How to add exception to random.randint in Python?

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Chapters
00:00 How To Add Exception To Random.Randint In Python?
00:35 Accepted Answer Score 8
00:49 Answer 2 Score 6
01:03 Answer 3 Score 1
01:36 Answer 4 Score 0
01:50 Thank you

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Full question
https://stackoverflow.com/questions/4210...

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Content licensed under CC BY-SA
https://meta.stackexchange.com/help/lice...

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Tags
#python #random

#avk47

ACCEPTED ANSWER

Score 8

You can do

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

ANSWER 2

Score 6

How about

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

for a one-liner

ANSWER 3

Score 1

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

ANSWER 4

Score 0

!= 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)