Python 2.7.3(默认，2014年2月27日，19:58:35)- IPython 1.1.0 - timeit:(github gist)

def diff(a, b):
  b = set(b)
  return [aa for aa in a if aa not in b]

def set_diff(a, b):
  return list(set(a) - set(b))

diff_lamb_hension = lambda l1,l2: [x for x in l1 if x not in l2]

diff_lamb_filter = lambda l1,l2: filter(lambda x: x not in l2, l1)

from difflib import SequenceMatcher
def squeezer(a, b):
  squeeze = SequenceMatcher(None, a, b)
  return reduce(lambda p,q: p+q, map(
    lambda t: squeeze.a[t[1]:t[2]],
      filter(lambda x:x[0]!='equal',
        squeeze.get_opcodes())))

结果:

# Small
a = range(10)
b = range(10/2)

timeit[diff(a, b)]
100000 loops, best of 3: 1.97 µs per loop

timeit[set_diff(a, b)]
100000 loops, best of 3: 2.71 µs per loop

timeit[diff_lamb_hension(a, b)]
100000 loops, best of 3: 2.1 µs per loop

timeit[diff_lamb_filter(a, b)]
100000 loops, best of 3: 3.58 µs per loop

timeit[squeezer(a, b)]
10000 loops, best of 3: 36 µs per loop

# Medium
a = range(10**4)
b = range(10**4/2)

timeit[diff(a, b)]
1000 loops, best of 3: 1.17 ms per loop

timeit[set_diff(a, b)]
1000 loops, best of 3: 1.27 ms per loop

timeit[diff_lamb_hension(a, b)]
1 loops, best of 3: 736 ms per loop

timeit[diff_lamb_filter(a, b)]
1 loops, best of 3: 732 ms per loop

timeit[squeezer(a, b)]
100 loops, best of 3: 12.8 ms per loop

# Big
a = xrange(10**7)
b = xrange(10**7/2)

timeit[diff(a, b)]
1 loops, best of 3: 1.74 s per loop

timeit[set_diff(a, b)]
1 loops, best of 3: 2.57 s per loop

timeit[diff_lamb_filter(a, b)]
# too long to wait for

timeit[diff_lamb_filter(a, b)]
# too long to wait for

timeit[diff_lamb_filter(a, b)]
# TypeError: sequence index must be integer, not 'slice'

@roman-bodnarchuk列表推导函数def diff(a, b)似乎更快。

Python 2.7.3(默认，2014年2月27日，19:58:35)- IPython 1.1.0 - timeit:(github gist)

def diff(a, b):
  b = set(b)
  return [aa for aa in a if aa not in b]

def set_diff(a, b):
  return list(set(a) - set(b))

diff_lamb_hension = lambda l1,l2: [x for x in l1 if x not in l2]

diff_lamb_filter = lambda l1,l2: filter(lambda x: x not in l2, l1)

from difflib import SequenceMatcher
def squeezer(a, b):
  squeeze = SequenceMatcher(None, a, b)
  return reduce(lambda p,q: p+q, map(
    lambda t: squeeze.a[t[1]:t[2]],
      filter(lambda x:x[0]!='equal',
        squeeze.get_opcodes())))

结果:

# Small
a = range(10)
b = range(10/2)

timeit[diff(a, b)]
100000 loops, best of 3: 1.97 µs per loop

timeit[set_diff(a, b)]
100000 loops, best of 3: 2.71 µs per loop

timeit[diff_lamb_hension(a, b)]
100000 loops, best of 3: 2.1 µs per loop

timeit[diff_lamb_filter(a, b)]
100000 loops, best of 3: 3.58 µs per loop

timeit[squeezer(a, b)]
10000 loops, best of 3: 36 µs per loop

# Medium
a = range(10**4)
b = range(10**4/2)

timeit[diff(a, b)]
1000 loops, best of 3: 1.17 ms per loop

timeit[set_diff(a, b)]
1000 loops, best of 3: 1.27 ms per loop

timeit[diff_lamb_hension(a, b)]
1 loops, best of 3: 736 ms per loop

timeit[diff_lamb_filter(a, b)]
1 loops, best of 3: 732 ms per loop

timeit[squeezer(a, b)]
100 loops, best of 3: 12.8 ms per loop

# Big
a = xrange(10**7)
b = xrange(10**7/2)

timeit[diff(a, b)]
1 loops, best of 3: 1.74 s per loop

timeit[set_diff(a, b)]
1 loops, best of 3: 2.57 s per loop

timeit[diff_lamb_filter(a, b)]
# too long to wait for

timeit[diff_lamb_filter(a, b)]
# too long to wait for

timeit[diff_lamb_filter(a, b)]
# TypeError: sequence index must be integer, not 'slice'

@roman-bodnarchuk列表推导函数def diff(a, b)似乎更快。

您可能希望使用集合而不是列表。

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])

在字典列表的情况下，当集合解引发时，完整列表理解解工作

TypeError: unhashable type: 'dict'

测试用例

def diff(a, b):
    return [aa for aa in a if aa not in b]

d1 = {"a":1, "b":1}
d2 = {"a":2, "b":2}
d3 = {"a":3, "b":3}

>>> diff([d1, d2, d3], [d2, d3])
[{'a': 1, 'b': 1}]
>>> diff([d1, d2, d3], [d1])
[{'a': 2, 'b': 2}, {'a': 3, 'b': 3}]

如果顺序无关紧要，你可以简单地计算集合差值:

>>> set([1,2,3,4]) - set([2,5])
set([1, 4, 3])
>>> set([2,5]) - set([1,2,3,4])
set([5])