Itemgetter

使用 itemgetter 将值分组为字典的键值对:

from itertools import groupby
from operator import itemgetter
adict = {'a': 1, 'b': 5, 'c': 1}

dict((i, dict(v)) for i, v in groupby(adict.items(), itemgetter(1)))
# Output: {1: {'a': 1, 'c': 1}, 5: {'b': 5}}

这对于像这样的 lambda 函数是等效的(但更快):

dict((i, dict(v)) for i, v in groupby(adict.items(), lambda x: x[1]))

或者按第二个元素排序元组列表首先将第一个元素排序为次要元素:

alist_of_tuples = [(5,2), (1,3), (2,2)]
sorted(alist_of_tuples, key=itemgetter(1,0))
# Output: [(2, 2), (5, 2), (1, 3)]