199

Is it even possible to concatenate vectors in Rust? If so, is there an elegant way to do so? I have something like this:

let mut a = vec![1, 2, 3];
let b = vec![4, 5, 6];

for val in &b {
    a.push(val);
}

Does anyone know of a better way?

nbro
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Joe Thomas
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6 Answers6

247

The structure std::vec::Vec has method append():

fn append(&mut self, other: &mut Vec<T>)

Moves all the elements of other into Self, leaving other empty.

From your example, the following code will concatenate two vectors by mutating a and b:

fn main() {
    let mut a = vec![1, 2, 3];
    let mut b = vec![4, 5, 6];

    a.append(&mut b);

    assert_eq!(a, [1, 2, 3, 4, 5, 6]);
    assert_eq!(b, []);
}

Alternatively, you can use Extend::extend() to append all elements of something that can be turned into an iterator (like Vec) to a given vector:

let mut a = vec![1, 2, 3];
let b = vec![4, 5, 6];

a.extend(b);
assert_eq!(a, [1, 2, 3, 4, 5, 6]);
// b is moved and can't be used anymore

Note that the vector b is moved instead of emptied. If your vectors contain elements that implement Copy, you can pass an immutable reference to one vector to extend() instead in order to avoid the move. In that case the vector b is not changed:

let mut a = vec![1, 2, 3];
let b = vec![4, 5, 6];

a.extend(&b);
assert_eq!(a, [1, 2, 3, 4, 5, 6]);
assert_eq!(b, [4, 5, 6]);
Lukas Kalbertodt
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77

I can't make it in one line. Damian Dziaduch

It is possible to do it in one line by using chain():

let c: Vec<i32> = a.into_iter().chain(b.into_iter()).collect(); // Consumed
let c: Vec<&i32> = a.iter().chain(b.iter()).collect(); // Referenced
let c: Vec<i32> = a.iter().cloned().chain(b.iter().cloned()).collect(); // Cloned
let c: Vec<i32> = a.iter().copied().chain(b.iter().copied()).collect(); // Copied

There are infinite ways.

Shepmaster
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Stargateur
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    What is the difference between consuming, cloning and coppying? I thought that there are only references (borrowing, yes?) and clones – Damian Dziaduch Jun 07 '19 at 10:37
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    @DamianDziaduch wow, that ***broad***, you ask me to explain Rust ;) If you have experience you could understand the following: basically consume move the data so `a` and `b` are **not** available anymore, the result is free to do what it want. Reference just borrow both vector, so you need to keep their alive as long the result want to live. Clone and Copy are very close, the first one can be expensive, the second one should be cheap. They just use both vector as a source without need them latter, so the result become free to live as long as it want as well that `a` and `b`. Hope it's clear. – Stargateur Jun 07 '19 at 10:53
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    How does the performance compare to equivalent versions of `extend`? – User Jul 19 '21 at 07:49
  • @Stargateur sorry what do you mean? Your reply appears at the bottom to me. And a nice result with what? – User Jul 21 '21 at 05:58
  • How is cloning cheaper than copying? How are they different? – Michael Dorst Dec 07 '21 at 07:38
  • @MichaelDorst https://doc.rust-lang.org/std/marker/trait.Copy.html#whats-the-difference-between-copy-and-clone – Stargateur Dec 07 '21 at 08:56
  • @Stargateur Based on that it seems to me that cloning could be either cheaper or more expensive than copying, depending on the type. Pretty sure for i32 it should be exactly the same. – Michael Dorst Dec 08 '21 at 11:01
32

Regarding the performance, slice::concat, append and extend are about the same. If you don't need the results immediately, making it a chained iterator is the fastest; if you need to collect(), it is the slowest:

#![feature(test)]

extern crate test;

use test::Bencher;

#[bench]
fn bench_concat___init__(b: &mut Bencher) {
    b.iter(|| {
        let mut x = vec![1i32; 100000];
        let mut y = vec![2i32; 100000];
    });
}

#[bench]
fn bench_concat_append(b: &mut Bencher) {
    b.iter(|| {
        let mut x = vec![1i32; 100000];
        let mut y = vec![2i32; 100000];
        x.append(&mut y)
    });
}

#[bench]
fn bench_concat_extend(b: &mut Bencher) {
    b.iter(|| {
        let mut x = vec![1i32; 100000];
        let mut y = vec![2i32; 100000];
        x.extend(y)
    });
}

#[bench]
fn bench_concat_concat(b: &mut Bencher) {
    b.iter(|| {
        let mut x = vec![1i32; 100000];
        let mut y = vec![2i32; 100000];
        [x, y].concat()
    });
}

#[bench]
fn bench_concat_iter_chain(b: &mut Bencher) {
    b.iter(|| {
        let mut x = vec![1i32; 100000];
        let mut y = vec![2i32; 100000];
        x.into_iter().chain(y.into_iter())
    });
}

#[bench]
fn bench_concat_iter_chain_collect(b: &mut Bencher) {
    b.iter(|| {
        let mut x = vec![1i32; 100000];
        let mut y = vec![2i32; 100000];
        x.into_iter().chain(y.into_iter()).collect::<Vec<i32>>()
    });
}
running 6 tests
test bench_concat___init__           ... bench:      27,261 ns/iter (+/- 3,129)
test bench_concat_append             ... bench:      52,820 ns/iter (+/- 9,243)
test bench_concat_concat             ... bench:      53,566 ns/iter (+/- 5,748)
test bench_concat_extend             ... bench:      53,920 ns/iter (+/- 7,329)
test bench_concat_iter_chain         ... bench:      26,901 ns/iter (+/- 1,306)
test bench_concat_iter_chain_collect ... bench:     190,334 ns/iter (+/- 16,107)
Tianyi Shi
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    I see chain is still broken on this... that sad. your `bench_concat_iter_chain` do nothing, iterator are lazy – Stargateur Nov 19 '20 at 07:01
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    running this on `rustc 1.55.0-nightly (885399992 2021-07-06)` have much better result, houra ! – Stargateur Jul 19 '21 at 08:32
  • In case others had the same hiccup as me: to run the above place in a `benches` folder in your crate. Ensure the nightly toolchain is installed with `rustup install nightly` and you can run the benchmarks with `cargo +nightly bench` – lukecampbell Feb 01 '23 at 12:55
18

I think the best method to concatenate one or more vector is this:

let first_number: Vec<usize> = Vec::from([0]);
let final_number: Vec<usize> = Vec::from([3]);
let middle_numbers: Vec<usize> = Vec::from([1,2]);

let numbers = [input_layer, middle_layers, output_layer].concat();
Mattia Samiolo
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  • This won’t matter for integers, but for some types it may be important that `slice::concat` needs to `clone` each element. – Anders Kaseorg Dec 27 '22 at 00:16
2

One option is to use the extend method, which allows you to append the elements of one vector to another. Like so:

let mut a = vec![1, 2, 3];
let b = vec![4, 5, 6];

a.extend(b);

This will append the elements of b to the end of a, resulting in a vector a with the elements [1, 2, 3, 4, 5, 6].

Another way is you can use the concat function from the std::iter module to concatenate two vectors. This function takes two vectors as arguments and returns a new vector that is the concatenation of the two input vectors. Like so:

use std::iter;

let a = vec![1, 2, 3];
let b = vec![4, 5, 6];

let c = iter::concat(a, b);

This will create a new vector c with the elements [1, 2, 3, 4, 5, 6].

You can also use the [..] operator to concatenate two vectors. This operator allows you to create a new vector that is the concatenation of two input vectors. Like so:

 let a = vec![1, 2, 3];
 let b = vec![4, 5, 6];

 let c = [&a[..], &b[..]].concat();

This will create a new vector c with the elements [1, 2, 3, 4, 5, 6].

AlexaP
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0

made some fix on Mattia Samiolo's answer:

            let first_number: Vec<usize> = Vec::from([0]);
            let final_number: Vec<usize> = Vec::from([3]);
            let middle_numbers: Vec<usize> = Vec::from([1, 2]);

            let numbers = [first_number, middle_numbers, final_number].concat();
            println!("{:?}", numbers);