有三种选择，其中两种是可以接受的，另一种不应该这样做。

在较高的级别上，这3个选项是:

减去两组(有时最好) 检查每个列表项是否存在于一个集合中(大多数情况下最好) 检查每个列表项是否存在于列表中(不做)

选项3)永远不应该超过选项2)。根据应用程序的需要，您可能更喜欢选项1)或2)，而在大多数用例中，2)可能是首选方法。2)与1)的性能非常相似，因为两者都具有O(m + n)个时间复杂度。相比之下，2)在空间复杂度上比1)有边际优势，并且既保持了原始列表的顺序，又保持了原始列表中的任何重复。

如果你想删除重复，不关心顺序，那么1)可能是最适合你的。

import time

def fun1(l1, l2):
    # Order and duplications in l1 are both lost, O(m) + O(n)
    return set(l1) - set(l2)

def fun2(l1, l2):
    # Order and duplications in l1 are both preserved, O(m) + O(n)
    l2_set = set(l2)
    return [item for item in l1 if item not in l2_set]

def fun3(l1, l2):
    # Order and duplications in l1 are both preserved, O(m * n)
    # Don't do
    return [item for item in l1 if item not in l2]

A = list(range(7500))
B = list(range(5000, 10000))

loops = 100

start = time.time()
for _ in range(loops):
    fun1(A, B)
print(f"fun1 time: {time.time() - start}")

start = time.time()
for _ in range(loops):
    fun2(A, B)
print(f"fun2 time: {time.time() - start}")

start = time.time()
for _ in range(loops):
    fun3(A, B)
print(f"fun3 time: {time.time() - start}")

fun1 time: 0.03749704360961914
fun2 time: 0.04109621047973633
fun3 time: 32.55076885223389

一个衬套:

diff = lambda l1,l2: [x for x in l1 if x not in l2]
diff(A,B)
diff(B,A)

Or:

diff = lambda l1,l2: filter(lambda x: x not in l2, l1)
diff(A,B)
diff(B,A)

当查看in -operator的TimeComplexity时，在最坏的情况下它与O(n)一起工作。即使是集合。

因此，当比较两个数组时，最好情况下的TimeComplexity为O(n)，最坏情况下为O(n²)。

另一种(但不幸的是更复杂)解决方案，在最好和最坏的情况下都适用于O(n):

# Compares the difference of list a and b
# uses a callback function to compare items
def diff(a, b, callback):
  a_missing_in_b = []
  ai = 0
  bi = 0

  a = sorted(a, callback)
  b = sorted(b, callback)

  while (ai < len(a)) and (bi < len(b)):

    cmp = callback(a[ai], b[bi])
    if cmp < 0:
      a_missing_in_b.append(a[ai])
      ai += 1
    elif cmp > 0:
      # Item b is missing in a
      bi += 1
    else:
      # a and b intersecting on this item
      ai += 1
      bi += 1

  # if a and b are not of same length, we need to add the remaining items
  for ai in xrange(ai, len(a)):
    a_missing_in_b.append(a[ai])


  return a_missing_in_b

e.g.

>>> a=[1,2,3]
>>> b=[2,4,6]
>>> diff(a, b, cmp)
[1, 3]

添加一个答案来处理我们想要严格区别重复的情况，也就是说，在第一个列表中有我们想要保留在结果中的重复。例如，得到，

[1, 1, 1, 2] - [1, 1] --> [1, 2]

我们可以用一个额外的计数器来得到一个优雅的差分函数。

from collections import Counter

def diff(first, second):
    secondCntr = Counter(second)
    second = set(second)
    res = []
    for i in first:
        if i not in second:
            res.append(i)
        elif i in secondCntr:
            if secondCntr[i] > 0:
                secondCntr[i] -= 1
            else:
                res.append(i)        
    return res

最简单的方法，

使用设置().difference(设置())

list_a = [1,2,3]
list_b = [2,3]
print set(list_a).difference(set(list_b))

答案设置([1])

你可以做一个

list(set(A)-set(B))

and

list(set(B)-set(A))