在Python中,计算两个列表之间的差值的最佳方法是什么?
例子
A = [1,2,3,4]
B = [2,5]
A - B = [1,3,4]
B - A = [5]
在Python中,计算两个列表之间的差值的最佳方法是什么?
例子
A = [1,2,3,4]
B = [2,5]
A - B = [1,3,4]
B - A = [5]
当前回答
上面的例子使计算差异的问题变得微不足道。假设排序或重复数据删除确实使计算差异变得更容易,但如果您的比较无法承担这些假设,那么您将需要一个diff算法的重要实现。请参阅python标准库中的difflib。
#! /usr/bin/python2
from difflib import SequenceMatcher
A = [1,2,3,4]
B = [2,5]
squeeze=SequenceMatcher( None, A, B )
print "A - B = [%s]"%( reduce( lambda p,q: p+q,
map( lambda t: squeeze.a[t[1]:t[2]],
filter(lambda x:x[0]!='equal',
squeeze.get_opcodes() ) ) ) )
或Python3…
#! /usr/bin/python3
from difflib import SequenceMatcher
from functools import reduce
A = [1,2,3,4]
B = [2,5]
squeeze=SequenceMatcher( None, A, B )
print( "A - B = [%s]"%( reduce( lambda p,q: p+q,
map( lambda t: squeeze.a[t[1]:t[2]],
filter(lambda x:x[0]!='equal',
squeeze.get_opcodes() ) ) ) ) )
输出:
A - B = [[1, 3, 4]]
其他回答
您可能希望使用集合而不是列表。
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。
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
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])
你可以做一个
list(set(A)-set(B))
and
list(set(B)-set(A))