Logo

Programming-Idioms

# 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.
New implementation

Be concise.

Be useful.

All contributions dictatorially edited by webmasters to match personal tastes.

Please do not paste any copyright violating material.

Please try to avoid dependencies to third-party libraries and frameworks.

Other implementations 
(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)
import "slices"
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)
import java.util.Collections;
items.removeAll(Collections.singleton(x));
items.removeIf(t -> t.equals(x));
import java.util.Iterator;
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 ]

Preview