In Python 3.x, I need to choose an integer at random among the indices of a given list. In other words, given
mylist = [0, 5, 6, 8, -10] I want to return a number between 0 and 4. What's the most Pythonic way to do it? I tried
import numpy as np my_list = [0, 5, 6, 8, -10] def choose_at_random(a_list): choice = np.random.randint(0, len(a_list)) return choice This works, but is this the Pythonic way to do it?
If you just want to use something from Python's standard library (and don't need anything vectorised like numpy) randrange is generally the easiest method for accomplishing this.
You'd use it something like:
from random import randrange from typing import Sized def random_index_from_sized(a_list: Sized) -> int: return randrange(len(a_list)) my_list = [0, 5, 6, 8, -10] random_index_from_sized(my_list) which would return an integer value in [0, 4].
numpy's randint is similarly defined, so could be used in the above definition as:
from numpy.random import randint def random_sized_index(a_list: Sized) -> int: return randint(len(a_list)) Returning a single value from numpy is kind of pointless, i.e. numpy is designed for returning large arrays. A quick timeit test says that randrange(5) takes ~0.3µs while randint(5) takes ~2µs (for a single value). If you want, e.g., 1000 values then [randrange(5) for _ in range(1000)] takes ~300µs, while randint(5, size=1000) only takes ~20µs.
import random #random is in the standard library(so no need to "pip") my_list = [0, 5, 6, 8, -10] def choose_at_random(a_list): choice = (random.choice(a_list)) return choice 
my_list (mostly, because they also seem to want an extra out-of-bounds index at the end as well). you just return uniform draws from the list
my_list. Thanks for making me notice my mistake. The question is now fixed.
random.randintis inclusive in both ends.+1, right?np.random.randintbut for builtinrandom.randintyou don't need the+1at the end.