Python中重復的兩個列表之間的區別

Question

我有兩個列表，其中包含許多相同的項目，包括重復的項目。 我想檢查第一個列表中的哪些項目不在第二個列表中。 例如，我可能有一個這樣的列表：

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

和一個這樣的列表：

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

比較這兩個列表，我想返回這樣的第三個列表：

l3 = ['c']

我目前正在使用我不久前編寫的一些糟糕的代碼，我相當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

我怎樣才能更好地完成這項任務？

Answer 1

您沒有說明訂單是否重要。 如果沒有，您可以在 >= Python 2.7 中執行此操作：

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

為兩個列表創建Counter ，然后subtract另一個列表中subtract一個。

from collections import Counter

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

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

Answer 3

要同時考慮重復項和元素的順序：

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

計數器是 Python 2.7 中的新功能。 對於從 b 中減去 a 的一般解決方案：

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

你可以試試這個

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

Answer 6

試試這個：

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)