def listsubtraction(parent,child):
    answer=[]
    for element in parent:
        if element not in child:
            answer.append(element)
    return answer

我认为这应该可行。我是初学者，所以请原谅我的错误

如果重复和订购项目是问题:

[i为a中的i，如果不是b中的i或b中的i，删除(i)]

a = [1,2,3,3,3,3,4]
b = [1,3]
result: [2, 3, 3, 3, 4]

试试这个。

def subtract_lists(a, b):
    """ Subtracts two lists. Throws ValueError if b contains items not in a """
    # Terminate if b is empty, otherwise remove b[0] from a and recurse
    return a if len(b) == 0 else [a[:i] + subtract_lists(a[i+1:], b[1:]) 
                                  for i in [a.index(b[0])]][0]

>>> x = [1,2,3,4,5,6,7,8,9,0]
>>> y = [1,3,5,7,9]
>>> subtract_lists(x,y)
[2, 4, 6, 8, 0]
>>> x = [1,2,3,4,5,6,7,8,9,0,9]
>>> subtract_lists(x,y)
[2, 4, 6, 8, 0, 9]     #9 is only deleted once
>>>

如果列表允许重复元素，你可以使用Counter from collections:

from collections import Counter
result = list((Counter(x)-Counter(y)).elements())

如果你需要保留x中元素的顺序:

result = [ v for c in [Counter(y)] for v in x if not c[v] or c.subtract([v]) ]

list1 = ['a', 'c', 'a', 'b', 'k'] 
list2 = ['a', 'a', 'a', 'a', 'b', 'c', 'c', 'd', 'e', 'f'] 
for e in list1: 
    try: 
        list2.remove(e) 
    except ValueError: 
        print(f'{e} not in list') 
list2 
# ['a', 'a', 'c', 'd', 'e', 'f']

这将改变list2。如果你想保护list2，只需复制它，并在这段代码中使用list2的副本。

使用一个列表推导式来计算差值，同时保持x的原始顺序:

[item for item in x if item not in y]

如果你不需要列表属性(例如，排序)，使用一个集差异，正如其他答案所建议的:

list(set(x) - set(y))

为了允许x - y中缀语法，在从list继承的类上重写__sub__:

class MyList(list):
    def __init__(self, *args):
        super(MyList, self).__init__(args)

    def __sub__(self, other):
        return self.__class__(*[item for item in self if item not in other])

用法:

x = MyList(1, 2, 3, 4)
y = MyList(2, 5, 2)
z = x - y