Kotlin: How to flatten list of Hashmaps - kotlin

How can I flatten a list of HashMaps in Kotlin?
var listOfMaps: List<Map<String, String>> = listOf(mapOf("test" to "test1"), mapOf("test2" to "test3"), mapOf("test4" to "test5"))
I would like to get:Map<String,String> with all key value paires

val map:Map<String, String> = listOfMaps
.flatMap { it.entries }
.associate { it.key to it.value }

you can do something like this if you don't know if the list can be empty.
val map = listOfMaps.fold(mapOf<String, String>()) {acc, value -> acc + value }
If the list never will be empty you can use reduce instead.
Thank you Demigod for the comments

You could use fold:
listOfMaps.fold(
mutableMapOf<String, String>(),
{ acc, item -> acc.also { it.putAll(item) } }
)
The first parameter mutableMapOf<String, String>() creates an empty map to put the values into. This is called the accumulator
The second parameter is a function which takes two arguments
The accumulator
An item from the original list
This function is run sequentially against all items in the list. In our case it adds all the items from each map to the accumulator.
Note: This function does not account for duplicate keys. If a later map has the same key as an earlier one then the value just gets overridden.
Also note (pun intended): We use acc.also {} as we want to return the actual map, not the return value from the addAll method

Well... seeing lots of solutions, I will add my two cents here:
If you don't mind losing the values of duplicated keys you can use something as follows:
listOfMaps.flatMap { it.entries }.associate{ it.key to it.value } // or: it.toPair() if you will
// depending on how large those lists can become, you may want to consider also using asSequence
If you instead want to collect all entries including duplicate keys (i.e. saving all the values), use the following instead (which then gives you a Map<String, List<String>>):
listOfMaps.flatMap { it.entries }.groupBy({ it.key }) { it.value }
Also here the comment regarding asSequence holds...
Finally if you can omit those maps within the list and just use a Pair instead, that will spare you the flatMap { it.entries }-call and make things even easier, e.g. you could just call .toMap() then for the first case and groupBy directly for the second and the question regarding asSequence no longer arises.

An extra addition to this, if you have single value maps, maybe you want to switch to a List<Prair<String, String>>. In that case, the solution is straight forward:
You would have something like:
var listOfMaps: List<Pair<String, String>> = listOf("test" to "test1", "test2" to "test3", "test4" to "test5")
and toMap() would dsolve it all:
listOfMaps.toMap()

If you don't have duplicate keys or don't care for them, you can do it like this:
val flattenedMap = listOfMaps.flatMap { it.toList() }.toMap()
If you have duplicate keys and care for them, you can do it like this:
val flattenedMap = mutableMapOf<String, MutableList<String>>().apply {
listOfMaps.flatMap { it.toList() }.forEach {
getOrPut(it.first) {
mutableListOf()
}.add(it.second)
}
}
The result will be Map<String, List<String>> then of course.

This solution works with same keys (and different collection as values) by merging them together instead of overwriting them
/**
* Merge two maps with lists
*/
fun <K,V>Map<K,Collection<V>>.mergeLists(other: Map<K,Collection<V>>) =
(this.keys + other.keys).associateWith {key ->
setOf(this[key], other[key]).filterNotNull()
}.mapValues { (_,b) -> b.flatten().distinct() }
/**
* Merge two maps with sets
*/
fun <K,V>Map<K,Collection<V>>.mergeSets(other: Map<K,Collection<V>>) =
(this.keys + other.keys).associateWith {key ->
setOf(this[key], other[key]).filterNotNull()
}.mapValues { (_,b) -> b.flatten().toSet() }
Then use like e.g. listOfMaps.reduce { a, b -> a.mergeSets(b) }
Test:
#Test
fun `should merge two maps with as lists or sets`() {
// GIVEN
val map1 = mapOf(
"a" to listOf(1, 2, 3),
"b" to listOf(4, 5, 6),
"c" to listOf(10),
"e" to emptyList()
)
val map2 = mapOf(
"a" to listOf(1, 9),
"b" to listOf(7),
"d" to listOf(null)
)
// WHEN
val mergedAsLists = map1.mergeLists(map2)
val mergedAsSets = map1.mergeSets(map2)
// THEN
listOf(mergedAsLists, mergedAsSets).forEach { merged ->
assertThat(merged.keys).containsOnly("a", "b", "c", "d", "e")
assertThat(merged["a"]).containsOnly(1,2,3,9)
assertThat(merged["b"]).containsOnly(4,5,6,7)
assertThat(merged["c"]).containsOnly(10)
assertThat(merged["d"]).containsOnly(null)
assertThat(merged["e"]).isEmpty()
}
}

Related

How to use destructuring declarations to get multiple value from single map iteration?

data class Person(val name: String, val age: String, val color: String)
val persons = listOf(Person("john", "20", "green"), Person("Mary", "20", "blue"))
val names = persons.map {it.name}
val ages = persons.map {it.age}
val colors = persons.map {it.color}
// I will need 3 iteration to get my names and ages and colors
val personsMap = persons.map { it.name to it.age }
val (names2, ages) = personsMap.unzip()
// I am using 1 iteration but with Map and uzip(), I can get names2,
// ages within a single iteration
// Now if I wish to get color
// By using map, or alternatives call helper.. can I retrieve names,
// ages and colors at one go
// I do know that we can use iteration forEachloop and add into a
// set/list, but
// are there better way of doing it
// val (names, ages, colors)
What are the alternative way to write to get multiple value in a single iteration to get its list from destructing.
In order to collect lists of different properties, at least one iteration is required. this can be done by defining an extension function as
fun List<Person>.collectProperties(): Triple<List<String>, List<String>, List<String>>{
val (names, ages, colors) = listOf(mutableListOf<String>(), mutableListOf(), mutableListOf())
this.forEach { person ->
names.add(person.name)
ages.add(person.age)
colors.add(person.color)
}
return Triple(names, ages, colors)
}
now you can use this as
val persons = listOf(Person("john", "20", "green"), Person("Mary", "20", "blue"))
val(names, ages, colors) = persons.collectProperties()
Your toList() call is redundant, so you're iterating 3 times instead of twice.
I can't think of a way to do this with a single iteration using standard library higher-order functions. You could write a function like this that maps and unzips in one step so it only iterates once:
inline fun <T, U, V> Iterable<T>.doubleMap(transform: (T) -> Pair<U, V>): Pair<List<U>, List<V>> {
val size = if (this is Collection<*>) size else 10
val listU = ArrayList<U>(size)
val listV = ArrayList<V>(size)
for (t in this) {
with (transform(t)) {
listU += first
listV += second
}
}
return listU to listV
}

How to sort two arrays in the same way but leave unsortable elements in their places?

I use this code to do that:
var letters = arrayOf("one", "two", "three")
var digits = arrayOf(2.36, 1.1, 3.0)
val x = digits.zip(letters).toMap().toSortedMap { o1, o2 -> o2.compareTo(o1)}
val keys = x.keys.toList() // [3.0, 2.36, 1.1]
val values = x.values.toList() // [three, one, two]
But noticed one thing which is not very good. If array of digits contains more than one the same element, only the last of them is visible in keys. Is there any way to put all the same elements and leave letters elements in it's places in such case? For example: if there is [3.0, 2.36, 1.1, 1.1] in array. This code turns it to [3.0, 2.36, 1.1]. How can I solve this problem?
You should not convert it to a map. Map is a data structure where keys are unique and you don't really use any features of the map here.
You can do this with the following code:
val lettersSorted = digits.zip(letters)
.sortedByDescending { it.first }
.map { it.second }
I'm not sure if you need a sorted list of letters only or both. And if both then whether you prefer to keep them in separate lists or in a single list. You can acquire all of these lists like this:
val sorted = digits.zip(letters)
.sortedByDescending { it.first } // sorted list of digits and letters
val sortedDigits = sorted.map { it.first }
val sortedLetters = sorted.map { it.second }

Removing elements from a list which are not in another list - Kotlin

I have two mutableLists, listOfA has so many objects including duplicates while listOfB has fewer. So I want to use listOfB to filter similar objects in listOfA so all list will have equal number of objects with equivalent keys at the end. Code below could explain more.
fun main() {
test()
}
data class ObjA(val key: String, val value: String)
data class ObjB(val key: String, val value: String, val ref: Int)
fun test() {
val listOfA = mutableListOf(
ObjA("one", ""),
ObjA("one", "o"),
ObjA("one", "on"),
ObjA("one", "one"),
ObjA("two", ""),
ObjA("two", "2"),
ObjA("two", "two"),
ObjA("three", "3"),
ObjA("four", "4"),
ObjA("five", "five")
)
//Use this list's object keys to get object with similar keys in above array.
val listOfB = mutableListOf(
ObjB("one", "i", 2),
ObjB("two", "ii", 5)
)
val distinctListOfA = listOfA.distinctBy { it.key } //Remove duplicates in listOfA
/*
val desiredList = doSomething to compare keys in distinctListOfA and listOfB
for (o in desiredList) {
println("key: ${o.key}, value: ${o.value}")
}
*/
/* I was hoping to get this kind of output with duplicates removed and comparison made.
key: one, value: one
key: two, value: two
*/
}
If you want to operate directly on that distinctListOfA you may want to use removeAll to remove all the matching entries from it. Just be sure that you initialize the keys of B only once so that it doesn't get evaluated every time the predicate is applied:
val keysOfB = listOfB.map { it.key } // or listOfB.map { it.key }.also { keysOfB ->
distinctListOfA.removeAll {
it.key !in keysOfB
}
//} // if "also" was used you need it
If you have a MutableMap<String, ObjA> in place after evaluating your unique values (and I think it may make more sense to operate on a Map here), the following might be what you are after:
val map : MutableMap<String, ObjA> = ...
map.keys.retainAll(listOfB.map { it.key })
retainAll just keeps those values that are matching the given collection entries and after applying it the map now contains only the keys one and two.
In case you want to keep your previous lists/maps and rather want a new list/map instead, you may just call something like the following before operating on it:
val newList = distinctListOfA.toList() // creates a new list with the same entries
val newMap = yourPreviousMap.toMutableMap() // create a new map with the same entries
I tried this
primaryList.removeAll { primaryItem ->
secondaryList.any { it.equals(primary.id, true) }
}
PrimaryList here is a list of objects
SecondaryList here is a list of strings

Mapping an iterable 1-to-n in kotlin

What I am trying to do is use the Iterable.map but instead of transforming every one value to one new value, I want to transform one value to multiple new values.
For example:
val myList = listOf("test", "123", "another.test", "test2")
val result = myList.map {
if(it.contains(".")) {
return#map it.split(".")
} else {
return#map it
}
}
//desired output: ["test", "123", "another", "test", "test2"]
This code would result in a List which contains both strings and lists of strings (type Any).
How could I most elegantly implement this?
One quick way to do this is using flatMap.
val output = myList.flatMap { if(it.contains(".")) it.split(".") else listOf(it) }
The flatMap method transforms the each element using given function and then flatten the result to a single list.
https://kotlinlang.org/api/latest/jvm/stdlib/kotlin.collections/flat-map.html

Kotlin - from a list of Maps, to a map grouped by key

I have a List<Map<Branch,Pair<String, Any>>> that I would like to convert in a single Map<Branch,List<Pair<String, Any>>> .
So if I have an initial list with simply 2 elements :
List
1. branch1 -> Pair(key1,value1)
branch2 -> Pair(key2,value2)
2. branch1 -> Pair(key1a,value1a)
I want to end up with :
Map
branch1 -> Pair(key1,value1)
Pair(key1a,value1a)
branch2 -> Pair(key2,value2)
so a kind of groupBy, using all the values of the keys in the initially nested maps..
I have tried with
list.groupBy{it-> it.keys.first()}
but obviously it doesn't work, as it uses only the first key. I want the same, but using all keys as individual values.
What is the most idiomatic way of doing this in Kotlin ? I have an ugly looking working version in Java, but I am quite sure Kotlin has a nice way of doing it.. it's just that I am not finding it so far !
Any idea ?
Thanks
The following:
val result =
listOfMaps.asSequence()
.flatMap {
it.asSequence()
}.groupBy({ it.key }, { it.value })
will give you the result of type Map<Branch,List<Pair<String, Any>>> with the contents you requested.
val list: List<Map<Branch, Pair<String, Any>>> = listOf()
val map = list
.flatMap { it.entries }
.groupBy { it.key }
.mapValues { entry -> entry.value.map { it.value } }
I've managed to write this.
data class Branch(val name: String)
data class Key(val name: String)
data class Value(val name: String)
val sharedBranch = Branch("1")
val listOfMaps: List<Map<Branch, Pair<Key, Value>>> = listOf(
mapOf(sharedBranch to Pair(Key("1"), Value("1")),
Branch("2") to Pair(Key("2"), Value("2"))),
mapOf(sharedBranch to Pair(Key("1a"), Value("1a")))
)
val mapValues: Map<Branch, List<Pair<Key, Value>>> = listOfMaps.asSequence()
.flatMap { map -> map.entries.asSequence() }
.groupBy(Map.Entry<Branch, Pair<Key, Value>>::key)
.mapValues { it.value.map(Map.Entry<Branch, Pair<Key, Value>>::value) }
println(mapValues)
Is it appliable for your needs?
Everyone else is using flatMap, but you can also consider using fold, which is a common operation for reducing a larger collection into a smaller one. (For example, you can fold a list of integers into a single sum; here, a list of maps into a single map).
Perhaps others will find this easier to read than the flatMap versions above:
val listOfMaps: List<Map<Key, Value>> = ...
val mergedMaps: Map<Key, List<Value>> =
listOfMaps
.fold(mutableMapOf()) { acc, curr ->
curr.forEach { entry -> acc.merge(entry.key, listOf(entry.value)) { new, old -> new + old } }
acc
}
What the above code is doing:
Create a new, empty map. This will be acc (that is, the accumulator).
Iterate through our list of maps.
Work on one map (curr) at a time.
For the current map, run over each of its key/value pairs.
For each key/value, call merge on acc, passing in a list of size one (wrapping the value). If nothing is associated with the key yet, that list is added; otherwise, it is appended to the list already there.
Return the accumulating map, so it's used again in the next step.
Surprised nobody has mentioned the associate function.
val listy: List<Map<String, Int>> =
listOf(mapOf("A" to 1, "B" to 2), mapOf("C" to 3, "D" to 4))
val flattened = listy
.flatMap { it.asSequence() }
.associate { it.key to it.value }
println(flattened)
will print out {A=1, B=2, C=3, D=4}
Extract it to an extension function
private fun <K, V> List<Map<K, V>>.group(): Map<K, List<V>> =
asSequence().flatMap { it.asSequence() }.groupBy({ it.key }, { it.value })
Use it like so:
val list = yourListOfMaps
val grouped = list.group()