Check the keys in the map matching with the List content in Java












13















I have a List of Strings and a Map. Every key in the map needs to present in the list else I need to throw an exception. As of now I am looping the list and checking the key and throw exception if the map doesn't contains the key. Below is the sample code is what I am doing. IS there any other way in Java8 we can do it in one line or something using streams and filters ?



And also the contents in the list and keys in the map should match. That I am already handling in the separate if condition.



import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

public class TestClass {

public static void main(String args) {

List<String> ll = new ArrayList<>();
Map<String, Integer> m = new HashMap<>();
ll.add("a");
ll.add("b");
ll.add("d");

m.put("a", 1);
m.put("b", 1);
m.put("c", 1);

if(ll.size() != m.size){
System.out.println("Throw Exception");
}

for(String s : ll) {

if(!m.containsKey(s)) {
System.out.println("Throw Exception");
}
}
}
}









share|improve this question




















  • 2





    You are saying different things in your question and in your code. In your question you said you want to check if every key is in the list, but in the code, you are checking if everything in the list is present in the map.

    – Sweeper
    Jan 11 at 7:02











  • How about this ll.stream().filter(s -> !m.containsKey(s)).forEach(s -> { throw new RuntimeException("Not found"); }); ?

    – manfromnowhere
    Jan 11 at 7:06













  • @Sweeper Sorry , I will modify the question. Technically, the contents in the list and keys in the map should match.

    – sparker
    2 days ago






  • 2





    If they really need to match, then test something like m.keySet().equals(new HashSet<>(ll)). Or maybe keep the required keys in a set in the first place.

    – Stuart Marks
    2 days ago






  • 1





    @sparker Seems like what you are expecting your code does is not what it actually is doing. Maybe think about it and rephrase the question to clear out the doubt.

    – nullpointer
    2 days ago
















13















I have a List of Strings and a Map. Every key in the map needs to present in the list else I need to throw an exception. As of now I am looping the list and checking the key and throw exception if the map doesn't contains the key. Below is the sample code is what I am doing. IS there any other way in Java8 we can do it in one line or something using streams and filters ?



And also the contents in the list and keys in the map should match. That I am already handling in the separate if condition.



import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

public class TestClass {

public static void main(String args) {

List<String> ll = new ArrayList<>();
Map<String, Integer> m = new HashMap<>();
ll.add("a");
ll.add("b");
ll.add("d");

m.put("a", 1);
m.put("b", 1);
m.put("c", 1);

if(ll.size() != m.size){
System.out.println("Throw Exception");
}

for(String s : ll) {

if(!m.containsKey(s)) {
System.out.println("Throw Exception");
}
}
}
}









share|improve this question




















  • 2





    You are saying different things in your question and in your code. In your question you said you want to check if every key is in the list, but in the code, you are checking if everything in the list is present in the map.

    – Sweeper
    Jan 11 at 7:02











  • How about this ll.stream().filter(s -> !m.containsKey(s)).forEach(s -> { throw new RuntimeException("Not found"); }); ?

    – manfromnowhere
    Jan 11 at 7:06













  • @Sweeper Sorry , I will modify the question. Technically, the contents in the list and keys in the map should match.

    – sparker
    2 days ago






  • 2





    If they really need to match, then test something like m.keySet().equals(new HashSet<>(ll)). Or maybe keep the required keys in a set in the first place.

    – Stuart Marks
    2 days ago






  • 1





    @sparker Seems like what you are expecting your code does is not what it actually is doing. Maybe think about it and rephrase the question to clear out the doubt.

    – nullpointer
    2 days ago














13












13








13


1






I have a List of Strings and a Map. Every key in the map needs to present in the list else I need to throw an exception. As of now I am looping the list and checking the key and throw exception if the map doesn't contains the key. Below is the sample code is what I am doing. IS there any other way in Java8 we can do it in one line or something using streams and filters ?



And also the contents in the list and keys in the map should match. That I am already handling in the separate if condition.



import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

public class TestClass {

public static void main(String args) {

List<String> ll = new ArrayList<>();
Map<String, Integer> m = new HashMap<>();
ll.add("a");
ll.add("b");
ll.add("d");

m.put("a", 1);
m.put("b", 1);
m.put("c", 1);

if(ll.size() != m.size){
System.out.println("Throw Exception");
}

for(String s : ll) {

if(!m.containsKey(s)) {
System.out.println("Throw Exception");
}
}
}
}









share|improve this question
















I have a List of Strings and a Map. Every key in the map needs to present in the list else I need to throw an exception. As of now I am looping the list and checking the key and throw exception if the map doesn't contains the key. Below is the sample code is what I am doing. IS there any other way in Java8 we can do it in one line or something using streams and filters ?



And also the contents in the list and keys in the map should match. That I am already handling in the separate if condition.



import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

public class TestClass {

public static void main(String args) {

List<String> ll = new ArrayList<>();
Map<String, Integer> m = new HashMap<>();
ll.add("a");
ll.add("b");
ll.add("d");

m.put("a", 1);
m.put("b", 1);
m.put("c", 1);

if(ll.size() != m.size){
System.out.println("Throw Exception");
}

for(String s : ll) {

if(!m.containsKey(s)) {
System.out.println("Throw Exception");
}
}
}
}






java lambda java-8 java-stream






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share|improve this question













share|improve this question




share|improve this question








edited 2 days ago







sparker

















asked Jan 11 at 6:53









sparkersparker

331414




331414








  • 2





    You are saying different things in your question and in your code. In your question you said you want to check if every key is in the list, but in the code, you are checking if everything in the list is present in the map.

    – Sweeper
    Jan 11 at 7:02











  • How about this ll.stream().filter(s -> !m.containsKey(s)).forEach(s -> { throw new RuntimeException("Not found"); }); ?

    – manfromnowhere
    Jan 11 at 7:06













  • @Sweeper Sorry , I will modify the question. Technically, the contents in the list and keys in the map should match.

    – sparker
    2 days ago






  • 2





    If they really need to match, then test something like m.keySet().equals(new HashSet<>(ll)). Or maybe keep the required keys in a set in the first place.

    – Stuart Marks
    2 days ago






  • 1





    @sparker Seems like what you are expecting your code does is not what it actually is doing. Maybe think about it and rephrase the question to clear out the doubt.

    – nullpointer
    2 days ago














  • 2





    You are saying different things in your question and in your code. In your question you said you want to check if every key is in the list, but in the code, you are checking if everything in the list is present in the map.

    – Sweeper
    Jan 11 at 7:02











  • How about this ll.stream().filter(s -> !m.containsKey(s)).forEach(s -> { throw new RuntimeException("Not found"); }); ?

    – manfromnowhere
    Jan 11 at 7:06













  • @Sweeper Sorry , I will modify the question. Technically, the contents in the list and keys in the map should match.

    – sparker
    2 days ago






  • 2





    If they really need to match, then test something like m.keySet().equals(new HashSet<>(ll)). Or maybe keep the required keys in a set in the first place.

    – Stuart Marks
    2 days ago






  • 1





    @sparker Seems like what you are expecting your code does is not what it actually is doing. Maybe think about it and rephrase the question to clear out the doubt.

    – nullpointer
    2 days ago








2




2





You are saying different things in your question and in your code. In your question you said you want to check if every key is in the list, but in the code, you are checking if everything in the list is present in the map.

– Sweeper
Jan 11 at 7:02





You are saying different things in your question and in your code. In your question you said you want to check if every key is in the list, but in the code, you are checking if everything in the list is present in the map.

– Sweeper
Jan 11 at 7:02













How about this ll.stream().filter(s -> !m.containsKey(s)).forEach(s -> { throw new RuntimeException("Not found"); }); ?

– manfromnowhere
Jan 11 at 7:06







How about this ll.stream().filter(s -> !m.containsKey(s)).forEach(s -> { throw new RuntimeException("Not found"); }); ?

– manfromnowhere
Jan 11 at 7:06















@Sweeper Sorry , I will modify the question. Technically, the contents in the list and keys in the map should match.

– sparker
2 days ago





@Sweeper Sorry , I will modify the question. Technically, the contents in the list and keys in the map should match.

– sparker
2 days ago




2




2





If they really need to match, then test something like m.keySet().equals(new HashSet<>(ll)). Or maybe keep the required keys in a set in the first place.

– Stuart Marks
2 days ago





If they really need to match, then test something like m.keySet().equals(new HashSet<>(ll)). Or maybe keep the required keys in a set in the first place.

– Stuart Marks
2 days ago




1




1





@sparker Seems like what you are expecting your code does is not what it actually is doing. Maybe think about it and rephrase the question to clear out the doubt.

– nullpointer
2 days ago





@sparker Seems like what you are expecting your code does is not what it actually is doing. Maybe think about it and rephrase the question to clear out the doubt.

– nullpointer
2 days ago












6 Answers
6






active

oldest

votes


















9















Every key in the map needs to present in the list else I need to throw
an exception




You could do it using Stream.anyMatch and iterating on the keyset of the map instead as (variable names updated for readability purpose) :



if(map.keySet().stream().anyMatch(key -> !list.contains(key))) {
throw new CustomException("");
}


Better and as simple as it gets, use List.containsAll :



if(!list.containsAll(map.keySet())) {
throw new CustomException("");
}


Important: If you can trade for O(n) space to reduce the runtime complexity, you can create a HashSet out of your List and then perform the lookups. It would reduce the runtime complexity from O(n^2) to O(n) and the implementation would look like:



Set<String> allUniqueElementsInList = new HashSet<>(list);
if(!allUniqueElementsInList.containsAll(map.keySet())) {
throw new CustomException("");
}





share|improve this answer





















  • 1





    Or if(! map.keySet().stream().allMatch(list::contains)), but of course, if(!list.containsAll(map.keySet())) is the canonical solution.

    – Holger
    2 days ago



















2














Try this:



if ((ll == null && m == null) ||                            // if both are null
((ll.size() == m.size() && m.keySet().containsAll(ll))) // or contain the same elements
) {
System.out.println("Collections contain the same elements");
} else {
throw new CustomException("Collections don't match!");
}





share|improve this answer































    2














    We can try adding the list to a set, then comparing that set with the keyset from your hashmap:



    List<String> ll = new ArrayList<>();
    ll.add("a");
    ll.add("b");
    ll.add("d");

    Map<String, Integer> m = new HashMap<>();
    m.put("a", 1);
    m.put("b", 1);
    m.put("c", 1);

    Set<String> set = new HashSet<String>(ll);

    if (Objects.equals(set, m.keySet())) {
    System.out.println("sets match");
    }
    else {
    System.out.println("sets do not match");
    }





    share|improve this answer





















    • 1





      I think OP is looking for an answer wrt to java-8 (something using streams and filters) here.

      – Nicholas K
      2 days ago








    • 1





      @NicholasK well there is nothing in the code here, that doesn't work with java8., right? But yeah the throwing of exception is definitely missing and also an unwanted creation of Set which doesn't seem to be required for the purpose of OP.

      – nullpointer
      2 days ago








    • 1





      Ummm... if (...m.keySet() == null ...) ... what is that? The keySet can never be null, and in any case: the comparison of the size() and the containsAll call boil down to a simple if (Objects.equals(m.keySet(), set)) ...

      – Marco13
      2 days ago











    • @MarcoPolo Thanks for the feedback, answer updated.

      – Tim Biegeleisen
      2 days ago



















    1














    Simply use the following :-



    m.keySet().stream().filter(e -> !ll.contains(e)).findAny()
    .ifPresent(e -> throwException("Key Not found : " + e));


    And define the throwException method below :



    public static void throwException(String msg) {
    throw new RuntimeException(msg);
    }





    share|improve this answer

































      0














      You can simply change your existing code to -



      if(!m.keySet().containsAll(ll)) {
      System.out.println("Throws Exception");
      }


      This will solve your problem. :)






      share|improve this answer





















      • 2





        This logic has a problem. Should the hashmap's keyset contain extra items, your code would still return true. Also, if null on both sides of the comparison means equal, then you would have to cover that case as well.

        – Tim Biegeleisen
        Jan 11 at 7:03











      • @p.bansal - The same check performed in reverse was what OP is mostly trying to look forward to.

        – nullpointer
        2 days ago






      • 1





        @nullpointer This matches the OP’s code but not the text in the question, which requests the reverse. So +1 from me on this answer. But then the OP said in comments that the map keys must match the list, a yet again different check. (Sigh.)

        – Stuart Marks
        2 days ago











      • @StuartMarks well, if they must match, I’d use if(m.size() != ll.size() || !m.keySet().containsAll(ll)), as I suppose, m.keySet().containsAll(ll) is potentially faster than ll.containsAll(m.keySet())

        – Holger
        2 days ago











      • @Holger At the time this answer was written, it matched the code in the question. The OP subsequently changed the requirements.

        – Stuart Marks
        2 days ago



















      0














      Here is another solution:



          if (ll  .parallelStream()
      .filter(v -> !m.containsKey(v)) // Filter alle values not contained in the map
      .count() == 0) { // If no values are left then every key was present
      // do something
      } else {
      throw new RuntimeException("hello");
      }


      Just wanted to show a different approach






      share|improve this answer

























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        6 Answers
        6






        active

        oldest

        votes








        6 Answers
        6






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        9















        Every key in the map needs to present in the list else I need to throw
        an exception




        You could do it using Stream.anyMatch and iterating on the keyset of the map instead as (variable names updated for readability purpose) :



        if(map.keySet().stream().anyMatch(key -> !list.contains(key))) {
        throw new CustomException("");
        }


        Better and as simple as it gets, use List.containsAll :



        if(!list.containsAll(map.keySet())) {
        throw new CustomException("");
        }


        Important: If you can trade for O(n) space to reduce the runtime complexity, you can create a HashSet out of your List and then perform the lookups. It would reduce the runtime complexity from O(n^2) to O(n) and the implementation would look like:



        Set<String> allUniqueElementsInList = new HashSet<>(list);
        if(!allUniqueElementsInList.containsAll(map.keySet())) {
        throw new CustomException("");
        }





        share|improve this answer





















        • 1





          Or if(! map.keySet().stream().allMatch(list::contains)), but of course, if(!list.containsAll(map.keySet())) is the canonical solution.

          – Holger
          2 days ago
















        9















        Every key in the map needs to present in the list else I need to throw
        an exception




        You could do it using Stream.anyMatch and iterating on the keyset of the map instead as (variable names updated for readability purpose) :



        if(map.keySet().stream().anyMatch(key -> !list.contains(key))) {
        throw new CustomException("");
        }


        Better and as simple as it gets, use List.containsAll :



        if(!list.containsAll(map.keySet())) {
        throw new CustomException("");
        }


        Important: If you can trade for O(n) space to reduce the runtime complexity, you can create a HashSet out of your List and then perform the lookups. It would reduce the runtime complexity from O(n^2) to O(n) and the implementation would look like:



        Set<String> allUniqueElementsInList = new HashSet<>(list);
        if(!allUniqueElementsInList.containsAll(map.keySet())) {
        throw new CustomException("");
        }





        share|improve this answer





















        • 1





          Or if(! map.keySet().stream().allMatch(list::contains)), but of course, if(!list.containsAll(map.keySet())) is the canonical solution.

          – Holger
          2 days ago














        9












        9








        9








        Every key in the map needs to present in the list else I need to throw
        an exception




        You could do it using Stream.anyMatch and iterating on the keyset of the map instead as (variable names updated for readability purpose) :



        if(map.keySet().stream().anyMatch(key -> !list.contains(key))) {
        throw new CustomException("");
        }


        Better and as simple as it gets, use List.containsAll :



        if(!list.containsAll(map.keySet())) {
        throw new CustomException("");
        }


        Important: If you can trade for O(n) space to reduce the runtime complexity, you can create a HashSet out of your List and then perform the lookups. It would reduce the runtime complexity from O(n^2) to O(n) and the implementation would look like:



        Set<String> allUniqueElementsInList = new HashSet<>(list);
        if(!allUniqueElementsInList.containsAll(map.keySet())) {
        throw new CustomException("");
        }





        share|improve this answer
















        Every key in the map needs to present in the list else I need to throw
        an exception




        You could do it using Stream.anyMatch and iterating on the keyset of the map instead as (variable names updated for readability purpose) :



        if(map.keySet().stream().anyMatch(key -> !list.contains(key))) {
        throw new CustomException("");
        }


        Better and as simple as it gets, use List.containsAll :



        if(!list.containsAll(map.keySet())) {
        throw new CustomException("");
        }


        Important: If you can trade for O(n) space to reduce the runtime complexity, you can create a HashSet out of your List and then perform the lookups. It would reduce the runtime complexity from O(n^2) to O(n) and the implementation would look like:



        Set<String> allUniqueElementsInList = new HashSet<>(list);
        if(!allUniqueElementsInList.containsAll(map.keySet())) {
        throw new CustomException("");
        }






        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 2 days ago

























        answered 2 days ago









        nullpointernullpointer

        45.1k1096184




        45.1k1096184








        • 1





          Or if(! map.keySet().stream().allMatch(list::contains)), but of course, if(!list.containsAll(map.keySet())) is the canonical solution.

          – Holger
          2 days ago














        • 1





          Or if(! map.keySet().stream().allMatch(list::contains)), but of course, if(!list.containsAll(map.keySet())) is the canonical solution.

          – Holger
          2 days ago








        1




        1





        Or if(! map.keySet().stream().allMatch(list::contains)), but of course, if(!list.containsAll(map.keySet())) is the canonical solution.

        – Holger
        2 days ago





        Or if(! map.keySet().stream().allMatch(list::contains)), but of course, if(!list.containsAll(map.keySet())) is the canonical solution.

        – Holger
        2 days ago













        2














        Try this:



        if ((ll == null && m == null) ||                            // if both are null
        ((ll.size() == m.size() && m.keySet().containsAll(ll))) // or contain the same elements
        ) {
        System.out.println("Collections contain the same elements");
        } else {
        throw new CustomException("Collections don't match!");
        }





        share|improve this answer




























          2














          Try this:



          if ((ll == null && m == null) ||                            // if both are null
          ((ll.size() == m.size() && m.keySet().containsAll(ll))) // or contain the same elements
          ) {
          System.out.println("Collections contain the same elements");
          } else {
          throw new CustomException("Collections don't match!");
          }





          share|improve this answer


























            2












            2








            2







            Try this:



            if ((ll == null && m == null) ||                            // if both are null
            ((ll.size() == m.size() && m.keySet().containsAll(ll))) // or contain the same elements
            ) {
            System.out.println("Collections contain the same elements");
            } else {
            throw new CustomException("Collections don't match!");
            }





            share|improve this answer













            Try this:



            if ((ll == null && m == null) ||                            // if both are null
            ((ll.size() == m.size() && m.keySet().containsAll(ll))) // or contain the same elements
            ) {
            System.out.println("Collections contain the same elements");
            } else {
            throw new CustomException("Collections don't match!");
            }






            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered 2 days ago









            ETOETO

            2,163423




            2,163423























                2














                We can try adding the list to a set, then comparing that set with the keyset from your hashmap:



                List<String> ll = new ArrayList<>();
                ll.add("a");
                ll.add("b");
                ll.add("d");

                Map<String, Integer> m = new HashMap<>();
                m.put("a", 1);
                m.put("b", 1);
                m.put("c", 1);

                Set<String> set = new HashSet<String>(ll);

                if (Objects.equals(set, m.keySet())) {
                System.out.println("sets match");
                }
                else {
                System.out.println("sets do not match");
                }





                share|improve this answer





















                • 1





                  I think OP is looking for an answer wrt to java-8 (something using streams and filters) here.

                  – Nicholas K
                  2 days ago








                • 1





                  @NicholasK well there is nothing in the code here, that doesn't work with java8., right? But yeah the throwing of exception is definitely missing and also an unwanted creation of Set which doesn't seem to be required for the purpose of OP.

                  – nullpointer
                  2 days ago








                • 1





                  Ummm... if (...m.keySet() == null ...) ... what is that? The keySet can never be null, and in any case: the comparison of the size() and the containsAll call boil down to a simple if (Objects.equals(m.keySet(), set)) ...

                  – Marco13
                  2 days ago











                • @MarcoPolo Thanks for the feedback, answer updated.

                  – Tim Biegeleisen
                  2 days ago
















                2














                We can try adding the list to a set, then comparing that set with the keyset from your hashmap:



                List<String> ll = new ArrayList<>();
                ll.add("a");
                ll.add("b");
                ll.add("d");

                Map<String, Integer> m = new HashMap<>();
                m.put("a", 1);
                m.put("b", 1);
                m.put("c", 1);

                Set<String> set = new HashSet<String>(ll);

                if (Objects.equals(set, m.keySet())) {
                System.out.println("sets match");
                }
                else {
                System.out.println("sets do not match");
                }





                share|improve this answer





















                • 1





                  I think OP is looking for an answer wrt to java-8 (something using streams and filters) here.

                  – Nicholas K
                  2 days ago








                • 1





                  @NicholasK well there is nothing in the code here, that doesn't work with java8., right? But yeah the throwing of exception is definitely missing and also an unwanted creation of Set which doesn't seem to be required for the purpose of OP.

                  – nullpointer
                  2 days ago








                • 1





                  Ummm... if (...m.keySet() == null ...) ... what is that? The keySet can never be null, and in any case: the comparison of the size() and the containsAll call boil down to a simple if (Objects.equals(m.keySet(), set)) ...

                  – Marco13
                  2 days ago











                • @MarcoPolo Thanks for the feedback, answer updated.

                  – Tim Biegeleisen
                  2 days ago














                2












                2








                2







                We can try adding the list to a set, then comparing that set with the keyset from your hashmap:



                List<String> ll = new ArrayList<>();
                ll.add("a");
                ll.add("b");
                ll.add("d");

                Map<String, Integer> m = new HashMap<>();
                m.put("a", 1);
                m.put("b", 1);
                m.put("c", 1);

                Set<String> set = new HashSet<String>(ll);

                if (Objects.equals(set, m.keySet())) {
                System.out.println("sets match");
                }
                else {
                System.out.println("sets do not match");
                }





                share|improve this answer















                We can try adding the list to a set, then comparing that set with the keyset from your hashmap:



                List<String> ll = new ArrayList<>();
                ll.add("a");
                ll.add("b");
                ll.add("d");

                Map<String, Integer> m = new HashMap<>();
                m.put("a", 1);
                m.put("b", 1);
                m.put("c", 1);

                Set<String> set = new HashSet<String>(ll);

                if (Objects.equals(set, m.keySet())) {
                System.out.println("sets match");
                }
                else {
                System.out.println("sets do not match");
                }






                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 2 days ago

























                answered Jan 11 at 7:02









                Tim BiegeleisenTim Biegeleisen

                220k1388141




                220k1388141








                • 1





                  I think OP is looking for an answer wrt to java-8 (something using streams and filters) here.

                  – Nicholas K
                  2 days ago








                • 1





                  @NicholasK well there is nothing in the code here, that doesn't work with java8., right? But yeah the throwing of exception is definitely missing and also an unwanted creation of Set which doesn't seem to be required for the purpose of OP.

                  – nullpointer
                  2 days ago








                • 1





                  Ummm... if (...m.keySet() == null ...) ... what is that? The keySet can never be null, and in any case: the comparison of the size() and the containsAll call boil down to a simple if (Objects.equals(m.keySet(), set)) ...

                  – Marco13
                  2 days ago











                • @MarcoPolo Thanks for the feedback, answer updated.

                  – Tim Biegeleisen
                  2 days ago














                • 1





                  I think OP is looking for an answer wrt to java-8 (something using streams and filters) here.

                  – Nicholas K
                  2 days ago








                • 1





                  @NicholasK well there is nothing in the code here, that doesn't work with java8., right? But yeah the throwing of exception is definitely missing and also an unwanted creation of Set which doesn't seem to be required for the purpose of OP.

                  – nullpointer
                  2 days ago








                • 1





                  Ummm... if (...m.keySet() == null ...) ... what is that? The keySet can never be null, and in any case: the comparison of the size() and the containsAll call boil down to a simple if (Objects.equals(m.keySet(), set)) ...

                  – Marco13
                  2 days ago











                • @MarcoPolo Thanks for the feedback, answer updated.

                  – Tim Biegeleisen
                  2 days ago








                1




                1





                I think OP is looking for an answer wrt to java-8 (something using streams and filters) here.

                – Nicholas K
                2 days ago







                I think OP is looking for an answer wrt to java-8 (something using streams and filters) here.

                – Nicholas K
                2 days ago






                1




                1





                @NicholasK well there is nothing in the code here, that doesn't work with java8., right? But yeah the throwing of exception is definitely missing and also an unwanted creation of Set which doesn't seem to be required for the purpose of OP.

                – nullpointer
                2 days ago







                @NicholasK well there is nothing in the code here, that doesn't work with java8., right? But yeah the throwing of exception is definitely missing and also an unwanted creation of Set which doesn't seem to be required for the purpose of OP.

                – nullpointer
                2 days ago






                1




                1





                Ummm... if (...m.keySet() == null ...) ... what is that? The keySet can never be null, and in any case: the comparison of the size() and the containsAll call boil down to a simple if (Objects.equals(m.keySet(), set)) ...

                – Marco13
                2 days ago





                Ummm... if (...m.keySet() == null ...) ... what is that? The keySet can never be null, and in any case: the comparison of the size() and the containsAll call boil down to a simple if (Objects.equals(m.keySet(), set)) ...

                – Marco13
                2 days ago













                @MarcoPolo Thanks for the feedback, answer updated.

                – Tim Biegeleisen
                2 days ago





                @MarcoPolo Thanks for the feedback, answer updated.

                – Tim Biegeleisen
                2 days ago











                1














                Simply use the following :-



                m.keySet().stream().filter(e -> !ll.contains(e)).findAny()
                .ifPresent(e -> throwException("Key Not found : " + e));


                And define the throwException method below :



                public static void throwException(String msg) {
                throw new RuntimeException(msg);
                }





                share|improve this answer






























                  1














                  Simply use the following :-



                  m.keySet().stream().filter(e -> !ll.contains(e)).findAny()
                  .ifPresent(e -> throwException("Key Not found : " + e));


                  And define the throwException method below :



                  public static void throwException(String msg) {
                  throw new RuntimeException(msg);
                  }





                  share|improve this answer




























                    1












                    1








                    1







                    Simply use the following :-



                    m.keySet().stream().filter(e -> !ll.contains(e)).findAny()
                    .ifPresent(e -> throwException("Key Not found : " + e));


                    And define the throwException method below :



                    public static void throwException(String msg) {
                    throw new RuntimeException(msg);
                    }





                    share|improve this answer















                    Simply use the following :-



                    m.keySet().stream().filter(e -> !ll.contains(e)).findAny()
                    .ifPresent(e -> throwException("Key Not found : " + e));


                    And define the throwException method below :



                    public static void throwException(String msg) {
                    throw new RuntimeException(msg);
                    }






                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited 2 days ago

























                    answered Jan 11 at 7:03









                    Nicholas KNicholas K

                    6,21051031




                    6,21051031























                        0














                        You can simply change your existing code to -



                        if(!m.keySet().containsAll(ll)) {
                        System.out.println("Throws Exception");
                        }


                        This will solve your problem. :)






                        share|improve this answer





















                        • 2





                          This logic has a problem. Should the hashmap's keyset contain extra items, your code would still return true. Also, if null on both sides of the comparison means equal, then you would have to cover that case as well.

                          – Tim Biegeleisen
                          Jan 11 at 7:03











                        • @p.bansal - The same check performed in reverse was what OP is mostly trying to look forward to.

                          – nullpointer
                          2 days ago






                        • 1





                          @nullpointer This matches the OP’s code but not the text in the question, which requests the reverse. So +1 from me on this answer. But then the OP said in comments that the map keys must match the list, a yet again different check. (Sigh.)

                          – Stuart Marks
                          2 days ago











                        • @StuartMarks well, if they must match, I’d use if(m.size() != ll.size() || !m.keySet().containsAll(ll)), as I suppose, m.keySet().containsAll(ll) is potentially faster than ll.containsAll(m.keySet())

                          – Holger
                          2 days ago











                        • @Holger At the time this answer was written, it matched the code in the question. The OP subsequently changed the requirements.

                          – Stuart Marks
                          2 days ago
















                        0














                        You can simply change your existing code to -



                        if(!m.keySet().containsAll(ll)) {
                        System.out.println("Throws Exception");
                        }


                        This will solve your problem. :)






                        share|improve this answer





















                        • 2





                          This logic has a problem. Should the hashmap's keyset contain extra items, your code would still return true. Also, if null on both sides of the comparison means equal, then you would have to cover that case as well.

                          – Tim Biegeleisen
                          Jan 11 at 7:03











                        • @p.bansal - The same check performed in reverse was what OP is mostly trying to look forward to.

                          – nullpointer
                          2 days ago






                        • 1





                          @nullpointer This matches the OP’s code but not the text in the question, which requests the reverse. So +1 from me on this answer. But then the OP said in comments that the map keys must match the list, a yet again different check. (Sigh.)

                          – Stuart Marks
                          2 days ago











                        • @StuartMarks well, if they must match, I’d use if(m.size() != ll.size() || !m.keySet().containsAll(ll)), as I suppose, m.keySet().containsAll(ll) is potentially faster than ll.containsAll(m.keySet())

                          – Holger
                          2 days ago











                        • @Holger At the time this answer was written, it matched the code in the question. The OP subsequently changed the requirements.

                          – Stuart Marks
                          2 days ago














                        0












                        0








                        0







                        You can simply change your existing code to -



                        if(!m.keySet().containsAll(ll)) {
                        System.out.println("Throws Exception");
                        }


                        This will solve your problem. :)






                        share|improve this answer















                        You can simply change your existing code to -



                        if(!m.keySet().containsAll(ll)) {
                        System.out.println("Throws Exception");
                        }


                        This will solve your problem. :)







                        share|improve this answer














                        share|improve this answer



                        share|improve this answer








                        edited 2 days ago









                        nullpointer

                        45.1k1096184




                        45.1k1096184










                        answered Jan 11 at 6:59









                        p.bansalp.bansal

                        954




                        954








                        • 2





                          This logic has a problem. Should the hashmap's keyset contain extra items, your code would still return true. Also, if null on both sides of the comparison means equal, then you would have to cover that case as well.

                          – Tim Biegeleisen
                          Jan 11 at 7:03











                        • @p.bansal - The same check performed in reverse was what OP is mostly trying to look forward to.

                          – nullpointer
                          2 days ago






                        • 1





                          @nullpointer This matches the OP’s code but not the text in the question, which requests the reverse. So +1 from me on this answer. But then the OP said in comments that the map keys must match the list, a yet again different check. (Sigh.)

                          – Stuart Marks
                          2 days ago











                        • @StuartMarks well, if they must match, I’d use if(m.size() != ll.size() || !m.keySet().containsAll(ll)), as I suppose, m.keySet().containsAll(ll) is potentially faster than ll.containsAll(m.keySet())

                          – Holger
                          2 days ago











                        • @Holger At the time this answer was written, it matched the code in the question. The OP subsequently changed the requirements.

                          – Stuart Marks
                          2 days ago














                        • 2





                          This logic has a problem. Should the hashmap's keyset contain extra items, your code would still return true. Also, if null on both sides of the comparison means equal, then you would have to cover that case as well.

                          – Tim Biegeleisen
                          Jan 11 at 7:03











                        • @p.bansal - The same check performed in reverse was what OP is mostly trying to look forward to.

                          – nullpointer
                          2 days ago






                        • 1





                          @nullpointer This matches the OP’s code but not the text in the question, which requests the reverse. So +1 from me on this answer. But then the OP said in comments that the map keys must match the list, a yet again different check. (Sigh.)

                          – Stuart Marks
                          2 days ago











                        • @StuartMarks well, if they must match, I’d use if(m.size() != ll.size() || !m.keySet().containsAll(ll)), as I suppose, m.keySet().containsAll(ll) is potentially faster than ll.containsAll(m.keySet())

                          – Holger
                          2 days ago











                        • @Holger At the time this answer was written, it matched the code in the question. The OP subsequently changed the requirements.

                          – Stuart Marks
                          2 days ago








                        2




                        2





                        This logic has a problem. Should the hashmap's keyset contain extra items, your code would still return true. Also, if null on both sides of the comparison means equal, then you would have to cover that case as well.

                        – Tim Biegeleisen
                        Jan 11 at 7:03





                        This logic has a problem. Should the hashmap's keyset contain extra items, your code would still return true. Also, if null on both sides of the comparison means equal, then you would have to cover that case as well.

                        – Tim Biegeleisen
                        Jan 11 at 7:03













                        @p.bansal - The same check performed in reverse was what OP is mostly trying to look forward to.

                        – nullpointer
                        2 days ago





                        @p.bansal - The same check performed in reverse was what OP is mostly trying to look forward to.

                        – nullpointer
                        2 days ago




                        1




                        1





                        @nullpointer This matches the OP’s code but not the text in the question, which requests the reverse. So +1 from me on this answer. But then the OP said in comments that the map keys must match the list, a yet again different check. (Sigh.)

                        – Stuart Marks
                        2 days ago





                        @nullpointer This matches the OP’s code but not the text in the question, which requests the reverse. So +1 from me on this answer. But then the OP said in comments that the map keys must match the list, a yet again different check. (Sigh.)

                        – Stuart Marks
                        2 days ago













                        @StuartMarks well, if they must match, I’d use if(m.size() != ll.size() || !m.keySet().containsAll(ll)), as I suppose, m.keySet().containsAll(ll) is potentially faster than ll.containsAll(m.keySet())

                        – Holger
                        2 days ago





                        @StuartMarks well, if they must match, I’d use if(m.size() != ll.size() || !m.keySet().containsAll(ll)), as I suppose, m.keySet().containsAll(ll) is potentially faster than ll.containsAll(m.keySet())

                        – Holger
                        2 days ago













                        @Holger At the time this answer was written, it matched the code in the question. The OP subsequently changed the requirements.

                        – Stuart Marks
                        2 days ago





                        @Holger At the time this answer was written, it matched the code in the question. The OP subsequently changed the requirements.

                        – Stuart Marks
                        2 days ago











                        0














                        Here is another solution:



                            if (ll  .parallelStream()
                        .filter(v -> !m.containsKey(v)) // Filter alle values not contained in the map
                        .count() == 0) { // If no values are left then every key was present
                        // do something
                        } else {
                        throw new RuntimeException("hello");
                        }


                        Just wanted to show a different approach






                        share|improve this answer






























                          0














                          Here is another solution:



                              if (ll  .parallelStream()
                          .filter(v -> !m.containsKey(v)) // Filter alle values not contained in the map
                          .count() == 0) { // If no values are left then every key was present
                          // do something
                          } else {
                          throw new RuntimeException("hello");
                          }


                          Just wanted to show a different approach






                          share|improve this answer




























                            0












                            0








                            0







                            Here is another solution:



                                if (ll  .parallelStream()
                            .filter(v -> !m.containsKey(v)) // Filter alle values not contained in the map
                            .count() == 0) { // If no values are left then every key was present
                            // do something
                            } else {
                            throw new RuntimeException("hello");
                            }


                            Just wanted to show a different approach






                            share|improve this answer















                            Here is another solution:



                                if (ll  .parallelStream()
                            .filter(v -> !m.containsKey(v)) // Filter alle values not contained in the map
                            .count() == 0) { // If no values are left then every key was present
                            // do something
                            } else {
                            throw new RuntimeException("hello");
                            }


                            Just wanted to show a different approach







                            share|improve this answer














                            share|improve this answer



                            share|improve this answer








                            edited 2 days ago

























                            answered 2 days ago









                            user489872user489872

                            1,36031335




                            1,36031335






























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