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Idiom #136 Remove all occurrences of a value from a list

Remove all occurrences of the value x from list items.
This will alter the original list or return a new list, depending on which is more idiomatic.

import std.algorithm.iteration;
import std.array;
items = items.filter!(a => a != x).array;

The result of .filter is not evaluated until it's yielded by .array

This solution works for any input range, not only arrays.
(remove #{x} items)

Puts the x value in a set that serves as simple predicate
New implementation...
< > ⌨
programming-idioms.org 
items = items.filter!(a => a != x).array;
(remove #{x} items)
items.remove(x);
items.RemoveAll(r => r == x);

items.removeWhere((y)=>y==x);
Enum.filter(items, fn v -> v != x end)
items = pack (items,items != x)
items = slices.DeleteFunc(items, func(e T) bool {
	return e == x
})
items2 := make([]T, 0, len(items))
for _, v := range items {
	if v != x {
		items2 = append(items2, v)
	}
}
func removeAll[S ~[]T, T comparable](items *S, x T) {
	j := 0
	for i, v := range *items {
		if v != x {
			(*items)[j] = (*items)[i]
			j++
		}
	}
	var zero T
	for k := j; k < len(*items); k++ {
		(*items)[k] = zero
	}
	*items = (*items)[:j]
}
j := 0
for i, v := range items {
	if v != x {
		items[j] = items[i]
		j++
	}
}
for k := j; k < len(items); k++ {
	items[k] = nil
}
items = items[:j]
j := 0
for i, v := range items {
	if v != x {
		items[j] = items[i]
		j++
	}
}
items = items[:j]
filter (/= x) items
const newlist = items.filter(y => x !== y)
items.removeAll(Collections.singleton(x));
items.removeIf(t -> t.equals(x));
Iterator<T> i = items.listIterator();
while (i.hasNext())
    if (i.next().equals(x)) i.remove();
(remove-if (lambda (val) (= val x)) items)
$newItems = array_diff($items, [$x]);
var
  i: integer;

for i:= items.count-1 downto 0 do
  if items[i] = x then
    items.delete(i);
my @filtered = grep { $x ne $_ } @items;
items = [a for a in items if a != x]
newlist = [item for item in items if item != x]
items = list(a for a in items if a != x)
while items.count(x):
    items.remove(x)
f = lambda a: a != x
items = list(filter(f, items))
i, n = 0, len(items)
while i != n:
    if items[i] == x:
        del items[i]
        n = n - 1
    else:
        i = i + 1
items.delete(x)
items.retain(|&item| item != x);
items = items.into_iter().filter(|&item| item != x).collect();
items.filter(_ != x)
items reject: [: y | y = x ]