如何将一个字典根据 key 值拆分成多个字典？ pytthon

现在有一个字典 a： a = { '0106': '50', '0117': '35', '0203': '30', '0205': '30', '3302': '45', '3305': '45' } 如何根据 key 值前面的两位数拆分成字典 b： b = { '01': { '0106': '50', '0117': '35', }, '02': { '0203': '30', '0205': '30', }, '33': { '3302': '45', '3305': '45', } }

请问大神用 python 代码如何优雅的实现？

字典

key

拆分

pytthon

10 条回复 • 2018-05-17 10:25:09 +08:00

aaronzjw

2018-05-11 17:44:54 +08:00 via iPhone

collection defaultdict 了解一下

alvin666

2018-05-11 17:49:31 +08:00 via Android

b={{}}
for key,value in a.items():
b[key[:2][key]]=value

xpresslink

2018-05-11 17:51:35 +08:00

>>> a = { '0106': '50', '0117': '35', '0203': '30', '0205': '30', '3302': '45', '3305': '45' }
>>> from itertools import groupby
>>> {k:dict(g) for k,g in groupby(sorted(a.items(), key=lambda x:x[0]), key=lambda x:x[0][:2])}
{'01': {'0106': '50', '0117': '35'}, '02': {'0203': '30', '0205': '30'}, '33': {'3302': '45', '3305': '45'}}

wrq

2018-05-11 18:02:43 +08:00

>>> a
{'0117': '35', '0106': '50', '0205': '30', '0203': '30', '3305': '45', '3302': '45'}
>>> {k[:2]: {i:j for i, j in a.items() if i.startswith(k[:2])} for k, _ in a.items()}
{'02': {'0205': '30', '0203': '30'}, '33': {'3305': '45', '3302': '45'}, '01': {'0106': '50', '0117': '35'}}

nullcc

2018-05-11 19:00:34 +08:00

讲真写出来以后我并不觉得优雅：
>>> groupby_res = groupby(my_dict, lambda e: e[0:2])
>>>res = dict([(mini_key, dict([(key, my_dict[key]) for key in my_dict.keys() if key.startswith(mini_key)])) for mini_key, group in groupby_res for key in group])
>>> print(res)
>>> {'01': {'0106': '50', '0117': '35'}, '02': {'0203': '30', '0205': '30'}, '33': {'3302': '45', '3305': '45'}}

个人感觉最好还是写得明确一点，分多步构造也没关系。

kethylar

2018-05-11 19:32:56 +08:00

from collections import defaultdict
ret = defaultdict(dict)
for k,v in a.items():
ret[k[:2]].update({k:v})

xuegj1010

2018-05-11 20:13:59 +08:00

@nullcc 能写出来都是大神，至少我自己写不出来

ericls

2018-05-11 23:42:13 +08:00

from itertools import group_by
b = {k: dict(v) for k, v in groupby(list(a.items()), lambda item: item[0][:2])

ericls

2018-05-11 23:43:16 +08:00

from itertools import group_by
{k: dict(v) for k, v in groupby(a.items(), lambda item: item[0][:2])}

seabottom1978

2018-05-17 10:25:09 +08:00

result = {}
for i in a:
if i[:2] not in result:
result[i[:2]] = {i : a[i]}
else:
result[i[:2]][i] = a[i]