combine two lists and map to new class object with rxjava









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Consider the following input : Calling network call will retrieve client object



data class Client(
val name: String,
val phoneNumber: String,
val frequentContacts: List<String>,
val allContacts: List<String>
)


what i need to do is to map similar names in frequentContacts and allContacts list in new object and subscribe on the output.



assume the response from network call will return this Client Object




"name": "Jack",
"phoneNumber": "90284302424",
"frequentContacts": [
"John",
"Sam"
],
"allContacts": [
"John",
"Adam",
"Peter",
"Kim",
"Sam"
]



what i need receive in subscribe newly create object .



data class clientViewModel(val name: String,val isFrequent: Boolean)


so in onSuccess i should have instance from clientViewModel



Expected output :



("John", true")



("Adam", false")



("Peter", false")



("Kim", false")



("Sam", true")



here what i am up to



clientRepository.getClientById(clientId)
.subscribeOn(Schedulers.io())
.observeOn(AndroidSchedulers.mainThread())
.doAfterTerminate view.hideProgress()
.flatMapit.frequentContacts
.subscribe
onSuccess(item: ClientViewModel)
onError()
onFinish()



but this not work because once i used flat map i lose the allContacts list
any help ?



I read about GroupBy operator but i am using Single...










share|improve this question























  • What do you think of: .flatMapit.allContacts.mapc->Pair(c,it.frequentContacts.contains(c))
    – Raphael
    Nov 9 at 19:22











  • @Raphael i tried it shows me type miss match error Required :SingleSource <out(???)> Found : List<Pair(String, Boolean)>
    – Joe Adams
    Nov 9 at 19:51














up vote
0
down vote

favorite












Consider the following input : Calling network call will retrieve client object



data class Client(
val name: String,
val phoneNumber: String,
val frequentContacts: List<String>,
val allContacts: List<String>
)


what i need to do is to map similar names in frequentContacts and allContacts list in new object and subscribe on the output.



assume the response from network call will return this Client Object




"name": "Jack",
"phoneNumber": "90284302424",
"frequentContacts": [
"John",
"Sam"
],
"allContacts": [
"John",
"Adam",
"Peter",
"Kim",
"Sam"
]



what i need receive in subscribe newly create object .



data class clientViewModel(val name: String,val isFrequent: Boolean)


so in onSuccess i should have instance from clientViewModel



Expected output :



("John", true")



("Adam", false")



("Peter", false")



("Kim", false")



("Sam", true")



here what i am up to



clientRepository.getClientById(clientId)
.subscribeOn(Schedulers.io())
.observeOn(AndroidSchedulers.mainThread())
.doAfterTerminate view.hideProgress()
.flatMapit.frequentContacts
.subscribe
onSuccess(item: ClientViewModel)
onError()
onFinish()



but this not work because once i used flat map i lose the allContacts list
any help ?



I read about GroupBy operator but i am using Single...










share|improve this question























  • What do you think of: .flatMapit.allContacts.mapc->Pair(c,it.frequentContacts.contains(c))
    – Raphael
    Nov 9 at 19:22











  • @Raphael i tried it shows me type miss match error Required :SingleSource <out(???)> Found : List<Pair(String, Boolean)>
    – Joe Adams
    Nov 9 at 19:51












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Consider the following input : Calling network call will retrieve client object



data class Client(
val name: String,
val phoneNumber: String,
val frequentContacts: List<String>,
val allContacts: List<String>
)


what i need to do is to map similar names in frequentContacts and allContacts list in new object and subscribe on the output.



assume the response from network call will return this Client Object




"name": "Jack",
"phoneNumber": "90284302424",
"frequentContacts": [
"John",
"Sam"
],
"allContacts": [
"John",
"Adam",
"Peter",
"Kim",
"Sam"
]



what i need receive in subscribe newly create object .



data class clientViewModel(val name: String,val isFrequent: Boolean)


so in onSuccess i should have instance from clientViewModel



Expected output :



("John", true")



("Adam", false")



("Peter", false")



("Kim", false")



("Sam", true")



here what i am up to



clientRepository.getClientById(clientId)
.subscribeOn(Schedulers.io())
.observeOn(AndroidSchedulers.mainThread())
.doAfterTerminate view.hideProgress()
.flatMapit.frequentContacts
.subscribe
onSuccess(item: ClientViewModel)
onError()
onFinish()



but this not work because once i used flat map i lose the allContacts list
any help ?



I read about GroupBy operator but i am using Single...










share|improve this question















Consider the following input : Calling network call will retrieve client object



data class Client(
val name: String,
val phoneNumber: String,
val frequentContacts: List<String>,
val allContacts: List<String>
)


what i need to do is to map similar names in frequentContacts and allContacts list in new object and subscribe on the output.



assume the response from network call will return this Client Object




"name": "Jack",
"phoneNumber": "90284302424",
"frequentContacts": [
"John",
"Sam"
],
"allContacts": [
"John",
"Adam",
"Peter",
"Kim",
"Sam"
]



what i need receive in subscribe newly create object .



data class clientViewModel(val name: String,val isFrequent: Boolean)


so in onSuccess i should have instance from clientViewModel



Expected output :



("John", true")



("Adam", false")



("Peter", false")



("Kim", false")



("Sam", true")



here what i am up to



clientRepository.getClientById(clientId)
.subscribeOn(Schedulers.io())
.observeOn(AndroidSchedulers.mainThread())
.doAfterTerminate view.hideProgress()
.flatMapit.frequentContacts
.subscribe
onSuccess(item: ClientViewModel)
onError()
onFinish()



but this not work because once i used flat map i lose the allContacts list
any help ?



I read about GroupBy operator but i am using Single...







kotlin rx-java2






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edited Nov 12 at 8:26









Jayson Minard

35.2k13103170




35.2k13103170










asked Nov 9 at 18:56









Joe Adams

11




11











  • What do you think of: .flatMapit.allContacts.mapc->Pair(c,it.frequentContacts.contains(c))
    – Raphael
    Nov 9 at 19:22











  • @Raphael i tried it shows me type miss match error Required :SingleSource <out(???)> Found : List<Pair(String, Boolean)>
    – Joe Adams
    Nov 9 at 19:51
















  • What do you think of: .flatMapit.allContacts.mapc->Pair(c,it.frequentContacts.contains(c))
    – Raphael
    Nov 9 at 19:22











  • @Raphael i tried it shows me type miss match error Required :SingleSource <out(???)> Found : List<Pair(String, Boolean)>
    – Joe Adams
    Nov 9 at 19:51















What do you think of: .flatMapit.allContacts.mapc->Pair(c,it.frequentContacts.contains(c))
– Raphael
Nov 9 at 19:22





What do you think of: .flatMapit.allContacts.mapc->Pair(c,it.frequentContacts.contains(c))
– Raphael
Nov 9 at 19:22













@Raphael i tried it shows me type miss match error Required :SingleSource <out(???)> Found : List<Pair(String, Boolean)>
– Joe Adams
Nov 9 at 19:51




@Raphael i tried it shows me type miss match error Required :SingleSource <out(???)> Found : List<Pair(String, Boolean)>
– Joe Adams
Nov 9 at 19:51












1 Answer
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up vote
0
down vote













Honestly I'd just do this:



clientRepository.getClientById(clientId)
.subscribeOn(Schedulers.io())
.observeOn(AndroidSchedulers.mainThread())
.doFinally view.hideProgress()
.map client ->
client.allContacts.map contact ->
clientViewModel(name = contact, isFrequent = client.frequentContacts.contains(contact))


.subscribeBy(onSuccess = list: List<clientViewModel> ->
...
, onError = err -> ... )





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    up vote
    0
    down vote













    Honestly I'd just do this:



    clientRepository.getClientById(clientId)
    .subscribeOn(Schedulers.io())
    .observeOn(AndroidSchedulers.mainThread())
    .doFinally view.hideProgress()
    .map client ->
    client.allContacts.map contact ->
    clientViewModel(name = contact, isFrequent = client.frequentContacts.contains(contact))


    .subscribeBy(onSuccess = list: List<clientViewModel> ->
    ...
    , onError = err -> ... )





    share|improve this answer
























      up vote
      0
      down vote













      Honestly I'd just do this:



      clientRepository.getClientById(clientId)
      .subscribeOn(Schedulers.io())
      .observeOn(AndroidSchedulers.mainThread())
      .doFinally view.hideProgress()
      .map client ->
      client.allContacts.map contact ->
      clientViewModel(name = contact, isFrequent = client.frequentContacts.contains(contact))


      .subscribeBy(onSuccess = list: List<clientViewModel> ->
      ...
      , onError = err -> ... )





      share|improve this answer






















        up vote
        0
        down vote










        up vote
        0
        down vote









        Honestly I'd just do this:



        clientRepository.getClientById(clientId)
        .subscribeOn(Schedulers.io())
        .observeOn(AndroidSchedulers.mainThread())
        .doFinally view.hideProgress()
        .map client ->
        client.allContacts.map contact ->
        clientViewModel(name = contact, isFrequent = client.frequentContacts.contains(contact))


        .subscribeBy(onSuccess = list: List<clientViewModel> ->
        ...
        , onError = err -> ... )





        share|improve this answer












        Honestly I'd just do this:



        clientRepository.getClientById(clientId)
        .subscribeOn(Schedulers.io())
        .observeOn(AndroidSchedulers.mainThread())
        .doFinally view.hideProgress()
        .map client ->
        client.allContacts.map contact ->
        clientViewModel(name = contact, isFrequent = client.frequentContacts.contains(contact))


        .subscribeBy(onSuccess = list: List<clientViewModel> ->
        ...
        , onError = err -> ... )






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 9 at 19:39









        EpicPandaForce

        46.1k14121239




        46.1k14121239



























             

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