Remove nth character from string

Question

I have seen many methods for removing the last character from a string. Is there however a way to remove any old character based on its index?

Answer 1

Swift 3.2

let str = "hello"
let position = 2
let subStr = str.prefix(upTo: str.index(str.startIndex, offsetBy: position)) + str.suffix(from: str.index(str.startIndex, offsetBy: (position + 1)))
print(subStr)

"helo"

Answer 2

var hello = "hello world!"

Let's say we want to remove the "w". (It's at the 6th index position.)

First: Create an Index for that position. (I'm making the return type Index explicit; it's not required).

let index:Index = hello.startIndex.advancedBy(6)

Second: Call removeAtIndex() and pass it our just-made index. (Notice it returns the character in question)

let choppedChar:Character = hello.removeAtIndex(index)

print(hello) // prints hello orld!

print(choppedChar) // prints w

Answer 3

While string indices aren't random-access and aren't numbers, you can advance them by a number in order to access the nth character:

var s = "Hello, I must be going"

s.removeAtIndex(advance(s.startIndex, 5))

println(s) // prints "Hello I must be going"

Of course, you should always check the string is at least 5 in length before doing this!

edit: as @MartinR points out, you can use the with-end-index version of advance to avoid the risk of running past the end:

let index = advance(s.startIndex, 5, s.endIndex)
if index != s.endIndex { s.removeAtIndex(index) }

As ever, optionals are your friend:

// find returns index of first match,
// as an optional with nil for no match
if let idx = s.characters.index(of:",") {
    // this will only be executed if non-nil,
    // idx will be the unwrapped result of find
    s.removeAtIndex(idx)
}

Answer 4

Here is a safe Swift 4 implementation.

var s = "Hello, I must be going"
var n = 5
if let index = s.index(s.startIndex, offsetBy: n, limitedBy: s.endIndex) {
    s.remove(at: index)
    print(s) // prints "Hello I must be going"
} else {
    print("\(n) is out of range")
}