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.
std::list<Foo> x;
x.remove_if(std::not_fn(p));j := 0
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]func Filter[S ~[]T, T any](x *S, p func(T) bool) {
j := 0
for i, v := range *x {
if p(v) {
(*x)[j] = (*x)[i]
j++
}
}
var zero T
for k := j; k < len(*x); k++ {
(*x)[k] = zero
}
*x = (*x)[:j]
}j := 0
for i, v := range x {
if p(v) {
x[j] = x[i]
j++
}
}
x = x[:j]del_count = 0
for i in range(len(x)):
if not p(x[i - del_count]):
del x[i - del_count]
del_count += 1std::erase_if(x, std::not_fn(p));
std::list<Foo> x; x.remove_if(std::not_fn(p));
x.RemoveAll(item => !p(item));
x.retainWhere((e) => p);
x = pack(x,.not.p())
j := 0
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]
del := func(t *T) bool { return !p(t) }
x = slices.DeleteFunc(x, del)
func Filter[S ~[]T, T any](x *S, p func(T) bool) {
j := 0
for i, v := range *x {
if p(v) {
(*x)[j] = (*x)[i]
j++
}
}
var zero T
for k := j; k < len(*x); k++ {
(*x)[k] = zero
}
*x = (*x)[:j]
}
j := 0
for i, v := range x {
if p(v) {
x[j] = x[i]
j++
}
}
x = x[:j]
for (const [key, value] of x.entries()) {
if (!p(value)) x.splice(key, 1);
}
x = x.filter((e) => p(e));
x.removeIf(p.negate());
for i := x.count-1 downto 0 do if not p(x.items[i]) then x.delete(i);
@x = grep { p($_) } @x;
del_count = 0
for i in range(len(x)):
if not p(x[i - del_count]):
del x[i - del_count]
del_count += 1
i, n = 0, len(x)
while i != n:
if not p(x[i]):
del x[i]
n = n - 1
else:
i = i + 1
x.select!(&:p)
x.retain(p);
let mut j = 0;
for i in 0..x.len() {
if p(x[i]) {
x[j] = x[i];
j += 1;
}
}
x.truncate(j);