Java stream group by and sum multiple fields

up vote
7
down vote

favorite

I have a List fooList

class Foo 
 private String category;
 private int amount;
 private int price;

 ... constructor, getters & setters

I would like to group by category and then sum amount aswell as price.

The result will be stored in a map:

Map<Foo, List<Foo>> map = new HashMap<>();

The key is the Foo holding the summarized amount and price, with a list as value for all the objects with the same category.

So far I've tried the following:

Map<String, List<Foo>> map = fooList.stream().collect(groupingBy(Foo::getCategory()));

Now I only need to replace the String key with a Foo object holding the summarized amount and price. Here is where I'm stuck. I can't seem to find any way of doing this.

edited Aug 28 at 12:52

YCF_L

31.5k103273

asked Aug 28 at 12:44

MatMat

299214

have you looked at Collectors.summarizingLong?
â€“Â dehasi
Aug 28 at 12:49

Yes but how do you summarize on multiple fields?
â€“Â MatMat
Aug 28 at 12:52

1

well only recently I've answered something very close to this... stackoverflow.com/a/52041895/1059372 in jdk-12 seems like there will be a BiCollector that will make your life easier
â€“Â Eugene
Aug 28 at 12:58

3

Do I understand you correctly, that the KEY shall hold the sums? And that the List<Foo> shall hold all original instances?
â€“Â jokster
Aug 28 at 12:59

@jokster yes, that is correct!
â€“Â MatMat
Aug 28 at 13:01

add a commentÂ |Â

up vote
7
down vote

favorite

I have a List fooList

class Foo 
 private String category;
 private int amount;
 private int price;

 ... constructor, getters & setters

I would like to group by category and then sum amount aswell as price.

The result will be stored in a map:

Map<Foo, List<Foo>> map = new HashMap<>();

The key is the Foo holding the summarized amount and price, with a list as value for all the objects with the same category.

So far I've tried the following:

Map<String, List<Foo>> map = fooList.stream().collect(groupingBy(Foo::getCategory()));

Now I only need to replace the String key with a Foo object holding the summarized amount and price. Here is where I'm stuck. I can't seem to find any way of doing this.

edited Aug 28 at 12:52

YCF_L

31.5k103273

asked Aug 28 at 12:44

MatMat

299214

have you looked at Collectors.summarizingLong?
â€“Â dehasi
Aug 28 at 12:49

Yes but how do you summarize on multiple fields?
â€“Â MatMat
Aug 28 at 12:52

1

well only recently I've answered something very close to this... stackoverflow.com/a/52041895/1059372 in jdk-12 seems like there will be a BiCollector that will make your life easier
â€“Â Eugene
Aug 28 at 12:58

3

Do I understand you correctly, that the KEY shall hold the sums? And that the List<Foo> shall hold all original instances?
â€“Â jokster
Aug 28 at 12:59

@jokster yes, that is correct!
â€“Â MatMat
Aug 28 at 13:01

add a commentÂ |Â

up vote
7
down vote

favorite

I have a List fooList

class Foo 
 private String category;
 private int amount;
 private int price;

 ... constructor, getters & setters

I would like to group by category and then sum amount aswell as price.

The result will be stored in a map:

Map<Foo, List<Foo>> map = new HashMap<>();

The key is the Foo holding the summarized amount and price, with a list as value for all the objects with the same category.

So far I've tried the following:

Map<String, List<Foo>> map = fooList.stream().collect(groupingBy(Foo::getCategory()));

Now I only need to replace the String key with a Foo object holding the summarized amount and price. Here is where I'm stuck. I can't seem to find any way of doing this.

edited Aug 28 at 12:52

YCF_L

31.5k103273

asked Aug 28 at 12:44

MatMat

299214

I have a List fooList

class Foo 
 private String category;
 private int amount;
 private int price;

 ... constructor, getters & setters

I would like to group by category and then sum amount aswell as price.

The result will be stored in a map:

Map<Foo, List<Foo>> map = new HashMap<>();

The key is the Foo holding the summarized amount and price, with a list as value for all the objects with the same category.

So far I've tried the following:

Map<String, List<Foo>> map = fooList.stream().collect(groupingBy(Foo::getCategory()));

Now I only need to replace the String key with a Foo object holding the summarized amount and price. Here is where I'm stuck. I can't seem to find any way of doing this.

edited Aug 28 at 12:52

YCF_L

31.5k103273

asked Aug 28 at 12:44

MatMat

299214

edited Aug 28 at 12:52

YCF_L

31.5k103273

edited Aug 28 at 12:52

YCF_L

31.5k103273

edited Aug 28 at 12:52

YCF_L

31.5k103273

asked Aug 28 at 12:44

MatMat

299214

asked Aug 28 at 12:44

MatMat

299214

asked Aug 28 at 12:44

MatMat

299214

have you looked at Collectors.summarizingLong?
â€“Â dehasi
Aug 28 at 12:49

Yes but how do you summarize on multiple fields?
â€“Â MatMat
Aug 28 at 12:52

1

well only recently I've answered something very close to this... stackoverflow.com/a/52041895/1059372 in jdk-12 seems like there will be a BiCollector that will make your life easier
â€“Â Eugene
Aug 28 at 12:58

3

Do I understand you correctly, that the KEY shall hold the sums? And that the List<Foo> shall hold all original instances?
â€“Â jokster
Aug 28 at 12:59

@jokster yes, that is correct!
â€“Â MatMat
Aug 28 at 13:01

add a commentÂ |Â

have you looked at Collectors.summarizingLong?
â€“Â dehasi
Aug 28 at 12:49

Yes but how do you summarize on multiple fields?
â€“Â MatMat
Aug 28 at 12:52

1

well only recently I've answered something very close to this... stackoverflow.com/a/52041895/1059372 in jdk-12 seems like there will be a BiCollector that will make your life easier
â€“Â Eugene
Aug 28 at 12:58

3

Do I understand you correctly, that the KEY shall hold the sums? And that the List<Foo> shall hold all original instances?
â€“Â jokster
Aug 28 at 12:59

@jokster yes, that is correct!
â€“Â MatMat
Aug 28 at 13:01

have you looked at Collectors.summarizingLong?
â€“Â dehasi
Aug 28 at 12:49

Yes but how do you summarize on multiple fields?
â€“Â MatMat
Aug 28 at 12:52

well only recently I've answered something very close to this... stackoverflow.com/a/52041895/1059372 in jdk-12 seems like there will be a BiCollector that will make your life easier
â€“Â Eugene
Aug 28 at 12:58

Do I understand you correctly, that the KEY shall hold the sums? And that the List<Foo> shall hold all original instances?
â€“Â jokster
Aug 28 at 12:59

@jokster yes, that is correct!
â€“Â MatMat
Aug 28 at 13:01

add a commentÂ |Â

5 Answers
5

active

oldest

votes

up vote
9
down vote

accepted

A bit ugly, but it should work:

list.stream().collect(Collectors.groupingBy(Foo::getCategory))
 .entrySet().stream()
 .collect(Collectors.toMap(x -> 
 int sumAmount = x.getValue().stream().mapToInt(Foo::getAmount).sum();
 int sumPrice= x.getValue().stream().mapToInt(Foo::getPrice).sum();
 return new Foo(x.getKey(), sumAmount, sumPrice);
 , Map.Entry::getValue));

answered Aug 28 at 13:21

Sweeper

54.7k960117

add a commentÂ |Â

up vote
4
down vote

My variation of Sweepers answer uses a reducing Collector instead of streaming twice to sum the individual fields:

 Map<Foo, List<Foo>> map = fooList.stream()
 .collect(Collectors.groupingBy(Foo::getCategory))
 .entrySet().stream()
 .collect(Collectors.toMap(e -> e.getValue().stream().collect(
 Collectors.reducing((l, r) -> new Foo(l.getCategory(),
 l.getAmount() + r.getAmount(),
 l.getPrice() + r.getPrice())))
 .get(), 
 e -> e.getValue()));

It is not really better though, as it creates a lot of short-lived Foos.

Note however that Foo is required to provide hashCode- and equals-implementations that take only category into account for the resulting map to work correctly. This would probably not be what you want for Foos in general. I would prefer defining a separate FooSummary class to contain the aggregated data.

edited Aug 29 at 6:37

answered Aug 28 at 13:45

Hulk

2,13411431

add a commentÂ |Â

up vote
3
down vote

If you have a special, dedicated constructor and hashCode and equals methods consistently implemented in Foo as follows:

public Foo(Foo that) // not a copy constructor!!!
 this.category = that.category;
 this.amount = 0;
 this.price = 0;


public int hashCode() 
 return Objects.hashCode(category);


public boolean equals(Object another) 
 if (another == this) return true;
 if (!(another instanceof Foo)) return false;
 Foo that = (Foo) another;
 return Objects.equals(this.category, that.category);

The hashCode and equals implementations above allow you to use Foo as a meaningful key in the map (otherwise your map would be broken).

Now, with the help of a new method in Foo that performs the aggregation of the amount and price attributes at the same time, you can do what you want in 2 steps. First the method:

public void aggregate(Foo that) 
 this.amount += that.amount;
 this.price += that.price;

Now the final solution:

Map<Foo, List<Foo>> result = fooList.stream().collect(
 Collectors.collectingAndThen(
 Collectors.groupingBy(Foo::new), // works: special ctor, hashCode & equals
 m -> m.forEach((k, v) -> v.forEach(k::aggregate)); return m; ));

EDIT: a few observations were missing...

On one hand, this solution forces you to use an implementation of hashCode and equals that considers two different Foo instances as equal if they belong to the same category. Maybe this is not desired, or you already have an implementation that takes more or other attributes into account.

On the other hand, using Foo as the key of a map which is used to group instances by one of its attributes is quite an uncommon use case. I think it would be better to just use the category attribute to group by category and have two maps: Map<String, List<Foo>> to keep the groups and Map<String, Foo> to keep the aggregated price and amount, with the key being the category in both cases.

Besides this, this solution mutates the keys of the map after the entries are put into it. This is dangerous, because this could break the map. However, here I'm only mutating attributes of Foo that don't participate in neither hashCode nor equals Foo's implementation. I think that this risk is acceptable in this case, due to the unusuality of the requirement.

edited Aug 29 at 13:35

answered Aug 28 at 15:29

Federico Peralta Schaffner

18.5k32862

add a commentÂ |Â

up vote
0
down vote

I suggest you create a helper class, which will hold amount and price

final class Pair 
 final int amount;
 final int price;

 Pair(int amount, int price) 
 this.amount = amount;
 this.price = price;

And then just collect list to map:

List<Foo> list =//....;

Map<Foo, Pair> categotyPrise = list.stream().collect(Collectors.toMap(foo -> foo,
 foo -> new Pair(foo.getAmount(), foo.getPrice()),
 (o, n) -> new Pair(o.amount + n.amount, o.price + n.price)));

answered Aug 28 at 13:05

dehasi

539615

that is a good start, but not what the OP needs... read the comments under the question
â€“Â Eugene
Aug 28 at 13:06

add a commentÂ |Â

up vote
0
down vote

My take on a solution :)

public static void main(String args) 
 List<Foo> foos = new ArrayList<>();
 foos.add(new Foo("A", 1, 10));
 foos.add(new Foo("A", 2, 10));
 foos.add(new Foo("A", 3, 10));
 foos.add(new Foo("B", 1, 10));
 foos.add(new Foo("C", 1, 10));
 foos.add(new Foo("C", 5, 10));

 List<Foo> summarized = new ArrayList<>();
 Map<Foo, List<Foo>> collect = foos.stream().collect(Collectors.groupingBy(new Function<Foo, Foo>() 
 @Override
 public Foo apply(Foo t) 
 Optional<Foo> fOpt = summarized.stream().filter(e -> e.getCategory().equals(t.getCategory())).findFirst();
 Foo f;
 if (!fOpt.isPresent()) 
 f = new Foo(t.getCategory(), 0, 0);
 summarized.add(f);
 else 
 f = fOpt.get();
 
 f.setAmount(f.getAmount() + t.getAmount());
 f.setPrice(f.getPrice() + t.getPrice());
 return f;
 
 ));
 System.out.println(collect);

edited Aug 28 at 13:19

answered Aug 28 at 13:14

KDM

3,7011921

add a commentÂ |Â

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5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

up vote
9
down vote

accepted

A bit ugly, but it should work:

list.stream().collect(Collectors.groupingBy(Foo::getCategory))
 .entrySet().stream()
 .collect(Collectors.toMap(x -> 
 int sumAmount = x.getValue().stream().mapToInt(Foo::getAmount).sum();
 int sumPrice= x.getValue().stream().mapToInt(Foo::getPrice).sum();
 return new Foo(x.getKey(), sumAmount, sumPrice);
 , Map.Entry::getValue));

answered Aug 28 at 13:21

Sweeper

54.7k960117

add a commentÂ |Â

up vote
9
down vote

accepted

A bit ugly, but it should work:

list.stream().collect(Collectors.groupingBy(Foo::getCategory))
 .entrySet().stream()
 .collect(Collectors.toMap(x -> 
 int sumAmount = x.getValue().stream().mapToInt(Foo::getAmount).sum();
 int sumPrice= x.getValue().stream().mapToInt(Foo::getPrice).sum();
 return new Foo(x.getKey(), sumAmount, sumPrice);
 , Map.Entry::getValue));

answered Aug 28 at 13:21

Sweeper

54.7k960117

add a commentÂ |Â

up vote
9
down vote

accepted

A bit ugly, but it should work:

list.stream().collect(Collectors.groupingBy(Foo::getCategory))
 .entrySet().stream()
 .collect(Collectors.toMap(x -> 
 int sumAmount = x.getValue().stream().mapToInt(Foo::getAmount).sum();
 int sumPrice= x.getValue().stream().mapToInt(Foo::getPrice).sum();
 return new Foo(x.getKey(), sumAmount, sumPrice);
 , Map.Entry::getValue));

answered Aug 28 at 13:21

Sweeper

54.7k960117

A bit ugly, but it should work:

list.stream().collect(Collectors.groupingBy(Foo::getCategory))
 .entrySet().stream()
 .collect(Collectors.toMap(x -> 
 int sumAmount = x.getValue().stream().mapToInt(Foo::getAmount).sum();
 int sumPrice= x.getValue().stream().mapToInt(Foo::getPrice).sum();
 return new Foo(x.getKey(), sumAmount, sumPrice);
 , Map.Entry::getValue));

answered Aug 28 at 13:21

Sweeper

54.7k960117

answered Aug 28 at 13:21

Sweeper

54.7k960117

answered Aug 28 at 13:21

Sweeper

54.7k960117

answered Aug 28 at 13:21

Sweeper

54.7k960117

add a commentÂ |Â

up vote
4
down vote

My variation of Sweepers answer uses a reducing Collector instead of streaming twice to sum the individual fields:

 Map<Foo, List<Foo>> map = fooList.stream()
 .collect(Collectors.groupingBy(Foo::getCategory))
 .entrySet().stream()
 .collect(Collectors.toMap(e -> e.getValue().stream().collect(
 Collectors.reducing((l, r) -> new Foo(l.getCategory(),
 l.getAmount() + r.getAmount(),
 l.getPrice() + r.getPrice())))
 .get(), 
 e -> e.getValue()));

It is not really better though, as it creates a lot of short-lived Foos.

edited Aug 29 at 6:37

answered Aug 28 at 13:45

Hulk

2,13411431

add a commentÂ |Â

up vote
4
down vote

My variation of Sweepers answer uses a reducing Collector instead of streaming twice to sum the individual fields:

 Map<Foo, List<Foo>> map = fooList.stream()
 .collect(Collectors.groupingBy(Foo::getCategory))
 .entrySet().stream()
 .collect(Collectors.toMap(e -> e.getValue().stream().collect(
 Collectors.reducing((l, r) -> new Foo(l.getCategory(),
 l.getAmount() + r.getAmount(),
 l.getPrice() + r.getPrice())))
 .get(), 
 e -> e.getValue()));

It is not really better though, as it creates a lot of short-lived Foos.

edited Aug 29 at 6:37

answered Aug 28 at 13:45

Hulk

2,13411431

add a commentÂ |Â

up vote
4
down vote

My variation of Sweepers answer uses a reducing Collector instead of streaming twice to sum the individual fields:

 Map<Foo, List<Foo>> map = fooList.stream()
 .collect(Collectors.groupingBy(Foo::getCategory))
 .entrySet().stream()
 .collect(Collectors.toMap(e -> e.getValue().stream().collect(
 Collectors.reducing((l, r) -> new Foo(l.getCategory(),
 l.getAmount() + r.getAmount(),
 l.getPrice() + r.getPrice())))
 .get(), 
 e -> e.getValue()));

It is not really better though, as it creates a lot of short-lived Foos.

edited Aug 29 at 6:37

answered Aug 28 at 13:45

Hulk

2,13411431

My variation of Sweepers answer uses a reducing Collector instead of streaming twice to sum the individual fields:

 Map<Foo, List<Foo>> map = fooList.stream()
 .collect(Collectors.groupingBy(Foo::getCategory))
 .entrySet().stream()
 .collect(Collectors.toMap(e -> e.getValue().stream().collect(
 Collectors.reducing((l, r) -> new Foo(l.getCategory(),
 l.getAmount() + r.getAmount(),
 l.getPrice() + r.getPrice())))
 .get(), 
 e -> e.getValue()));

It is not really better though, as it creates a lot of short-lived Foos.

edited Aug 29 at 6:37

answered Aug 28 at 13:45

Hulk

2,13411431

edited Aug 29 at 6:37

answered Aug 28 at 13:45

Hulk

2,13411431

answered Aug 28 at 13:45

Hulk

2,13411431

answered Aug 28 at 13:45

Hulk

2,13411431

add a commentÂ |Â

up vote
3
down vote

If you have a special, dedicated constructor and hashCode and equals methods consistently implemented in Foo as follows:

public Foo(Foo that) // not a copy constructor!!!
 this.category = that.category;
 this.amount = 0;
 this.price = 0;


public int hashCode() 
 return Objects.hashCode(category);


public boolean equals(Object another) 
 if (another == this) return true;
 if (!(another instanceof Foo)) return false;
 Foo that = (Foo) another;
 return Objects.equals(this.category, that.category);

The hashCode and equals implementations above allow you to use Foo as a meaningful key in the map (otherwise your map would be broken).

Now, with the help of a new method in Foo that performs the aggregation of the amount and price attributes at the same time, you can do what you want in 2 steps. First the method:

public void aggregate(Foo that) 
 this.amount += that.amount;
 this.price += that.price;

Now the final solution:

Map<Foo, List<Foo>> result = fooList.stream().collect(
 Collectors.collectingAndThen(
 Collectors.groupingBy(Foo::new), // works: special ctor, hashCode & equals
 m -> m.forEach((k, v) -> v.forEach(k::aggregate)); return m; ));

EDIT: a few observations were missing...

edited Aug 29 at 13:35

answered Aug 28 at 15:29

Federico Peralta Schaffner

18.5k32862

add a commentÂ |Â

up vote
3
down vote

If you have a special, dedicated constructor and hashCode and equals methods consistently implemented in Foo as follows:

public Foo(Foo that) // not a copy constructor!!!
 this.category = that.category;
 this.amount = 0;
 this.price = 0;


public int hashCode() 
 return Objects.hashCode(category);


public boolean equals(Object another) 
 if (another == this) return true;
 if (!(another instanceof Foo)) return false;
 Foo that = (Foo) another;
 return Objects.equals(this.category, that.category);

The hashCode and equals implementations above allow you to use Foo as a meaningful key in the map (otherwise your map would be broken).

Now, with the help of a new method in Foo that performs the aggregation of the amount and price attributes at the same time, you can do what you want in 2 steps. First the method:

public void aggregate(Foo that) 
 this.amount += that.amount;
 this.price += that.price;

Now the final solution:

Map<Foo, List<Foo>> result = fooList.stream().collect(
 Collectors.collectingAndThen(
 Collectors.groupingBy(Foo::new), // works: special ctor, hashCode & equals
 m -> m.forEach((k, v) -> v.forEach(k::aggregate)); return m; ));

EDIT: a few observations were missing...

edited Aug 29 at 13:35

answered Aug 28 at 15:29

Federico Peralta Schaffner

18.5k32862

add a commentÂ |Â

up vote
3
down vote

If you have a special, dedicated constructor and hashCode and equals methods consistently implemented in Foo as follows:

public Foo(Foo that) // not a copy constructor!!!
 this.category = that.category;
 this.amount = 0;
 this.price = 0;


public int hashCode() 
 return Objects.hashCode(category);


public boolean equals(Object another) 
 if (another == this) return true;
 if (!(another instanceof Foo)) return false;
 Foo that = (Foo) another;
 return Objects.equals(this.category, that.category);

The hashCode and equals implementations above allow you to use Foo as a meaningful key in the map (otherwise your map would be broken).

Now, with the help of a new method in Foo that performs the aggregation of the amount and price attributes at the same time, you can do what you want in 2 steps. First the method:

public void aggregate(Foo that) 
 this.amount += that.amount;
 this.price += that.price;

Now the final solution:

Map<Foo, List<Foo>> result = fooList.stream().collect(
 Collectors.collectingAndThen(
 Collectors.groupingBy(Foo::new), // works: special ctor, hashCode & equals
 m -> m.forEach((k, v) -> v.forEach(k::aggregate)); return m; ));

EDIT: a few observations were missing...

edited Aug 29 at 13:35

answered Aug 28 at 15:29

Federico Peralta Schaffner

18.5k32862

If you have a special, dedicated constructor and hashCode and equals methods consistently implemented in Foo as follows:

public Foo(Foo that) // not a copy constructor!!!
 this.category = that.category;
 this.amount = 0;
 this.price = 0;


public int hashCode() 
 return Objects.hashCode(category);


public boolean equals(Object another) 
 if (another == this) return true;
 if (!(another instanceof Foo)) return false;
 Foo that = (Foo) another;
 return Objects.equals(this.category, that.category);

The hashCode and equals implementations above allow you to use Foo as a meaningful key in the map (otherwise your map would be broken).

Now, with the help of a new method in Foo that performs the aggregation of the amount and price attributes at the same time, you can do what you want in 2 steps. First the method:

public void aggregate(Foo that) 
 this.amount += that.amount;
 this.price += that.price;

Now the final solution:

Map<Foo, List<Foo>> result = fooList.stream().collect(
 Collectors.collectingAndThen(
 Collectors.groupingBy(Foo::new), // works: special ctor, hashCode & equals
 m -> m.forEach((k, v) -> v.forEach(k::aggregate)); return m; ));

EDIT: a few observations were missing...

edited Aug 29 at 13:35

answered Aug 28 at 15:29

Federico Peralta Schaffner

18.5k32862

edited Aug 29 at 13:35

answered Aug 28 at 15:29

Federico Peralta Schaffner

18.5k32862

answered Aug 28 at 15:29

Federico Peralta Schaffner

18.5k32862

answered Aug 28 at 15:29

Federico Peralta Schaffner

18.5k32862

add a commentÂ |Â

up vote
0
down vote

I suggest you create a helper class, which will hold amount and price

final class Pair 
 final int amount;
 final int price;

 Pair(int amount, int price) 
 this.amount = amount;
 this.price = price;

And then just collect list to map:

List<Foo> list =//....;

Map<Foo, Pair> categotyPrise = list.stream().collect(Collectors.toMap(foo -> foo,
 foo -> new Pair(foo.getAmount(), foo.getPrice()),
 (o, n) -> new Pair(o.amount + n.amount, o.price + n.price)));

answered Aug 28 at 13:05

dehasi

539615

that is a good start, but not what the OP needs... read the comments under the question
â€“Â Eugene
Aug 28 at 13:06

add a commentÂ |Â

up vote
0
down vote

I suggest you create a helper class, which will hold amount and price

final class Pair 
 final int amount;
 final int price;

 Pair(int amount, int price) 
 this.amount = amount;
 this.price = price;

And then just collect list to map:

List<Foo> list =//....;

Map<Foo, Pair> categotyPrise = list.stream().collect(Collectors.toMap(foo -> foo,
 foo -> new Pair(foo.getAmount(), foo.getPrice()),
 (o, n) -> new Pair(o.amount + n.amount, o.price + n.price)));

answered Aug 28 at 13:05

dehasi

539615

that is a good start, but not what the OP needs... read the comments under the question
â€“Â Eugene
Aug 28 at 13:06

add a commentÂ |Â

up vote
0
down vote

I suggest you create a helper class, which will hold amount and price

final class Pair 
 final int amount;
 final int price;

 Pair(int amount, int price) 
 this.amount = amount;
 this.price = price;

And then just collect list to map:

List<Foo> list =//....;

Map<Foo, Pair> categotyPrise = list.stream().collect(Collectors.toMap(foo -> foo,
 foo -> new Pair(foo.getAmount(), foo.getPrice()),
 (o, n) -> new Pair(o.amount + n.amount, o.price + n.price)));

answered Aug 28 at 13:05

dehasi

539615

I suggest you create a helper class, which will hold amount and price

final class Pair 
 final int amount;
 final int price;

 Pair(int amount, int price) 
 this.amount = amount;
 this.price = price;

And then just collect list to map:

List<Foo> list =//....;

Map<Foo, Pair> categotyPrise = list.stream().collect(Collectors.toMap(foo -> foo,
 foo -> new Pair(foo.getAmount(), foo.getPrice()),
 (o, n) -> new Pair(o.amount + n.amount, o.price + n.price)));

answered Aug 28 at 13:05

dehasi

539615

answered Aug 28 at 13:05

dehasi

539615

answered Aug 28 at 13:05

dehasi

539615

answered Aug 28 at 13:05

dehasi

539615

that is a good start, but not what the OP needs... read the comments under the question
â€“Â Eugene
Aug 28 at 13:06

add a commentÂ |Â

that is a good start, but not what the OP needs... read the comments under the question
â€“Â Eugene
Aug 28 at 13:06

that is a good start, but not what the OP needs... read the comments under the question
â€“Â Eugene
Aug 28 at 13:06

add a commentÂ |Â

up vote
0
down vote

My take on a solution :)

public static void main(String args) 
 List<Foo> foos = new ArrayList<>();
 foos.add(new Foo("A", 1, 10));
 foos.add(new Foo("A", 2, 10));
 foos.add(new Foo("A", 3, 10));
 foos.add(new Foo("B", 1, 10));
 foos.add(new Foo("C", 1, 10));
 foos.add(new Foo("C", 5, 10));

 List<Foo> summarized = new ArrayList<>();
 Map<Foo, List<Foo>> collect = foos.stream().collect(Collectors.groupingBy(new Function<Foo, Foo>() 
 @Override
 public Foo apply(Foo t) 
 Optional<Foo> fOpt = summarized.stream().filter(e -> e.getCategory().equals(t.getCategory())).findFirst();
 Foo f;
 if (!fOpt.isPresent()) 
 f = new Foo(t.getCategory(), 0, 0);
 summarized.add(f);
 else 
 f = fOpt.get();
 
 f.setAmount(f.getAmount() + t.getAmount());
 f.setPrice(f.getPrice() + t.getPrice());
 return f;
 
 ));
 System.out.println(collect);

edited Aug 28 at 13:19

answered Aug 28 at 13:14

KDM

3,7011921

add a commentÂ |Â

up vote
0
down vote

My take on a solution :)

public static void main(String args) 
 List<Foo> foos = new ArrayList<>();
 foos.add(new Foo("A", 1, 10));
 foos.add(new Foo("A", 2, 10));
 foos.add(new Foo("A", 3, 10));
 foos.add(new Foo("B", 1, 10));
 foos.add(new Foo("C", 1, 10));
 foos.add(new Foo("C", 5, 10));

 List<Foo> summarized = new ArrayList<>();
 Map<Foo, List<Foo>> collect = foos.stream().collect(Collectors.groupingBy(new Function<Foo, Foo>() 
 @Override
 public Foo apply(Foo t) 
 Optional<Foo> fOpt = summarized.stream().filter(e -> e.getCategory().equals(t.getCategory())).findFirst();
 Foo f;
 if (!fOpt.isPresent()) 
 f = new Foo(t.getCategory(), 0, 0);
 summarized.add(f);
 else 
 f = fOpt.get();
 
 f.setAmount(f.getAmount() + t.getAmount());
 f.setPrice(f.getPrice() + t.getPrice());
 return f;
 
 ));
 System.out.println(collect);

edited Aug 28 at 13:19

answered Aug 28 at 13:14

KDM

3,7011921

add a commentÂ |Â

up vote
0
down vote

My take on a solution :)

public static void main(String args) 
 List<Foo> foos = new ArrayList<>();
 foos.add(new Foo("A", 1, 10));
 foos.add(new Foo("A", 2, 10));
 foos.add(new Foo("A", 3, 10));
 foos.add(new Foo("B", 1, 10));
 foos.add(new Foo("C", 1, 10));
 foos.add(new Foo("C", 5, 10));

 List<Foo> summarized = new ArrayList<>();
 Map<Foo, List<Foo>> collect = foos.stream().collect(Collectors.groupingBy(new Function<Foo, Foo>() 
 @Override
 public Foo apply(Foo t) 
 Optional<Foo> fOpt = summarized.stream().filter(e -> e.getCategory().equals(t.getCategory())).findFirst();
 Foo f;
 if (!fOpt.isPresent()) 
 f = new Foo(t.getCategory(), 0, 0);
 summarized.add(f);
 else 
 f = fOpt.get();
 
 f.setAmount(f.getAmount() + t.getAmount());
 f.setPrice(f.getPrice() + t.getPrice());
 return f;
 
 ));
 System.out.println(collect);

edited Aug 28 at 13:19

answered Aug 28 at 13:14

KDM

3,7011921

My take on a solution :)

public static void main(String args) 
 List<Foo> foos = new ArrayList<>();
 foos.add(new Foo("A", 1, 10));
 foos.add(new Foo("A", 2, 10));
 foos.add(new Foo("A", 3, 10));
 foos.add(new Foo("B", 1, 10));
 foos.add(new Foo("C", 1, 10));
 foos.add(new Foo("C", 5, 10));

 List<Foo> summarized = new ArrayList<>();
 Map<Foo, List<Foo>> collect = foos.stream().collect(Collectors.groupingBy(new Function<Foo, Foo>() 
 @Override
 public Foo apply(Foo t) 
 Optional<Foo> fOpt = summarized.stream().filter(e -> e.getCategory().equals(t.getCategory())).findFirst();
 Foo f;
 if (!fOpt.isPresent()) 
 f = new Foo(t.getCategory(), 0, 0);
 summarized.add(f);
 else 
 f = fOpt.get();
 
 f.setAmount(f.getAmount() + t.getAmount());
 f.setPrice(f.getPrice() + t.getPrice());
 return f;
 
 ));
 System.out.println(collect);

edited Aug 28 at 13:19

answered Aug 28 at 13:14

KDM

3,7011921

edited Aug 28 at 13:19

answered Aug 28 at 13:14

KDM

3,7011921

answered Aug 28 at 13:14

KDM

3,7011921

answered Aug 28 at 13:14

KDM

3,7011921

add a commentÂ |Â

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