Making a random int just be itself once in loop

Question

So.. im working on this loop:

stuff_so_far = [intl_pt]

for i in range(0, num_pts - 1):
    rdm = random.randint(0, len(points_in_code) - 1)
    a = (stuff_so_far[i][0] + points_in_code[rdm][0]) // 2
    b = (stuff_so_far[i][1] + points_in_code[rdm][1]) // 2
    stuff_so_far.append((a, b))

Basically what i want to achive is to get a random index for "points_in_code" every time the code loops. It is doing that now, but what i want to know is, how do i make it not randomly repeat a number? As in, if in the first iteration of the loop rdm gets set to 1, and then in the second iteration of the loop rdm gets set to 3, and in some cases, rdm can be set to 1 again in the third itertion. How do i make it not be 1 again (as long as the loop is still going)?

Ive tried everything i know and searched online but i found nothing, how do i make that happen without altering my code too much? (im new to programming)
I know each time i call random.randint(), i am creating a single random number, it does not magically change to a new random not used before number everytime the loop iterates.

Asked By: new_programmer_python

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Source

Answer 1

You can use random.sample:

import random

points_in_code = [11, 4, 13, 18, 7, 12] # Just a example
num_pts = 4

indexes = random.sample(range(len(points_in_code)), num_pts - 1)
for rdm, i in zip(indexes, range(0, num_pts - 1)):
    pass # rdm will be a random unique index
    # i will increase 1 each iteration

You could also use enumerate(indexes) instead of zip(indexes, range(0, num_pts - 1)) but then you would need to reverse i and rdm.

See the documentation for random.sample for more info. See also info on zip and enumerate

Answered By: mousetail

Answer 2

Try below approach

import random as rnd

## Lets say you want at max 100
MAX=100

arr = [i for i in range(101)]

while True:
    curr = rnd.choice(arr)
    arr.remove(curr)
    print(curr)
    if len(arr)==0: break

Your code :

stuff_so_far = [intl_pt]
choice_arr = [i for i in range(len(points_in_code))]
for i in range(0, num_pts - 1):
    rdm = random.choice(choice_arr)
    choice_arr.remove(rdm)
    if len(choice_arr)==0:
        print("No more choices available")
        break

    a = (stuff_so_far[i][0] + points_in_code[rdm][0]) // 2
    b = (stuff_so_far[i][1] + points_in_code[rdm][1]) // 2
    stuff_so_far.append((a, b))

Answered By: Anubhav Sharma

Answer 3

You can use an Array for keep track of which random number you already used. You only need to regenerate the random number if you get one which you already had.

stuff_so_far = [intl_pt]
tracking_num = [] # keeps track of the random number that was alreadey used.


for i in range(num_pts - 1): # small tip you can create a for-loop without the zero in the beginning.
    # while loop in which a number will be regenerated if the current random number
    # is contained in the 'tracking_num' array.
    rdm = random.randint(0, len(points_in_code) - 1)
    while tracking_num.__contains__(rdm):
        rdm = random.randint(0, len(points_in_code) - 1)

    a = (stuff_so_far[i][0] + points_in_code[rdm][0]) // 2
    b = (stuff_so_far[i][1] + points_in_code[rdm][1]) // 2
    stuff_so_far.append((a, b))

Answered By: JPudel

Making a random int just be itself once in loop

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