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 ？
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aaronzjw Nov 13, 2016 via Android
sort+lambda
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justou Nov 13, 2016  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 |
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3
rogwan OP |
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4
pupboss Nov 13, 2016
def score(s):
return s['student_score'] bar = sorted(foo, key = score) print(bar) |
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aaronzjw Nov 13, 2016
@rogwan print sorted(list, key=lambda student: student['student_score'])[-3:]
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Mutoo Nov 13, 2016 2
[:3] 这个表情好搞笑
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ipconfiger Nov 13, 2016
@rogwan 居然还有这种书
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9
triostones Nov 13, 2016
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}] |
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rogwan OP @ipconfiger 就是 Python 核心编程那本书啊
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11
rogwan OP @triostones 谢谢，这个方法也 OK ^-^
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12
staticor Nov 13, 2016
python cookbook 里肯定是提了 第 1 章.
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14
timeship Nov 13, 2016
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
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15
gemini Nov 14, 2016
top3 = sorted(list, key=lambda stu:int(stu['student_score']), reverse=True)[0:3]
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ericls Nov 14, 2016
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 |
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18
hl Nov 14, 2016
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 #################### |
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19
jayyjh Nov 14, 2016
sorted_dict = sorted(dict.items(), key=lambda item: item[1])
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