为什么扩展功能在 ArrayList 上不起作用<arraylist<double> &gt; </arraylist<double>

Question

I am very new to Android development in Kotlin. I have an ArrayList consisting two ArrayLists of Double type. I want to discard/ slice everything after the first element inside the Big ArrayList. On coming across the properties described here on Kotlin's page, I found some functions like dropLast , take etc. However, they don't perform on implementation also there is no error. I am still getting the same output as input having the same length. Although the functions like add , get under Functions column are working fine. I am surely missing something here. What would be the way for achieving this?

Below is a dummy code:-

fun padding(tokenizedinput : ArrayList<ArrayList<Double>>) : ArrayList<ArrayList<Double>> {
    var temp_storage = tokenizedinput // of size 2
    temp_storage.take(1) // OPTION 1. Only want to have first element in this ArrayList
    temp_storage.dropLast(1) // OPTION 2. Only want to drop 1 element from the end
    println("FInal size: "+ temp_storage.size) //still size 2. Why not 1!?
    return temp_storage
}

Answer 1

temp_storage.take(1)

This returns a revised List . It does not modify the List on which you call it. You are ignoring the returned value.

temp_storage.dropLast(1)

Same — you are ignoring the work that the function is doing.

println("FInal size: "+ temp_storage.size) //still size 2. Why not 1!?

It is the same size because nothing that you did modified it.

What is the way of achieving this?

If I am understanding what you want, use:

fun padding(tokenizedinput : ArrayList<ArrayList<Double>>) = arrayListOf(tokenizedinput[0])

Here, we:

• Get the first element of tokenizedinput

• Wrap that in an ArrayList , since you wanted an ArrayList<ArrayList<Double>> response

Answer 2

List.take(n) or List.dropLast(n) would return a new list with the operation. It would NOT MODIFY the existing list. Try logging or printing this way:-

println(temp_storage.take(1).size) // would be 1
println(temp_storage.dropLast(1).size) // would be 1

The above outputs would be 1 , iff the size of the List is 2

To convert to the existing list use:-

temp_storage = ArrayList(temp_storage.dropLast(1)) // need to cast it to ArrayList<T> before assigning

Answer 3

To add to what the other answers have already said, from the actual class which contains this method:

The take method:

/**
 * Returns a list containing first [n] elements.
 * 
 * @throws IllegalArgumentException if [n] is negative.
 * 
 * @sample samples.collections.Collections.Transformations.take
 */
public fun <T> Iterable<T>.take(n: Int): List<T> {
    require(n >= 0) { "Requested element count $n is less than zero." }
    if (n == 0) return emptyList()
    if (this is Collection<T>) {
        if (n >= size) return toList()
        if (n == 1) return listOf(first())
    }
    var count = 0
    val list = ArrayList<T>(n)
    for (item in this) {
        if (count++ == n)
            break
        list.add(item)
    }
    return list.optimizeReadOnlyList()
} 

and also the dropLast :

/**
 * Returns a list containing all elements except last [n] elements.
 * 
 * @throws IllegalArgumentException if [n] is negative.
 * 
 * @sample samples.collections.Collections.Transformations.drop
 */
public fun <T> List<T>.dropLast(n: Int): List<T> {
    require(n >= 0) { "Requested element count $n is less than zero." }
    return take((size - n).coerceAtLeast(0))
}

which can be found in _Collections.kt

This means that it returns a list, it doesn't modify the original collection