获取列表中的元组数，而与元素顺序无关 [英] Get count of tuples in list regardless of elements orders

问题描述

我有一个看起来像这样的元组列表:

I have a list of tuples that looks something like this:

my_list = [(1,12),(12,1),(12,1),(20,15),(7,8),(15,20)]

无论顺序如何，我都希望对数字组合进行计数.例如，如果要简单打印，我希望输出为:

I want to get a count of the number combinations regardless of the order. For example, if it were to be simply printed, I'd like the output to be:

1,12 = 3
20,15 = 2
7,8 = 1

基本上，我有一个连接列表，但方向无关紧要，所以1到12与12到1相同.

Basically, I have a list of connections but the direction doesn't matter, so 1 to 12 is the same as 12 to 1.

我已经为此工作了一段时间，无法提出一个干净的解决方案.我想出的一切都必须检查两个方向，但是元组的列表是随机的，因此涵盖所有可能性将是荒谬的.我可以很容易地用set或其他东西来计算唯一的元组，但是同样，我想到的双向"计数的每个解决方案都是草率的.

I have been working on this for a while and can't come up with a clean solution. Everything I've come up with would have to check for both directions, but the list of tuples is random, so covering every possibility would be ridiculous. I could easily count the unique tuples with set or something, but again, every solution for the "bi-directional" count I've thought about is sloppy.

我觉得这应该不是很难，但是我花了这么长时间来研究，对此我感到非常困惑.任何帮助将不胜感激！

I feel like this shouldn't be very hard, but I am thoroughly at brain fry from working on this so long. Any help would be greatly appreciated!

推荐答案

您可以按大小交换所有职位，然后使用collections.counter:

You can swap all positions by size and then use collections.counter:

from collections import Counter
l = [(1,12),(12,1),(12,1),(20,15),(7,8),(15,20)]
new_l = Counter([[(b, a), (a, b)][a < b] for a, b in l])
for a, b in new_l.items():
  print('{},{}:{}'.format(*(list(a)+[b])))

输出:

15,20:2
7,8:1
1,12:3

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