我如何比较python中的两个列表，并返回第二个需要具有相同值而不管顺序？

Question

a = [1, 2, 3, 4]
b = [2, 4, 3, 1]
c = [2, 3]

When comparing a to b, should return True : all items in a are presented in b , and all items in b are presented in a .

When comparing a to c , should return False : there are items in a that don't exist on c .

What is the pythonic way to do it?

Answer 1

排序，然后比较。

sorted(a) == sorted(b)

Answer 2

Use sets or frozensets.

set_a = {1, 2, 3, 4} #python 2.7 or higher set literal, use the set(iter) syntax for older versions
set_b = {2, 4, 4, 1}

set_a == set_b

set_a - set_b == set_b - set_a

The biggest advantage of using sets over any list method is that it's highly readable, you haven't mutated your original iterable, it can perform well even in cases where a is huge and b is tiny (checking whether a and b have the same length first is a good optimization if you expect this case often, though), and using the right data structure for the job is pythonic.

Answer 3

Use set s:

In [4]: set(a) == set(b)
Out[4]: True

In [5]: set(a) == set(c)
Out[5]: False

Answer 4

Turn them into sets:

>>> set([1,2,3,4]) == set([2,4,3,1])
True

>>> set([2, 3]) == set([1,2,3,4])
False

If your lists contain duplicate items, you'll have to compare their lengths too. Sets collapse duplicates.