Python 使用理解合并字典键和值

Question

我如何使用 python COMPREHENSION 巩固以下内容

FROM(字典列表)

[
 {'server':'serv1','os':'Linux','archive':'/my/folder1'}
 ,{'server':'serv2','os':'Linux','archive':'/my/folder1'}
 ,{'server':'serv3','os':'Linux','archive':'/my/folder2'}
 ,{'server':'serv4','os':'AIX','archive':'/my/folder1'}
 ,{'server':'serv5','os':'AIX','archive':'/my/folder1'}
]

TO(以元组为键的字典列表和以“服务器编号”为值的列表

[
 {('Linux','/my/folder1'):['serv1','serv2']}
 ,('Linux','/my/folder2'):['serv3']}
 .{('AIX','/my/folder1'):['serv4','serv5']}
]

Answer 1

需要能够为您的字典设置默认值并多次使用相同的键可能会使字典理解有点笨拙。我更喜欢这样的东西:

一个defaultdict可能有帮助:

from collections import defaultdict

lst = [
 {'server':'serv1','os':'Linux','archive':'/my/folder1'},
 {'server':'serv2','os':'Linux','archive':'/my/folder1'},
 {'server':'serv3','os':'Linux','archive':'/my/folder2'},
 {'server':'serv4','os':'AIX','archive':'/my/folder1'},
 {'server':'serv5','os':'AIX','archive':'/my/folder1'}
]

dct = defaultdict(list)

for d in lst:
    key = d['os'], d['archive']
    dct[key].append(d['server'])

如果你希望最后有一个标准的字典(实际上我真的没有看到一个很好的理由)你可以使用 dict.setdefault为了创建一个空列表，其中键尚不存在:

dct = {}

for d in lst:
    key = d['os'], d['archive']
    dct.setdefault(key, []).append(d['server'])

documentation on defaultdict (vs. setdefault) :

This technique is simpler and faster than an equivalent technique using dict.setdefault()