index number from Array of Strings Swift

Question

I need to check if the index number of a string is divisible by 2.

Here is what I need to do:

I have a string 0BCB7A0D87AD101B500B I need to remove all "0" characters from the string, but only if they are at an odd number in the string. I need to break it down as follows

0B|CB|7A|0D|87|AD|10|1B|50|0B

and only remove the "0" if it is the first character in the pair.

B|CB|7A|D|87|AD|10|1B|50|B

Then put the string back together

BCB7AD87AD101B50B

This code doesn't work, but this is how I am looking at doing it. I'm breaking up the string into an array:

   var characters = Array("0BCB7A0D87AD101B500B")

    for letter in characters {
        if characters(index: Int) % 2 != 0 { // if index has a remainder after being divided by 2
           if letter == "0"{ 
           characters.removeAtIndex(index: Int)
            }
        }

    }

I just can't divide or get the index number of the letter/string in the array.

Glorfindel · Accepted Answer · 2015-08-12 15:28:23Z

2

This would work, it uses a separate variable (index) to keep track of the index:

var characters = Array("0BCB7A0D87AD101B500B")
var newCharacters = Array<Character>()
var index = 0
for letter in characters {
    if index % 2 == 1 || letter != "0" {
        newCharacters.append(letter)
    }
    index++
}
var newString = String(newCharacters)

edited Aug 12, 2015 at 15:28

answered Aug 12, 2015 at 15:11

Glorfindel

22.8k13 gold badges97 silver badges124 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

ThundercatChris Over a year ago

Hmmm, this produces BCB7AD87AD181B53B349E51E9688B0000000. It has removed all the 0s.

Glorfindel Over a year ago

I think you are using another input string, then.

Gabriele Petronella · Accepted Answer · 2015-08-12 15:25:05Z

1

A functional approach to the problem

// zip each character with the index (enumerate)
let charsWithIndex = enumerate("0BCB7A0D87AD101B500B")
// filter away the unwanted characters (filter) and get rid of the indexes (map)
let filteredCharacters = filter(charsWithIndex) { index, char in !(char == "0" && index % 2 == 0) }.map { String($0.1) }
// put the string back together (join)
let filteredString = "".join(filteredCharacters)

edited Aug 12, 2015 at 15:25

answered Aug 12, 2015 at 15:20

Gabriele Petronella

108k21 gold badges223 silver badges238 bronze badges

Comments

ThundercatChris · Accepted Answer · 2015-08-12 15:27:18Z

0

Thank ou for your code. I was able to adjust it to remove at the right points.

var characters = Array("0BCB7A0D87AD101B500B")

    var newCharacters = Array<Character>()
    var index = 1
    for letter in characters {
        if index % 2 == 0 {
            newCharacters.append(letter)
        }
        if index % 2 != 0 && letter != "0" {
            newCharacters.append(letter)
        }
        index++
    }
    var newString = String(newCharacters)

answered Aug 12, 2015 at 15:27

ThundercatChris

4817 silver badges25 bronze badges

1 Comment

Glorfindel Over a year ago

That's exactly what my code is doing, though. There should be no difference. I let index begin at 0, because string positions start with 0. And you have split the if statement.

Collectives™ on Stack Overflow

index number from Array of Strings Swift

3 Answers 3

2 Comments

Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

2 Comments

Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related