Reputation: 38500
How do I cycle through an iterator a finite number of times?
I would expect the output of something like this to be 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3
and then stop:
vec![1, 2, 3].iter().cycle(4)
// ^ but .cycle() doesn't take an argument...
I don't know the length of the iterator to begin with.
Upvotes: 10
Views: 5324
Reputation: 38500
One could leverage slice::repeat
, but I can't imagine this is very efficient:
let v = vec![1, 2, 3];
let it = v.iter();
println!("{:?}", it.map(|x| *x).collect::<Vec<i32>>().repeat(4).iter());
Upvotes: 0
Reputation: 13942
One simple way is to repeat the iterator itself, take the first 4 and flatten:
fn main() {
let v = vec![1, 2, 3];
let res = std::iter::repeat(v.iter())
.take(4)
.flatten()
.collect::<Vec<_>>();
dbg!(res);
}
Some micro-benchmark result using code in this gist comparing 3 different approaches:
cycle_n
implementation mimicking Iterator::cycle
.Kudos to rustc
, cycle_n
consistently outperforms the other two when the input is reasonably large whereas repeat-take-flatten performs the best for small input.
Upvotes: 15
Reputation: 42739
There is no such an iterator in the std lib.
If you know the iterator size, you can take
your number times the size of the iterator:
fn cycle_n_times<T: Clone>(slice: &[T], count: usize) -> impl Iterator<Item = &T> {
slice.iter().cycle().take(slice.len() * count)
}
Or you can write your own that is more general:
pub struct Ncycles<I> {
orig: I,
iter: I,
count: usize,
}
impl<I: Clone> Ncycles<I> {
fn new(iter: I, count: usize) -> Ncycles<I> {
Ncycles {
orig: iter.clone(),
iter,
count,
}
}
}
impl<I> Iterator for Ncycles<I>
where
I: Clone + Iterator,
{
type Item = <I as Iterator>::Item;
#[inline]
fn next(&mut self) -> Option<<I as Iterator>::Item> {
match self.iter.next() {
None if self.count == 0 => None,
None => {
self.iter = self.orig.clone();
self.count -= 1;
self.iter.next()
}
y => y,
}
}
}
#[test]
fn it_work() {
Ncycles::new(vec![1, 2, 3].iter(), 4).eq(&[1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3]);
}
Upvotes: 4