Use all but the last element from an iterator

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Solution 1

As Sebastian Redl points out, checking the length of each chunk is the better solution for your specific case.

To answer the question you asked ("Use all but the last element from an iterator"), you can use Iterator::peekable to look ahead one. That will tell you if you are on the last item or not and you can decide to skip processing it if so.

let things = [0, 1, 2, 3, 4];

let mut chunks = things.chunks(2).peekable();
while let Some(chunk) = chunks.next() {
    if chunks.peek().is_some() {
        print!("Not the last: ");
    } else {
        print!("This is the last: ")
    }

    println!("{:?}", chunk);
}

To be sure that all parts have equal length, I just want to drop that last element

Always dropping the last element won't do this. For example, if you evenly chunk up your input, then always dropping the last element would lose a full chunk. You'd have to do some pre-calculation to decide if you need to drop it or not.

Solution 2

You can filter() the chunks iterator on the slice's len() being the amount you passed to chunks():

let things = [0, 1, 2, 3, 4];

for chunk in things.chunks(2).filter(|c| c.len() == 2) {
    println!("{:?}", chunk);
}

As of Rust 1.31, you can use the chunks_exact method as well:

let things = [0, 1, 2, 3, 4];

for chunk in things.chunks_exact(2) {
    println!("{:?}", chunk);
}

Note that the returned iterator also has the method remainder if you need to get the uneven amount of items at the very end.

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Innot Kauker
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Innot Kauker

Updated on September 14, 2022

Comments

  • Innot Kauker
    Innot Kauker over 1 year

    I want to split a Vec into some parts of equal length, and then map over them. I have an iterator resulting from a call to Vec's chunks() method. This may leave me with a part that will be smaller than other parts, which will be the last element generated by it.

    To be sure that all parts have equal length, I just want to drop that last element and then call map() on what's left.

    • JHBonarius
      JHBonarius over 5 years
      although there's a better solution for your case, note the existence of std::iter::take
  • Innot Kauker
    Innot Kauker over 6 years
    That's even more explicit. Thanks!
  • Shepmaster
    Shepmaster over 6 years
    You could also slice the input to a multiple of the chunk size and avoid a check at each step.