use of if else statement in python getting runtime error

Question

n=range(101)
if n%2==0:
     print("weird")
if range(2,6):
     print("not weird")
if range(6,21):
     print("weird")
if n>20:
     print("not weird")
else :
     print("weird")

Given an integer, , perform the following conditional actions:

If is odd, print Weird If is even and in the inclusive range of 2 to 5, print Not Weird If is even and in the inclusive range of 6 to 20 , print Weird If is even and greater than 20, print Not Weird

Please include the error that you're getting

John M
– John M

2018-08-30 14:43:21 +00:00
Commented Aug 30, 2018 at 14:43 — John M
– John M, Commented Aug 30, 2018 at 14:43
it's weird indeed

fantaghirocco
– fantaghirocco

2018-08-30 15:08:03 +00:00
Commented Aug 30, 2018 at 15:08 — fantaghirocco
– fantaghirocco, Commented Aug 30, 2018 at 15:08

blhsing · Accepted Answer · 2018-08-30 14:31:15Z

1

You should use a for loop to iterate through the range generator, and you should use the in operator to test if n is in a range, and if you mean to print only upon the first matching rule for each number, you should use elif statements instead of if for the conditions that follows the first. You might also want to print the number itself so that you know which numbers are "weird" and which are "not weird":

for n in range(101):
    print(n, end=' ')
    if n%2==0:
         print("weird")
    elif n in range(2,6):
         print("not weird")
    elif n in range(6,21):
         print("weird")
    elif n>20:
         print("not weird")
    else:
         print("weird")

answered Aug 30, 2018 at 14:31

blhsing

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