how to get the index of a reoccurring character in swift

-2

var string = "HELLO WOLRD"

I want to get all of the indexes of the letter "L" in the string (which in this case would be 2,3 and 8). How do I do that in swift?

edited Nov 15 '18 at 16:07

Dávid Pásztor

22.4k82850

asked Nov 15 '18 at 15:47

Orane Pereira

3

It would be useful to show what you have already tried

– Matt
Nov 15 '18 at 15:59

add a comment |

-2

var string = "HELLO WOLRD"

I want to get all of the indexes of the letter "L" in the string (which in this case would be 2,3 and 8). How do I do that in swift?

edited Nov 15 '18 at 16:07

Dávid Pásztor

22.4k82850

asked Nov 15 '18 at 15:47

Orane Pereira

3

It would be useful to show what you have already tried

– Matt
Nov 15 '18 at 15:59

add a comment |

-2

var string = "HELLO WOLRD"

I want to get all of the indexes of the letter "L" in the string (which in this case would be 2,3 and 8). How do I do that in swift?

edited Nov 15 '18 at 16:07

Dávid Pásztor

22.4k82850

asked Nov 15 '18 at 15:47

Orane Pereira

var string = "HELLO WOLRD"

I want to get all of the indexes of the letter "L" in the string (which in this case would be 2,3 and 8). How do I do that in swift?

swift string

edited Nov 15 '18 at 16:07

Dávid Pásztor

22.4k82850

asked Nov 15 '18 at 15:47

Orane Pereira

edited Nov 15 '18 at 16:07

Dávid Pásztor

22.4k82850

asked Nov 15 '18 at 15:47

Orane Pereira

edited Nov 15 '18 at 16:07

Dávid Pásztor

22.4k82850

edited Nov 15 '18 at 16:07

Dávid Pásztor

22.4k82850

edited Nov 15 '18 at 16:07

Dávid Pásztor

22.4k82850

asked Nov 15 '18 at 15:47

Orane Pereira

asked Nov 15 '18 at 15:47

Orane Pereira

asked Nov 15 '18 at 15:47

Orane Pereira

3

It would be useful to show what you have already tried

– Matt
Nov 15 '18 at 15:59

add a comment |

3

It would be useful to show what you have already tried

– Matt
Nov 15 '18 at 15:59

It would be useful to show what you have already tried

– Matt
Nov 15 '18 at 15:59

add a comment |

2 Answers
2

active

oldest

votes

You could use enumerated() along with compactMap like this:

var string = "HELLO WOLRD"



let indices = string.enumerated().compactMap { $0.element == "L" ? $0.offset : nil }

print(indices)

[2, 3, 8]

Explanation:

string.enumerated() generates a sequence of tuple pairs where the first item is the offset of the character, and the second is the character.

The closure used by compactMap takes each tuple in turn and returns the offset of the character if it matches L or returns nil if it does not match. compactMap leaves out the nil values and just returns the offsets.

edited Nov 15 '18 at 16:44

answered Nov 15 '18 at 16:08

vacawama

98.7k15177203

add a comment |

You can use reduce(into:) on the enumerated String.

let helloWorld = "HELLO WORLD"



let charsWithIndices = helloWorld.enumerated().reduce(into: [Character:[Int]](), { result, current in

    if result[current.element] != nil {

        result[current.element]!.append(current.offset)

    } else {

        result[current.element] = [current.offset]

    }

})



print(charsWithIndices)

["W": [6], "D": [10], "R": [8], "E": [1], "H": [0], " ": [5], "O": [4, 7], "L": [2, 3, 9]]

Or if you want to use the indices straight away for indexing your String, you can change make charsWithIndices of type [Character:[String.Index]]:

You can use reduce(into:) on the enumerated String.

let helloWorld = "HELLO WORLD"

let charsWithIndices = helloWorld.enumerated().reduce(into: [Character:[String.Index]](), { result, current in

    let index = helloWorld.index(helloWorld.startIndex, offsetBy: current.offset)

    if result[current.element] != nil {

        result[current.element]!.append(index)

    } else {

        result[current.element] = [index]

    }

})

answered Nov 15 '18 at 16:03

Dávid Pásztor

22.4k82850

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

You could use enumerated() along with compactMap like this:

var string = "HELLO WOLRD"



let indices = string.enumerated().compactMap { $0.element == "L" ? $0.offset : nil }

print(indices)

[2, 3, 8]

Explanation:

string.enumerated() generates a sequence of tuple pairs where the first item is the offset of the character, and the second is the character.

The closure used by compactMap takes each tuple in turn and returns the offset of the character if it matches L or returns nil if it does not match. compactMap leaves out the nil values and just returns the offsets.

edited Nov 15 '18 at 16:44

answered Nov 15 '18 at 16:08

vacawama

98.7k15177203

add a comment |

You could use enumerated() along with compactMap like this:

var string = "HELLO WOLRD"



let indices = string.enumerated().compactMap { $0.element == "L" ? $0.offset : nil }

print(indices)

[2, 3, 8]

Explanation:

string.enumerated() generates a sequence of tuple pairs where the first item is the offset of the character, and the second is the character.

The closure used by compactMap takes each tuple in turn and returns the offset of the character if it matches L or returns nil if it does not match. compactMap leaves out the nil values and just returns the offsets.

edited Nov 15 '18 at 16:44

answered Nov 15 '18 at 16:08

vacawama

98.7k15177203

add a comment |

You could use enumerated() along with compactMap like this:

var string = "HELLO WOLRD"



let indices = string.enumerated().compactMap { $0.element == "L" ? $0.offset : nil }

print(indices)

[2, 3, 8]

Explanation:

string.enumerated() generates a sequence of tuple pairs where the first item is the offset of the character, and the second is the character.

The closure used by compactMap takes each tuple in turn and returns the offset of the character if it matches L or returns nil if it does not match. compactMap leaves out the nil values and just returns the offsets.

edited Nov 15 '18 at 16:44

answered Nov 15 '18 at 16:08

vacawama

98.7k15177203

You could use enumerated() along with compactMap like this:

var string = "HELLO WOLRD"



let indices = string.enumerated().compactMap { $0.element == "L" ? $0.offset : nil }

print(indices)

[2, 3, 8]

Explanation:

string.enumerated() generates a sequence of tuple pairs where the first item is the offset of the character, and the second is the character.

The closure used by compactMap takes each tuple in turn and returns the offset of the character if it matches L or returns nil if it does not match. compactMap leaves out the nil values and just returns the offsets.

edited Nov 15 '18 at 16:44

answered Nov 15 '18 at 16:08

vacawama

98.7k15177203

edited Nov 15 '18 at 16:44

answered Nov 15 '18 at 16:08

vacawama

98.7k15177203

answered Nov 15 '18 at 16:08

vacawama

98.7k15177203

answered Nov 15 '18 at 16:08

vacawama

98.7k15177203

add a comment |

You can use reduce(into:) on the enumerated String.

let helloWorld = "HELLO WORLD"



let charsWithIndices = helloWorld.enumerated().reduce(into: [Character:[Int]](), { result, current in

    if result[current.element] != nil {

        result[current.element]!.append(current.offset)

    } else {

        result[current.element] = [current.offset]

    }

})



print(charsWithIndices)

["W": [6], "D": [10], "R": [8], "E": [1], "H": [0], " ": [5], "O": [4, 7], "L": [2, 3, 9]]

Or if you want to use the indices straight away for indexing your String, you can change make charsWithIndices of type [Character:[String.Index]]:

You can use reduce(into:) on the enumerated String.

let helloWorld = "HELLO WORLD"

let charsWithIndices = helloWorld.enumerated().reduce(into: [Character:[String.Index]](), { result, current in

    let index = helloWorld.index(helloWorld.startIndex, offsetBy: current.offset)

    if result[current.element] != nil {

        result[current.element]!.append(index)

    } else {

        result[current.element] = [index]

    }

})

answered Nov 15 '18 at 16:03

Dávid Pásztor

22.4k82850

add a comment |

You can use reduce(into:) on the enumerated String.

let helloWorld = "HELLO WORLD"



let charsWithIndices = helloWorld.enumerated().reduce(into: [Character:[Int]](), { result, current in

    if result[current.element] != nil {

        result[current.element]!.append(current.offset)

    } else {

        result[current.element] = [current.offset]

    }

})



print(charsWithIndices)

["W": [6], "D": [10], "R": [8], "E": [1], "H": [0], " ": [5], "O": [4, 7], "L": [2, 3, 9]]

Or if you want to use the indices straight away for indexing your String, you can change make charsWithIndices of type [Character:[String.Index]]:

You can use reduce(into:) on the enumerated String.

let helloWorld = "HELLO WORLD"

let charsWithIndices = helloWorld.enumerated().reduce(into: [Character:[String.Index]](), { result, current in

    let index = helloWorld.index(helloWorld.startIndex, offsetBy: current.offset)

    if result[current.element] != nil {

        result[current.element]!.append(index)

    } else {

        result[current.element] = [index]

    }

})

answered Nov 15 '18 at 16:03

Dávid Pásztor

22.4k82850

add a comment |

You can use reduce(into:) on the enumerated String.

let helloWorld = "HELLO WORLD"



let charsWithIndices = helloWorld.enumerated().reduce(into: [Character:[Int]](), { result, current in

    if result[current.element] != nil {

        result[current.element]!.append(current.offset)

    } else {

        result[current.element] = [current.offset]

    }

})



print(charsWithIndices)

["W": [6], "D": [10], "R": [8], "E": [1], "H": [0], " ": [5], "O": [4, 7], "L": [2, 3, 9]]

Or if you want to use the indices straight away for indexing your String, you can change make charsWithIndices of type [Character:[String.Index]]:

You can use reduce(into:) on the enumerated String.

let helloWorld = "HELLO WORLD"

let charsWithIndices = helloWorld.enumerated().reduce(into: [Character:[String.Index]](), { result, current in

    let index = helloWorld.index(helloWorld.startIndex, offsetBy: current.offset)

    if result[current.element] != nil {

        result[current.element]!.append(index)

    } else {

        result[current.element] = [index]

    }

})

answered Nov 15 '18 at 16:03

Dávid Pásztor

22.4k82850

You can use reduce(into:) on the enumerated String.

let helloWorld = "HELLO WORLD"



let charsWithIndices = helloWorld.enumerated().reduce(into: [Character:[Int]](), { result, current in

    if result[current.element] != nil {

        result[current.element]!.append(current.offset)

    } else {

        result[current.element] = [current.offset]

    }

})



print(charsWithIndices)

["W": [6], "D": [10], "R": [8], "E": [1], "H": [0], " ": [5], "O": [4, 7], "L": [2, 3, 9]]

Or if you want to use the indices straight away for indexing your String, you can change make charsWithIndices of type [Character:[String.Index]]:

You can use reduce(into:) on the enumerated String.

let helloWorld = "HELLO WORLD"

let charsWithIndices = helloWorld.enumerated().reduce(into: [Character:[String.Index]](), { result, current in

    let index = helloWorld.index(helloWorld.startIndex, offsetBy: current.offset)

    if result[current.element] != nil {

        result[current.element]!.append(index)

    } else {

        result[current.element] = [index]

    }

})

answered Nov 15 '18 at 16:03

Dávid Pásztor

22.4k82850

answered Nov 15 '18 at 16:03

Dávid Pásztor

22.4k82850

answered Nov 15 '18 at 16:03

Dávid Pásztor

22.4k82850

answered Nov 15 '18 at 16:03

Dávid Pásztor

22.4k82850

add a comment |

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