Explain this code please

Question

x, y, z = 0, 1, 0
if x == 1 or y == 1 or z == 1: print('passed')
if 1 in (x, y, z):
print('passed')
if x or y or z:
print('passed')
if any((x, y, z)):
print('passed')
#output
passed
passed
passed
passed

Answer 1

Ramya Thiurmalisamy any() returns true if any of the element in the iterable is true, else it returns false... So, any((0,1,0)) returns true... Coz, y=1 (ie., true)

Answer 2

The conditions you specified all are true... So, each if condition gets executed separately one after the other (since it's not nested if) and prints passed 4 times...
The condition of or is : if any of the two conditions is tfue then it returns true..
So, in 1st condition( false or true or false) == true... Prints passed 1 time
2nd condition : 1 is in (0,1,0) so, true... Prints passed 2nd time...
3rd condition : 0 or 1 or 0 ==True.. So, prints passed 3rd time
4th condition : any((0,10)) gives true... So prints passed 4th time....
Hope you understand.. 👍👍

Answer 3

any(0,1,0) here, any one of from argument list, otherthan zero treated as true.. In 3rd condition also x or y or z if any one of x, y, z not zero results true..
any(0,2,0) =>true
(0 or 5 or 0) results true..

Answer 4

sarada lakshmi Tq

Answer 5

sarada lakshmi if you please explain how the 4 th condition becomes true?

Answer 6

Jayakrishna🇮🇳 Tq.