I am using numpy's random.choise to get n integers in range(0,1000), where the same integer can't be chosen twice (replace=false).

But since each integer is a starting point of a sublist with a certain length (e.g.- 10), I want to have the random function not choosing any other integer that is - 10 from already chosen ones.

using the p= parameter is not possible because i do not know in advance which n's will be chosen. (It could be possible to use a loop - where each iteration the -10 integers from the newly chosen one are added to the list of probabilities with assigned probability 0, but that seems to me like a not optimal solution...)

for example:

myseries = list(range(1000))
n = 30 
blockRange = 10
np.random.choice(myseries, n, replace=False)

This returns 30 numbers, two of them are 33 and 37 - but I want to 'forbid' this (if 33 is there, no number between 24 to 42 should be allowed!)

thanks

CodePudding user response：

A solution is to build it iteratively, it will be slower than a numpy method for sure but I can't find a way to do your particular random distribution with numpy.

I create a res list and a forbidden set.

I iterate n times (here n=30), try to compute a random number in specific bounds. If it's not forbidden, meaning that this number is not in the forbidden set I can add it to my res list.

And I also add to my forbidden set every int in this range : [rand_num-9; rand_numb[.

But if it is in my forbidden set I retry thanks to a while loop to find a suitable random number.

code :

import random 

n, lower_bound, upper_bound = 30, 0, 1000
res = []
forbidden = set()
for _ in range(n):
    rand_numb = random.randint(lower_bound, upper_bound)
    while rand_numb in forbidden:
        rand_numb = random.randint(0,1000)
    res.append(rand_numb)
    forbidden = forbidden.union(set(range(rand_numb-9, rand_numb 10)))