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

asked Nov 22 at 14:28

Vincent F

1,9181229

add a comment |

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.

Any idea ?

Thanks

asked Nov 22 at 14:28

Vincent F

1,9181229

add a comment |

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.

Any idea ?

Thanks

asked Nov 22 at 14:28

Vincent F

1,9181229

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.

Any idea ?

Thanks

kotlin

asked Nov 22 at 14:28

Vincent F

1,9181229

asked Nov 22 at 14:28

Vincent F

1,9181229

asked Nov 22 at 14:28

Vincent F

1,9181229

asked Nov 22 at 14:28

Vincent F

1,9181229

asked Nov 22 at 14:28

Vincent F

1,9181229

add a comment |

4 Answers
4

active

oldest

votes

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 at 15:21

Roland

9,39711141

Thanks, it works. FYI I've made the change here : github.com/societe-generale/github-crawler/blob/…
– Vincent F
Nov 22 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 at 16:43

add a comment |

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 at 14:52

Andrey Ilyunin

1,259220

what do you make of marko's comment?
– Tim Castelijns
Nov 22 at 15:06

@TimCastelijns, Kotlin's Map and MutableMap interfaces do not contain this method
– Andrey Ilyunin
Nov 22 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 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 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 at 16:09

|
show 7 more comments

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 at 16:08

Frank Neblung

919417

add a comment |

val list: List>> = listOf()
val map = list.flatMap { it.entries }.groupBy { it.key }.mapValues { entry -> entry.value.map { it.value } }

answered Nov 23 at 8:27

Constantin Chernishov

2047

add a comment |

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

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 at 15:21

Roland

9,39711141

Thanks, it works. FYI I've made the change here : github.com/societe-generale/github-crawler/blob/…
– Vincent F
Nov 22 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 at 16:43

add a comment |

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 at 15:21

Roland

9,39711141

Thanks, it works. FYI I've made the change here : github.com/societe-generale/github-crawler/blob/…
– Vincent F
Nov 22 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 at 16:43

add a comment |

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 at 15:21

Roland

9,39711141

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 at 15:21

Roland

9,39711141

answered Nov 22 at 15:21

Roland

9,39711141

answered Nov 22 at 15:21

Roland

9,39711141

answered Nov 22 at 15:21

Roland

9,39711141

Thanks, it works. FYI I've made the change here : github.com/societe-generale/github-crawler/blob/…
– Vincent F
Nov 22 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 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 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 at 16:43

Thanks, it works. FYI I've made the change here : github.com/societe-generale/github-crawler/blob/…
– Vincent F
Nov 22 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 at 16:43

add a comment |

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 at 14:52

Andrey Ilyunin

1,259220

what do you make of marko's comment?
– Tim Castelijns
Nov 22 at 15:06

@TimCastelijns, Kotlin's Map and MutableMap interfaces do not contain this method
– Andrey Ilyunin
Nov 22 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 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 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 at 16:09

|
show 7 more comments

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 at 14:52

Andrey Ilyunin

1,259220

what do you make of marko's comment?
– Tim Castelijns
Nov 22 at 15:06

@TimCastelijns, Kotlin's Map and MutableMap interfaces do not contain this method
– Andrey Ilyunin
Nov 22 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 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 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 at 16:09

|
show 7 more comments

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 at 14:52

Andrey Ilyunin

1,259220

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 at 14:52

Andrey Ilyunin

1,259220

answered Nov 22 at 14:52

Andrey Ilyunin

1,259220

answered Nov 22 at 14:52

Andrey Ilyunin

1,259220

answered Nov 22 at 14:52

Andrey Ilyunin

1,259220

what do you make of marko's comment?
– Tim Castelijns
Nov 22 at 15:06

@TimCastelijns, Kotlin's Map and MutableMap interfaces do not contain this method
– Andrey Ilyunin
Nov 22 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 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 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 at 16:09

|
show 7 more comments

what do you make of marko's comment?
– Tim Castelijns
Nov 22 at 15:06

@TimCastelijns, Kotlin's Map and MutableMap interfaces do not contain this method
– Andrey Ilyunin
Nov 22 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 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 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 at 16:09

what do you make of marko's comment?
– Tim Castelijns
Nov 22 at 15:06

@TimCastelijns, Kotlin's Map and MutableMap interfaces do not contain this method
– Andrey Ilyunin
Nov 22 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 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 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 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 at 16:09

|
show 7 more comments

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 at 16:08

Frank Neblung

919417

add a comment |

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 at 16:08

Frank Neblung

919417

add a comment |

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 at 16:08

Frank Neblung

919417

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 at 16:08

Frank Neblung

919417

answered Nov 22 at 16:08

Frank Neblung

919417

answered Nov 22 at 16:08

Frank Neblung

919417

answered Nov 22 at 16:08

Frank Neblung

919417

add a comment |

val list: List>> = listOf()
val map = list.flatMap { it.entries }.groupBy { it.key }.mapValues { entry -> entry.value.map { it.value } }

answered Nov 23 at 8:27

Constantin Chernishov

2047

add a comment |

val list: List>> = listOf()
val map = list.flatMap { it.entries }.groupBy { it.key }.mapValues { entry -> entry.value.map { it.value } }

answered Nov 23 at 8:27

Constantin Chernishov

2047

add a comment |

val list: List>> = listOf()
val map = list.flatMap { it.entries }.groupBy { it.key }.mapValues { entry -> entry.value.map { it.value } }

answered Nov 23 at 8:27

Constantin Chernishov

2047

val list: List>> = listOf()
val map = list.flatMap { it.entries }.groupBy { it.key }.mapValues { entry -> entry.value.map { it.value } }

answered Nov 23 at 8:27

Constantin Chernishov

2047

answered Nov 23 at 8:27

Constantin Chernishov

2047

answered Nov 23 at 8:27

Constantin Chernishov

2047

answered Nov 23 at 8:27

Constantin Chernishov

2047

add a comment |

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