forming a specific list with Java 8 streams
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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
edited Nov 22 '18 at 13:47
Naman
46k12102206
asked Nov 22 '18 at 11:50
Wassim MakniWassim Makni
186114
add a comment |
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
edited Nov 22 '18 at 13:47
Naman
46k12102206
asked Nov 22 '18 at 11:50
Wassim MakniWassim Makni
186114
Are the elements at index 3, 7 and 11 alwaysand, or could they also be i.e.or, or maybe another boolean operation?
– Federico Peralta Schaffner
Nov 22 '18 at 12:38
2
@FedericoPeraltaSchaffner they could beand / orbut for the moment i don't need it
– Wassim Makni
Nov 22 '18 at 13:24
add a comment |
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
edited Nov 22 '18 at 13:47
Naman
46k12102206
asked Nov 22 '18 at 11:50
Wassim MakniWassim Makni
186114
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
edited Nov 22 '18 at 13:47
Naman
46k12102206
asked Nov 22 '18 at 11:50
Wassim MakniWassim Makni
186114
edited Nov 22 '18 at 13:47
Naman
46k12102206
asked Nov 22 '18 at 11:50
Wassim MakniWassim Makni
186114
edited Nov 22 '18 at 13:47
Naman
46k12102206
edited Nov 22 '18 at 13:47
Naman
46k12102206
edited Nov 22 '18 at 13:47
Naman
46k12102206
46k12102206
asked Nov 22 '18 at 11:50
Wassim MakniWassim Makni
186114
asked Nov 22 '18 at 11:50
Wassim MakniWassim Makni
186114
asked Nov 22 '18 at 11:50
Wassim MakniWassim Makni
186114
186114
Are the elements at index 3, 7 and 11 alwaysand, or could they also be i.e.or, or maybe another boolean operation?
– Federico Peralta Schaffner
Nov 22 '18 at 12:38
2
@FedericoPeraltaSchaffner they could beand / orbut for the moment i don't need it
– Wassim Makni
Nov 22 '18 at 13:24
add a comment |
Are the elements at index 3, 7 and 11 alwaysand, or could they also be i.e.or, or maybe another boolean operation?
– Federico Peralta Schaffner
Nov 22 '18 at 12:38
2
@FedericoPeraltaSchaffner they could beand / orbut for the moment i don't need it
– Wassim Makni
Nov 22 '18 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 '18 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 '18 at 12:38
2
2
@FedericoPeraltaSchaffner they could be
and / or but for the moment i don't need it– Wassim Makni
Nov 22 '18 at 13:24
@FedericoPeraltaSchaffner they could be
and / or but for the moment i don't need it– Wassim Makni
Nov 22 '18 at 13:24
add a comment |
2 Answers
2
active
oldest
votes
4
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());
answered Nov 22 '18 at 13:44
NamanNaman
46k12102206
2
By the way, since the placement oflimitis irrelevant here, I’d chain it immediately toiterate, as that’s semantically closer and makes it easier to understand the iteration logic.
– Holger
Nov 22 '18 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.
– Naman
Nov 22 '18 at 14:04
1
@nullpointer right, as I said, the placement oflimitis 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 the4).
– Holger
Nov 22 '18 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 '18 at 14:14
2
@WassimMakni true, also a cleaner way could be as suggested by Holger as well usingrange..API
– Naman
Nov 22 '18 at 14:15
|
show 1 more comment
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());
answered Nov 22 '18 at 12:35
AomineAomine
42.8k74678
2
@WassimMakni it’s not working for Java 8, but you can replaceiterate(0, i -> i < source.size(), i -> i + 4)withrange(0, source.size()/4).map(i -> i*4), which might even turn out to be more efficient.
– Holger
Nov 22 '18 at 13:50
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
4
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());
answered Nov 22 '18 at 13:44
NamanNaman
46k12102206
2
By the way, since the placement oflimitis irrelevant here, I’d chain it immediately toiterate, as that’s semantically closer and makes it easier to understand the iteration logic.
– Holger
Nov 22 '18 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.
– Naman
Nov 22 '18 at 14:04
1
@nullpointer right, as I said, the placement oflimitis 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 the4).
– Holger
Nov 22 '18 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 '18 at 14:14
2
@WassimMakni true, also a cleaner way could be as suggested by Holger as well usingrange..API
– Naman
Nov 22 '18 at 14:15
|
show 1 more comment
4
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());
answered Nov 22 '18 at 13:44
NamanNaman
46k12102206
2
By the way, since the placement oflimitis irrelevant here, I’d chain it immediately toiterate, as that’s semantically closer and makes it easier to understand the iteration logic.
– Holger
Nov 22 '18 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.
– Naman
Nov 22 '18 at 14:04
1
@nullpointer right, as I said, the placement oflimitis 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 the4).
– Holger
Nov 22 '18 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 '18 at 14:14
2
@WassimMakni true, also a cleaner way could be as suggested by Holger as well usingrange..API
– Naman
Nov 22 '18 at 14:15
|
show 1 more comment
4
4
4
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());
answered Nov 22 '18 at 13:44
NamanNaman
46k12102206
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());
answered Nov 22 '18 at 13:44
NamanNaman
46k12102206
edited Nov 22 '18 at 14:14
answered Nov 22 '18 at 13:44
NamanNaman
46k12102206
answered Nov 22 '18 at 13:44
NamanNaman
46k12102206
answered Nov 22 '18 at 13:44
NamanNaman
46k12102206
46k12102206
2
By the way, since the placement oflimitis irrelevant here, I’d chain it immediately toiterate, as that’s semantically closer and makes it easier to understand the iteration logic.
– Holger
Nov 22 '18 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.
– Naman
Nov 22 '18 at 14:04
1
@nullpointer right, as I said, the placement oflimitis 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 the4).
– Holger
Nov 22 '18 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 '18 at 14:14
2
@WassimMakni true, also a cleaner way could be as suggested by Holger as well usingrange..API
– Naman
Nov 22 '18 at 14:15
|
show 1 more comment
2
By the way, since the placement oflimitis irrelevant here, I’d chain it immediately toiterate, as that’s semantically closer and makes it easier to understand the iteration logic.
– Holger
Nov 22 '18 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.
– Naman
Nov 22 '18 at 14:04
1
@nullpointer right, as I said, the placement oflimitis 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 the4).
– Holger
Nov 22 '18 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 '18 at 14:14
2
@WassimMakni true, also a cleaner way could be as suggested by Holger as well usingrange..API
– Naman
Nov 22 '18 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 '18 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 '18 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.
– Naman
Nov 22 '18 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.
– Naman
Nov 22 '18 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 '18 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 '18 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 '18 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 '18 at 14:14
2
2
@WassimMakni true, also a cleaner way could be as suggested by Holger as well using
range.. API– Naman
Nov 22 '18 at 14:15
@WassimMakni true, also a cleaner way could be as suggested by Holger as well using
range.. API– Naman
Nov 22 '18 at 14:15
|
show 1 more comment
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());
answered Nov 22 '18 at 12:35
AomineAomine
42.8k74678
2
@WassimMakni it’s not working for Java 8, but you can replaceiterate(0, i -> i < source.size(), i -> i + 4)withrange(0, source.size()/4).map(i -> i*4), which might even turn out to be more efficient.
– Holger
Nov 22 '18 at 13:50
add a comment |
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());
answered Nov 22 '18 at 12:35
AomineAomine
42.8k74678
2
@WassimMakni it’s not working for Java 8, but you can replaceiterate(0, i -> i < source.size(), i -> i + 4)withrange(0, source.size()/4).map(i -> i*4), which might even turn out to be more efficient.
– Holger
Nov 22 '18 at 13:50
add a comment |
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());
answered Nov 22 '18 at 12:35
AomineAomine
42.8k74678
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());
answered Nov 22 '18 at 12:35
AomineAomine
42.8k74678
answered Nov 22 '18 at 12:35
AomineAomine
42.8k74678
answered Nov 22 '18 at 12:35
AomineAomine
42.8k74678
answered Nov 22 '18 at 12:35
AomineAomine
42.8k74678
42.8k74678
2
@WassimMakni it’s not working for Java 8, but you can replaceiterate(0, i -> i < source.size(), i -> i + 4)withrange(0, source.size()/4).map(i -> i*4), which might even turn out to be more efficient.
– Holger
Nov 22 '18 at 13:50
add a comment |
2
@WassimMakni it’s not working for Java 8, but you can replaceiterate(0, i -> i < source.size(), i -> i + 4)withrange(0, source.size()/4).map(i -> i*4), which might even turn out to be more efficient.
– Holger
Nov 22 '18 at 13:50
2
2
@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 '18 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 '18 at 13:50
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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 '18 at 12:38
2
@FedericoPeraltaSchaffner they could be
and / orbut for the moment i don't need it– Wassim Makni
Nov 22 '18 at 13:24