Convert Map to Map<String, Set> with filter and streams












6














I would like to convert my map which looks like this:



{

  key="someKey1", value=Apple(id="1", color="green"),

  key="someKey2", value=Apple(id="2", color="red"),

  key="someKey3", value=Apple(id="3", color="green"),

  key="someKey4", value=Apple(id="4", color="red"),

}


to another map which puts all apples of the same color into the same list:



{

  key="red", value=list={apple1, apple3},

  key="green", value=list={apple2, apple4},  

}


I tried the following:



Map<String, Set<Apple>> sortedApples = appleMap.entrySet()

    .stream()

    .collect(Collectors.toMap(l -> l.getColour, ???));


Am I on the right track? Should I use filters for this task? Is there an easier way?






java lambda java-8 java-stream







share|improve this question





























    6














    I would like to convert my map which looks like this:



    {
    
  key="someKey1", value=Apple(id="1", color="green"),
    
  key="someKey2", value=Apple(id="2", color="red"),
    
  key="someKey3", value=Apple(id="3", color="green"),
    
  key="someKey4", value=Apple(id="4", color="red"),
    
}


    to another map which puts all apples of the same color into the same list:



    {
    
  key="red", value=list={apple1, apple3},
    
  key="green", value=list={apple2, apple4},  
    
}


    I tried the following:



    Map<String, Set<Apple>> sortedApples = appleMap.entrySet()
    
    .stream()
    
    .collect(Collectors.toMap(l -> l.getColour, ???));


    Am I on the right track? Should I use filters for this task? Is there an easier way?






    java lambda java-8 java-stream







    share|improve this question



























      6












      6








      6







      I would like to convert my map which looks like this:



      {
      
  key="someKey1", value=Apple(id="1", color="green"),
      
  key="someKey2", value=Apple(id="2", color="red"),
      
  key="someKey3", value=Apple(id="3", color="green"),
      
  key="someKey4", value=Apple(id="4", color="red"),
      
}


      to another map which puts all apples of the same color into the same list:



      {
      
  key="red", value=list={apple1, apple3},
      
  key="green", value=list={apple2, apple4},  
      
}


      I tried the following:



      Map<String, Set<Apple>> sortedApples = appleMap.entrySet()
      
    .stream()
      
    .collect(Collectors.toMap(l -> l.getColour, ???));


      Am I on the right track? Should I use filters for this task? Is there an easier way?






      java lambda java-8 java-stream







      share|improve this question















      I would like to convert my map which looks like this:



      {
      
  key="someKey1", value=Apple(id="1", color="green"),
      
  key="someKey2", value=Apple(id="2", color="red"),
      
  key="someKey3", value=Apple(id="3", color="green"),
      
  key="someKey4", value=Apple(id="4", color="red"),
      
}


      to another map which puts all apples of the same color into the same list:



      {
      
  key="red", value=list={apple1, apple3},
      
  key="green", value=list={apple2, apple4},  
      
}


      I tried the following:



      Map<String, Set<Apple>> sortedApples = appleMap.entrySet()
      
    .stream()
      
    .collect(Collectors.toMap(l -> l.getColour, ???));


      Am I on the right track? Should I use filters for this task? Is there an easier way?






      java lambda java-8 java-stream




      java lambda java-8 java-stream






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 10 mins ago









      Aomine

      39.8k73770




      39.8k73770










      asked 34 mins ago









      M4V3N

      528




      528
























          3 Answers
          3






          active

          oldest

          votes


















          4














          Collectors.groupingBy is more suitable than Collectors.toMap for this task (though both can be used).



          Map<String, List<Apple>> sortedApples = 
          
    appleMap.values()
          
            .stream()
          
            .collect(Collectors.groupingBy(Apple::getColour));


          Or, to group them into Sets use:



          Map<String, Set<Apple>> sortedApples = 
          
    appleMap.values()
          
            .stream()
          
            .collect(Collectors.groupingBy(Apple::getColour,
          
                                           Collectors.mapping(Function.identity(),
          
                                                              Collectors.toSet())));





          share|improve this answer





























            4














            You can use Collectors.groupingBy and Collectors.toSet()



            Map<String, Set<Apple>> sortedApples = appleMap.values() // Collection<Apple>
            
        .stream() // Stream<Apple>
            
        .collect(Collectors.groupingBy(Apple::getColour, // groupBy colour
            
                Collectors.mapping(a -> a, Collectors.toSet()))); // collect to Set





            share|improve this answer































              1














              if you want to proceed with toMap you can get the result as follows:



              map.entrySet()
              
   .stream()
              
   .collect(toMap(s -> s.getValue().getColour(), 
              
                    v ->  new HashSet<>(singleton(v.getValue())), 
              
            (l, r) -> {l.addAll(r); return l;}));


              But I would not suggest it because as you can see this it is not very readable and not the idiomatic approach.



              This is a job for the groupingBy collector shown in the other answers or do it shorter without a stream:



              Map<String, Set<Apple>> res = new HashMap<>();
              
map.forEach((k, v) -> res.computeIfAbsent(v.getColour(), e -> new HashSet<>()).add(v));





              share|improve this answer























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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                4














                Collectors.groupingBy is more suitable than Collectors.toMap for this task (though both can be used).



                Map<String, List<Apple>> sortedApples = 
                
    appleMap.values()
                
            .stream()
                
            .collect(Collectors.groupingBy(Apple::getColour));


                Or, to group them into Sets use:



                Map<String, Set<Apple>> sortedApples = 
                
    appleMap.values()
                
            .stream()
                
            .collect(Collectors.groupingBy(Apple::getColour,
                
                                           Collectors.mapping(Function.identity(),
                
                                                              Collectors.toSet())));





                share|improve this answer


























                  4














                  Collectors.groupingBy is more suitable than Collectors.toMap for this task (though both can be used).



                  Map<String, List<Apple>> sortedApples = 
                  
    appleMap.values()
                  
            .stream()
                  
            .collect(Collectors.groupingBy(Apple::getColour));


                  Or, to group them into Sets use:



                  Map<String, Set<Apple>> sortedApples = 
                  
    appleMap.values()
                  
            .stream()
                  
            .collect(Collectors.groupingBy(Apple::getColour,
                  
                                           Collectors.mapping(Function.identity(),
                  
                                                              Collectors.toSet())));





                  share|improve this answer
























                    4












                    4








                    4






                    Collectors.groupingBy is more suitable than Collectors.toMap for this task (though both can be used).



                    Map<String, List<Apple>> sortedApples = 
                    
    appleMap.values()
                    
            .stream()
                    
            .collect(Collectors.groupingBy(Apple::getColour));


                    Or, to group them into Sets use:



                    Map<String, Set<Apple>> sortedApples = 
                    
    appleMap.values()
                    
            .stream()
                    
            .collect(Collectors.groupingBy(Apple::getColour,
                    
                                           Collectors.mapping(Function.identity(),
                    
                                                              Collectors.toSet())));





                    share|improve this answer












                    Collectors.groupingBy is more suitable than Collectors.toMap for this task (though both can be used).



                    Map<String, List<Apple>> sortedApples = 
                    
    appleMap.values()
                    
            .stream()
                    
            .collect(Collectors.groupingBy(Apple::getColour));


                    Or, to group them into Sets use:



                    Map<String, Set<Apple>> sortedApples = 
                    
    appleMap.values()
                    
            .stream()
                    
            .collect(Collectors.groupingBy(Apple::getColour,
                    
                                           Collectors.mapping(Function.identity(),
                    
                                                              Collectors.toSet())));






                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered 31 mins ago









                    Eran

                    279k37452537




                    279k37452537

























                        4














                        You can use Collectors.groupingBy and Collectors.toSet()



                        Map<String, Set<Apple>> sortedApples = appleMap.values() // Collection<Apple>
                        
        .stream() // Stream<Apple>
                        
        .collect(Collectors.groupingBy(Apple::getColour, // groupBy colour
                        
                Collectors.mapping(a -> a, Collectors.toSet()))); // collect to Set





                        share|improve this answer




























                          4














                          You can use Collectors.groupingBy and Collectors.toSet()



                          Map<String, Set<Apple>> sortedApples = appleMap.values() // Collection<Apple>
                          
        .stream() // Stream<Apple>
                          
        .collect(Collectors.groupingBy(Apple::getColour, // groupBy colour
                          
                Collectors.mapping(a -> a, Collectors.toSet()))); // collect to Set





                          share|improve this answer


























                            4












                            4








                            4






                            You can use Collectors.groupingBy and Collectors.toSet()



                            Map<String, Set<Apple>> sortedApples = appleMap.values() // Collection<Apple>
                            
        .stream() // Stream<Apple>
                            
        .collect(Collectors.groupingBy(Apple::getColour, // groupBy colour
                            
                Collectors.mapping(a -> a, Collectors.toSet()))); // collect to Set





                            share|improve this answer














                            You can use Collectors.groupingBy and Collectors.toSet()



                            Map<String, Set<Apple>> sortedApples = appleMap.values() // Collection<Apple>
                            
        .stream() // Stream<Apple>
                            
        .collect(Collectors.groupingBy(Apple::getColour, // groupBy colour
                            
                Collectors.mapping(a -> a, Collectors.toSet()))); // collect to Set






                            share|improve this answer














                            share|improve this answer



                            share|improve this answer








                            edited 16 mins ago

























                            answered 30 mins ago









                            nullpointer

                            42.5k1090175




                            42.5k1090175























                                1














                                if you want to proceed with toMap you can get the result as follows:



                                map.entrySet()
                                
   .stream()
                                
   .collect(toMap(s -> s.getValue().getColour(), 
                                
                    v ->  new HashSet<>(singleton(v.getValue())), 
                                
            (l, r) -> {l.addAll(r); return l;}));


                                But I would not suggest it because as you can see this it is not very readable and not the idiomatic approach.



                                This is a job for the groupingBy collector shown in the other answers or do it shorter without a stream:



                                Map<String, Set<Apple>> res = new HashMap<>();
                                
map.forEach((k, v) -> res.computeIfAbsent(v.getColour(), e -> new HashSet<>()).add(v));





                                share|improve this answer




























                                  1














                                  if you want to proceed with toMap you can get the result as follows:



                                  map.entrySet()
                                  
   .stream()
                                  
   .collect(toMap(s -> s.getValue().getColour(), 
                                  
                    v ->  new HashSet<>(singleton(v.getValue())), 
                                  
            (l, r) -> {l.addAll(r); return l;}));


                                  But I would not suggest it because as you can see this it is not very readable and not the idiomatic approach.



                                  This is a job for the groupingBy collector shown in the other answers or do it shorter without a stream:



                                  Map<String, Set<Apple>> res = new HashMap<>();
                                  
map.forEach((k, v) -> res.computeIfAbsent(v.getColour(), e -> new HashSet<>()).add(v));





                                  share|improve this answer


























                                    1












                                    1








                                    1






                                    if you want to proceed with toMap you can get the result as follows:



                                    map.entrySet()
                                    
   .stream()
                                    
   .collect(toMap(s -> s.getValue().getColour(), 
                                    
                    v ->  new HashSet<>(singleton(v.getValue())), 
                                    
            (l, r) -> {l.addAll(r); return l;}));


                                    But I would not suggest it because as you can see this it is not very readable and not the idiomatic approach.



                                    This is a job for the groupingBy collector shown in the other answers or do it shorter without a stream:



                                    Map<String, Set<Apple>> res = new HashMap<>();
                                    
map.forEach((k, v) -> res.computeIfAbsent(v.getColour(), e -> new HashSet<>()).add(v));





                                    share|improve this answer














                                    if you want to proceed with toMap you can get the result as follows:



                                    map.entrySet()
                                    
   .stream()
                                    
   .collect(toMap(s -> s.getValue().getColour(), 
                                    
                    v ->  new HashSet<>(singleton(v.getValue())), 
                                    
            (l, r) -> {l.addAll(r); return l;}));


                                    But I would not suggest it because as you can see this it is not very readable and not the idiomatic approach.



                                    This is a job for the groupingBy collector shown in the other answers or do it shorter without a stream:



                                    Map<String, Set<Apple>> res = new HashMap<>();
                                    
map.forEach((k, v) -> res.computeIfAbsent(v.getColour(), e -> new HashSet<>()).add(v));






                                    share|improve this answer














                                    share|improve this answer



                                    share|improve this answer








                                    edited 11 mins ago

























                                    answered 17 mins ago









                                    Aomine

                                    39.8k73770




                                    39.8k73770






























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