Question

我是编程的新手，我正在创建一个配对计划，比较我使用列表存储的个性分数，这是我的代码：

`ps = int(personality_score)
potential_partners = partners.Partners()
while potential_partners.available():
partner = []
personality_scores = []
a = potential_partners.get_name()
f = int(potential_partners.get_personality_score())
if ps == f:
print("This is your match" + a)
else:
g = abs(int(ps - f))
h = int(g)
personality_scores.append(h)
partner.append(a)
partner_compatability =list(zip(personality_scores, partner))
partner_compatability.sort(key = operator.itemgetter(0))
for sub in partner_compatability:
print(sub)`

我已经查看了与此相关的多个问题和答案，但没有一个对我有用，我的列表输出是：

`[['Mary Smith', 1]]
[['Juan Lopez', 5]]
[['Leslie Liu', 11]]
[['Tatiana Ivanov', 15]]
[['Andre Leroy', 11]]
[['Sam Augusta', 7]]
[['Adalbert Weber', 1]]`

但应从最低分到最高分：

`[['Mary Smith', 1]] 
[['Adalbert Weber', 1]]
[['Juan Lopez', 5]]
[['Sam Augusta', 7]]
[['Leslie Liu', 11]]
[['Andre Leroy', 11]]
[['Tatiana Ivanov', 15]]`

Answer 1

（我也是新手程序员，直到有更好的答案）你可以做我做的事情：只需颠倒顺序然后排序。

lst = [['Mary Smith', 1],
['Juan Lopez', 5],
['Leslie Liu', 11],
['Tatiana Ivanov', 15],
['Andre Leroy', 11],
['Sam Augusta', 7],
['Adalbert Weber', 1]]

然后：

sorted_lst = [[k,v] for v,k in lst]
sorted_lst.sort()
print(sorted_lst)

Answer 2

您可以直接对partner_compatability

进行排序

sorted_list = partner_compatability.sort(key=lambda x: x[1])

我试过这样，因为没有提供列表值

In [67]: a
Out[67]: [8, 9, 3]
In [68]: b
Out[68]: [5, 6, 3]

In [69]: partner_compatability = list(zip(a, b))

In [70]: partner_compatability
Out[70]: [(8, 5), (9, 6), (3, 3)]

In [71]: partner_compatability.sort(key=lambda x: x[1])
In [72]: partner_compatability
Out[72]: [(3, 3), (8, 5), (9, 6)]

根据您的评论，

partner_compatability = [[(1, 'Mary Smith')],
 [(5, 'Juan Lopez')],
 [(11, 'Leslie Liu')],
 [(15, 'Tatiana Ivanov')],
 [(11, 'Andre Leroy')],
 [(7, 'Sam Augusta')],
 [(1, 'Adalbert Weber')]]

partner_compatability.sort(key=lambda x: x[0][0])

In [71]: partner_compatability
Out[71]:
[[(1, 'Mary Smith')],
 [(1, 'Adalbert Weber')],
 [(5, 'Juan Lopez')],
 [(7, 'Sam Augusta')],
 [(11, 'Leslie Liu')],
 [(11, 'Andre Leroy')],
 [(15, 'Tatiana Ivanov')]]

Answer 3

import operator

data = [
    ['Mary Smith', 1],
    ['Juan Lopez', 5],
    ['Leslie Liu', 11],
    ['Tatiana Ivanov', 15],
    ['Andre Leroy', 11],
    ['Sam Augusta', 7],
    ['Adalbert Weber', 1],
]

data.sort(key = operator.itemgetter(1))
for sub in data:
    print(sub)

打印

['Mary Smith', 1]
['Adalbert Weber', 1]
['Juan Lopez', 5]
['Sam Augusta', 7]
['Leslie Liu', 11]
['Andre Leroy', 11]
['Tatiana Ivanov', 15]

虽然不为人所知，但操作符函数应该比等效的lambda快。还有attrgetter。当密钥应该是多个项目的元组时，两者都可以使用。

python

3 个答案: