按值（列表）对字典应用功能对其排序

Question

我有一本字典：

students = {
    '1': [32, 14, 31, 23],
    '2': [32, 8, 36.5, 22],
    '3': [26, 11, 39, 15.5],
    '4': [34, 15, 44, 25],
    '5': [30, 6, 25, 24],
    '6': [26, 8, 31, 13],
    '7': [24, 4, 16, 17],
    '8': [22, 2, 0, 11.2],
    '9': [22, 4, 15, 10],
    '10': [31, 4, 16, 4.2],
    '11': [18, 3.5, 0, 0],
    '12': [28, 5, 30, 18.5],
    '13': [34, 13, 23, 13],
}

我想按每个列表的总和对字典进行排序。 所以我可以这样：

import operator

students = {
    '1': sum([32, 14, 31, 23]),
    '2': sum([32, 8, 36.5, 22]),
    '3': sum([26, 11, 39, 15.5]),
    '4': sum([34, 15, 44, 25]),
    '5': sum([30, 6, 25, 24]),
    '6': sum([26, 8, 31, 13]),
    '7': sum([24, 4, 16, 17]),
    '8': sum([22, 2, 0, 11.2]),
    '9': sum([22, 4, 15, 10]),
    '10': sum([31, 4, 16, 4.2]),
    '11': sum([18, 3.5, 0, 0]),
    '12': sum([28, 5, 30, 18.5]),
    '13': sum([34, 13, 23, 13]),
}

sorted_students = sorted(
    students.items(), key=operator.itemgetter(1), reverse=True)

但：

1. 我不想更改字典，我想将列表作为值
2. 我不想写多次sum() 。

我尝试了key=sum(operator.itemgetter(1))但是很明显它不起作用，我得到了：

TypeError: 'operator.itemgetter' object is not iterable

有什么更有效的方式来做我想做的事？

Answer 1

字典未排序，因此无法对字典进行“排序”。 但是，如果需要一个单独的已排序键列表，则可以这样做：

>>> students = {
    '1': [32, 14, 31, 23],
    '2': [32, 8, 36.5, 22],
    '3': [26, 11, 39, 15.5],
    '4': [34, 15, 44, 25],
    '5': [30, 6, 25, 24],
    '6': [26, 8, 31, 13],
    '7': [24, 4, 16, 17],
    '8': [22, 2, 0, 11.2],
    '9': [22, 4, 15, 10],
    '10': [31, 4, 16, 4.2],
    '11': [18, 3.5, 0, 0],
    '12': [28, 5, 30, 18.5],
    '13': [34, 13, 23, 13],
}
>>> sortedKeys = sorted(students.keys(), key = lambda x: sum(students[x]))
>>> sortedKeys
['11', '8', '9', '10', '7', '6', '12', '13', '5', '3', '2', '1', '4']

那么你可以遍历列表，你会通过遍历dict ，因为iter(dict)通过对按键进行迭代dict反正。

>>> for key in sortedKeys:
    print(key,sum(students[key]))


11 21.5
8 35.2
9 51
10 55.2
7 61
6 78
12 81.5
13 83
5 85
3 91.5
2 98.5
1 100
4 118

Answer 2

它们的键必须是可操作的，因此您需要将操作包装在一个函数中：

key = lambda x: sum(x[1])

Answer 3

如果要基于最高总和订购，请在按总和排序后使用OrderedDict ：

from collections import OrderedDict

students = OrderedDict(sorted(students.items(),key=lambda x: sum(x[1])))


OrderedDict([('11', [18, 3.5, 0, 0]), ('8', [22, 2, 0, 11.2]),
           ('9', [22, 4, 15, 10]), ('10', [31, 4, 16, 4.2]), 
           ('7', [24, 4, 16, 17]), ('6', [26, 8, 31, 13]),
          ('12', [28, 5, 30, 18.5]), ('13', [34, 13, 23, 13]), 
          ('5', [30, 6, 25, 24]), ('3', [26, 11, 39, 15.5]), 
          ('2', [32, 8, 36.5, 22]), ('1', [32, 14, 31, 23]), 
         ('4', [34, 15, 44, 25])])

如果要从最高和最低sorted ，请在sorted使用-sum(...

students = OrderedDict(sorted(students.items(),key=lambda x: -sum(x[1])))

print(students)
OrderedDict([('4', [34, 15, 44, 25]), ('1', [32, 14, 31, 23]),
        ('2', [32, 8, 36.5, 22]), ('3', [26, 11, 39, 15.5]), 
        ('5', [30, 6, 25, 24]), ('13', [34, 13, 23, 13]), 
        ('12', [28, 5, 30, 18.5]), ('6', [26, 8, 31, 13]),
        ('7', [24, 4, 16, 17]), ('10', [31, 4, 16, 4.2]), 
        ('9', [22, 4, 15, 10]), ('8', [22, 2, 0, 11.2]), 
        ('11', [18, 3.5, 0, 0])])

Answer 4

由于无法对字典进行排序，因此我们只能得到排序后的字典的表示形式，该字典将是一个列表，实际上是一个元组列表。

d = {k: sum(i for i in v) for k,v in students.items()}
print (sorted(d.items(), key=operator.itemgetter(1)))

输出：

[('11', 21.5), ('8', 35.2), ('9', 51), ('10', 55.2), ('7', 61), ('6', 78), ('12', 81.5), ('13', 83), ('5', 85), ('3', 91.5), ('2', 98.5), ('1', 100), ('4', 118)]