Programming-Idioms

Idiom

# 247 Filter list in-place

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Lead paragraph

Remove all the elements from list _x that _don'_t satisfy the predicate _p, without allocating a new list.
Keep all the elements that _do satisfy _p.

For languages that don't have mutable lists, refer to idiom #57 instead.

Implementation

Language

Rust

Added by

Last edited by

Imports

Code

let mut j = 0;
for i in 0..x.len() {
    if p(x[i]) {
        x[j] = x[i];
        j += 1;
    }
}
x.truncate(j);

Documentation URL

Original attribution URL

Online demo

Edit summary

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Implementation edit is for fixing errors and enhancing with metadata.

Instead of changing the code of the snippet, consider creating another Rust implementation.

Other implementations

j := 0
for i, v := range x {
	if p(v) {
		x[j] = x[i]
		j++
	}
}
x = x[:j]

Demo

for i, v := range x {
	if p(v) {
		x[j] = x[i]
		j++
	}
}
for k := j; k < len(x); k++ {
	x[k] = nil
}
x = x[:j]

Demo

classes

for i := x.count-1 downto 0 do
  if not p(x.items[i]) then x.delete(i);

#include <list>
#include <functional>

std::list<Foo> x;
x.remove_if(std::not_fn(p));

#include <functional>

std::erase_if(x, std::not_fn(p));

x.select!(&:p)

for (const [key, value] of x.entries()) {
	if (!p(value)) x.splice(key, 1);
}

Preview