有这样一个 list，怎么把列表中的字典进行排序？

list = [ {'student_name': zhangsan, 'student_score': 65}, {'student_name': lisi, 'student_score': 95}, {'student_name': wangwu, 'student_score': 80}, {'student_name': maliu, 'student_score': 75}, {'student_name': zhuqi, 'student_score': 88} ]

把 5 个学生成绩从高到低排序，取前三名，要怎么处理这样的 list ？

list

student_score'

student_name'

排序

20 条回复 • 2016-11-14 17:57:38 +08:00

aaronzjw

2016-11-13 15:06:20 +08:00 via Android

sort+lambda

justou

2016-11-13 15:12:43 +08:00

from operator import itemgetter

lst = [ {'student_name': 'zhangsan', 'student_score': 65},
{'student_name': 'lisi', 'student_score': 95},
{'student_name': 'wangwu', 'student_score': 80},
{'student_name': 'maliu', 'student_score': 75},
{'student_name': 'zhuqi', 'student_score': 88} ]

top3 = sorted(lst, key=itemgetter('student_score'), reverse=True)[:3]

print top3

rogwan

2016-11-13 15:12:48 +08:00

@aaronzjw
sorted(list, key=lambda student_score:student_score[1], reverse=True)[0:3]

这样解决，出错，结果不对。。。

pupboss

2016-11-13 15:14:01 +08:00

def score(s):
return s['student_score']

bar = sorted(foo, key = score)

print(bar)

rogwan

2016-11-13 15:20:55 +08:00

@justou 赞！刚查书去，二楼的方法也是书上推荐的标准做法

aaronzjw

2016-11-13 17:30:27 +08:00

@rogwan print sorted(list, key=lambda student: student['student_score'])[-3:]

Mutoo

2016-11-13 17:34:34 +08:00

[:3] 这个表情好搞笑

ipconfiger

2016-11-13 17:48:56 +08:00

@rogwan 居然还有这种书

triostones

2016-11-13 18:01:40 +08:00

In [12]: import heapq

In [13]: from operator import itemgetter

In [14]: scores = [ {'student_name': 'zhangsan', 'student_score': 65}, {'student_name': 'lisi', 'student_score': 95}, {'student_name':'wangwu', 'student_score': 80}, {'student_name': 'maliu', 'student_sco
...: re': 75}, {'student_name': 'zhuqi', 'student_score': 88} ]

In [15]: heapq.nlargest(3, scores, key=itemgetter('student_score'))
Out[15]:
[{'student_name': 'lisi', 'student_score': 95},
{'student_name': 'zhuqi', 'student_score': 88},
{'student_name': 'wangwu', 'student_score': 80}]

rogwan

2016-11-13 18:47:49 +08:00

@ipconfiger 就是 Python 核心编程那本书啊

rogwan

2016-11-13 18:53:57 +08:00

@triostones 谢谢，这个方法也 OK ^-^

staticor

2016-11-13 19:33:46 +08:00

python cookbook 里肯定是提了第 1 章.

rogwan

2016-11-13 20:19:07 +08:00

@staticor 是，去翻了， 1.13 节有讲

timeship

2016-11-13 21:51:29 +08:00

楼上正确， sorted （ list, key=itemgetter )然后取前三个，好像很多书里都有讲过类似的 QAQ

gemini

2016-11-14 00:00:26 +08:00

top3 = sorted(list, key=lambda stu:int(stu['student_score']), reverse=True)[0:3]

ericls

2016-11-14 00:07:08 +08:00

lst = [ {'student_name': 'zhangsan', 'student_score': 65},
{'student_name': 'lisi', 'student_score': 95},
{'student_name': 'wangwu', 'student_score': 80},
{'student_name': 'maliu', 'student_score': 75},
{'student_name': 'zhuqi', 'student_score': 88} ]

sorted(lst, key=lambda stu: stu["student_score"], reverse=True)[:3]

明明就可以

rogwan

2016-11-14 11:07:54 +08:00

@ericls 确实可以 lol

2016-11-14 13:59:07 +08:00

请大家参考 python 官方高级数据结构 heapq 章节，可以使用内置高效的方法来最简单的实现

students_list = [ {'student_name': 'zhangsan', 'student_score': 65},
{'student_name': 'lisi', 'student_score': 95},
{'student_name': 'wangwu', 'student_score': 80},
{'student_name': 'maliu', 'student_score': 75},
{'student_name': 'zhuqi', 'student_score': 88} ]

####################
import heapq

score_first_3 = heapq.nlargest(3,students_list,key=lambda student:student["student_score"])

print score_first_3
####################

jayyjh

2016-11-14 15:55:29 +08:00

sorted_dict = sorted(dict.items(), key=lambda item: item[1])

shyrock

2016-11-14 17:57:38 +08:00

@hl 语法上看不是最简单的，难道用 heapq 的效率更高？