我有两个列表，每个列表都有非唯一的数字，这意味着它们可以多次具有相同的值。
我需要找到两者之间的区别，因为相同的值可能会出现多次(所以我无法区分每个集合之间的区别)。因此，我需要检查某个值在第一个列表中出现的次数是否多于第二个。
列表是:
l1 = [1, 2, 5, 3, 3, 4, 9, 8, 2]
l2 = [1, 1, 3, 2, 4, 8, 9]
# Sorted and justified
l1 = [1, 2, 2, 3, 3, 4, 5, 8, 9]
l2 = [1, 1, 2, 3, 4, 8, 9]
列表元素可以是字符串或整数或 float 。 所以结果列表应该是:
difference(l1, l2) == [3, 5, 2]
# There is an extra 2 and 3 in l1 that is not in l2, and a 5 in l1 but not l2.
difference(l2, l1) == [1]
# The extra 1 is the only value in l2 but not in l1.
我已经尝试过列表理解[x for x in l1 if x not in l2] 但这不起作用，因为它没有考虑到两者中的重复值。
请您参考如下方法:
如果顺序不重要，您可以使用计数器(参见标准库的collections 模块):
from collections import Counter
l1 = [1,2,5,3,3,4,9,8,2]
l2 = [1,1,3,2,4,8,9]
c1 = Counter(l1) # Counter({2: 2, 3: 2, 1: 1, 5: 1, 4: 1, 9: 1, 8: 1})
c2 = Counter(l2) # Counter({1: 2, 3: 1, 2: 1, 4: 1, 8: 1, 9: 1})
diff1 = list((c1-c2).keys()) # [2, 5, 3]
diff2 = list((c2-c1).keys()) # [1]
这是相当通用的，也适用于字符串:
...
l1 = ['foo', 'foo', 'bar']
l2 = ['foo', 'bar', 'bar', 'baz']
...
# diff1 == ['foo']
# diff2 == ['bar', 'baz']
