comparing two lists elements in python

Question

I have 2 lists which consider versions '2.0.0', Im comparing their element and get the list of [True, True, True] meanings, how can i take from list of [True, True, True] meanings, only one meaning True, or if it will be 1 False in that list, how can i get False, globally, i need to override magic methor eq and firstly Im trying to make it simple functionally. I expect to see True if they are equil, and False if something is different in those to lists.

import numpy

a = ['2','0','0']
b = ['2','0','0']

print(numpy.in1d(a, b))
if numpy.in1d(a, b) == [True, True, True]:
    print('equils')

Answer 1

您正在寻找numpy.all

numpy.all(a == b)

Answer 2

Just use all()

if all(numpy.in1d(a, b)):
    print('equils')

Answer 3

I'm not sure why you would use numpy for that when a simple comparison of tuples will work out of the box

a = ('2','0','0')
b = ('2','0','0')

a == b # --> True


a = ('2','1','0')
b = ('2','0')

a == b # --> False