有这样一个 list，怎么把列表中的字典进行排序？

[rogwan]

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 ？

[aaronzjw]

sort+lambda

[justou]

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]

@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]

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

bar = sorted(foo, key = score)

print(bar)

[rogwan]

@justou 赞！刚查书去，二楼的方法也是书上推荐的标准做法

[aaronzjw]

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

[Mutoo]

[:3] 这个表情好搞笑

[ipconfiger]

@rogwan 居然还有这种书

[triostones]

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]

@ipconfiger 就是 Python 核心编程那本书啊

[rogwan]

@triostones 谢谢，这个方法也 OK ^-^

[staticor]

python cookbook 里肯定是提了 第 1 章.

[rogwan]

@staticor 是，去翻了， 1.13 节有讲

[timeship]

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]

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

[ericls]

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]

@ericls 确实可以 lol

[hl]

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]

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

[shyrock]

@hl 语法上看不是最简单的，难道用 heapq 的效率更高？