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
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]
>>>
其他回答
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)
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
您可能希望使用集合而不是列表。
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])
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。