I can not find within the documentation of Vec<T> how to retrieve a slice from a specified range.
Is there something like this in the standard library:
let a = vec![1, 2, 3, 4];
let suba = a.subvector(0, 2); // Contains [1, 2];
The documentation for Vec covers this in the section titled "slicing".
You can create a slice of a Vec or array by indexing it with a Range (or RangeInclusive, RangeFrom, RangeTo, RangeToInclusive, or RangeFull), for example:
fn main() {
let a = vec![1, 2, 3, 4, 5];
// With a start and an end
println!("{:?}", &a[1..4]);
// With a start and an end, inclusive
println!("{:?}", &a[1..=3]);
// With just a start
println!("{:?}", &a[2..]);
// With just an end
println!("{:?}", &a[..3]);
// With just an end, inclusive
println!("{:?}", &a[..=2]);
// All elements
println!("{:?}", &a[..]);
}
If you wish to convert the entire Vec to a slice, you can use deref coercion:
fn main() {
let a = vec![1, 2, 3, 4, 5];
let b: &[i32] = &a;
println!("{:?}", b);
}
This coercion is automatically applied when calling a function:
fn print_it(b: &[i32]) {
println!("{:?}", b);
}
fn main() {
let a = vec![1, 2, 3, 4, 5];
print_it(&a);
}
You can also call Vec::as_slice, but it's a bit less common:
fn main() {
let a = vec![1, 2, 3, 4, 5];
let b = a.as_slice();
println!("{:?}", b);
}
See also:
Why is it discouraged to accept a reference to a String (&String), Vec (&Vec), or Box (&Box) as a function argument?
Related
I have a mutable reference of vector of mutable references of vectors I want to iterate over the vector and then change the value of its elements.
let mut a = vec![1, 2, 3];
let mut b = vec![4, 5, 6];
let mut c = vec![&mut a , &mut b];
let mut v = &mut c;
I want to iterate over v and do operations on its elements
preferably using an iterator and not by indexing.(I specifically want to iterate over a reference of a vector of references of vector and not just a 2d vector)
fn main() {
let mut a = vec![1, 2, 3];
let mut b = vec![4, 5, 6];
let mut c = vec![&mut a, &mut b];
let v = &mut c;
for row in v.iter_mut() {
for el in row.iter_mut() {
*el *= 2;
}
}
println!("{:?}", v);
}
[[2, 4, 6], [8, 10, 12]]
Or if you need the indices for the computation:
fn main() {
let mut a = vec![1, 2, 3];
let mut b = vec![4, 5, 6];
let mut c = vec![&mut a, &mut b];
let v = &mut c;
for (y, row) in v.iter_mut().enumerate() {
for (x, el) in row.iter_mut().enumerate() {
*el *= 2 * x + y;
}
}
println!("{:?}", v);
}
[[0, 4, 12], [4, 15, 30]]
Is there any straightforward way to insert or replace multiple elements from &[T] and/or Vec<T> in the middle or at the beginning of a Vec in linear time?
I could only find std::vec::Vec::insert, but that's only for inserting a single element in O(n) time, so I obviously cannot call that in a loop.
I could do a split_off at that index, extend the new elements into the left half of the split, and then extend the second half into the first, but is there a better way?
As of Rust 1.21.0, Vec::splice is available and allows inserting at any point, including fully prepending:
let mut vec = vec![1, 5];
let slice = &[2, 3, 4];
vec.splice(1..1, slice.iter().cloned());
println!("{:?}", vec); // [1, 2, 3, 4, 5]
The docs state:
Note 4: This is optimal if:
The tail (elements in the vector after range) is empty
or replace_with yields fewer elements than range’s length
or the lower bound of its size_hint() is exact.
In this case, the lower bound of the slice's iterator should be exact, so it should perform one memory move.
splice is a bit more powerful in that it allows you to remove a range of values (the first argument), insert new values (the second argument), and optionally get the old values (the result of the call).
Replacing a set of items
let mut vec = vec![0, 1, 5];
let slice = &[2, 3, 4];
vec.splice(..2, slice.iter().cloned());
println!("{:?}", vec); // [2, 3, 4, 5]
Getting the previous values
let mut vec = vec![0, 1, 2, 3, 4];
let slice = &[9, 8, 7];
let old: Vec<_> = vec.splice(3.., slice.iter().cloned()).collect();
println!("{:?}", vec); // [0, 1, 2, 9, 8, 7]
println!("{:?}", old); // [3, 4]
Okay, there is no appropriate method in Vec interface (as I can see). But we can always implement the same thing ourselves.
memmove
When T is Copy, probably the most obvious way is to move the memory, like this:
fn push_all_at<T>(v: &mut Vec<T>, offset: usize, s: &[T]) where T: Copy {
match (v.len(), s.len()) {
(_, 0) => (),
(current_len, _) => {
v.reserve_exact(s.len());
unsafe {
v.set_len(current_len + s.len());
let to_move = current_len - offset;
let src = v.as_mut_ptr().offset(offset as isize);
if to_move > 0 {
let dst = src.offset(s.len() as isize);
std::ptr::copy_memory(dst, src, to_move);
}
std::ptr::copy_nonoverlapping_memory(src, s.as_ptr(), s.len());
}
},
}
}
shuffle
If T is not copy, but it implements Clone, we can append given slice to the end of the Vec, and move it to the required position using swaps in linear time:
fn push_all_at<T>(v: &mut Vec<T>, mut offset: usize, s: &[T]) where T: Clone + Default {
match (v.len(), s.len()) {
(_, 0) => (),
(0, _) => { v.push_all(s); },
(_, _) => {
assert!(offset <= v.len());
let pad = s.len() - ((v.len() - offset) % s.len());
v.extend(repeat(Default::default()).take(pad));
v.push_all(s);
let total = v.len();
while total - offset >= s.len() {
for i in 0 .. s.len() { v.swap(offset + i, total - s.len() + i); }
offset += s.len();
}
v.truncate(total - pad);
},
}
}
iterators concat
Maybe the best choice will be to not modify Vec at all. For example, if you are going to access the result via iterator, we can just build iterators chain from our chunks:
let v: &[usize] = &[0, 1, 2];
let s: &[usize] = &[3, 4, 5, 6];
let offset = 2;
let chain = v.iter().take(offset).chain(s.iter()).chain(v.iter().skip(offset));
let result: Vec<_> = chain.collect();
println!("Result: {:?}", result);
I was trying to prepend to a vector in rust and found this closed question that was linked here, (despite this question being both prepend and insert AND efficiency. I think my answer would be better as an answer for that other, more precises question because I can't attest to the efficiency), but the following code helped me prepend, (and the opposite.) [I'm sure that the other two answers are more efficient, but the way that I learn, I like having answers that can be cut-n-pasted with examples that demonstrate an application of the answer.]
pub trait Unshift<T> { fn unshift(&mut self, s: &[T]) -> (); }
pub trait UnshiftVec<T> { fn unshift_vec(&mut self, s: Vec<T>) -> (); }
pub trait UnshiftMemoryHog<T> { fn unshift_memory_hog(&mut self, s: Vec<T>) -> (); }
pub trait Shift<T> { fn shift(&mut self) -> (); }
pub trait ShiftN<T> { fn shift_n(&mut self, s: usize) -> (); }
impl<T: std::clone::Clone> ShiftN<T> for Vec<T> {
fn shift_n(&mut self, s: usize) -> ()
// where
// T: std::clone::Clone,
{
self.drain(0..s);
}
}
impl<T: std::clone::Clone> Shift<T> for Vec<T> {
fn shift(&mut self) -> ()
// where
// T: std::clone::Clone,
{
self.drain(0..1);
}
}
impl<T: std::clone::Clone> Unshift<T> for Vec<T> {
fn unshift(&mut self, s: &[T]) -> ()
// where
// T: std::clone::Clone,
{
self.splice(0..0, s.to_vec());
}
}
impl<T: std::clone::Clone> UnshiftVec<T> for Vec<T> {
fn unshift_vec(&mut self, s: Vec<T>) -> ()
where
T: std::clone::Clone,
{
self.splice(0..0, s);
}
}
impl<T: std::clone::Clone> UnshiftMemoryHog<T> for Vec<T> {
fn unshift_memory_hog(&mut self, s: Vec<T>) -> ()
where
T: std::clone::Clone,
{
let mut tmp: Vec<_> = s.to_owned();
//let mut tmp: Vec<_> = s.clone(); // this also works for some data types
/*
let local_s: Vec<_> = self.clone(); // explicit clone()
tmp.extend(local_s); // to vec is possible
*/
tmp.extend(self.clone());
*self = tmp;
//*self = (*tmp).to_vec(); // Just because it compiles, doesn't make it right.
}
}
// this works for: v = unshift(v, &vec![8]);
// (If you don't want to impl Unshift for Vec<T>)
#[allow(dead_code)]
fn unshift_fn<T>(v: Vec<T>, s: &[T]) -> Vec<T>
where
T: Clone,
{
// create a mutable vec and fill it
// with a clone of the array that we want
// at the start of the vec.
let mut tmp: Vec<_> = s.to_owned();
// then we add the existing vector to the end
// of the temporary vector.
tmp.extend(v);
// return the tmp vec that is identitcal
// to unshift-ing the original vec.
tmp
}
/*
N.B. It is sometimes (often?) more memory efficient to reverse
the vector and use push/pop, rather than splice/drain;
Especially if you create your vectors in "stack order" to begin with.
*/
fn main() {
let mut v: Vec<usize> = vec![1, 2, 3];
println!("Before push:\t {:?}", v);
v.push(0);
println!("After push:\t {:?}", v);
v.pop();
println!("popped:\t\t {:?}", v);
v.drain(0..1);
println!("drain(0..1)\t {:?}", v);
/*
// We could use a function
let c = v.clone();
v = unshift_fn(c, &vec![0]);
*/
v.splice(0..0, vec![0]);
println!("splice(0..0, vec![0]) {:?}", v);
v.shift_n(1);
println!("shift\t\t {:?}", v);
v.unshift_memory_hog(vec![8, 16, 31, 1]);
println!("MEMORY guzzler unshift {:?}", v);
//v.drain(0..3);
v.drain(0..=2);
println!("back to the start: {:?}", v);
v.unshift_vec(vec![0]);
println!("zerothed with unshift: {:?}", v);
let mut w = vec![4, 5, 6];
/*
let prepend_this = &[1, 2, 3];
w.unshift_vec(prepend_this.to_vec());
*/
w.unshift(&[1, 2, 3]);
assert_eq!(&w, &[1, 2, 3, 4, 5, 6]);
println!("{:?} == {:?}", &w, &[1, 2, 3, 4, 5, 6]);
}
I have a 2-dimensional vector:
let vec2d = vec![
vec![1, 1, 1],
vec![1, 1, 1],
];
I can yield the total of elements stored this way:
let mut n_vec_element: i32 = 0;
for i in vec2d.iter() {
n_vec_element += i.len() as i32;
}
println!("2D vector elements :{}", n_vec_element); // prints 6
When I increase the dimensions, the loop gets longer:
let mut n_vec_element: i32 = 0;
let vec3d = vec![
vec![
vec![1, 3, 5 as i32],
vec![2, 4, 6 as i32],
vec![3, 5, 7 as i32],
],
vec![
vec![1, 3, 5 as i32],
vec![2, 4, 6 as i32],
vec![3, 5, 7 as i32],
]
];
for i in vec3d.iter() {
// I must add another iter everytime I increment the dimension by 1.
// Else, it returns the number of stored vector instead of the vector
// elements.
for j in i.iter() {
n_vec_size += j.len() as i32;
}
};
println!("3D vector elements :{}", n_vec_element); // prints 18
There must be a more concise way to do this, but I haven't figured it out yet. Initially, I tried using vector's len() function but as I said above, it returns number of vectors stored instead of its elements.
You do not need an explicit loop to do so:
let vec2d = vec![
vec![1, 1, 1],
vec![1, 1, 1],
];
let n_vec_element: usize = vec2d.iter().map(Vec::len).sum();
assert_eq!(n_vec_element, 6);
For a 3d vector, you can do the same:
let vec3d = vec![
vec![
vec![1, 3, 5 as i32],
vec![2, 4, 6 as i32],
vec![3, 5, 7 as i32],
],
vec![
vec![1, 3, 5 as i32],
vec![2, 4, 6 as i32],
vec![3, 5, 7 as i32],
]
];
let n_vec_element: usize = vec3d.iter().flatten().map(Vec::len).sum();
assert_eq!(n_vec_element, 18);
With a 4D vector, you can put 2 flatten, etc.
With the specialization feature (i.e., with the nightly compiler), you can generalize this with an unique method:
#![feature(specialization)]
trait Count {
fn count(self) -> usize;
}
impl<T> Count for T {
default fn count(self) -> usize {
1
}
}
impl<T> Count for T
where
T: IntoIterator,
T::Item: Count,
{
fn count(self) -> usize {
self.into_iter().map(|x| x.count()).sum()
}
}
fn main() {
let v = vec![1, 2, 3];
assert_eq!(v.count(), 3);
let v = vec![vec![1, 2, 3], vec![4, 5, 6]];
assert_eq!(v.count(), 6);
}
This code is supposed to increment each value in a vector by 1:
fn main() {
let mut v = vec![2, 3, 1, 4, 2, 5];
let i = v.iter_mut();
for j in i {
*j += 1;
println!("{}", j);
}
println!("{:?}", &mut v);
}
It doesn't work because of the borrowing rules of Rust:
error[E0499]: cannot borrow `v` as mutable more than once at a time
--> src/main.rs:8:27
|
3 | let i = v.iter_mut();
| - first mutable borrow occurs here
...
8 | println!("{:?}", &mut v);
| ^ second mutable borrow occurs here
9 | }
| - first borrow ends here
How can I accomplish this task?
Don't store the mutable iterator; use it directly in the loop instead:
fn main() {
let mut v = vec![2, 3, 1, 4, 2, 5];
for j in v.iter_mut() { // or for j in &mut v
*j += 1;
println!("{}", j);
}
println!("{:?}", &v); // note that I dropped mut here; it's not needed
}
Your code will work as-is in a future version of Rust thanks to non-lexical lifetimes:
#![feature(nll)]
fn main() {
let mut v = vec![2, 3, 1, 4, 2, 5];
let i = v.iter_mut();
for j in i {
*j += 1;
println!("{}", j);
}
println!("{:?}", &mut v);
}
playground
You call also just use map and collect like,
>> let mut v = vec![5,1,4,2,3];
>> v.iter_mut().map(|x| *x += 1).collect::<Vec<_>>();
>> v
[6, 2, 5, 3, 4]
In my opinion the simplest and most readable solution:
#![feature(nll)]
fn main() {
let mut v = vec![2, 3, 1, 4, 2, 5];
for i in 0..v.len() {
v[i] += 1;
println!("{}", j);
}
println!("{:?}", v);
}
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?
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]);
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.
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)
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();
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].
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);