How to remove character from odd index in Swift String - ios

Question

I am new to swift & I have to remove characters from odd index from a given String
I did try with following code
var myStringObject = "HelloTestString"
myStringObject.enumerated().filter({ index, char in !(char == "1" && index % 2 == 0) })
but I am unable to find the desired result string. Can you please guide me how to remove characters from odd index in String.

Answer 1

You can't filter by two filter types. So you have to move to the old approach.
var myStringObject = "HelloTestString"
var newString = ""
var index = 0
while index < myStringObject.count {
if index % 2 != 0 {
let firstIndex: String.Index = myStringObject.startIndex
let desiredChar: Character = myStringObject[myStringObject.index(firstIndex, offsetBy: index)]
newString = newString + "\(desiredChar)"
}
index = index + 1
}
print(newString) //elTsSrn

Answer 2

Keeping your starting solution:
let couples = myStringObject.enumerated().filter { (arg0) -> Bool in
let (offset, _) = arg0
return offset % 2 == 0
}
print(couples)
or
let couples = myStringObject.enumerated().filter { (offset, _) -> Bool in
return offset % 2 == 0
} //Which is more similar to last version
You have then, an array of Tuples where first element is the offset, and the second the element.
$>[(offset: 0, element: "H"), (offset: 2, element: "l"), (offset: 4, element: "o"), (offset: 6, element: "e"), (offset: 8, element: "t"), (offset: 10, element: "t"), (offset: 12, element: "i"), (offset: 14, element: "g")]
Let's keep only the letters (and back to String, not a String.Element):
let onlyletters = couples.map({ String($0.element) })
Let's get it back into a String.
let result = onlyletters.joined()
In one line:
let oneLine = myStringObject.enumerated().filter({ $0.0 % 2 == 0 }).map({ String($0.element) }).joined()

Answer 3

You can create an auxiliary index and use defer to increase it after each iteration on your collection, this way you don't need to enumerate your string:
let string = "HelloTestString"
var index = 0
let filtered = string.filter { _ in
defer { index += 1 }
return index % 2 == 1
}
print(filtered) // "elTsSrn"
If you need to mutate your original string you can use removeAll with the same approach:
var string = "HelloTestString"
var index = 0
string.removeAll { _ in
defer { index += 1 }
return index % 2 == 0
}
print(string)
Implementing your own method
extending RangeReplaceableCollection to filter or removeAll elements that are in odd or even positions:
extension RangeReplaceableCollection {
var oddIndicesElements: Self {
var position = 0
return filter { _ in
defer { position += 1 }
return position % 2 == 1
}
}
var evenIndicesElements: Self {
var position = 0
return filter { _ in
defer { position += 1 }
return position % 2 == 0
}
}
mutating func removeAllEvenIndicesElements() {
var position = 0
removeAll { _ in
defer { position += 1 }
return position % 2 == 0
}
}
mutating func removeAllOddIndicesElements() {
var position = 0
removeAll { _ in
defer { position += 1 }
return position % 2 == 1
}
}
}
var myStringObject = "HelloTestString"
print(myStringObject.oddIndicesElements) // "elTsSrn"
myStringObject.removeAllEvenIndicesElements()
print(myStringObject) // "elTsSrn"