转化列表

Question

我有一个这样的名单：

[['0', '10'], ['0', '11'], ['0', '12'], ['1', '10'], ['1', '11']] 

哪能组在列表中的元素，例如像上面？

['0': ['10','11','12']],['1': ['10','11']] 

Answer 1

迭代 - 放入字典。

d = {} 
l = [['0', '10'], ['0', '11'], ['0', '12'], ['1', '10'], ['1', '11']] 
for p in l: 
    if p[0] in d: 
     d[p[0]].append(p[1]) 
    else: 
     d[p[0]] = [p[1]] 

>>> d 
{'1': ['10', '11'], '0': ['10', '11', '12']} 

使用defaultdict：

from collections import defaultdict 

d = defaultdict(list) 
l = [['0', '10'], ['0', '11'], ['0', '12'], ['1', '10'], ['1', '11']] 
for p in l: 
    d[p[0]].append(p[1]) 

一行代码：使用dict comprehension（有点浪费，但没有进口，并要求2.7+）

>>> dd = {key: [i[1] for i in l if i[0] == key] for (key, value) in l} 
>>> dd 
{'1': ['10', '11'], '0': ['10', '11', '12']} 

Answer 2

你可以使用collections.defaultdict：

import collections 

l = [['0', '10'], ['0', '11'], ['0', '12'], ['1', '10'], ['1', '11']] 

d = collections.defaultdict(list) 
for k, v in l: 
    d[k].append(v) 
print(d) 

Answer 3

您可以使用dict.setdefault()：

In [16]: lis=[['0', '10'], ['0', '11'], ['0', '12'], ['1', '10'], ['1', '11']] 

In [17]: dic={} 

In [18]: for x,y in lis: 
    ....:  dic.setdefault(x,[]).append(y) 
    ....:  

In [19]: dic 
Out[19]: {'0': ['10', '11', '12'], '1': ['10', '11']} 

和名单也甚至itertools.groupby将正常工作：

In [39]: from operator import itemgetter 

In [40]: from itertools import groupby 

In [34]: lis=[['0', '10'], ['0', '11'], ['0', '12'], ['1', '10'], ['1', '11']] 

In [35]: {k:[x[1] for x in g] for k,g in groupby(lis,key=itemgetter(0))} 
Out[35]: {'0': ['10', '11', '12'], '1': ['10', '11']}