Dynamic groupingBy keyselector in Kotlin
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I would like to pass a dynamic keySelector to my groupingBy function. Is this possible?
private fun hasNumberOfIdenticalValues(hand: ArrayList<Card>, values: ArrayList<String>, limit: Int) : Boolean {
var numberOfOccurrences : Map<String, Int> = hashMapOf()
for(value in values) {
numberOfOccurrences = hand.groupingBy{it.suit}.eachCount()
}
return true
}
java kotlin
asked Nov 10 at 19:28
Paul Mcloughlin
1,2782917
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0
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favorite
I would like to pass a dynamic keySelector to my groupingBy function. Is this possible?
private fun hasNumberOfIdenticalValues(hand: ArrayList<Card>, values: ArrayList<String>, limit: Int) : Boolean {
var numberOfOccurrences : Map<String, Int> = hashMapOf()
for(value in values) {
numberOfOccurrences = hand.groupingBy{it.suit}.eachCount()
}
return true
}
java kotlin
asked Nov 10 at 19:28
Paul Mcloughlin
1,2782917
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I would like to pass a dynamic keySelector to my groupingBy function. Is this possible?
private fun hasNumberOfIdenticalValues(hand: ArrayList<Card>, values: ArrayList<String>, limit: Int) : Boolean {
var numberOfOccurrences : Map<String, Int> = hashMapOf()
for(value in values) {
numberOfOccurrences = hand.groupingBy{it.suit}.eachCount()
}
return true
}
java kotlin
asked Nov 10 at 19:28
Paul Mcloughlin
1,2782917
I would like to pass a dynamic keySelector to my groupingBy function. Is this possible?
private fun hasNumberOfIdenticalValues(hand: ArrayList<Card>, values: ArrayList<String>, limit: Int) : Boolean {
var numberOfOccurrences : Map<String, Int> = hashMapOf()
for(value in values) {
numberOfOccurrences = hand.groupingBy{it.suit}.eachCount()
}
return true
}
java kotlin
java kotlin
asked Nov 10 at 19:28
Paul Mcloughlin
1,2782917
asked Nov 10 at 19:28
Paul Mcloughlin
1,2782917
asked Nov 10 at 19:28
Paul Mcloughlin
1,2782917
asked Nov 10 at 19:28
Paul Mcloughlin
1,2782917
asked Nov 10 at 19:28
Paul Mcloughlin
1,2782917
1,2782917
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
You can, you just need to return the string that will classify the interation. For instance, to separete the cards by color of the suit:
fun Card.color(): String {
return if(listof("diamonds","hearts").contains(this.suit))
"red"
else
"black"
}
for(value in values) {
numberOfOccurrences = hand.groupingBy{it.color()}.eachCount()
}
answered Nov 10 at 20:08
Raphael
1,30511020
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
You can, you just need to return the string that will classify the interation. For instance, to separete the cards by color of the suit:
fun Card.color(): String {
return if(listof("diamonds","hearts").contains(this.suit))
"red"
else
"black"
}
for(value in values) {
numberOfOccurrences = hand.groupingBy{it.color()}.eachCount()
}
answered Nov 10 at 20:08
Raphael
1,30511020
add a comment |
up vote
0
down vote
You can, you just need to return the string that will classify the interation. For instance, to separete the cards by color of the suit:
fun Card.color(): String {
return if(listof("diamonds","hearts").contains(this.suit))
"red"
else
"black"
}
for(value in values) {
numberOfOccurrences = hand.groupingBy{it.color()}.eachCount()
}
answered Nov 10 at 20:08
Raphael
1,30511020
add a comment |
up vote
0
down vote
up vote
0
down vote
You can, you just need to return the string that will classify the interation. For instance, to separete the cards by color of the suit:
fun Card.color(): String {
return if(listof("diamonds","hearts").contains(this.suit))
"red"
else
"black"
}
for(value in values) {
numberOfOccurrences = hand.groupingBy{it.color()}.eachCount()
}
answered Nov 10 at 20:08
Raphael
1,30511020
You can, you just need to return the string that will classify the interation. For instance, to separete the cards by color of the suit:
fun Card.color(): String {
return if(listof("diamonds","hearts").contains(this.suit))
"red"
else
"black"
}
for(value in values) {
numberOfOccurrences = hand.groupingBy{it.color()}.eachCount()
}
answered Nov 10 at 20:08
Raphael
1,30511020
answered Nov 10 at 20:08
Raphael
1,30511020
answered Nov 10 at 20:08
Raphael
1,30511020
answered Nov 10 at 20:08
Raphael
1,30511020
1,30511020
add a comment |
add a comment |
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