Remove i-th item from list items.
This will alter the original list or return a new list, depending on which is more idiomatic.
Note that in most languages, the smallest valid value for i is 0.
copy(items[i:], items[i+1:])
items[len(items)-1] = nil
items = items[:len(items)-1]items.remove(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))(define (removeElementByIndex L i)
(if (null? L)
null
(if (= i 0)
(cdr L)
(cons (car L) (removeElementByIndex (cdr L) (- i 1)))
)
)
)
Items.Delete (I);
(defn remove-idx [i items]
(keep-indexed #(when-not (= i %1) %2) items))
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)
copy(items[i:], items[i+1:]) items[len(items)-1] = nil items = items[:len(items)-1]
items = append(items[:i], items[i+1:]...)
take i items ++ drop (1 + i) items
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];
unset($items[$i]);
list[i] := list[high(list)]; setlength(list,high(list));
$removed_element = splice @items, $i, 1;
items.pop(i)
del items[i]
items.delete_at(i)
items.remove(i)
val without = {
val (left, right) = items.splitAt(i)
left ++ right.drop(1)
}
items removeAt: i.