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rogwan
V2EX  ›  Python

有这样一个 list,怎么把列表中的字典进行排序?

  •  
  •   rogwan · 2016-11-13 15:02:37 +08:00 · 4852 次点击
    这是一个创建于 2725 天前的主题,其中的信息可能已经有所发展或是发生改变。

    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 ?

    20 条回复    2016-11-14 17:57:38 +08:00
    aaronzjw
        1
    aaronzjw  
       2016-11-13 15:06:20 +08:00 via Android
    sort+lambda
    justou
        2
    justou  
       2016-11-13 15:12:43 +08:00   ❤️ 1
    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
        3
    rogwan  
    OP
       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
        4
    pupboss  
       2016-11-13 15:14:01 +08:00
    def score(s):
    return s['student_score']

    bar = sorted(foo, key = score)

    print(bar)
    rogwan
        5
    rogwan  
    OP
       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
        6
    aaronzjw  
       2016-11-13 17:30:27 +08:00
    @rogwan print sorted(list, key=lambda student: student['student_score'])[-3:]
    Mutoo
        7
    Mutoo  
       2016-11-13 17:34:34 +08:00   ❤️ 2
    [:3] 这个表情好搞笑
    ipconfiger
        8
    ipconfiger  
       2016-11-13 17:48:56 +08:00
    @rogwan 居然还有这种书
    triostones
        9
    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
        10
    rogwan  
    OP
       2016-11-13 18:47:49 +08:00
    @ipconfiger 就是 Python 核心编程那本书啊
    rogwan
        11
    rogwan  
    OP
       2016-11-13 18:53:57 +08:00
    @triostones 谢谢,这个方法也 OK ^-^
    staticor
        12
    staticor  
       2016-11-13 19:33:46 +08:00
    python cookbook 里肯定是提了 第 1 章.
    rogwan
        13
    rogwan  
    OP
       2016-11-13 20:19:07 +08:00
    @staticor 是,去翻了, 1.13 节有讲
    timeship
        14
    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
        15
    gemini  
       2016-11-14 00:00:26 +08:00
    top3 = sorted(list, key=lambda stu:int(stu['student_score']), reverse=True)[0:3]
    ericls
        16
    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
        17
    rogwan  
    OP
       2016-11-14 11:07:54 +08:00
    @ericls 确实可以 lol
    hl
        18
    hl  
       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
        19
    jayyjh  
       2016-11-14 15:55:29 +08:00
    sorted_dict = sorted(dict.items(), key=lambda item: item[1])
    shyrock
        20
    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?
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