Compare to multiple values in an if statement [duplicate]

Answer 1

You can use in, or check for the index and catch the error:

num in a will check if the item num is in the list a.

>>> 1 in [1, 2, 5]
True
>>> 3 in [1, 2, 5]
False
>>> 100 in range(101)
True

try getting the index, and except to catch the IndexError:

def isIn(item, lst):
    try:
        lst.index(item)
        return True
    except ValueError:
        return False
    return False

>>> isIn(5, [1, 2, 5])
True
>>> isIn(5, [1, 2, 3])
False

Answer 2

Use the in operator

if num in a : 

as in

def test(num): 
    a = [1, 2, 3]
    if num in a : 
        return True
    else : 
        return False

a work around would be (as suggested by Padraic)

 def test(num): 
        a = [1, 2, 3]
        return num in a

This would work because, The in operator compares if the LHS is present in the RHS and returns a boolean value respectively.

Also this is possible

test = lambda x:  num in [1, 2, 3]

That is all in a single line!