How do I generate non-repeating random numbers in a while loop? (Python 3)

Question

So far what I have is

import random
def go():
    rounds = 0
    while rounds < 5:
        number = random.randint(1, 5)
        if number == 1:
            print('a')
        elif number == 2:
            print('b')
        elif number == 3:
            print('c')
        elif number == 4:
            print('d')
        elif number == 5:
            print('e')
        rounds = rounds + 1
go()

and the output ends up beings something along the lines of

e
e
c
b
e

How do I make it so a number is only used once and the letters do not repeat? (ex. something like)

a
e
b
c
d

Thanks in advance

you can try random.sample("abcde", 5) random library python 3 — Jose Ricardo Bustos M.
– Jose Ricardo Bustos M., Commented Sep 15, 2015 at 22:37
your specific code example suggests that you want L = list("abcde"); random.shuffle(L) i.e., get a random permutation of "abcde" letters. — jfs
– jfs, Commented Sep 15, 2015 at 23:37

asdf · Accepted Answer · 2015-09-15 22:43:53Z

9

The random.sample(population, k) method returns a sample of unique values from the specified population of length k.

r = random.sample("abcde", 5)
for element in r:
    print(element)

edited Sep 15, 2015 at 22:43

answered Sep 15, 2015 at 22:39

asdf

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5 Comments

asdf Over a year ago

@zeurosis if it worked for you please select it as the accepted answer

zeurosis Over a year ago

sorry how do I do that? I haven't used this website a whole lot

asdf Over a year ago

@zeurosis click the check mark under the upvote/downvote icons on the left of the post

asdf Over a year ago

@zeurosis you might want to do that on your previous questions as well ;)

jfs Over a year ago

random.sample("abcde", 5) is a weird way to emulate random.shuffle()

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