Python中重复的两个列表之间的区别

Question

我有两个列表，其中包含许多相同的项目，包括重复的项目。 我想检查第一个列表中的哪些项目不在第二个列表中。 例如，我可能有一个这样的列表：

l1 = ['a', 'b', 'c', 'b', 'c']

和一个这样的列表：

l2 = ['a', 'b', 'c', 'b']

比较这两个列表，我想返回这样的第三个列表：

l3 = ['c']

我目前正在使用我不久前编写的一些糟糕的代码，我相当肯定它甚至不能正常工作，如下所示。

def list_difference(l1,l2):
    for i in range(0, len(l1)):
        for j in range(0, len(l2)):
            if l1[i] == l1[j]:
                l1[i] = 'damn'
                l2[j] = 'damn'
    l3 = []
    for item in l1:
        if item!='damn':
            l3.append(item)
    return l3

我怎样才能更好地完成这项任务？

Answer 1

您没有说明订单是否重要。 如果没有，您可以在 >= Python 2.7 中执行此操作：

l1 = ['a', 'b', 'c', 'b', 'c']
l2 = ['a', 'b', 'c', 'b']

from collections import Counter

c1 = Counter(l1)
c2 = Counter(l2)

diff = c1-c2
print list(diff.elements())

Answer 2

为两个列表创建Counter ，然后subtract另一个列表中subtract一个。

from collections import Counter

a = [1,2,3,1,2]
b = [1,2,3,1]

c = Counter(a)
c.subtract(Counter(b))

Answer 3

要同时考虑重复项和元素的顺序：

from collections import Counter

def list_difference(a, b):
    count = Counter(a) # count items in a
    count.subtract(b)  # subtract items that are in b
    diff = []
    for x in a:
        if count[x] > 0:
           count[x] -= 1
           diff.append(x)
    return diff

例子

print(list_difference("z y z x v x y x u".split(), "x y z w z".split()))
# -> ['y', 'x', 'v', 'x', 'u']

Python 2.5 版本：

from collections import defaultdict 

def list_difference25(a, b):
    # count items in a
    count = defaultdict(int) # item -> number of occurrences
    for x in a:
        count[x] += 1

    # subtract items that are in b
    for x in b: 
        count[x] -= 1

    diff = []
    for x in a:
        if count[x] > 0:
           count[x] -= 1
           diff.append(x)
    return diff

Answer 4

计数器是 Python 2.7 中的新功能。 对于从 b 中减去 a 的一般解决方案：

def list_difference(b, a):
    c = list(b)
    for item in a:
       try:
           c.remove(item)
       except ValueError:
           pass            #or maybe you want to keep a values here
    return c

Answer 5

你可以试试这个

list(filter(lambda x:l1.remove(x),li2)) print(l1)

Answer 6

试试这个：

from collections import Counter
from typing import Sequence

def duplicates_difference(a: Sequence, b: Sequence) -> Counter:
    """
    >>> duplicates_difference([1,2],[1,2,2,3])
    Counter({2: 1, 3: 1})
    """
    shorter, longer = sorted([a, b], key=len)
    return Counter(longer) - Counter(shorter)