making the loop to continue to produce for outputs

Question

Here is the code I have:

def generate(x)
    two = {}
    for x in range(1,7315):
        two.update({vowels[random.randint(0,4)] + alpha[random.randint(0,21)]:0})  
        return two      
generate(x)

this only returns a single value, how could I make it return multiple values?

`return` causes function to exit and stop. You can't make a function return multiple values. — Akshat Mahajan, Apr 05 '16 at 03:34
Maybe you want to use `yield` keyword. Have a look at this [answer](http://stackoverflow.com/a/231855/2932244). — JRodDynamite, Apr 05 '16 at 03:53

score 0 · Answer 1 · answered Apr 05 '16 at 03:36

return a tuple with your values

def returnTwoNumbers():
    return (1, 0)

print(returnTwoNumbers()[0])
print(returnTwoNumbers()[1])

#output:
#1
#0

It also looks like you're trying to get a random vowel from your list of vowels. Using random.choice is a much more readable way to get a random item from a list:

import random
vowelList = ['a', 'e', 'i', 'o', 'u']

print (random.choice(vowelList))

score 0 · Answer 2 · answered Apr 05 '16 at 03:37

0

You can use a tuple to return multiple values from a function e.g.:

return (one, two, three)

answered Apr 05 '16 at 03:37

heemayl

39,294
7
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score 0 · Answer 3 · answered Apr 05 '16 at 03:42

0

You have wrong indentation

def generate():
    two = {}
    for x in range(1,7315):
        two.update({vowels[random.randint(0,4)] + alpha[random.randint(0,21)]:0})  
    return two      

twos = generate()

answered Apr 05 '16 at 03:42

Sergey Gornostaev

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3
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making the loop to continue to produce for outputs

3 Answers3