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).