Question

我正在使用两个列表列表：

list1 = [[],["1","3","4"],["0"],["1","2","5","4"]]
list2 = [["1","2","3"],["8","7","6"],[],["9","8","6","4"]]

for (v, c) in zip(list1, list2):
    joined_list =sorted(set(v + c))

这当前将值放到列表的末尾，我希望它进入与他们目前最新的相同的索引吗？

编辑：忘了他们是字符串，代码就在这里 https://github.com/ishikawa-rei/EPRO-CALC/blob/master/massaging.py 它是第一个函数，并获得两个列表的输入

Answer 1

您在每次循环迭代中重新定位joined_list。您必须在循环之前定义列表并在迭代时收集项目：

joined_list = []
for v, c in zip(list1, list2):
    joined_list.append(sorted(set(v + c)))

或者您可以使用list comprehension：

joined_list = [sorted(set(v + c)) for v, c in zip(list1, list2)]

Answer 2

您应该使用append方法。

joined_list = []
for (v, c) in zip(list1, list2):
    joined.append(sorted(set(v+c)))

print(joined)

这输出以下内容：

[[1, 2, 3], [1, 3, 4, 6, 7, 8], [0], [1, 2, 4, 5, 6, 8, 9]]

将列表列表加入相同的索引

2 个答案: