here is example of the list. I want to make dynamic where maybe the the value will become more.
val list = arrayListOf("A", "B", "C", "A", "A", "B") //Maybe they will be more
I want the output like:-
val result = list[i] + " size: " + list[i].size
So the output will display every String with the size.
A size: 3
B size: 2
C size: 1
If I add more value, so the result will increase also.
You can use groupBy in this way:
val result = list.groupBy { it }.map { it.key to it.value.size }.toMap()
Jeoffrey's way is better actually, since he is using .mapValues() directly, instead of an extra call to .toMap(). I'm just leaving this answer her since
I believe that the other info I put is relevant.
This will give a Map<String, Int>, where the Int is the count of the occurences.
This result will not change when you change the original list. That is not how the language works. If you want something like that, you'd need quite a bit of work, like overwriting the add function from your collection to refresh the result map.
Also, I see no reason for you to use an ArrayList, especially since you are expecting to increase the size of that collection, I'd stick with MutableList if I were you.
I think the terminology you're looking for is "frequency" here: the number of times an element appears in a list.
You can usually count elements in a list using the count method like this:
val numberOfAs = list.count { it == "A" }
This approach is pretty inefficient if you need to count all elements though, in which case you can create a map of frequencies the following way:
val freqs = list.groupBy { it }.mapValues { (_, g) -> g.size }
freqs here will be a Map where each key is a unique element from the original list, and the value is the corresponding frequency of that element in the list.
This works by first grouping elements that are equal to each other via groupBy, which returns a Map<String, List<String>> where each key is a unique element from the original list, and each value is the group of all elements in the list that were equal to the key.
Then mapValues will transform that map so that the values are the sizes of the groups instead of the groups themselves.
An improved approach, as suggested by #broot is to make use of Kotlin's Grouping class which has a built-in eachCount method:
val freqs = list.groupingBy { it }.eachCount()
Related
In Kotlin you can extract list items using a step like this:
val numbers = listOf("one", "two", "three", "four", "five", "six")
println(numbers.slice(0..4 step 2))
Is there a similar way to do this for a map? So instead of listOf, I'm using mapOf.
First you need to think about whether it makes sense to use slice on a Map. The Map interface does not guarantee an order, so it does not have the concept of indices that you can select from, which is what slice does.
If your Map is backed by a specific implementation that guarantees an iteration order, such as LinkedHashMap, then it can make sense to convert the Map's entries into a List so you can slice them, and then you can convert the result back to a Map.
val result = someLinkedHashMap
.entries
.toList()
.slice(someRange)
.associate { it.key to it.value } // change the filtered entries list into a new map
The associate function's documentation guarantees a Map implementation that has consistent execution order.
slice() is an extension function for Arrays and Lists.
To convert a Map to a List, first get the entries Set, and then convert the Set into a List.
Run in Kotlin Playground
fun main() {
val numbers = mapOf(
"one" to 1,
"two" to 2,
"three" to 3,
"four" to 4,
"five" to 5,
"six" to 6,
)
// get the Map entries, and convert them to a List - then slice
val sliced = numbers.entries.toList().slice(0..4 step 2)
println(sliced)
}
Output:
[one=1, three=3, five=5]
This question already has answers here:
Split a list into groups of consecutive elements based on a condition in Kotlin
(4 answers)
Closed 7 months ago.
I want to groupBy a list of items by its value, but only if subsequent, and ignore grouping otherwise:
input:
val values = listOf("Apple", "Apple", "Grape", "Grape", "Apple", "Cherry", "Cherry", "Grape")
output: {"Apple"=2, "Grape"=2, "Apple"=1, "Cherry"=2, "Grape"=1}
There's no built in option for this in Kotlin - it has to be custom, so there are many different options.
Because you need to keep track of the previous element, to compare the current one against, you need to have some sort of state. To achieve this you could use zipWithNext or windowed to group elements. Or use fold and accumulate the values into a list - removing and adding the last element depending on whether there's a break in the sequence.
To try and keep things a bit more clearer (even if it breaks the norms a bit) I recommend using vars and a single loop. I used the buildList { } DSL, which creates a clear scope for the operation.
val result: List<Pair<String, Int>> = buildList {
var previousElement: String? = null
var currentCount: Int = 0
// iterate over each incoming value
values.forEach { currentElement: String ->
// currentElement is new - so increment the count
currentCount++
// if we have a break in the sequence...
if (currentElement != previousElement) {
// then add the current element and count to our output
add(currentElement to currentCount)
// reset the count
currentCount = 0
}
// end this iteration - update 'previous'
previousElement = currentElement
}
}
Note that result will match the order of your initial list.
You cloud use MultiValueMap which can has duplicated keys. Since there is no native model you should implement yourself or use the open-source library.
Here is a reference.
Map implementation with duplicate keys
For comparison purposes, here's a short but inefficient solution written in the functional style using fold():
fun <E> List<E>.mergeConsecutive(): List<Pair<E, Int>>
= fold(listOf()) { acc, e ->
if (acc.isNotEmpty() && acc.last().first == e) {
val currentTotal = acc.last().second
acc.dropLast(1) + (e to currentTotal + 1)
} else
acc + (e to 1)
}
The accumulator builds up the list of pairs, incrementing its last entry when we get a duplicate, or appending a new entry when there's a different item. (You could make it slightly shorter by replacing the currentTotal with a call to let(), but that would be even harder to read.)
It uses immutable Lists and Pairs, and so has to create a load of temporary ones as it goes — which makes this pretty inefficient (𝒪(𝑛²)), and I wouldn't recommend it for production code. But hopefully it's instructive.
There is a data class as fruits.
data class Fruits(
val code: String, //Unique
val name: String
)
The base list indexed items with boolean variable is as below.
val indexList: MutableList<Boolean> = MutableList(baseFruitList.size) { false }
Now the Favourite Indexed list is as below
val favList: MutableList<Boolean> = MutableList(favFruitList.size) { true}
I want a combined full list which basically has the fav item indicated as true.
Ex:
baseFruitList = {[FT1,apple],[FT2,grapes],[FT3,banana],[FT4,mango],[FT5,pears]}
favList = {[FT2,grapes],[FT4,mango]}
The final index list should have
finalIndexed = {false,true,false,true,false}
How can we achieve in Kotlin, without iterating through each element.
You can do
val finalIndexed = baseFruitList.map { it in favList }
assuming, like #Tenfour04 is asking, that name is guaranteed to be a specific value (including matching case) for a specific code (since that combination is how a data class matches another, e.g. for checking if it's in another list)
If you can't guarantee that, this is safer:
val finalIndexed = baseFruitList.map { fruit ->
favList.any { fav.code == fruit.code }
}
but here you have to iterate over all the favs (at least until you find a match) looking to see if one has the code.
But really, if code is the unique identifier here, why not just store those in your favList?
favList = listOf("FT2", "FT4") // or a Set would be more efficient, and more correct!
val finalIndexed = baseFruitList.map { it.code in favList }
I don't know what you mean about "without iterating through each element" - if you mean without an explicit indexed for loop, then you can use these simple functions like I have here. But there's always some amount of iteration involved. Sets are always an option to help you minimise that
I have a list of ids. I want to filter my list and only keep the values in that list that match the id.
fun filterHelper(ids: List<Int>, list: List<People>) {
list.filter { ids.contains(it.id) }
}
But this is very inefficient. It is essentially traversing the list O(n^2). Does Kotlin let me do better?
I asked a similar question about slicing maps recently. The answer is that there is no good built-in function, but you can work around by using a Set instead of a List for your ids, which gets you O(1) lookup time for the comparisons, so O(n) in total.
data class People(val id: Int)
fun main() {
val people = listOf(People(1), People(2), People(3), People(4))
val ids = setOf(2, 4)
val filtered = people.filter { it.id in ids }
println(filtered)
}
Output:
[People(id=2), People(id=4)]
It's worth mentioning that if you already have a list, you can easily convert to a set with:
list.toSet()
In Kotlin, how can I take the first n elements of this array:
val allColours = arrayOf(
Pair(Color.RED, Color.WHITE),
Pair(Color.RED, Color.BLACK),
Pair(Color.YELLOW, Color.BLACK),
Pair(Color.GREEN, Color.WHITE),
Pair(Color.BLUE, Color.WHITE),
Pair(Color.BLUE, Color.WHITE),
Pair(Color.CYAN, Color.BLACK),
Pair(Color.WHITE, Color.BLACK))
So how can I fill pegColours with the first say 3 Pairs?
var pegColours: Array<Pair<Color,Color>> = //???
I tried allColours.take but it gave an error:
Expecting an element
You need to specify the number of items you want to take.
allColours.take(3)
For a random number of random indices, you can use the following:
val indexes = arrayOf(2, 4, 6)
allColours.filterIndexed { index, s -> indexes.contains(index) }
Note that you can write an extension method for this:
fun <T> Array<T>.filterByIndices(vararg indices: Int) = filterIndexed { index, _ -> indices.contains(index) }
Alternatively, if the indices are consecutive, you can use slice:
allColours.slice(1..3)
The problem with your code that you create pairs with color constants which are Ints (allColours has type Array<Pair<Int, Int>>), but you expect Array<Pair<Color, Color>>. What you have to do is change type pegColours type and use take:
var pegColours: Array<Pair<Int, Int>> = allColours.take(3).toTypedArray()
Also you have to call toTypedArray() cause Array.take returns List rather than Array. Or you can change pegColours type as following:
var pegColours: List<Pair<Int, Int>> = allColours.take(3)
I know you already proposed the usage of take, but alternatively ranges and a simple map also help to write idiomatic code as shown next:
var pegColours = (0 until 3)
.map { allColours[it] }
.toTypedArray()
You are very close :)
val allColours = arrayOf("red", "blue", "green")
kotlin.io.println(allColours.take(2))
Will give you first two elements ["red", "blue"]
You have to specify the number of elements you want to take from the array