你可以做一个

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

and

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

如果你不关心项目的顺序或重复，请使用set。使用列表推导式:

>>> def diff(first, second):
        second = set(second)
        return [item for item in first if item not in second]

>>> diff(A, B)
[1, 3, 4]
>>> diff(B, A)
[5]
>>> 

简单的代码，让你与多个项目的差异，如果你想要:

a=[1,2,3,3,4]
b=[2,4]
tmp = copy.deepcopy(a)
for k in b:
    if k in tmp:
        tmp.remove(k)
print(tmp)

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]

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))