CODEWITHSUNDEEP
pythonrandompython-itertools
maze
maze

Reputation: 940

How can I generate n random numbers without a for loop

I am trying to generate 4 random integers between 0-9. I can easily achieve this using a for-loop:

digits = [random.randrange(10) for i in range(4)]

Assuming it is possible, I was wondering about other ways to yield the same result, perhaps without directly using a for-loop. I was experimenting with itertools and this didn't work:

digits = itertools.repeat(random.randrange(10), 4)

The main purpose of my question is to expose myself to additional methods. Thanks.

Upvotes: 8

Views: 4923

Answers (3)

Chiheb Nexus
Chiheb Nexus

Reputation: 9257

What about using a generator and a while loop ? 

from random import randrange

def rand(val, repeat=4):
    j = 1
    while j <= repeat:
        yield randrange(val)
        j += 1

final = list(rand(10))

Output:

[2, 6, 8, 8]

Also, using a recursion:

from random import randrange

def rand(val, repeat=4, b=1, temp=[]):
    if b < repeat:
        temp.append(randrange(val))
        return rand(val, b= b+1, temp=temp)
    else:
        return temp + [randrange(val)]

final = rand(10)
print(final)

Output:

[1, 9, 3, 1]

Upvotes: 1

Leon
Leon

Reputation: 32484

This is not really something that I would do, but since you would like to expose yourself to additional methods, then taking advantage of the 0-9 range you can also obtain 4 such integers as follows:

map(int, "%04d"%random.randrange(10**4))

It actually generates a four digit string, and extracts the digits from it.

Upvotes: 0

codelessbugging
codelessbugging

Reputation: 2909

numpy.random.randint(0, 10, size=4)

or to use itertools

list(map(random.randrange, itertools.repeat(10, 4)))

or even

list(map(random.randrange, [10]*4))

Upvotes: 14

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