将列表列表的索引打印为主要列表（python）

Question

我想打印列表列表的索引元素。

我有如下列表

list1=[["a","b","c"],["a","b"],["a","b","c","d"]]

我所期望的

expectedlist=[[0,1,2],[0,1],[0,1,2,3]]

我尝试了这段代码

list1=[["a","b","c"],["a","b"],["a","b","c","d"]]
a=[]

for i,v in enumerate(list1):
    for k,r in enumerate(v):
        a+=[k]

print(a)

但它只打印了一个列表。

[0，1，2，0，1，0，1，2，3]

Expectedlist = [[0,1,2]，[0,1]，[0,1,2,3]]

Answer 1

您可以简单地获取子列表的len并使用range生成一个list 。 应该相当快。

>>> list1 = [["a", "b", "c"], ["a", "b"], ["a", "b", "c", "d"]]

>>> [list(range(len(sub))) for sub in list1]
[[0, 1, 2], [0, 1], [0, 1, 2, 3]]

Answer 2

尝试这个，

>>> ls = [["a","b","c"],["a","b"],["a","b","c","d"]]
>>> [[i for i,v in enumerate(el)] for el in ls]

[[0, 1, 2], [0, 1], [0, 1, 2, 3]]

Answer 3

代码的问题是您使用的是单个列表，而最终输出是嵌套列表。

因此，您需要两个列表。

a = []
for i, v in enumerate(list1): 
    b = []
    for k, r in enumerate(v): 
        b+=[k] # Also b.append(k)
    a.append(b)    

print(a)
# [[0, 1, 2], [0, 1], [0, 1, 2, 3]]

Answer 4

您可以使用列表推导，如下所示：

>>> list1=[["a","b","c"],["a","b"],["a","b","c","d"]]
>>> [[k.index(i) for i in k] for k in list1]
[[0, 1, 2], [0, 1], [0, 1, 2, 3]]

Answer 5

或使用map ：

>>> list(map(lambda x: list(range(len(x))), list1))
[[0, 1, 2], [0, 1], [0, 1, 2, 3]]
>>> 

如果值是唯一的：

>>> [list(map(i.index, i)) for i in list1]
[[0, 1, 2], [0, 1], [0, 1, 2, 3]]
>>>