Convert Map to Map<String, Set> with filter and streams
I would like to convert my map which looks like this:
{
key="someKey1", value=Apple(id="1", color="green"),
key="someKey2", value=Apple(id="2", color="red"),
key="someKey3", value=Apple(id="3", color="green"),
key="someKey4", value=Apple(id="4", color="red"),
}
to another map which puts all apples of the same color into the same list:
{
key="red", value=list={apple1, apple3},
key="green", value=list={apple2, apple4},
}
I tried the following:
Map<String, Set<Apple>> sortedApples = appleMap.entrySet()
.stream()
.collect(Collectors.toMap(l -> l.getColour, ???));
Am I on the right track? Should I use filters for this task? Is there an easier way?
java lambda java-8 java-stream
edited 10 mins ago
Aomine
39.8k73770
asked 34 mins ago
M4V3N
528
add a comment |
I would like to convert my map which looks like this:
{
key="someKey1", value=Apple(id="1", color="green"),
key="someKey2", value=Apple(id="2", color="red"),
key="someKey3", value=Apple(id="3", color="green"),
key="someKey4", value=Apple(id="4", color="red"),
}
to another map which puts all apples of the same color into the same list:
{
key="red", value=list={apple1, apple3},
key="green", value=list={apple2, apple4},
}
I tried the following:
Map<String, Set<Apple>> sortedApples = appleMap.entrySet()
.stream()
.collect(Collectors.toMap(l -> l.getColour, ???));
Am I on the right track? Should I use filters for this task? Is there an easier way?
java lambda java-8 java-stream
edited 10 mins ago
Aomine
39.8k73770
asked 34 mins ago
M4V3N
528
add a comment |
I would like to convert my map which looks like this:
{
key="someKey1", value=Apple(id="1", color="green"),
key="someKey2", value=Apple(id="2", color="red"),
key="someKey3", value=Apple(id="3", color="green"),
key="someKey4", value=Apple(id="4", color="red"),
}
to another map which puts all apples of the same color into the same list:
{
key="red", value=list={apple1, apple3},
key="green", value=list={apple2, apple4},
}
I tried the following:
Map<String, Set<Apple>> sortedApples = appleMap.entrySet()
.stream()
.collect(Collectors.toMap(l -> l.getColour, ???));
Am I on the right track? Should I use filters for this task? Is there an easier way?
java lambda java-8 java-stream
edited 10 mins ago
Aomine
39.8k73770
asked 34 mins ago
M4V3N
528
I would like to convert my map which looks like this:
{
key="someKey1", value=Apple(id="1", color="green"),
key="someKey2", value=Apple(id="2", color="red"),
key="someKey3", value=Apple(id="3", color="green"),
key="someKey4", value=Apple(id="4", color="red"),
}
to another map which puts all apples of the same color into the same list:
{
key="red", value=list={apple1, apple3},
key="green", value=list={apple2, apple4},
}
I tried the following:
Map<String, Set<Apple>> sortedApples = appleMap.entrySet()
.stream()
.collect(Collectors.toMap(l -> l.getColour, ???));
Am I on the right track? Should I use filters for this task? Is there an easier way?
java lambda java-8 java-stream
java lambda java-8 java-stream
edited 10 mins ago
Aomine
39.8k73770
asked 34 mins ago
M4V3N
528
edited 10 mins ago
Aomine
39.8k73770
asked 34 mins ago
M4V3N
528
edited 10 mins ago
Aomine
39.8k73770
edited 10 mins ago
Aomine
39.8k73770
edited 10 mins ago
Aomine
39.8k73770
39.8k73770
asked 34 mins ago
M4V3N
528
asked 34 mins ago
M4V3N
528
asked 34 mins ago
M4V3N
528
528
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
Collectors.groupingBy is more suitable than Collectors.toMap for this task (though both can be used).
Map<String, List<Apple>> sortedApples =
appleMap.values()
.stream()
.collect(Collectors.groupingBy(Apple::getColour));
Or, to group them into Sets use:
Map<String, Set<Apple>> sortedApples =
appleMap.values()
.stream()
.collect(Collectors.groupingBy(Apple::getColour,
Collectors.mapping(Function.identity(),
Collectors.toSet())));
answered 31 mins ago
Eran
279k37452537
add a comment |
You can use Collectors.groupingBy and Collectors.toSet()
Map<String, Set<Apple>> sortedApples = appleMap.values() // Collection<Apple>
.stream() // Stream<Apple>
.collect(Collectors.groupingBy(Apple::getColour, // groupBy colour
Collectors.mapping(a -> a, Collectors.toSet()))); // collect to Set
answered 30 mins ago
nullpointer
42.5k1090175
add a comment |
if you want to proceed with toMap you can get the result as follows:
map.entrySet()
.stream()
.collect(toMap(s -> s.getValue().getColour(),
v -> new HashSet<>(singleton(v.getValue())),
(l, r) -> {l.addAll(r); return l;}));
But I would not suggest it because as you can see this it is not very readable and not the idiomatic approach.
This is a job for the groupingBy collector shown in the other answers or do it shorter without a stream:
Map<String, Set<Apple>> res = new HashMap<>();
map.forEach((k, v) -> res.computeIfAbsent(v.getColour(), e -> new HashSet<>()).add(v));
answered 17 mins ago
Aomine
39.8k73770
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Collectors.groupingBy is more suitable than Collectors.toMap for this task (though both can be used).
Map<String, List<Apple>> sortedApples =
appleMap.values()
.stream()
.collect(Collectors.groupingBy(Apple::getColour));
Or, to group them into Sets use:
Map<String, Set<Apple>> sortedApples =
appleMap.values()
.stream()
.collect(Collectors.groupingBy(Apple::getColour,
Collectors.mapping(Function.identity(),
Collectors.toSet())));
answered 31 mins ago
Eran
279k37452537
add a comment |
Collectors.groupingBy is more suitable than Collectors.toMap for this task (though both can be used).
Map<String, List<Apple>> sortedApples =
appleMap.values()
.stream()
.collect(Collectors.groupingBy(Apple::getColour));
Or, to group them into Sets use:
Map<String, Set<Apple>> sortedApples =
appleMap.values()
.stream()
.collect(Collectors.groupingBy(Apple::getColour,
Collectors.mapping(Function.identity(),
Collectors.toSet())));
answered 31 mins ago
Eran
279k37452537
add a comment |
Collectors.groupingBy is more suitable than Collectors.toMap for this task (though both can be used).
Map<String, List<Apple>> sortedApples =
appleMap.values()
.stream()
.collect(Collectors.groupingBy(Apple::getColour));
Or, to group them into Sets use:
Map<String, Set<Apple>> sortedApples =
appleMap.values()
.stream()
.collect(Collectors.groupingBy(Apple::getColour,
Collectors.mapping(Function.identity(),
Collectors.toSet())));
answered 31 mins ago
Eran
279k37452537
Collectors.groupingBy is more suitable than Collectors.toMap for this task (though both can be used).
Map<String, List<Apple>> sortedApples =
appleMap.values()
.stream()
.collect(Collectors.groupingBy(Apple::getColour));
Or, to group them into Sets use:
Map<String, Set<Apple>> sortedApples =
appleMap.values()
.stream()
.collect(Collectors.groupingBy(Apple::getColour,
Collectors.mapping(Function.identity(),
Collectors.toSet())));
answered 31 mins ago
Eran
279k37452537
answered 31 mins ago
Eran
279k37452537
answered 31 mins ago
Eran
279k37452537
answered 31 mins ago
Eran
279k37452537
279k37452537
add a comment |
add a comment |
You can use Collectors.groupingBy and Collectors.toSet()
Map<String, Set<Apple>> sortedApples = appleMap.values() // Collection<Apple>
.stream() // Stream<Apple>
.collect(Collectors.groupingBy(Apple::getColour, // groupBy colour
Collectors.mapping(a -> a, Collectors.toSet()))); // collect to Set
answered 30 mins ago
nullpointer
42.5k1090175
add a comment |
You can use Collectors.groupingBy and Collectors.toSet()
Map<String, Set<Apple>> sortedApples = appleMap.values() // Collection<Apple>
.stream() // Stream<Apple>
.collect(Collectors.groupingBy(Apple::getColour, // groupBy colour
Collectors.mapping(a -> a, Collectors.toSet()))); // collect to Set
answered 30 mins ago
nullpointer
42.5k1090175
add a comment |
You can use Collectors.groupingBy and Collectors.toSet()
Map<String, Set<Apple>> sortedApples = appleMap.values() // Collection<Apple>
.stream() // Stream<Apple>
.collect(Collectors.groupingBy(Apple::getColour, // groupBy colour
Collectors.mapping(a -> a, Collectors.toSet()))); // collect to Set
answered 30 mins ago
nullpointer
42.5k1090175
You can use Collectors.groupingBy and Collectors.toSet()
Map<String, Set<Apple>> sortedApples = appleMap.values() // Collection<Apple>
.stream() // Stream<Apple>
.collect(Collectors.groupingBy(Apple::getColour, // groupBy colour
Collectors.mapping(a -> a, Collectors.toSet()))); // collect to Set
answered 30 mins ago
nullpointer
42.5k1090175
edited 16 mins ago
answered 30 mins ago
nullpointer
42.5k1090175
answered 30 mins ago
nullpointer
42.5k1090175
answered 30 mins ago
nullpointer
42.5k1090175
42.5k1090175
add a comment |
add a comment |
if you want to proceed with toMap you can get the result as follows:
map.entrySet()
.stream()
.collect(toMap(s -> s.getValue().getColour(),
v -> new HashSet<>(singleton(v.getValue())),
(l, r) -> {l.addAll(r); return l;}));
But I would not suggest it because as you can see this it is not very readable and not the idiomatic approach.
This is a job for the groupingBy collector shown in the other answers or do it shorter without a stream:
Map<String, Set<Apple>> res = new HashMap<>();
map.forEach((k, v) -> res.computeIfAbsent(v.getColour(), e -> new HashSet<>()).add(v));
answered 17 mins ago
Aomine
39.8k73770
add a comment |
if you want to proceed with toMap you can get the result as follows:
map.entrySet()
.stream()
.collect(toMap(s -> s.getValue().getColour(),
v -> new HashSet<>(singleton(v.getValue())),
(l, r) -> {l.addAll(r); return l;}));
But I would not suggest it because as you can see this it is not very readable and not the idiomatic approach.
This is a job for the groupingBy collector shown in the other answers or do it shorter without a stream:
Map<String, Set<Apple>> res = new HashMap<>();
map.forEach((k, v) -> res.computeIfAbsent(v.getColour(), e -> new HashSet<>()).add(v));
answered 17 mins ago
Aomine
39.8k73770
add a comment |
if you want to proceed with toMap you can get the result as follows:
map.entrySet()
.stream()
.collect(toMap(s -> s.getValue().getColour(),
v -> new HashSet<>(singleton(v.getValue())),
(l, r) -> {l.addAll(r); return l;}));
But I would not suggest it because as you can see this it is not very readable and not the idiomatic approach.
This is a job for the groupingBy collector shown in the other answers or do it shorter without a stream:
Map<String, Set<Apple>> res = new HashMap<>();
map.forEach((k, v) -> res.computeIfAbsent(v.getColour(), e -> new HashSet<>()).add(v));
answered 17 mins ago
Aomine
39.8k73770
if you want to proceed with toMap you can get the result as follows:
map.entrySet()
.stream()
.collect(toMap(s -> s.getValue().getColour(),
v -> new HashSet<>(singleton(v.getValue())),
(l, r) -> {l.addAll(r); return l;}));
But I would not suggest it because as you can see this it is not very readable and not the idiomatic approach.
This is a job for the groupingBy collector shown in the other answers or do it shorter without a stream:
Map<String, Set<Apple>> res = new HashMap<>();
map.forEach((k, v) -> res.computeIfAbsent(v.getColour(), e -> new HashSet<>()).add(v));
answered 17 mins ago
Aomine
39.8k73770
edited 11 mins ago
answered 17 mins ago
Aomine
39.8k73770
answered 17 mins ago
Aomine
39.8k73770
answered 17 mins ago
Aomine
39.8k73770
39.8k73770
add a comment |
add a comment |
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