Start learning free

LESSON 113 OF 678 · ARRAYS I

Insert and Remove at a Position in Swift Arrays

Lesson Read the full explanation

TL ; DR:

insert(x, at: i) puts x at index i. Use remove(at: i) to take out the item at index i. The other items move to make room or to close the gap.


var colors = ["red", "green", "blue"]

colors.insert("pink", at: 1)
print(colors)   // ["red", "pink", "green", "blue"]

colors.remove(at: 0)
print(colors)   // ["pink", "green", "blue"]

Full lesson:

append adds an item at the end. To put one anywhere else in an array, use insert(x, at: i). The items from index i on move one place to the right.

var pets = ["Rex", "Milo"]
pets.insert("Luna", at: 0)
print(pets)   // ["Luna", "Rex", "Milo"]

You met remove(at:) in the last lesson. It takes out the item at index i. The items after it move one place to the left. It also returns the item:

let first = pets.remove(at: 0)
print(first)   // Luna
print(pets)    // ["Rex", "Milo"]

Watch the indexes. After an insert or a removal, the items behind that spot have new indexes.

Note the order in insert. The value comes before the index.

Your task

pets holds the pets waiting at the vet, in order.

Bella has an appointment, so she goes second, at index 1. Then Milo goes home. Take him out of the line.

The code below prints how many pets wait once Bella is in. At the end, it prints the pets left.

Expected output:

4
Rex Bella Luna
STEP 1 OF 2
    Ctrl + Enter to check

    SWIFT · ARRAYS I

    Array Insert Remove in Swift, explained

    What these two methods do

    append only adds at the end. When an item has to go somewhere else, Swift gives you insert(_:at:). You write the value first and the index after at:. The items from that index on shift one place to the right to make room.

    var colors = ["red", "green", "blue"]
colors.insert("pink", at: 1)
print(colors)

    ["red", "pink", "green", "blue"]

    remove(at:) does the opposite. It takes out the item at an index, and the items after it shift one place to the left to close the gap.

    colors.remove(at: 0)
print(colors)

    ["pink", "green", "blue"]

    Details worth knowing

    remove(at:) hands the item back

    The removed item is the method's result, so you can keep it:

    var pets = ["Rex", "Milo", "Luna"]
let gone = pets.remove(at: 1)
print(gone)
print(pets)

    Milo
["Rex", "Luna"]

    If you don't need it, just ignore it: Swift doesn't complain.

    Valid indexes

    For insert, the index can be anything from 0 up to count. Inserting at count puts the item at the end. For remove(at:), the index must be an existing one, from 0 to count - 1. An index outside those limits stops the program with a run-time error.

    Here is insert at the very end next to insert at the front, using the line after Milo has gone home:

    pets.insert("Zed", at: pets.count)
print(pets)
pets.insert("Ada", at: 0)
print(pets)

    ["Rex", "Bella", "Luna", "Zed"]
["Ada", "Rex", "Bella", "Luna", "Zed"]

    Inserting at index 0 shifts every item. That is fine for a short list like this one, but it is the reason a long array is slower to change at the front than at the end: all the items behind the spot have to move.

    The array must be a var

    Both methods change the array, so it has to be declared with var. With let, the compiler refuses:

    error: cannot use mutating member on immutable value: 'pets' is a 'let' constant

    Indexes move

    After an insert or a removal, every item behind that spot has a new index. If you plan a second change, work out the indexes after the first one.

    The exercise: a vet's waiting line

    pets starts as ["Rex", "Milo", "Luna"]. Bella has an appointment, so she goes second, at index 1:

    pets.insert("Bella", at: 1)

    The line is now ["Rex", "Bella", "Milo", "Luna"], so pets.count prints 4. Milo, who used to be at index 1, moved to index 2. That is why the next step removes index 2 and not 1:

    pets.remove(at: 2)

    The line is now Rex, Bella and Luna. The finished program:

    var pets = ["Rex", "Milo", "Luna"]
pets.insert("Bella", at: 1)
print(pets.count)
pets.remove(at: 2)
print(pets[0], pets[1], pets[2])

    It prints:

    4
Rex Bella Luna

    Both calls look at the array as it is at that moment, line by line. That is why the order of the two lines matters: swap them, and index 2 would still hold Luna, because Bella wouldn't be in the line yet.

    Common mistakes

    • Writing the index first. It is insert("Bella", at: 1), with the value before the index.
    • Removing by the old index. Once Bella is in, Milo is at 2, not 1. Removing index 1 would send Bella home instead.
    • Declaring the array with let.

    Key takeaways

    • insert(x, at: i) puts x at index i and shifts the rest right.
    • remove(at: i) takes out the item at i, returns it, and shifts the rest left.
    • Both need a var array and an index in range.
    Keep going with Swift500+ step-by-step lessons, from your first line of code to coding interviews. Free to start.
    Start learning free