Quickest and most concise way to generate a list of random, sorted integers

Question

I want to generate a list of length n which contains n random unique integers, ordered by numerical value and with a difference of no more than 10 between any two elements.

I wrote this function:

import random
def make_list(n):
  list = []
  num = 0
  for i in range(30):
    num = num + random.randint(1,10)
    list.append(num)
  return(list)

It works but it seems verbose and I feel like there should be a better, more concise way using list comprehensions or other.

Answer 1

You could use a generator expression to generate the intervals, then call itertools.accumulate to calculate the running sums.

>>> from itertools import accumulate
>>> from random import randint

>>> list(accumulate(randint(1, 10) for _ in range(30)))
[4, 8, 12, 16, 20, 26, 31, 41, 49, 50, 58, 63, 72, 77, 81, 86, 88, 92, 101, 106, 109, 114, 116, 122, 123, 133, 143, 153, 159, 160]

Answer 2

import random

def make_list(n):
  num = 0
  return [num := num + random.randint(1, 10) for _ in range(n)]

You can use := operator after Python3.8

https://stackoverflow.com/a/54514656/17560347

Answer 3

Using islice and accumulate from the itertools module:

import random
import itertools as it

def make_list(n):
    i = iter(lambda: random.randint(1, 10), None)
    return list(it.islice(it.accumulate(i), n))

Answer 4

You can use random.choices with the desired n passed as the k argument:

import random
from itertools import accumulate

print(list(accumulate(random.choices(range(1, 11), k=5))))

Demo: https://replit.com/@blhsing/VerifiablePleasingSolidstatedrive