Programming-Idioms

unset($items[$i]);

take i items ++ drop (1 + i) items

(defn remove-idx [i items]
    (keep-indexed #(when-not (= i %1) %2) items))

Items.Delete (I);

items.erase (items.begin () + i);

items.RemoveAt(i);

items = items.remove(i);

items.removeAt(i);

List.delete_at(items, i)

{Left, [_|Right]} = lists:split(I-1, Items),
Left ++ Right.

integer, allocatable, dimension(:) :: items
items = [items(:i-1), items(i+1:)]

items = slices.Delete(items, i, i+1)

items = append(items[:i], items[i+1:]...)

copy(items[i:], items[i+1:])
items[len(items)-1] = nil
items = items[:len(items)-1]

let new_list = items.filter(function(val,idx,ary) { return idx != i });

items.splice(i,1);

items.remove(i);

items.remove(i);

items.removeAt(i)

(defun remove-item-by-index (index items)
  "Construct a fresh list which is a copy of ITEMS (a list) excluding the item at INDEX."
  (loop for item in items
          as i from 0
          unless (= index i) collect item))

local removedItem = table.remove(items,i)

[items removeObjectAtIndex:i];

list[i] := list[high(list)];
setlength(list,high(list));

$removed_element = splice @items, $i, 1;

del items[i]

items.pop(i)

items.delete_at(i) 

items.remove(i)

val without = {
  val (left, right) = items.splitAt(i)
  left ++ right.drop(1)
}

(define (removeElementByIndex L i)
  (if (null? L)
      null
      (if (= i 0)
          (cdr L)
          (cons (car L) (removeElementByIndex (cdr L) (- i 1)))
       )
    )
)

items removeAt: i.