根据python中的其他元素对列表进行排序

如果我有一个包含以下结构元素的列表：

[team1, points1, playedgames1, team2, points2, playedgames2, team3, points3, playedgames3]

等等。一个例子（有3个团队）：

希望它看起来像这样：

["Inter", 3, 2, "Juventus", 5, 2, "Milan", 6, 2]

等等，为更多的团队。如您所见，列表现在排序在最低点之后。基本上，现在是：

["team2, points2, playedgames2, team3, points3, playedgames3, team1, points2, playedgames2]

由于点S2的值最低。因此，我可以像这样对列表进行排序，根据积分，同时保留列表的结构（团队、积分、玩过的游戏）等等。这可能吗

元素从文本文件中检索。

这里有一种方法：

import operator

L = (team1, points1, playedgames1, team2, points2, playedgames2, team3, points3, playedgames3)
L = [tuple(L[i:i+3]) for i in range(None, len(L), 3)]
answer = sorted(L, key=operator.itemgetter(2))

ls = ["Milan", 6, 2, "Inter", 3, 2, "Juventus", 5, 2]
>>> sorted([ls[i:i+3] for i in range(0,len(ls),3)], key=lambda x:x[1])
[['Inter', 3, 2], ['Juventus', 5, 2], ['Milan', 6, 2]]

---

我建议您将记录从文件转换为元组列表。 之后，您可以通过以下简单方法解决问题：

ls.sort(key=operator.itemgetter(0))

或者您可以创建函数：

def my_sort(my_list):
   list_of_tuples = sorted([tuple(l[i:i+3]) for i in range(0, len(my_list), 3], key=operator.itemgetter(0))
   arr = []
   for i in b:
       arr += [i[0]] + [i[1]]
   return arr

另外，

operator.itemgetter

工作得更快

在[17]中：%%timeit
排序（a，key=operator.itemgetter（0））

10000个回路，最好为3:114µs/回路

在[18]中：%%timeit
已排序（a，key=lambda x:x[0]）

10000个回路，最好为3个：每个回路210µs

---

首先，您需要使用步骤3拆分列表，然后根据第二个值的排序列表进行比较，然后根据第二个值的排序列表创建最后一个列表：

>>> new=[ls[i:i+3] for i in range(0,len(ls),3)]
>>> new
[('Milan', 6, 2), ('Inter', 3, 2), ('Juventus', 5, 2)]
>>> val2_l=sorted(zip(*new)[1])
>>> val2_l
[3, 5, 6]
>>> v=zip(*new)[1]
>>> v
(6, 3, 5)
>>> ["team{}, points{}, playedgames{}".format(v.index(i)+1,v.index(i)+1,v.index(i)+1) for i in val2_l]
['team2, points2, playedgames2', 'team3, points3, playedgames3', 'team1, points1, playedgames1']
>>> [''.join(last)]
['team2, points2, playedgames2team3, points3, playedgames3team1, points1, playedgames1']

---

如果它应该是一个列表（我不小心写了括号而不是括号，现在编辑了），那也行吗？谢谢！这正是我想要的，也是可以理解的！

>>> new=[ls[i:i+3] for i in range(0,len(ls),3)]
>>> new
[('Milan', 6, 2), ('Inter', 3, 2), ('Juventus', 5, 2)]
>>> val2_l=sorted(zip(*new)[1])
>>> val2_l
[3, 5, 6]
>>> v=zip(*new)[1]
>>> v
(6, 3, 5)
>>> ["team{}, points{}, playedgames{}".format(v.index(i)+1,v.index(i)+1,v.index(i)+1) for i in val2_l]
['team2, points2, playedgames2', 'team3, points3, playedgames3', 'team1, points1, playedgames1']
>>> [''.join(last)]
['team2, points2, playedgames2team3, points3, playedgames3team1, points1, playedgames1']