Kotlin - from a list of Maps, to a map grouped by key
0
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
kotlin
asked Nov 22 '18 at 14:28
Vincent FVincent F
2,0181329
add a comment |
0
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
kotlin
asked Nov 22 '18 at 14:28
Vincent FVincent F
2,0181329
add a comment |
0
0
0
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
kotlin
asked Nov 22 '18 at 14:28
Vincent FVincent F
2,0181329
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
kotlin
kotlin
asked Nov 22 '18 at 14:28
Vincent FVincent F
2,0181329
asked Nov 22 '18 at 14:28
Vincent FVincent F
2,0181329
asked Nov 22 '18 at 14:28
Vincent FVincent F
2,0181329
asked Nov 22 '18 at 14:28
Vincent FVincent F
2,0181329
asked Nov 22 '18 at 14:28
Vincent FVincent F
2,0181329
2,0181329
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
2
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.
answered Nov 22 '18 at 15:21
RolandRoland
10k11241
Thanks, it works. FYI I've made the change here : github.com/societe-generale/github-crawler/blob/…
– Vincent F
Nov 22 '18 at 16:23
well... you could have said, that it actually ends up in aMap<Branch, Map<String, Any>>... that.mapValuescould also already be simplified to:.mapValues { it.value.toMap() }
– Roland
Nov 22 '18 at 16:43
add a comment |
1
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?
answered Nov 22 '18 at 14:52
Andrey IlyuninAndrey Ilyunin
1,286220
what do you make of marko's comment?
– Tim Castelijns
Nov 22 '18 at 15:06
@TimCastelijns, Kotlin'sMapandMutableMapinterfaces do not contain this method
– Andrey Ilyunin
Nov 22 '18 at 15:44
On the JVM, this is legal Kotlin:val map = mutableMapOf(1 to "a", 2 to "b"); map.merge(1, "c") { v0, v1 -> v0 + v1}
– Marko Topolnik
Nov 22 '18 at 16:07
@TimCastelijns I tried to write it, butmergeis not a good fit, you can do it withcompute. It's not a great win overgroupBysolutions, though.val result = mutableMapOf<Branch, MutableList<Pair<String, Any>>>(); listOfMaps.forEach { map -> map.forEach { k, v -> result.compute(k) { _, v0 -> v0?.also { it.add(v) } ?: mutableListOf(v) } } }
– Marko Topolnik
Nov 22 '18 at 16:08
@MarkoTopolnik yes, but it is if you have mutable map returned to you as an argument. otherwise you are forced to create an excessive copy. I didn't write is is illegal, I have wrote thatKotlin's Mapinterface does not have this method
– Andrey Ilyunin
Nov 22 '18 at 16:09
|
show 7 more comments
0
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()
answered Nov 22 '18 at 16:08
Frank NeblungFrank Neblung
956417
add a comment |
0
val list: List>> = listOf()
val map = list.flatMap { it.entries }.groupBy { it.key }.mapValues { entry -> entry.value.map { it.value } }
answered Nov 23 '18 at 8:27
Constantin ChernishovConstantin Chernishov
20417
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
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.
answered Nov 22 '18 at 15:21
RolandRoland
10k11241
Thanks, it works. FYI I've made the change here : github.com/societe-generale/github-crawler/blob/…
– Vincent F
Nov 22 '18 at 16:23
well... you could have said, that it actually ends up in aMap<Branch, Map<String, Any>>... that.mapValuescould also already be simplified to:.mapValues { it.value.toMap() }
– Roland
Nov 22 '18 at 16:43
add a comment |
2
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.
answered Nov 22 '18 at 15:21
RolandRoland
10k11241
Thanks, it works. FYI I've made the change here : github.com/societe-generale/github-crawler/blob/…
– Vincent F
Nov 22 '18 at 16:23
well... you could have said, that it actually ends up in aMap<Branch, Map<String, Any>>... that.mapValuescould also already be simplified to:.mapValues { it.value.toMap() }
– Roland
Nov 22 '18 at 16:43
add a comment |
2
2
2
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.
answered Nov 22 '18 at 15:21
RolandRoland
10k11241
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.
answered Nov 22 '18 at 15:21
RolandRoland
10k11241
answered Nov 22 '18 at 15:21
RolandRoland
10k11241
answered Nov 22 '18 at 15:21
RolandRoland
10k11241
answered Nov 22 '18 at 15:21
RolandRoland
10k11241
10k11241
Thanks, it works. FYI I've made the change here : github.com/societe-generale/github-crawler/blob/…
– Vincent F
Nov 22 '18 at 16:23
well... you could have said, that it actually ends up in aMap<Branch, Map<String, Any>>... that.mapValuescould also already be simplified to:.mapValues { it.value.toMap() }
– Roland
Nov 22 '18 at 16:43
add a comment |
Thanks, it works. FYI I've made the change here : github.com/societe-generale/github-crawler/blob/…
– Vincent F
Nov 22 '18 at 16:23
well... you could have said, that it actually ends up in aMap<Branch, Map<String, Any>>... that.mapValuescould also already be simplified to:.mapValues { it.value.toMap() }
– Roland
Nov 22 '18 at 16:43
Thanks, it works. FYI I've made the change here : github.com/societe-generale/github-crawler/blob/…
– Vincent F
Nov 22 '18 at 16:23
Thanks, it works. FYI I've made the change here : github.com/societe-generale/github-crawler/blob/…
– Vincent F
Nov 22 '18 at 16:23
well... you could have said, that it actually ends up in a
Map<Branch, Map<String, Any>>... that .mapValues could also already be simplified to: .mapValues { it.value.toMap() }– Roland
Nov 22 '18 at 16:43
well... you could have said, that it actually ends up in a
Map<Branch, Map<String, Any>>... that .mapValues could also already be simplified to: .mapValues { it.value.toMap() }– Roland
Nov 22 '18 at 16:43
add a comment |
1
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?
answered Nov 22 '18 at 14:52
Andrey IlyuninAndrey Ilyunin
1,286220
what do you make of marko's comment?
– Tim Castelijns
Nov 22 '18 at 15:06
@TimCastelijns, Kotlin'sMapandMutableMapinterfaces do not contain this method
– Andrey Ilyunin
Nov 22 '18 at 15:44
On the JVM, this is legal Kotlin:val map = mutableMapOf(1 to "a", 2 to "b"); map.merge(1, "c") { v0, v1 -> v0 + v1}
– Marko Topolnik
Nov 22 '18 at 16:07
@TimCastelijns I tried to write it, butmergeis not a good fit, you can do it withcompute. It's not a great win overgroupBysolutions, though.val result = mutableMapOf<Branch, MutableList<Pair<String, Any>>>(); listOfMaps.forEach { map -> map.forEach { k, v -> result.compute(k) { _, v0 -> v0?.also { it.add(v) } ?: mutableListOf(v) } } }
– Marko Topolnik
Nov 22 '18 at 16:08
@MarkoTopolnik yes, but it is if you have mutable map returned to you as an argument. otherwise you are forced to create an excessive copy. I didn't write is is illegal, I have wrote thatKotlin's Mapinterface does not have this method
– Andrey Ilyunin
Nov 22 '18 at 16:09
|
show 7 more comments
1
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?
answered Nov 22 '18 at 14:52
Andrey IlyuninAndrey Ilyunin
1,286220
what do you make of marko's comment?
– Tim Castelijns
Nov 22 '18 at 15:06
@TimCastelijns, Kotlin'sMapandMutableMapinterfaces do not contain this method
– Andrey Ilyunin
Nov 22 '18 at 15:44
On the JVM, this is legal Kotlin:val map = mutableMapOf(1 to "a", 2 to "b"); map.merge(1, "c") { v0, v1 -> v0 + v1}
– Marko Topolnik
Nov 22 '18 at 16:07
@TimCastelijns I tried to write it, butmergeis not a good fit, you can do it withcompute. It's not a great win overgroupBysolutions, though.val result = mutableMapOf<Branch, MutableList<Pair<String, Any>>>(); listOfMaps.forEach { map -> map.forEach { k, v -> result.compute(k) { _, v0 -> v0?.also { it.add(v) } ?: mutableListOf(v) } } }
– Marko Topolnik
Nov 22 '18 at 16:08
@MarkoTopolnik yes, but it is if you have mutable map returned to you as an argument. otherwise you are forced to create an excessive copy. I didn't write is is illegal, I have wrote thatKotlin's Mapinterface does not have this method
– Andrey Ilyunin
Nov 22 '18 at 16:09
|
show 7 more comments
1
1
1
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?
answered Nov 22 '18 at 14:52
Andrey IlyuninAndrey Ilyunin
1,286220
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?
answered Nov 22 '18 at 14:52
Andrey IlyuninAndrey Ilyunin
1,286220
answered Nov 22 '18 at 14:52
Andrey IlyuninAndrey Ilyunin
1,286220
answered Nov 22 '18 at 14:52
Andrey IlyuninAndrey Ilyunin
1,286220
answered Nov 22 '18 at 14:52
Andrey IlyuninAndrey Ilyunin
1,286220
1,286220
what do you make of marko's comment?
– Tim Castelijns
Nov 22 '18 at 15:06
@TimCastelijns, Kotlin'sMapandMutableMapinterfaces do not contain this method
– Andrey Ilyunin
Nov 22 '18 at 15:44
On the JVM, this is legal Kotlin:val map = mutableMapOf(1 to "a", 2 to "b"); map.merge(1, "c") { v0, v1 -> v0 + v1}
– Marko Topolnik
Nov 22 '18 at 16:07
@TimCastelijns I tried to write it, butmergeis not a good fit, you can do it withcompute. It's not a great win overgroupBysolutions, though.val result = mutableMapOf<Branch, MutableList<Pair<String, Any>>>(); listOfMaps.forEach { map -> map.forEach { k, v -> result.compute(k) { _, v0 -> v0?.also { it.add(v) } ?: mutableListOf(v) } } }
– Marko Topolnik
Nov 22 '18 at 16:08
@MarkoTopolnik yes, but it is if you have mutable map returned to you as an argument. otherwise you are forced to create an excessive copy. I didn't write is is illegal, I have wrote thatKotlin's Mapinterface does not have this method
– Andrey Ilyunin
Nov 22 '18 at 16:09
|
show 7 more comments
what do you make of marko's comment?
– Tim Castelijns
Nov 22 '18 at 15:06
@TimCastelijns, Kotlin'sMapandMutableMapinterfaces do not contain this method
– Andrey Ilyunin
Nov 22 '18 at 15:44
On the JVM, this is legal Kotlin:val map = mutableMapOf(1 to "a", 2 to "b"); map.merge(1, "c") { v0, v1 -> v0 + v1}
– Marko Topolnik
Nov 22 '18 at 16:07
@TimCastelijns I tried to write it, butmergeis not a good fit, you can do it withcompute. It's not a great win overgroupBysolutions, though.val result = mutableMapOf<Branch, MutableList<Pair<String, Any>>>(); listOfMaps.forEach { map -> map.forEach { k, v -> result.compute(k) { _, v0 -> v0?.also { it.add(v) } ?: mutableListOf(v) } } }
– Marko Topolnik
Nov 22 '18 at 16:08
@MarkoTopolnik yes, but it is if you have mutable map returned to you as an argument. otherwise you are forced to create an excessive copy. I didn't write is is illegal, I have wrote thatKotlin's Mapinterface does not have this method
– Andrey Ilyunin
Nov 22 '18 at 16:09
what do you make of marko's comment?
– Tim Castelijns
Nov 22 '18 at 15:06
what do you make of marko's comment?
– Tim Castelijns
Nov 22 '18 at 15:06
@TimCastelijns, Kotlin's
Map and MutableMap interfaces do not contain this method– Andrey Ilyunin
Nov 22 '18 at 15:44
@TimCastelijns, Kotlin's
Map and MutableMap interfaces do not contain this method– Andrey Ilyunin
Nov 22 '18 at 15:44
On the JVM, this is legal Kotlin:
val map = mutableMapOf(1 to "a", 2 to "b"); map.merge(1, "c") { v0, v1 -> v0 + v1}– Marko Topolnik
Nov 22 '18 at 16:07
On the JVM, this is legal Kotlin:
val map = mutableMapOf(1 to "a", 2 to "b"); map.merge(1, "c") { v0, v1 -> v0 + v1}– Marko Topolnik
Nov 22 '18 at 16:07
@TimCastelijns I tried to write it, but
merge is not a good fit, you can do it with compute. It's not a great win over groupBy solutions, though. val result = mutableMapOf<Branch, MutableList<Pair<String, Any>>>(); listOfMaps.forEach { map -> map.forEach { k, v -> result.compute(k) { _, v0 -> v0?.also { it.add(v) } ?: mutableListOf(v) } } }– Marko Topolnik
Nov 22 '18 at 16:08
@TimCastelijns I tried to write it, but
merge is not a good fit, you can do it with compute. It's not a great win over groupBy solutions, though. val result = mutableMapOf<Branch, MutableList<Pair<String, Any>>>(); listOfMaps.forEach { map -> map.forEach { k, v -> result.compute(k) { _, v0 -> v0?.also { it.add(v) } ?: mutableListOf(v) } } }– Marko Topolnik
Nov 22 '18 at 16:08
@MarkoTopolnik yes, but it is if you have mutable map returned to you as an argument. otherwise you are forced to create an excessive copy. I didn't write is is illegal, I have wrote that
Kotlin's Map interface does not have this method– Andrey Ilyunin
Nov 22 '18 at 16:09
@MarkoTopolnik yes, but it is if you have mutable map returned to you as an argument. otherwise you are forced to create an excessive copy. I didn't write is is illegal, I have wrote that
Kotlin's Map interface does not have this method– Andrey Ilyunin
Nov 22 '18 at 16:09
|
show 7 more comments
0
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()
answered Nov 22 '18 at 16:08
Frank NeblungFrank Neblung
956417
add a comment |
0
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()
answered Nov 22 '18 at 16:08
Frank NeblungFrank Neblung
956417
add a comment |
0
0
0
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()
answered Nov 22 '18 at 16:08
Frank NeblungFrank Neblung
956417
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()
answered Nov 22 '18 at 16:08
Frank NeblungFrank Neblung
956417
answered Nov 22 '18 at 16:08
Frank NeblungFrank Neblung
956417
answered Nov 22 '18 at 16:08
Frank NeblungFrank Neblung
956417
answered Nov 22 '18 at 16:08
Frank NeblungFrank Neblung
956417
956417
add a comment |
add a comment |
0
val list: List>> = listOf()
val map = list.flatMap { it.entries }.groupBy { it.key }.mapValues { entry -> entry.value.map { it.value } }
answered Nov 23 '18 at 8:27
Constantin ChernishovConstantin Chernishov
20417
add a comment |
0
val list: List>> = listOf()
val map = list.flatMap { it.entries }.groupBy { it.key }.mapValues { entry -> entry.value.map { it.value } }
answered Nov 23 '18 at 8:27
Constantin ChernishovConstantin Chernishov
20417
add a comment |
0
0
0
val list: List>> = listOf()
val map = list.flatMap { it.entries }.groupBy { it.key }.mapValues { entry -> entry.value.map { it.value } }
answered Nov 23 '18 at 8:27
Constantin ChernishovConstantin Chernishov
20417
val list: List>> = listOf()
val map = list.flatMap { it.entries }.groupBy { it.key }.mapValues { entry -> entry.value.map { it.value } }
answered Nov 23 '18 at 8:27
Constantin ChernishovConstantin Chernishov
20417
answered Nov 23 '18 at 8:27
Constantin ChernishovConstantin Chernishov
20417
answered Nov 23 '18 at 8:27
Constantin ChernishovConstantin Chernishov
20417
answered Nov 23 '18 at 8:27
Constantin ChernishovConstantin Chernishov
20417
20417
add a comment |
add a comment |
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