python之如果条目匹配，则减少列表列表

我在 python 中有一个列表，看起来像

[['boy','121','is a male child'],['boy','121','is male'],['boy','121','is a child'],['girl','122','is a female child'],['girl','122','is a child']] 

我想根据每个列表中的前 2 个条目减少列表，得到

[['boy','121',is a male child, is male, is a child'],['girl','122','is a female child','is a child']] 

有没有一种方法可以在不创建虚拟列表的情况下有效地做到这一点？

请您参考如下方法:

作为此类任务的更 pythonic 方式，您可以使用字典:

>>> li=[['boy','121','is a male child'],['boy','121','is male'],['boy','121','is a child'],['girl','122','is a female child'],['girl','122','is a child']] 
>>>  
>>> d={} 
>>>  
>>> for i,j,k in li: 
...   d.setdefault((i,j),[]).append(k) 
...  
>>> d 
{('boy', '121'): ['is a male child', 'is male', 'is a child'], ('girl', '122'): ['is a female child', 'is a child']} 

setdefault(key[, default])

If key is in the dictionary, return its value. If not, insert key with a value of default and return default. default defaults to None.

如果你想将元素放在 1 个容器中，你可以遍历这些项目并将值转换为 tuple 然后用键对其进行广告:

>>> [i+tuple(j) for i,j in d.items()] 
[('boy', '121', 'is a male child', 'is male', 'is a child'), ('girl', '122', 'is a female child', 'is a child')] 

正如@jonrsharpe 所说，作为一种更优雅的方式，您还可以使用 collections.defaultdict :

>>> from collections import defaultdict 
>>>  
>>> d=defaultdict(list) 
>>> for i,j,k in li: 
...   d[i,j].append(k) 
...  
>>> d 
defaultdict(<type 'list'>, {('boy', '121'): ['is a male child', 'is male', 'is a child'], ('girl', '122'): ['is a female child', 'is a child']})