How to replace string characters that are not in a reference list in kotlin

Question

I have a reference string on which the allowed characters are listed. Then I also have input strings, from which not allowed characters should be replaced with a fixed character, in this example "0".

I can use filter but it removes the characters altogether, does not offer a replacement. Please note that it is not about being alphanumerical, there are ALLOWED non-alphanumerical characters and there are not allowed alphanumerical characters, referenceStr happens to be arbitrary.

var referenceStr = "abcdefg"
var inputStr = "abcqwyzt"
inputStr = inputStr.filter{it in referenceStr}

This yields:

"abc"

But I need:

"abc00000"

I also considered replace but it looks more like when you have a complete reference list of characters that are NOT allowed. My case is the other way around.

Alternative quick solution: inputStr.replace("[^$referenceStr]".toRegex(), "0") (caveat: it only works with restrictions, e.g. referenceStr must not contain "]") — k314159
– k314159, Commented May 13, 2022 at 11:56
@uğur-dinç Are you sure? This solution replace characters on current iteration. For example if we have val referenceStr = "zbcdefg" the result will be 0bc000z0, It is correct approach for solving you issue? — Sky
– Sky, Commented May 13, 2022 at 11:58

Sweeper · Accepted Answer · 2022-05-13 12:10:28Z

1

Given:

val referenceStr = "abcd][efg"
val replacementChar = '0'
val inputStr = "abcqwyzt[]"

You can do this with a regex [^<referenceStr>], where <referenceStr> should be replaced with referenceStr:

val result = inputStr.replace("[^${Regex.escape(referenceStr)}]".toRegex(), replacementChar.toString())
println(result)

Note that Regex.escape is used to make sure that the characters in referenceStr are all interpreted literally.

Alternatively, use map:

val result = inputStr.map {
    if (it !in referenceStr) replacementChar else it
}.joinToString(separator = "")

In the lambda decide whether the current char "it" should be transformed to replacementChar, or itself. map creates a List<Char>, so you need to use joinToString to make the result a String again.

answered May 13, 2022 at 12:10

Sweeper

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2 Comments

k314159 Over a year ago

Is it just me or does the Kotlin stdlib seem a bit inconsistent that String.filter returns String, but String.map returns List?

Sweeper Over a year ago

@k314159 Note that this is using CharSequence.map and String.filter. String.map doesn't actually exist. But it is a bit inconsistent, isn't it? I think there should be a String.map extension that returns a String too.

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