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

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

(remove #{x} items)

items.remove(x);

using System.Collections.Generic;

items.RemoveAll(r => r == x);

import std.algorithm.iteration;
import std.array;

items = items.filter!(a => a != x).array;

items.removeWhere((y)=>y==x);

Enum.filter(items, fn v -> v != x end)

items = pack (items,items != x)

j := 0
for i, v := range items {
	if v != x {
		items[j] = items[i]
		j++
	}
}
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]

items2 := make([]T, 0, len(items))
for _, v := range items {
	if v != x {
		items2 = append(items2, v)
	}
}

filter (/= x) items

const newlist = items.filter(y => x !== y)

import java.util.Collections;

items.removeAll(Collections.singleton(x));

$newItems = array_diff($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;

newlist = [item for item in items if item != x]

items.delete(x)

items = items.into_iter().filter(|&item| item != x).collect();

items.filter(_ != x)

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Idiom created by programming-idioms.org

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