可能重复值的两个列表之间的Python差异

Question

我有两个列表，每个列表都具有非唯一编号，这意味着它们可以具有多次相同的值。

我需要找到两者之间的差异，考虑到相同的值可能会出现多次（所以我不能采取每组之间的差异）的事实。所以，我需要检查一个值是否在第一个列表中出现的次数多于第二个列表中的次数。

的列表是：

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] 

列表中的元件可以是字符串或整数或浮点数。 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. 

我已经试过列表理解[x for x in l1 if x not in l2]这是不行的，因为它没有考虑在这两个重复的值。

Answer 1

如果订单不重要的是，你可以使用一个Counter（见collections模块的标准库）：

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] 

这是相当普遍的，并与琴弦的作品，太：

... 
l1 = ['foo', 'foo', 'bar'] 
l2 = ['foo', 'bar', 'bar', 'baz'] 
... 
# diff1 == ['foo'] 
# diff2 == ['bar', 'baz'] 

Answer 2

我有一种感觉，很多人会来这里为multiset的差异（例如：[1, 1, 1, 2, 2, 2, 3, 3] - [1, 2, 2] == [1, 1, 2, 3, 3]），所以我也会在这里发布该答案：

import collections 

def multiset_difference(a, b): 
    """Compute a - b of two multisets a and b""" 
    a = collections.Counter(a) 
    b = collections.Counter(b) 

    difference = a - b 
    return difference # Remove this line if you want it as a list 

    as_list = [] 
    for item, count in difference.items(): 
     as_list.extend([item] * count) 
    return as_list 

def ordered_multiset_difference(a, b): 
    """As above, but preserves order and is O(ab) worst case""" 
    difference = list(a) 
    depleted = set() # Values that aren't in difference to prevent searching the list again 
    for i in b: 
     if i not in depleted: 
      try: 
       difference.remove(i) 
      except ValueError: 
       depleted.add(i) 
    return difference 

Answer 3

使用Counter可能是一个更好的选择，但要自己把它卷：

def diff(a, b): 
    result = [] 
    cpy = b[:] 
    for ele in a: 
     if ele in cpy: 
      cpy.remove(ele) 
     else: 
      result.append(ele) 
    return result 

或虐待的一行：

def diff(a, b): 
    return [ele for ele in a if ele not in b or b.remove(ele)] 

的一个衬垫的过程中破坏b的差异，所以你可能想通过它一个副本：diff(l1, l2[:])，或使用：

def diff(a, b): 
    cpy = b[:] 
    return [ele for ele in a if ele not in cpy or cpy.remove(ele)]