Java 8 Strem filter map in map — Map<String,Map>












5














How to filter a Map<String,Map<String,Employee>> using Java 8 Filter?



I have to filter only when any of employee in the list having a field value Gender = "M".



Input:




Map<String,Map<String,Employee>>




Output:




Map<String,Map<String,Employee>>




Filter criteria:




Employee.genter = "M"




Also i have to return empty map if the filtered result is empty.



I tried the below, but it is not working as expected. It is returning the only if all the Employees are with gender "M".



tempCollection.entrySet().stream()
.filter(i -> i.getValue().entrySet().stream().allMatch(e-> "M".equals(e.getValue().getGender())))
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));









share|improve this question
























  • Return all the employees whose gender is M.
    – user1578872
    2 days ago
















5














How to filter a Map<String,Map<String,Employee>> using Java 8 Filter?



I have to filter only when any of employee in the list having a field value Gender = "M".



Input:




Map<String,Map<String,Employee>>




Output:




Map<String,Map<String,Employee>>




Filter criteria:




Employee.genter = "M"




Also i have to return empty map if the filtered result is empty.



I tried the below, but it is not working as expected. It is returning the only if all the Employees are with gender "M".



tempCollection.entrySet().stream()
.filter(i -> i.getValue().entrySet().stream().allMatch(e-> "M".equals(e.getValue().getGender())))
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));









share|improve this question
























  • Return all the employees whose gender is M.
    – user1578872
    2 days ago














5












5








5


2





How to filter a Map<String,Map<String,Employee>> using Java 8 Filter?



I have to filter only when any of employee in the list having a field value Gender = "M".



Input:




Map<String,Map<String,Employee>>




Output:




Map<String,Map<String,Employee>>




Filter criteria:




Employee.genter = "M"




Also i have to return empty map if the filtered result is empty.



I tried the below, but it is not working as expected. It is returning the only if all the Employees are with gender "M".



tempCollection.entrySet().stream()
.filter(i -> i.getValue().entrySet().stream().allMatch(e-> "M".equals(e.getValue().getGender())))
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));









share|improve this question















How to filter a Map<String,Map<String,Employee>> using Java 8 Filter?



I have to filter only when any of employee in the list having a field value Gender = "M".



Input:




Map<String,Map<String,Employee>>




Output:




Map<String,Map<String,Employee>>




Filter criteria:




Employee.genter = "M"




Also i have to return empty map if the filtered result is empty.



I tried the below, but it is not working as expected. It is returning the only if all the Employees are with gender "M".



tempCollection.entrySet().stream()
.filter(i -> i.getValue().entrySet().stream().allMatch(e-> "M".equals(e.getValue().getGender())))
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));






java java-8 hashmap java-stream






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 2 days ago









Aomine

39.3k73669




39.3k73669










asked 2 days ago









user1578872

1,54452564




1,54452564












  • Return all the employees whose gender is M.
    – user1578872
    2 days ago


















  • Return all the employees whose gender is M.
    – user1578872
    2 days ago
















Return all the employees whose gender is M.
– user1578872
2 days ago




Return all the employees whose gender is M.
– user1578872
2 days ago












5 Answers
5






active

oldest

votes


















3














You could simply iterate on the key-value pairs and filter as:



Map<String, Map<String, Employee>> output = new HashMap<>();
tempCollection.forEach((k, v) -> {
if (v.values().stream().anyMatch(i -> "M".equals(i.getGender()))) {
output.put(k, v.entrySet()
.stream()
.filter(i -> "M".equals(i.getValue().getGender()))
.collect(toMap(Map.Entry::getKey, Map.Entry::getValue)));
}
});





share|improve this answer























  • anyMatch returns all the result even there is a single match, I want the key value to be removed if there is no match,
    – user1578872
    2 days ago










  • If none of the employee is with the Gender "M", it still returns Map<String,emptyMap()>
    – user1578872
    2 days ago










  • @user1578872 i have to return empty map if the filtered result is empty... that's what you meant, right?
    – nullpointer
    2 days ago












  • It should be completeEmpty map in this case. Like, return emptyMap(), not Map<String,emptyMap()>
    – user1578872
    2 days ago






  • 2




    I think put would be more ideal rather than putIfAbsent as the key will always be unique.
    – Aomine
    2 days ago





















1














The function allMatch only matches if every element in the stream matches the predicate; you can use anyMatch to match if any element matches the predicate:



tempCollection.entrySet().stream()
.filter(i -> i.getValue().entrySet().stream().anyMatch(e-> "M".equals(e.getValue().getGender())))
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));





share|improve this answer





























    1














    Seems like what you're after is given a Entry<String,Map<String,Employee>> if there's any employee who has a gender of "M" then filter the inner Map<String,Employee> to contain only entries with a gender "M".



    in which case you can filter along with anyMatch for the first criterion. Further, at the collecting phase, you can then apply the filtering on the inner map:



    tempCollection.entrySet().stream()
    .filter(i -> i.getValue().values().stream().anyMatch(e -> "M".equals(e.getGender())))
    .collect(toMap(Map.Entry::getKey,
    v -> v.getValue().entrySet().stream()
    .filter(i -> "M".equals(i.getValue().getGender()))
    .collect(toMap(Map.Entry::getKey, Map.Entry::getValue))));





    share|improve this answer























    • anyMatch returns all the values in the map even there is only one employee with Gender M.
      – user1578872
      2 days ago










    • @user1578872 right, thanks to the accepted answer, I can now understand your requirement. the description wasn't very clear IMO. anyway edited to accommodate.
      – Aomine
      2 days ago





















    1














    You may do it like so,



    Map<String, Map<String, Employee>> maleEmpMap = tempCollection.entrySet().stream()
    .map(e -> new AbstractMap.SimpleEntry<>(e.getKey(),
    e.getValue().entrySet().stream().filter(emp -> "M".equals(emp.getValue().getGender()))
    .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue))))
    .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));


    Take each map entry from the outer map and create a new map entry using the same key and the newly constructed nested map with the same key and which has only male employees as it's value. Finally collect all those matching entries into a new outer map. This will gracefully handle the empty scenario which you mentioned in the problem statement above.






    share|improve this answer































      0














      Other way would be like this:



      map.values()
      .removeIf(entry->entry.values().stream().anyMatch(e -> !"M".equals(e.getGender())));

      map.entrySet()
      .removeIf(entry->entry.getValue().size() == 0);





      share|improve this answer





















        Your Answer






        StackExchange.ifUsing("editor", function () {
        StackExchange.using("externalEditor", function () {
        StackExchange.using("snippets", function () {
        StackExchange.snippets.init();
        });
        });
        }, "code-snippets");

        StackExchange.ready(function() {
        var channelOptions = {
        tags: "".split(" "),
        id: "1"
        };
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function() {
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled) {
        StackExchange.using("snippets", function() {
        createEditor();
        });
        }
        else {
        createEditor();
        }
        });

        function createEditor() {
        StackExchange.prepareEditor({
        heartbeatType: 'answer',
        autoActivateHeartbeat: false,
        convertImagesToLinks: true,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: 10,
        bindNavPrevention: true,
        postfix: "",
        imageUploader: {
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        },
        onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        });


        }
        });














        draft saved

        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53927656%2fjava-8-strem-filter-map-in-map-mapstring-mapstring-employee%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3














        You could simply iterate on the key-value pairs and filter as:



        Map<String, Map<String, Employee>> output = new HashMap<>();
        tempCollection.forEach((k, v) -> {
        if (v.values().stream().anyMatch(i -> "M".equals(i.getGender()))) {
        output.put(k, v.entrySet()
        .stream()
        .filter(i -> "M".equals(i.getValue().getGender()))
        .collect(toMap(Map.Entry::getKey, Map.Entry::getValue)));
        }
        });





        share|improve this answer























        • anyMatch returns all the result even there is a single match, I want the key value to be removed if there is no match,
          – user1578872
          2 days ago










        • If none of the employee is with the Gender "M", it still returns Map<String,emptyMap()>
          – user1578872
          2 days ago










        • @user1578872 i have to return empty map if the filtered result is empty... that's what you meant, right?
          – nullpointer
          2 days ago












        • It should be completeEmpty map in this case. Like, return emptyMap(), not Map<String,emptyMap()>
          – user1578872
          2 days ago






        • 2




          I think put would be more ideal rather than putIfAbsent as the key will always be unique.
          – Aomine
          2 days ago


















        3














        You could simply iterate on the key-value pairs and filter as:



        Map<String, Map<String, Employee>> output = new HashMap<>();
        tempCollection.forEach((k, v) -> {
        if (v.values().stream().anyMatch(i -> "M".equals(i.getGender()))) {
        output.put(k, v.entrySet()
        .stream()
        .filter(i -> "M".equals(i.getValue().getGender()))
        .collect(toMap(Map.Entry::getKey, Map.Entry::getValue)));
        }
        });





        share|improve this answer























        • anyMatch returns all the result even there is a single match, I want the key value to be removed if there is no match,
          – user1578872
          2 days ago










        • If none of the employee is with the Gender "M", it still returns Map<String,emptyMap()>
          – user1578872
          2 days ago










        • @user1578872 i have to return empty map if the filtered result is empty... that's what you meant, right?
          – nullpointer
          2 days ago












        • It should be completeEmpty map in this case. Like, return emptyMap(), not Map<String,emptyMap()>
          – user1578872
          2 days ago






        • 2




          I think put would be more ideal rather than putIfAbsent as the key will always be unique.
          – Aomine
          2 days ago
















        3












        3








        3






        You could simply iterate on the key-value pairs and filter as:



        Map<String, Map<String, Employee>> output = new HashMap<>();
        tempCollection.forEach((k, v) -> {
        if (v.values().stream().anyMatch(i -> "M".equals(i.getGender()))) {
        output.put(k, v.entrySet()
        .stream()
        .filter(i -> "M".equals(i.getValue().getGender()))
        .collect(toMap(Map.Entry::getKey, Map.Entry::getValue)));
        }
        });





        share|improve this answer














        You could simply iterate on the key-value pairs and filter as:



        Map<String, Map<String, Employee>> output = new HashMap<>();
        tempCollection.forEach((k, v) -> {
        if (v.values().stream().anyMatch(i -> "M".equals(i.getGender()))) {
        output.put(k, v.entrySet()
        .stream()
        .filter(i -> "M".equals(i.getValue().getGender()))
        .collect(toMap(Map.Entry::getKey, Map.Entry::getValue)));
        }
        });






        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 2 days ago

























        answered 2 days ago









        nullpointer

        41.4k1087175




        41.4k1087175












        • anyMatch returns all the result even there is a single match, I want the key value to be removed if there is no match,
          – user1578872
          2 days ago










        • If none of the employee is with the Gender "M", it still returns Map<String,emptyMap()>
          – user1578872
          2 days ago










        • @user1578872 i have to return empty map if the filtered result is empty... that's what you meant, right?
          – nullpointer
          2 days ago












        • It should be completeEmpty map in this case. Like, return emptyMap(), not Map<String,emptyMap()>
          – user1578872
          2 days ago






        • 2




          I think put would be more ideal rather than putIfAbsent as the key will always be unique.
          – Aomine
          2 days ago




















        • anyMatch returns all the result even there is a single match, I want the key value to be removed if there is no match,
          – user1578872
          2 days ago










        • If none of the employee is with the Gender "M", it still returns Map<String,emptyMap()>
          – user1578872
          2 days ago










        • @user1578872 i have to return empty map if the filtered result is empty... that's what you meant, right?
          – nullpointer
          2 days ago












        • It should be completeEmpty map in this case. Like, return emptyMap(), not Map<String,emptyMap()>
          – user1578872
          2 days ago






        • 2




          I think put would be more ideal rather than putIfAbsent as the key will always be unique.
          – Aomine
          2 days ago


















        anyMatch returns all the result even there is a single match, I want the key value to be removed if there is no match,
        – user1578872
        2 days ago




        anyMatch returns all the result even there is a single match, I want the key value to be removed if there is no match,
        – user1578872
        2 days ago












        If none of the employee is with the Gender "M", it still returns Map<String,emptyMap()>
        – user1578872
        2 days ago




        If none of the employee is with the Gender "M", it still returns Map<String,emptyMap()>
        – user1578872
        2 days ago












        @user1578872 i have to return empty map if the filtered result is empty... that's what you meant, right?
        – nullpointer
        2 days ago






        @user1578872 i have to return empty map if the filtered result is empty... that's what you meant, right?
        – nullpointer
        2 days ago














        It should be completeEmpty map in this case. Like, return emptyMap(), not Map<String,emptyMap()>
        – user1578872
        2 days ago




        It should be completeEmpty map in this case. Like, return emptyMap(), not Map<String,emptyMap()>
        – user1578872
        2 days ago




        2




        2




        I think put would be more ideal rather than putIfAbsent as the key will always be unique.
        – Aomine
        2 days ago






        I think put would be more ideal rather than putIfAbsent as the key will always be unique.
        – Aomine
        2 days ago















        1














        The function allMatch only matches if every element in the stream matches the predicate; you can use anyMatch to match if any element matches the predicate:



        tempCollection.entrySet().stream()
        .filter(i -> i.getValue().entrySet().stream().anyMatch(e-> "M".equals(e.getValue().getGender())))
        .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));





        share|improve this answer


























          1














          The function allMatch only matches if every element in the stream matches the predicate; you can use anyMatch to match if any element matches the predicate:



          tempCollection.entrySet().stream()
          .filter(i -> i.getValue().entrySet().stream().anyMatch(e-> "M".equals(e.getValue().getGender())))
          .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));





          share|improve this answer
























            1












            1








            1






            The function allMatch only matches if every element in the stream matches the predicate; you can use anyMatch to match if any element matches the predicate:



            tempCollection.entrySet().stream()
            .filter(i -> i.getValue().entrySet().stream().anyMatch(e-> "M".equals(e.getValue().getGender())))
            .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));





            share|improve this answer












            The function allMatch only matches if every element in the stream matches the predicate; you can use anyMatch to match if any element matches the predicate:



            tempCollection.entrySet().stream()
            .filter(i -> i.getValue().entrySet().stream().anyMatch(e-> "M".equals(e.getValue().getGender())))
            .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));






            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered 2 days ago









            Tordek

            6,97622861




            6,97622861























                1














                Seems like what you're after is given a Entry<String,Map<String,Employee>> if there's any employee who has a gender of "M" then filter the inner Map<String,Employee> to contain only entries with a gender "M".



                in which case you can filter along with anyMatch for the first criterion. Further, at the collecting phase, you can then apply the filtering on the inner map:



                tempCollection.entrySet().stream()
                .filter(i -> i.getValue().values().stream().anyMatch(e -> "M".equals(e.getGender())))
                .collect(toMap(Map.Entry::getKey,
                v -> v.getValue().entrySet().stream()
                .filter(i -> "M".equals(i.getValue().getGender()))
                .collect(toMap(Map.Entry::getKey, Map.Entry::getValue))));





                share|improve this answer























                • anyMatch returns all the values in the map even there is only one employee with Gender M.
                  – user1578872
                  2 days ago










                • @user1578872 right, thanks to the accepted answer, I can now understand your requirement. the description wasn't very clear IMO. anyway edited to accommodate.
                  – Aomine
                  2 days ago


















                1














                Seems like what you're after is given a Entry<String,Map<String,Employee>> if there's any employee who has a gender of "M" then filter the inner Map<String,Employee> to contain only entries with a gender "M".



                in which case you can filter along with anyMatch for the first criterion. Further, at the collecting phase, you can then apply the filtering on the inner map:



                tempCollection.entrySet().stream()
                .filter(i -> i.getValue().values().stream().anyMatch(e -> "M".equals(e.getGender())))
                .collect(toMap(Map.Entry::getKey,
                v -> v.getValue().entrySet().stream()
                .filter(i -> "M".equals(i.getValue().getGender()))
                .collect(toMap(Map.Entry::getKey, Map.Entry::getValue))));





                share|improve this answer























                • anyMatch returns all the values in the map even there is only one employee with Gender M.
                  – user1578872
                  2 days ago










                • @user1578872 right, thanks to the accepted answer, I can now understand your requirement. the description wasn't very clear IMO. anyway edited to accommodate.
                  – Aomine
                  2 days ago
















                1












                1








                1






                Seems like what you're after is given a Entry<String,Map<String,Employee>> if there's any employee who has a gender of "M" then filter the inner Map<String,Employee> to contain only entries with a gender "M".



                in which case you can filter along with anyMatch for the first criterion. Further, at the collecting phase, you can then apply the filtering on the inner map:



                tempCollection.entrySet().stream()
                .filter(i -> i.getValue().values().stream().anyMatch(e -> "M".equals(e.getGender())))
                .collect(toMap(Map.Entry::getKey,
                v -> v.getValue().entrySet().stream()
                .filter(i -> "M".equals(i.getValue().getGender()))
                .collect(toMap(Map.Entry::getKey, Map.Entry::getValue))));





                share|improve this answer














                Seems like what you're after is given a Entry<String,Map<String,Employee>> if there's any employee who has a gender of "M" then filter the inner Map<String,Employee> to contain only entries with a gender "M".



                in which case you can filter along with anyMatch for the first criterion. Further, at the collecting phase, you can then apply the filtering on the inner map:



                tempCollection.entrySet().stream()
                .filter(i -> i.getValue().values().stream().anyMatch(e -> "M".equals(e.getGender())))
                .collect(toMap(Map.Entry::getKey,
                v -> v.getValue().entrySet().stream()
                .filter(i -> "M".equals(i.getValue().getGender()))
                .collect(toMap(Map.Entry::getKey, Map.Entry::getValue))));






                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 2 days ago

























                answered 2 days ago









                Aomine

                39.3k73669




                39.3k73669












                • anyMatch returns all the values in the map even there is only one employee with Gender M.
                  – user1578872
                  2 days ago










                • @user1578872 right, thanks to the accepted answer, I can now understand your requirement. the description wasn't very clear IMO. anyway edited to accommodate.
                  – Aomine
                  2 days ago




















                • anyMatch returns all the values in the map even there is only one employee with Gender M.
                  – user1578872
                  2 days ago










                • @user1578872 right, thanks to the accepted answer, I can now understand your requirement. the description wasn't very clear IMO. anyway edited to accommodate.
                  – Aomine
                  2 days ago


















                anyMatch returns all the values in the map even there is only one employee with Gender M.
                – user1578872
                2 days ago




                anyMatch returns all the values in the map even there is only one employee with Gender M.
                – user1578872
                2 days ago












                @user1578872 right, thanks to the accepted answer, I can now understand your requirement. the description wasn't very clear IMO. anyway edited to accommodate.
                – Aomine
                2 days ago






                @user1578872 right, thanks to the accepted answer, I can now understand your requirement. the description wasn't very clear IMO. anyway edited to accommodate.
                – Aomine
                2 days ago













                1














                You may do it like so,



                Map<String, Map<String, Employee>> maleEmpMap = tempCollection.entrySet().stream()
                .map(e -> new AbstractMap.SimpleEntry<>(e.getKey(),
                e.getValue().entrySet().stream().filter(emp -> "M".equals(emp.getValue().getGender()))
                .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue))))
                .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));


                Take each map entry from the outer map and create a new map entry using the same key and the newly constructed nested map with the same key and which has only male employees as it's value. Finally collect all those matching entries into a new outer map. This will gracefully handle the empty scenario which you mentioned in the problem statement above.






                share|improve this answer




























                  1














                  You may do it like so,



                  Map<String, Map<String, Employee>> maleEmpMap = tempCollection.entrySet().stream()
                  .map(e -> new AbstractMap.SimpleEntry<>(e.getKey(),
                  e.getValue().entrySet().stream().filter(emp -> "M".equals(emp.getValue().getGender()))
                  .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue))))
                  .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));


                  Take each map entry from the outer map and create a new map entry using the same key and the newly constructed nested map with the same key and which has only male employees as it's value. Finally collect all those matching entries into a new outer map. This will gracefully handle the empty scenario which you mentioned in the problem statement above.






                  share|improve this answer


























                    1












                    1








                    1






                    You may do it like so,



                    Map<String, Map<String, Employee>> maleEmpMap = tempCollection.entrySet().stream()
                    .map(e -> new AbstractMap.SimpleEntry<>(e.getKey(),
                    e.getValue().entrySet().stream().filter(emp -> "M".equals(emp.getValue().getGender()))
                    .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue))))
                    .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));


                    Take each map entry from the outer map and create a new map entry using the same key and the newly constructed nested map with the same key and which has only male employees as it's value. Finally collect all those matching entries into a new outer map. This will gracefully handle the empty scenario which you mentioned in the problem statement above.






                    share|improve this answer














                    You may do it like so,



                    Map<String, Map<String, Employee>> maleEmpMap = tempCollection.entrySet().stream()
                    .map(e -> new AbstractMap.SimpleEntry<>(e.getKey(),
                    e.getValue().entrySet().stream().filter(emp -> "M".equals(emp.getValue().getGender()))
                    .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue))))
                    .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));


                    Take each map entry from the outer map and create a new map entry using the same key and the newly constructed nested map with the same key and which has only male employees as it's value. Finally collect all those matching entries into a new outer map. This will gracefully handle the empty scenario which you mentioned in the problem statement above.







                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited 2 days ago

























                    answered 2 days ago









                    Ravindra Ranwala

                    8,31731634




                    8,31731634























                        0














                        Other way would be like this:



                        map.values()
                        .removeIf(entry->entry.values().stream().anyMatch(e -> !"M".equals(e.getGender())));

                        map.entrySet()
                        .removeIf(entry->entry.getValue().size() == 0);





                        share|improve this answer


























                          0














                          Other way would be like this:



                          map.values()
                          .removeIf(entry->entry.values().stream().anyMatch(e -> !"M".equals(e.getGender())));

                          map.entrySet()
                          .removeIf(entry->entry.getValue().size() == 0);





                          share|improve this answer
























                            0












                            0








                            0






                            Other way would be like this:



                            map.values()
                            .removeIf(entry->entry.values().stream().anyMatch(e -> !"M".equals(e.getGender())));

                            map.entrySet()
                            .removeIf(entry->entry.getValue().size() == 0);





                            share|improve this answer












                            Other way would be like this:



                            map.values()
                            .removeIf(entry->entry.values().stream().anyMatch(e -> !"M".equals(e.getGender())));

                            map.entrySet()
                            .removeIf(entry->entry.getValue().size() == 0);






                            share|improve this answer












                            share|improve this answer



                            share|improve this answer










                            answered 2 days ago









                            Hadi J

                            9,70731641




                            9,70731641






























                                draft saved

                                draft discarded




















































                                Thanks for contributing an answer to Stack Overflow!


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid



                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.


                                To learn more, see our tips on writing great answers.





                                Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                                Please pay close attention to the following guidance:


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid



                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.


                                To learn more, see our tips on writing great answers.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53927656%2fjava-8-strem-filter-map-in-map-mapstring-mapstring-employee%23new-answer', 'question_page');
                                }
                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown







                                Popular posts from this blog

                                "Incorrect syntax near the keyword 'ON'. (on update cascade, on delete cascade,)

                                Alcedinidae

                                RAC Tourist Trophy