Question

我是python和编程的新手，需要一些帮助：我有一个我之前在程序中创建的列表，通过循环附加（即我现在不能重新定义我的列表来解决我的问题），24个4元组：

elementary = [(23, 1, 18, 4), (23, 1, 6, 16), (23, 1, 4, 18), (23, 2, 18, 3), (23, 2, 12, 9), (23, 2, 9, 12), (23, 2, 3, 18), (23, 3, 18, 2), (23, 3, 2, 18), (23, 4, 18, 1), (23, 4, 1, 18), (23, 5, 14, 7), (23, 5, 7, 14), (23, 6, 16, 1), (23, 6, 9, 8), (23, 6, 8, 9), (23, 6, 1, 16), (23, 7, 14, 5), (23, 7, 5, 14), (23, 8, 9, 6), (23, 8, 6, 9), (23, 9, 12, 2), (23, 9, 8, 6), (23, 9, 6, 8), (23, 9, 2, 12), (23, 12, 9, 2), (23, 12, 2, 9), (23, 14, 7, 5), (23, 14, 5, 7), (23, 16, 1, 6), (23, 18, 4, 1), (23, 18, 3, 2), (23, 18, 2, 3), (23, 18, 1, 4)]

但现在想要摆脱刚刚重新安排的元组...换句话说，在第一个元组（(23,1,18,4)）之后我会放弃(23,1,4,18), (23,4,1,18)等等......，如果可能的话，我想在整个列表中这样做，这样我最终只会得到6个完全不同的4元组。有没有办法做到这一点，而不是在我的程序中先回去做一些不同的事情？任何帮助将不胜感激。谢谢！

Answer 1

怎么样：

{tuple(sorted(t)): t for t in elementary}.values()

Answer 2

作为1-liner，对每个4元组进行排序，然后创建一组结果，shich具有删除重复的效果。我假设你的4元组被允许改变元素的顺序。

set(tuple(sorted(i)) for i in elementary)

>>> set((5, 7, 14, 23), (6, 8, 9, 23), (2, 3, 18, 23), (1, 4, 18, 23), (1, 6, 16, 23), (2, 9, 12, 23))

Answer 3

你真的只有24岁吗？如果是这样，带有一些不必要的内存分配的慢速解决方案在这里可以正常工作，并节省您的编程时间：

elementary_unique = set(tuple(sorted(t)) for t in elementary)

现在elementary_unique是一个集合而不是列表 - 如果重要，可以使用

elementary_unique = list(set(tuple(sorted(t)) for t in elementary))

相反，但这会比第一个版本慢一点。

Answer 4

您可以在sorted(tuple1) == sorted(tuple2)时识别等效组合。

代码很简短：

>>> set(map(tuple, map(sorted, elementary)))
set([(5, 7, 14, 23), (6, 8, 9, 23), (2, 3, 18, 23), 
     (1, 4, 18, 23), (1, 6, 16, 23), (2, 9, 12, 23)])

如果你需要保留每个第一个不同元组的顺序，那么需要更多的工作：

>>> uniq = set()
>>> for t in elementary:
    s = tuple(sorted(t))
    if s not in uniq:
        uniq.add(s)
        print t

(23, 1, 18, 4)
(23, 1, 6, 16)
(23, 2, 18, 3)
(23, 2, 12, 9)
(23, 5, 14, 7)
(23, 6, 9, 8)

Answer 5

如果你真的想通过比较来解决这个问题：

In [1028]: elementary = [(23, 1, 18, 4), (23, 1, 6, 16), (23, 1, 4, 18), (23, 2, 18, 3), (23, 2, 12, 9), (23, 2, 9, 12), (23, 2, 3, 18), (23, 3, 18, 2), (23, 3, 2, 18), (23, 4, 18, 1), (23, 4, 1, 18), (23, 5, 14, 7), (23, 5, 7, 14), (23, 6, 16, 1), (23, 6, 9, 8), (23, 6, 8, 9), (23, 6, 1, 16), (23, 7, 14, 5), (23, 7, 5, 14), (23, 8, 9, 6), (23, 8, 6, 9), (23, 9, 12, 2), (23, 9, 8, 6), (23, 9, 6, 8), (23, 9, 2, 12), (23, 12, 9, 2), (23, 12, 2, 9), (23, 14, 7, 5), (23, 14, 5, 7), (23, 16, 1, 6), (23, 18, 4, 1), (23, 18, 3, 2), (23, 18, 2, 3), (23, 18, 1, 4)]

In [1029]: for e in elementary:                                                                                                                                  
               add = True                                                                                                                                                   
               for a in answer:                                                                                                                                                 
                   if all(_e in a and e.count(_e)==a.count(_e) and len(e)==len(a) for _e in e):
                       add = False
               if add:
                   answer.append(e)    

In [1030]: answer
Out[1030]: 
[(23, 1, 18, 4),
 (23, 1, 6, 16),
 (23, 2, 18, 3),
 (23, 2, 12, 9),
 (23, 5, 14, 7),
 (23, 6, 9, 8)]

Python元组列表

5 个答案: