forming a specific list with Java 8 streams












9














I'am currently working in a java project which I have a list of strings and I want them to have a specific format using streams .



For example



Input : [nom, contains, b, and, prenom, contains, y, and, age, >=, 1, and, age, <=, 100]
Ouput: 



[

{key:"nom",

operation:"contains",

value:"b"

},

{

key:"prenom",

operation:"contains",

value:"y"

},

{

key:"age",

operation:">=",

value: 1

},

{

key:"age",

operation:"<=",

value: 1000

}]


I wrote a very basic code without using streams:



List filter = [nom, contains, b, and, prenom, contains, y, and, age, >=, 1, and, age, <=, 100]

    List<SearchCriteria> formedFilter = new ArrayList<>();

    SearchCriteria sc = new SearchCriteria();

    if(filter != null){

        for(int i = 0 ;i< filter.size();i++){

            if(i % 4 ==0){

                sc.setKey((String) filter.get(i));

            }else if(i % 4 == 1){

                sc.setOperation((String) filter.get(i));



            }else if(i % 4 ==2){

                sc.setValue(filter.get(i));

                formedFilter.add(sc);



            }else{

                sc = new SearchCriteria();

            }

        }

    }


SearchCriteria Class



public class SearchCriteria {

    private String key;

    private String operation;

    private Object value;



    public SearchCriteria() {

    }



    public SearchCriteria(String key, String operation, Object value) {

        this.key = key;

        this.operation = operation;

        this.value = value;

    }



   // getters and setters

}





java java-8 java-stream







share|improve this question
























  • Are the elements at index 3, 7 and 11 always and, or could they also be i.e. or, or maybe another boolean operation?
    – Federico Peralta Schaffner
    Nov 22 at 12:38






  • 2




    @FedericoPeraltaSchaffner they could be and / or but for the moment i don't need it
    – Wassim Makni
    Nov 22 at 13:24
















9














I'am currently working in a java project which I have a list of strings and I want them to have a specific format using streams .



For example



Input : [nom, contains, b, and, prenom, contains, y, and, age, >=, 1, and, age, <=, 100]
Ouput: 



[

{key:"nom",

operation:"contains",

value:"b"

},

{

key:"prenom",

operation:"contains",

value:"y"

},

{

key:"age",

operation:">=",

value: 1

},

{

key:"age",

operation:"<=",

value: 1000

}]


I wrote a very basic code without using streams:



List filter = [nom, contains, b, and, prenom, contains, y, and, age, >=, 1, and, age, <=, 100]

    List<SearchCriteria> formedFilter = new ArrayList<>();

    SearchCriteria sc = new SearchCriteria();

    if(filter != null){

        for(int i = 0 ;i< filter.size();i++){

            if(i % 4 ==0){

                sc.setKey((String) filter.get(i));

            }else if(i % 4 == 1){

                sc.setOperation((String) filter.get(i));



            }else if(i % 4 ==2){

                sc.setValue(filter.get(i));

                formedFilter.add(sc);



            }else{

                sc = new SearchCriteria();

            }

        }

    }


SearchCriteria Class



public class SearchCriteria {

    private String key;

    private String operation;

    private Object value;



    public SearchCriteria() {

    }



    public SearchCriteria(String key, String operation, Object value) {

        this.key = key;

        this.operation = operation;

        this.value = value;

    }



   // getters and setters

}





java java-8 java-stream







share|improve this question
























  • Are the elements at index 3, 7 and 11 always and, or could they also be i.e. or, or maybe another boolean operation?
    – Federico Peralta Schaffner
    Nov 22 at 12:38






  • 2




    @FedericoPeraltaSchaffner they could be and / or but for the moment i don't need it
    – Wassim Makni
    Nov 22 at 13:24














9












9








9


1





I'am currently working in a java project which I have a list of strings and I want them to have a specific format using streams .



For example



Input : [nom, contains, b, and, prenom, contains, y, and, age, >=, 1, and, age, <=, 100]
Ouput: 



[

{key:"nom",

operation:"contains",

value:"b"

},

{

key:"prenom",

operation:"contains",

value:"y"

},

{

key:"age",

operation:">=",

value: 1

},

{

key:"age",

operation:"<=",

value: 1000

}]


I wrote a very basic code without using streams:



List filter = [nom, contains, b, and, prenom, contains, y, and, age, >=, 1, and, age, <=, 100]

    List<SearchCriteria> formedFilter = new ArrayList<>();

    SearchCriteria sc = new SearchCriteria();

    if(filter != null){

        for(int i = 0 ;i< filter.size();i++){

            if(i % 4 ==0){

                sc.setKey((String) filter.get(i));

            }else if(i % 4 == 1){

                sc.setOperation((String) filter.get(i));



            }else if(i % 4 ==2){

                sc.setValue(filter.get(i));

                formedFilter.add(sc);



            }else{

                sc = new SearchCriteria();

            }

        }

    }


SearchCriteria Class



public class SearchCriteria {

    private String key;

    private String operation;

    private Object value;



    public SearchCriteria() {

    }



    public SearchCriteria(String key, String operation, Object value) {

        this.key = key;

        this.operation = operation;

        this.value = value;

    }



   // getters and setters

}





java java-8 java-stream







share|improve this question















I'am currently working in a java project which I have a list of strings and I want them to have a specific format using streams .



For example



Input : [nom, contains, b, and, prenom, contains, y, and, age, >=, 1, and, age, <=, 100]
Ouput: 



[

{key:"nom",

operation:"contains",

value:"b"

},

{

key:"prenom",

operation:"contains",

value:"y"

},

{

key:"age",

operation:">=",

value: 1

},

{

key:"age",

operation:"<=",

value: 1000

}]


I wrote a very basic code without using streams:



List filter = [nom, contains, b, and, prenom, contains, y, and, age, >=, 1, and, age, <=, 100]

    List<SearchCriteria> formedFilter = new ArrayList<>();

    SearchCriteria sc = new SearchCriteria();

    if(filter != null){

        for(int i = 0 ;i< filter.size();i++){

            if(i % 4 ==0){

                sc.setKey((String) filter.get(i));

            }else if(i % 4 == 1){

                sc.setOperation((String) filter.get(i));



            }else if(i % 4 ==2){

                sc.setValue(filter.get(i));

                formedFilter.add(sc);



            }else{

                sc = new SearchCriteria();

            }

        }

    }


SearchCriteria Class



public class SearchCriteria {

    private String key;

    private String operation;

    private Object value;



    public SearchCriteria() {

    }



    public SearchCriteria(String key, String operation, Object value) {

        this.key = key;

        this.operation = operation;

        this.value = value;

    }



   // getters and setters

}





java java-8 java-stream




java java-8 java-stream






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 22 at 13:47









nullpointer

42.1k1089175




42.1k1089175










asked Nov 22 at 11:50









Wassim Makni

181113




181113












  • Are the elements at index 3, 7 and 11 always and, or could they also be i.e. or, or maybe another boolean operation?
    – Federico Peralta Schaffner
    Nov 22 at 12:38






  • 2




    @FedericoPeraltaSchaffner they could be and / or but for the moment i don't need it
    – Wassim Makni
    Nov 22 at 13:24


















  • Are the elements at index 3, 7 and 11 always and, or could they also be i.e. or, or maybe another boolean operation?
    – Federico Peralta Schaffner
    Nov 22 at 12:38






  • 2




    @FedericoPeraltaSchaffner they could be and / or but for the moment i don't need it
    – Wassim Makni
    Nov 22 at 13:24
















Are the elements at index 3, 7 and 11 always and, or could they also be i.e. or, or maybe another boolean operation?
– Federico Peralta Schaffner
Nov 22 at 12:38




Are the elements at index 3, 7 and 11 always and, or could they also be i.e. or, or maybe another boolean operation?
– Federico Peralta Schaffner
Nov 22 at 12:38




2




2




@FedericoPeraltaSchaffner they could be and / or but for the moment i don't need it
– Wassim Makni
Nov 22 at 13:24




@FedericoPeraltaSchaffner they could be and / or but for the moment i don't need it
– Wassim Makni
Nov 22 at 13:24












2 Answers
2






active

oldest

votes


















5














For a Java 8 version of Aomine's answer you can use:



List<SearchCriteria> formedFilter = IntStream.iterate(0, i -> i + 4)

            .limit(filter.size() / 4 + 1) // + 1 to consider the last group as well

            .mapToObj(i -> new SearchCriteria(filter.get(i), filter.get(i + 1), filter.get(i + 2)))

            .collect(Collectors.toList());



Alternatively, similar to a suggestion by Holger, you can use the rangeClosed API from IntStream as:



List<SearchCriteria> formedFilter2 = IntStream.rangeClosed(0, filter.size() / 4)

        .map(i -> i * 4)

        .mapToObj(i -> new SearchCriteria(filter.get(i), filter.get(i + 1), filter.get(i + 2)))

        .collect(Collectors.toList());





share|improve this answer



















  • 2




    By the way, since the placement of limit is irrelevant here, I’d chain it immediately to iterate, as that’s semantically closer and makes it easier to understand the iteration logic.
    – Holger
    Nov 22 at 14:01










  • @Holger I just kept the short-circuiting closer to the terminal operation and the order wouldn't really matter here is what my understanding is.
    – nullpointer
    Nov 22 at 14:04






  • 1




    @nullpointer right, as I said, the placement of limit is irrelevant here, so it’s just a matter of style. I see your point in emphasizing the result list’s size, but still think, the iteration logic is more important (consider the appearance of the 4).
    – Holger
    Nov 22 at 14:08






  • 2




    @nullpointer I think that replacing .limit(filter.size() / 4 + 1) with .limit(filter.size() / 4 + 1) do the job correctly
    – Wassim Makni
    Nov 22 at 14:14






  • 2




    @WassimMakni true, also a cleaner way could be as suggested by Holger as well using range.. API
    – nullpointer
    Nov 22 at 14:15



















2














Using JDK 9, you can do:



IntStream.iterate(0, i -> i < source.size(), i -> i + 4)

         .mapToObj(x -> new SearchCriteria(source.get(x), source.get(x + 1), source.get(x + 2)))

         .collect(toList());





share|improve this answer

















  • 1




    @WassimMakni it’s not working for Java 8, but you can replace iterate(0, i -> i < source.size(), i -> i + 4) with range(0, source.size()/4).map(i -> i*4), which might even turn out to be more efficient.
    – Holger
    Nov 22 at 13:50











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









5














For a Java 8 version of Aomine's answer you can use:



List<SearchCriteria> formedFilter = IntStream.iterate(0, i -> i + 4)

            .limit(filter.size() / 4 + 1) // + 1 to consider the last group as well

            .mapToObj(i -> new SearchCriteria(filter.get(i), filter.get(i + 1), filter.get(i + 2)))

            .collect(Collectors.toList());



Alternatively, similar to a suggestion by Holger, you can use the rangeClosed API from IntStream as:



List<SearchCriteria> formedFilter2 = IntStream.rangeClosed(0, filter.size() / 4)

        .map(i -> i * 4)

        .mapToObj(i -> new SearchCriteria(filter.get(i), filter.get(i + 1), filter.get(i + 2)))

        .collect(Collectors.toList());





share|improve this answer



















  • 2




    By the way, since the placement of limit is irrelevant here, I’d chain it immediately to iterate, as that’s semantically closer and makes it easier to understand the iteration logic.
    – Holger
    Nov 22 at 14:01










  • @Holger I just kept the short-circuiting closer to the terminal operation and the order wouldn't really matter here is what my understanding is.
    – nullpointer
    Nov 22 at 14:04






  • 1




    @nullpointer right, as I said, the placement of limit is irrelevant here, so it’s just a matter of style. I see your point in emphasizing the result list’s size, but still think, the iteration logic is more important (consider the appearance of the 4).
    – Holger
    Nov 22 at 14:08






  • 2




    @nullpointer I think that replacing .limit(filter.size() / 4 + 1) with .limit(filter.size() / 4 + 1) do the job correctly
    – Wassim Makni
    Nov 22 at 14:14






  • 2




    @WassimMakni true, also a cleaner way could be as suggested by Holger as well using range.. API
    – nullpointer
    Nov 22 at 14:15
















5














For a Java 8 version of Aomine's answer you can use:



List<SearchCriteria> formedFilter = IntStream.iterate(0, i -> i + 4)

            .limit(filter.size() / 4 + 1) // + 1 to consider the last group as well

            .mapToObj(i -> new SearchCriteria(filter.get(i), filter.get(i + 1), filter.get(i + 2)))

            .collect(Collectors.toList());



Alternatively, similar to a suggestion by Holger, you can use the rangeClosed API from IntStream as:



List<SearchCriteria> formedFilter2 = IntStream.rangeClosed(0, filter.size() / 4)

        .map(i -> i * 4)

        .mapToObj(i -> new SearchCriteria(filter.get(i), filter.get(i + 1), filter.get(i + 2)))

        .collect(Collectors.toList());





share|improve this answer



















  • 2




    By the way, since the placement of limit is irrelevant here, I’d chain it immediately to iterate, as that’s semantically closer and makes it easier to understand the iteration logic.
    – Holger
    Nov 22 at 14:01










  • @Holger I just kept the short-circuiting closer to the terminal operation and the order wouldn't really matter here is what my understanding is.
    – nullpointer
    Nov 22 at 14:04






  • 1




    @nullpointer right, as I said, the placement of limit is irrelevant here, so it’s just a matter of style. I see your point in emphasizing the result list’s size, but still think, the iteration logic is more important (consider the appearance of the 4).
    – Holger
    Nov 22 at 14:08






  • 2




    @nullpointer I think that replacing .limit(filter.size() / 4 + 1) with .limit(filter.size() / 4 + 1) do the job correctly
    – Wassim Makni
    Nov 22 at 14:14






  • 2




    @WassimMakni true, also a cleaner way could be as suggested by Holger as well using range.. API
    – nullpointer
    Nov 22 at 14:15














5












5








5






For a Java 8 version of Aomine's answer you can use:



List<SearchCriteria> formedFilter = IntStream.iterate(0, i -> i + 4)

            .limit(filter.size() / 4 + 1) // + 1 to consider the last group as well

            .mapToObj(i -> new SearchCriteria(filter.get(i), filter.get(i + 1), filter.get(i + 2)))

            .collect(Collectors.toList());



Alternatively, similar to a suggestion by Holger, you can use the rangeClosed API from IntStream as:



List<SearchCriteria> formedFilter2 = IntStream.rangeClosed(0, filter.size() / 4)

        .map(i -> i * 4)

        .mapToObj(i -> new SearchCriteria(filter.get(i), filter.get(i + 1), filter.get(i + 2)))

        .collect(Collectors.toList());





share|improve this answer














For a Java 8 version of Aomine's answer you can use:



List<SearchCriteria> formedFilter = IntStream.iterate(0, i -> i + 4)

            .limit(filter.size() / 4 + 1) // + 1 to consider the last group as well

            .mapToObj(i -> new SearchCriteria(filter.get(i), filter.get(i + 1), filter.get(i + 2)))

            .collect(Collectors.toList());



Alternatively, similar to a suggestion by Holger, you can use the rangeClosed API from IntStream as:



List<SearchCriteria> formedFilter2 = IntStream.rangeClosed(0, filter.size() / 4)

        .map(i -> i * 4)

        .mapToObj(i -> new SearchCriteria(filter.get(i), filter.get(i + 1), filter.get(i + 2)))

        .collect(Collectors.toList());






share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 22 at 14:14

























answered Nov 22 at 13:44









nullpointer

42.1k1089175




42.1k1089175








  • 2




    By the way, since the placement of limit is irrelevant here, I’d chain it immediately to iterate, as that’s semantically closer and makes it easier to understand the iteration logic.
    – Holger
    Nov 22 at 14:01










  • @Holger I just kept the short-circuiting closer to the terminal operation and the order wouldn't really matter here is what my understanding is.
    – nullpointer
    Nov 22 at 14:04






  • 1




    @nullpointer right, as I said, the placement of limit is irrelevant here, so it’s just a matter of style. I see your point in emphasizing the result list’s size, but still think, the iteration logic is more important (consider the appearance of the 4).
    – Holger
    Nov 22 at 14:08






  • 2




    @nullpointer I think that replacing .limit(filter.size() / 4 + 1) with .limit(filter.size() / 4 + 1) do the job correctly
    – Wassim Makni
    Nov 22 at 14:14






  • 2




    @WassimMakni true, also a cleaner way could be as suggested by Holger as well using range.. API
    – nullpointer
    Nov 22 at 14:15














  • 2




    By the way, since the placement of limit is irrelevant here, I’d chain it immediately to iterate, as that’s semantically closer and makes it easier to understand the iteration logic.
    – Holger
    Nov 22 at 14:01










  • @Holger I just kept the short-circuiting closer to the terminal operation and the order wouldn't really matter here is what my understanding is.
    – nullpointer
    Nov 22 at 14:04






  • 1




    @nullpointer right, as I said, the placement of limit is irrelevant here, so it’s just a matter of style. I see your point in emphasizing the result list’s size, but still think, the iteration logic is more important (consider the appearance of the 4).
    – Holger
    Nov 22 at 14:08






  • 2




    @nullpointer I think that replacing .limit(filter.size() / 4 + 1) with .limit(filter.size() / 4 + 1) do the job correctly
    – Wassim Makni
    Nov 22 at 14:14






  • 2




    @WassimMakni true, also a cleaner way could be as suggested by Holger as well using range.. API
    – nullpointer
    Nov 22 at 14:15








2




2




By the way, since the placement of limit is irrelevant here, I’d chain it immediately to iterate, as that’s semantically closer and makes it easier to understand the iteration logic.
– Holger
Nov 22 at 14:01




By the way, since the placement of limit is irrelevant here, I’d chain it immediately to iterate, as that’s semantically closer and makes it easier to understand the iteration logic.
– Holger
Nov 22 at 14:01












@Holger I just kept the short-circuiting closer to the terminal operation and the order wouldn't really matter here is what my understanding is.
– nullpointer
Nov 22 at 14:04




@Holger I just kept the short-circuiting closer to the terminal operation and the order wouldn't really matter here is what my understanding is.
– nullpointer
Nov 22 at 14:04




1




1




@nullpointer right, as I said, the placement of limit is irrelevant here, so it’s just a matter of style. I see your point in emphasizing the result list’s size, but still think, the iteration logic is more important (consider the appearance of the 4).
– Holger
Nov 22 at 14:08




@nullpointer right, as I said, the placement of limit is irrelevant here, so it’s just a matter of style. I see your point in emphasizing the result list’s size, but still think, the iteration logic is more important (consider the appearance of the 4).
– Holger
Nov 22 at 14:08




2




2




@nullpointer I think that replacing .limit(filter.size() / 4 + 1) with .limit(filter.size() / 4 + 1) do the job correctly
– Wassim Makni
Nov 22 at 14:14




@nullpointer I think that replacing .limit(filter.size() / 4 + 1) with .limit(filter.size() / 4 + 1) do the job correctly
– Wassim Makni
Nov 22 at 14:14




2




2




@WassimMakni true, also a cleaner way could be as suggested by Holger as well using range.. API
– nullpointer
Nov 22 at 14:15




@WassimMakni true, also a cleaner way could be as suggested by Holger as well using range.. API
– nullpointer
Nov 22 at 14:15













2














Using JDK 9, you can do:



IntStream.iterate(0, i -> i < source.size(), i -> i + 4)

         .mapToObj(x -> new SearchCriteria(source.get(x), source.get(x + 1), source.get(x + 2)))

         .collect(toList());





share|improve this answer

















  • 1




    @WassimMakni it’s not working for Java 8, but you can replace iterate(0, i -> i < source.size(), i -> i + 4) with range(0, source.size()/4).map(i -> i*4), which might even turn out to be more efficient.
    – Holger
    Nov 22 at 13:50
















2














Using JDK 9, you can do:



IntStream.iterate(0, i -> i < source.size(), i -> i + 4)

         .mapToObj(x -> new SearchCriteria(source.get(x), source.get(x + 1), source.get(x + 2)))

         .collect(toList());





share|improve this answer

















  • 1




    @WassimMakni it’s not working for Java 8, but you can replace iterate(0, i -> i < source.size(), i -> i + 4) with range(0, source.size()/4).map(i -> i*4), which might even turn out to be more efficient.
    – Holger
    Nov 22 at 13:50














2












2








2






Using JDK 9, you can do:



IntStream.iterate(0, i -> i < source.size(), i -> i + 4)

         .mapToObj(x -> new SearchCriteria(source.get(x), source.get(x + 1), source.get(x + 2)))

         .collect(toList());





share|improve this answer












Using JDK 9, you can do:



IntStream.iterate(0, i -> i < source.size(), i -> i + 4)

         .mapToObj(x -> new SearchCriteria(source.get(x), source.get(x + 1), source.get(x + 2)))

         .collect(toList());






share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 22 at 12:35









Aomine

39.5k73669




39.5k73669








  • 1




    @WassimMakni it’s not working for Java 8, but you can replace iterate(0, i -> i < source.size(), i -> i + 4) with range(0, source.size()/4).map(i -> i*4), which might even turn out to be more efficient.
    – Holger
    Nov 22 at 13:50














  • 1




    @WassimMakni it’s not working for Java 8, but you can replace iterate(0, i -> i < source.size(), i -> i + 4) with range(0, source.size()/4).map(i -> i*4), which might even turn out to be more efficient.
    – Holger
    Nov 22 at 13:50








1




1




@WassimMakni it’s not working for Java 8, but you can replace iterate(0, i -> i < source.size(), i -> i + 4) with range(0, source.size()/4).map(i -> i*4), which might even turn out to be more efficient.
– Holger
Nov 22 at 13:50




@WassimMakni it’s not working for Java 8, but you can replace iterate(0, i -> i < source.size(), i -> i + 4) with range(0, source.size()/4).map(i -> i*4), which might even turn out to be more efficient.
– Holger
Nov 22 at 13:50


















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