itertools.groupby()分组字典列表

## itertools.groupby()分组字典列表数据

from operator import itemgetter
from itertools import groupby


students = [
    {'name': 'Peter', 'age': 19, 'score': 95},
    {'name': 'Lily', 'age': 22, 'score': 90},
    {'name': 'Stanley', 'age': 22, 'score': 92},
    {'name': 'Bob', 'age': 20, 'score': 88},
    {'name': 'Well', 'age': 20, 'score': 82}
]

for key, group in groupby(sorted(students, key=itemgetter("age")), key=itemgetter("age")):
    # groupby()函数同时返回分组关键字和一个与关键字相对应的可迭代对象
    # itemgetter()同样可以接收多个关键字，也可以使用匿名函数代替此函数，但速度相比之下较慢
    print("Age: %s" % key)
    for g in group:
        print(g)
"""
Age: 19
{'name': 'Peter', 'age': 19, 'score': 95}
Age: 20
{'name': 'Bob', 'age': 20, 'score': 88}
{'name': 'Well', 'age': 20, 'score': 82}
Age: 22
{'name': 'Lily', 'age': 22, 'score': 90}
{'name': 'Stanley', 'age': 22, 'score': 92}

"""
# 也可以使用defaultdict()创建一个一对多字典方便进行随机访问
from collections import defaultdict


students_by_age = defaultdict(list)
for key, group in groupby(sorted(students, key=itemgetter("age")), key=itemgetter("age")):
    for i in group:
        students_by_age[key].append(i)

print(list(students_by_age[22]))
# [{'name': 'Lily', 'age': 22, 'score': 90}, {'name': 'Stanley', 'age': 22, 'score': 92}]

参考资料：
　　Python Cookbook, 3rd edition, by David Beazley and Brian K. Jones (O’Reilly).