我有一个关于在 Python 中计算列表中出现次数的问题。
这是列表:
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。
我正在尝试像这样使用 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，我的代码现在不这样做。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!
最佳答案
应该这样做:
counter = Counter(tuple(sorted(tup)) for tup in your_list)
print counter
关于python - 计算列表中的元组，无论元组中每个元素的顺序如何，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/31953807/