What is the easiest way to get value from `Map<...

# announcements

[Danish Ansari]

What is the easiest way to get value from Map<String,String> with case insensitive String key?

More context and detailed scenario in the thread...

[Danish Ansari]

I might have a Map<String,String> like this

val input = mutableMapOf<String, String>().apply {
    put("@Username", "John Doe")
    // OR
    // put("@USERNAME", "John Doe")
    // OR
    // put("@username", "John Doe")
}

And I have to use username at lots of different places so I have created a getter extension on Map<String,String> like this

interval val Map<String, String>.username: String?
    get() {
        val caseInsensitiveMap = TreeMap<String, String>(String.CASE_INSENSITIVE_ORDER)
        caseInsensitiveMap.putAll(this)
        return caseInsensitiveMap["@USERNAME"]
    }

I know this is not a very optimized way, that's why the question 😬


And I will access it like this

val username = input.username
println(username) // Must print "John Doe"

[Arkadii Ivanov]

Perhaps wrap keys into a class with custom equals/hashCode implementations?

[Danish Ansari]

Actually I can't change definition of Map because it's already in use at many place in the project, so I can't change the key now

[streetsofboston]

But you can change the definition of what you put as keys in these `Map`s?
Then @Arkadii Ivanov's is the way to go

[streetsofboston]

If it has to be a Map<*String*, String>, then Arkadii's suggestion won't work, since Strings are not extensible (you can't sub-class them)

[Arkadii Ivanov]

For JVM you can try TreeMap which accepts a Comparator. But it will give O(logN) time.

[Nikky]

can you call lowercase() on the key on insert ?
that would not chang the type.. but it would avoid duplicates

[Danish Ansari]

I don't do insertion, my project is a library used by other developers (that's why I'm worried about the casing).

I guess the best thing I think of is to change definition of Map to TreeMap

[Vadim Kapustin]

val Map<String, String>.username: String?
    get() = this.asSequence()
        .filter { it.key.uppercase() == "@USERNAME" }
        .firstOrNull()?.value

@Danish Ansari maybe like this...

[Vadim Kapustin]

in fact, your version will get the last matching value. To do this, use lastOrNull instead of firstOrNull, which is somewhat slower...

[Danish Ansari]

Thanks a lot @Vadim Kapustin I'll surely try this approach