Struct rand::distributions::slice::Slice
source · pub struct Slice<'a, T> {
slice: &'a [T],
range: Uniform<usize>,
}
Expand description
A distribution to sample items uniformly from a slice.
Slice::new
constructs a distribution referencing a slice and uniformly
samples references from the items in the slice. It may do extra work up
front to make sampling of multiple values faster; if only one sample from
the slice is required, SliceRandom::choose
can be more efficient.
Steps are taken to avoid bias which might be present in naive
implementations; for example slice[rng.gen() % slice.len()]
samples from
the slice, but may be more likely to select numbers in the low range than
other values.
This distribution samples with replacement; each sample is independent.
Sampling without replacement requires state to be retained, and therefore
cannot be handled by a distribution; you should instead consider methods
on SliceRandom
, such as SliceRandom::choose_multiple
.
§Example
use rand::Rng;
use rand::distributions::Slice;
let vowels = ['a', 'e', 'i', 'o', 'u'];
let vowels_dist = Slice::new(&vowels).unwrap();
let rng = rand::thread_rng();
// build a string of 10 vowels
let vowel_string: String = rng
.sample_iter(&vowels_dist)
.take(10)
.collect();
println!("{}", vowel_string);
assert_eq!(vowel_string.len(), 10);
assert!(vowel_string.chars().all(|c| vowels.contains(&c)));
For a single sample, SliceRandom::choose
may be preferred:
use rand::seq::SliceRandom;
let vowels = ['a', 'e', 'i', 'o', 'u'];
let mut rng = rand::thread_rng();
println!("{}", vowels.choose(&mut rng).unwrap())
Fields§
§slice: &'a [T]
§range: Uniform<usize>