How to add exception to random.randint in Python?

You can do

number = random.randint(0,10)
while number == 5:
   number = random.randint(0,10)

How about

random.choice([x for x in range(11) if x != 5])

for a one-liner

If you actually want an error to be raised then you can use assert

number = random.randint(0, 10)
assert number != 5

or raise an error if your condition is met.

number = random.randint(0, 10)
if number == 5:
    raise ValueError # or another Exception of choice

Or if you want to keep trying until you get a random number that isn't 5, then

while True:
    number = random.randint(0, 10)
    if number != 5:
        break

!= is actually an operator that returns true or false try this:

import random
l=[i for i in range(1,11)]
l.remove(5)
print random.choice(l)