Question

我要打印列表列表的索引元素。

我的列表如下

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

我期望什么

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list从中生成一个range。应该会更快。

>>> 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

尝试一下

>>> 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：

"activities": [                                             
    {                                                         
      "type": "message",                                      
      "id": "XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX",               
      "timestamp": "2019-07-01T15:18:56.8251462Z",            
      "serviceUrl": "XXXXXXXXXXXXXXXXXXXXXXXXX",   
      "channelId": "directline",                              
      "from": {                                               
        "id": "user1"                                         
      },                                                      
      "conversation": {                                       
        "id": "XXXXXXXXXXXXXXXXXXXXXXXXXXXXX"                      
      },                                                      
      "text": "the milk"                                      
    },                                                        
    {                                                         
      "type": "message",                                      
      "id": "XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX",               
      "timestamp": "2019-07-01T15:18:57.6856172Z",            
      "localTimestamp": "2019-07-01T15:18:57.5099359+00:00",  
      "channelId": "directline",                              
      "from": {                                               
        "id": "XXXXXX",                                        
        "name": "XXXXXX"                                       
      },                                                      
      "conversation": {                                       
        "id": "XXXXXXXXXXXXXXXXXX"                      
      },                                                      
      "text": "The product is in second line 3rd row",        
      "attachments": [],                                      
      "entities": [],                                         
      "replyToId": "XXXXXXXXXXXXXXXXXXXXXXXXXXXXX"         
    }                                                         
  ],                                                          
  "watermark": "1"                                            
}

Answer 5

或使用let tmp = Int64(10) print(String(tmp))：

map

如果值是唯一的：

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

将列表列表的索引打印为主要列表（python）

5 个答案: