转换列表

Question

我有一个这样的列表:

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

如何对列表中的元素进行分组,例如上面的内容？

['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])

one-liner:使用dict理解(有点浪费,但没有进口,需要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)