计算列表中的元组,无论元组中每个元素的顺序如何

Question

我有一个关于在Python中列出列表中的事件的问题.

这是清单:

aList = [(11, 0), (9, 7), (23, 9), (25, 3), (9, 9), (21, 2), (10, 9), (14, 14), (8, 13), (14, 9), (11, 4), (1, 14), (3, 9), (3, 1), (11, 9), (9, 1), (7, 0), (9, 3), (9, 3), (16, 11), (9, 7), (9, 13), (11, 9), (26, 18), (18, 9), (11, 14), (9, 9), (24, 26), (12, 21), (1, 14), (3, 14), (15, 14), (26, 9), (11, 3), (4, 14), (9, 14), (26, 4), (7, 26), (9, 3), (13, 3), (9, 24), (14, 9), (3, 26), (7, 25), (5, 9), (9, 5), (14, 4), (9, 0), (4, 26), (4, 26), (9, 9), (18, 18), (9, 7), (7, 6), (9, 9), (14, 13), (11, 9), (3, 9), (9, 15), (25, 9), (10, 24), (0, 4), (10, 3), (8, 12), (9, 4), (20, 9), (9, 9), (6, 9), (9, 8), (9, 9), (24, 16), (9, 11), (14, 9), (7, 12), (1, 9), (9, 13), (5, 13), (9, 9), (25, 9), (4, 9), (1, 3), (10, 9), (12, 9), (9, 9), (11, 11), (14, 3), (9, 25), (16, 1), (25, 16), (25, 16), (5, 9), (9, 3), (5, 1), (10, 24), (10, 25), (24, 1), (9, 1), (24, 9), (9, 7), (25, 3)]

如您所见,这是一个元组列表.无论元组中每个元素的顺序如何,我都需要计算每个元组的出现次数.

我正试图像这样使用Collections的计数器:

cnt = Counter()
for l in aList:
    cnt[l] += 1

print cnt

像(3,9)和(9,3)这样的元组应该算作相同的,我的代码现在不做.有没有办法用计数器实现这种计数？或者我应该使用其他方法？

谢谢!

Answer 1

这应该这样做:

counter = Counter(tuple(sorted(tup)) for tup in your_list)
print counter