) for x in izip_longest(l1,l2,fillvalue="No data")]
[['a', [1, 2]], ['b', [1]], ['c', [1, 2, 3]], ['d', 'No data']]