python 之 列表中字典的分组(三种写法，总有一款适合你)

一、使用场景

在工作中有时我们需要进行对数据进行分组，统计每组下的用户

users = [
    {"username": "zhangsan", "age": 18},
    {"username": "lisi", "age": 20},
    {"username": "wanger", "age": 23},
    {"username": "mazi", "age": 19},
    {"username": "zhaowu", "age": 18},
    {"username": "maliu", "age": 22},
    {"username": "guiqi", "age": 18}
]

• ​​temp1比如按年龄分组，得出每组年龄下的用户信息​​

{
    18: [
        {'username': 'zhangsan', 'age': 18}, 
        {'username': 'zhaowu', 'age': 18},
        {'username': 'guiqi', 'age': 18}
    ], 
    20: [
        {'username': 'lisi', 'age': 20}
    ], 
    23: [{'username': 'wanger', 'age': 23}], 
    19: [{'username': 'mazi', 'age': 19}], 
    22: [{'username': 'maliu', 'age': 22}]
}

• ​​temp2 获取按照年龄分组，分组下为用户名​​

{
  18: ['zhangsan', 'zhaowu', 'guiqi'], 
  20: ['lisi'], 
  23: ['wanger'], 
  19: ['mazi'], 
  22: ['maliu']
}

二、解法三种，终有一款适合你

• ​​解法一​​

temp1 = {}
temp2 = {}
for user in users:
    temp1.setdefault(user.get("age"), []).append(user)
    temp2.setdefault(user.get("age"), []).append(user.get("username"))
print(temp1)
print(temp2)

• ​​解法二​​

from collections import defaultdict

temp1 = defaultdict(list)
temp2 = defaultdict(list)

for user in users:
    temp1[user['age']].append(user)
    temp2[user['age']].append(user.get("username"))
print(temp1)
print(temp2)

• ​​解法三​​

from operator import itemgetter
from itertools import groupby

temp1 = {}
temp2 = {}
users.sort(key=itemgetter('age'))  排序必须排序
for age, items in groupby(users, key=itemgetter('age')):
    temp1[age] = [item for item in items]   # 与下冲突注销一行及成功
    temp2[age] = [item.get('username') for item in items]

print(temp1)
print(temp2)