How to get a List from a HashMap<String,List>
I want to extract a List<E>
from a Map<String,List<E>>
(E
is a random Class) using stream()
.
I want a simple one-line method using java 8's stream.
What I have tried until now :
HashMap<String,List<E>> map = new HashMap<>();
List<E> list = map.values(); // does not compile
list = map.values().stream().collect(Collectors.toList()); // does not compile
java collections java-8 java-stream collectors
add a comment |
I want to extract a List<E>
from a Map<String,List<E>>
(E
is a random Class) using stream()
.
I want a simple one-line method using java 8's stream.
What I have tried until now :
HashMap<String,List<E>> map = new HashMap<>();
List<E> list = map.values(); // does not compile
list = map.values().stream().collect(Collectors.toList()); // does not compile
java collections java-8 java-stream collectors
add a comment |
I want to extract a List<E>
from a Map<String,List<E>>
(E
is a random Class) using stream()
.
I want a simple one-line method using java 8's stream.
What I have tried until now :
HashMap<String,List<E>> map = new HashMap<>();
List<E> list = map.values(); // does not compile
list = map.values().stream().collect(Collectors.toList()); // does not compile
java collections java-8 java-stream collectors
I want to extract a List<E>
from a Map<String,List<E>>
(E
is a random Class) using stream()
.
I want a simple one-line method using java 8's stream.
What I have tried until now :
HashMap<String,List<E>> map = new HashMap<>();
List<E> list = map.values(); // does not compile
list = map.values().stream().collect(Collectors.toList()); // does not compile
java collections java-8 java-stream collectors
java collections java-8 java-stream collectors
edited 19 mins ago
Nicholas K
5,81951031
5,81951031
asked 11 hours ago
Yassine Ben Hamida
13111
13111
add a comment |
add a comment |
6 Answers
6
active
oldest
votes
map.values()
returns a Collection<List<E>>
not a List<E>
, if you want the latter then you're required to flatten the nested List<E>
into a single List<E>
as follows:
List<E> result = map.values()
.stream()
.flatMap(List::stream)
.collect(Collectors.toList());
add a comment |
Here's the Java9 solution,
List<E> result = map.values().stream()
.collect(Collectors.flatMapping(List::stream, Collectors.toList()));
6
Just the apiNote :- TheflatMapping()
collectors are most useful when used in a multi-level reduction, such as downstream of agroupingBy
orpartitioningBy
.
– nullpointer
11 hours ago
3
It decreases indeed the readability here. The Java 8 way is clearly more relevant here.
– davidxxx
11 hours ago
2
In fact it is not the Java 9 solution. It is a way possible with Java 9. But the Java 9 way to favor is of course the same as with Java 8 : stackoverflow.com/a/54054456/270371
– davidxxx
10 hours ago
add a comment |
In addition to other answers:
List<E> result = map.values()
.stream()
.collect(ArrayList::new, List::addAll, List::addAll);
This could also do the trick.
add a comment |
Or use forEach
map.forEach((k,v)->list.addAll(v));
or as Aomine commented use this
map.values().forEach(list::addAll);
3
good idea in showing a non-stream version. btw it would be better to iterate over thevalues
since you're not doing anything with thek
i.e. you can doList<String> result = new ArrayList<>(); map.values().forEach(result::addAll);
– Aomine
11 hours ago
add a comment |
You can use Collection.stream
with flatMap
as:
Map<String, List<E>> map = new HashMap<>(); // program to interface
List<E> list = map.values()
.stream()
.flatMap(Collection::stream)
.collect(Collectors.toList());
or use a non-stream version as:
List<E> list = new ArrayList<>();
map.values().forEach(list::addAll)
add a comment |
Simply use :-
map.values().stream().flatMap(List::stream).collect(Collectors.toList());
1
This wont' work, rather it will give youList<List<E>>
– Ravindra Ranwala
11 hours ago
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
map.values()
returns a Collection<List<E>>
not a List<E>
, if you want the latter then you're required to flatten the nested List<E>
into a single List<E>
as follows:
List<E> result = map.values()
.stream()
.flatMap(List::stream)
.collect(Collectors.toList());
add a comment |
map.values()
returns a Collection<List<E>>
not a List<E>
, if you want the latter then you're required to flatten the nested List<E>
into a single List<E>
as follows:
List<E> result = map.values()
.stream()
.flatMap(List::stream)
.collect(Collectors.toList());
add a comment |
map.values()
returns a Collection<List<E>>
not a List<E>
, if you want the latter then you're required to flatten the nested List<E>
into a single List<E>
as follows:
List<E> result = map.values()
.stream()
.flatMap(List::stream)
.collect(Collectors.toList());
map.values()
returns a Collection<List<E>>
not a List<E>
, if you want the latter then you're required to flatten the nested List<E>
into a single List<E>
as follows:
List<E> result = map.values()
.stream()
.flatMap(List::stream)
.collect(Collectors.toList());
answered 11 hours ago
Aomine
40.7k73870
40.7k73870
add a comment |
add a comment |
Here's the Java9 solution,
List<E> result = map.values().stream()
.collect(Collectors.flatMapping(List::stream, Collectors.toList()));
6
Just the apiNote :- TheflatMapping()
collectors are most useful when used in a multi-level reduction, such as downstream of agroupingBy
orpartitioningBy
.
– nullpointer
11 hours ago
3
It decreases indeed the readability here. The Java 8 way is clearly more relevant here.
– davidxxx
11 hours ago
2
In fact it is not the Java 9 solution. It is a way possible with Java 9. But the Java 9 way to favor is of course the same as with Java 8 : stackoverflow.com/a/54054456/270371
– davidxxx
10 hours ago
add a comment |
Here's the Java9 solution,
List<E> result = map.values().stream()
.collect(Collectors.flatMapping(List::stream, Collectors.toList()));
6
Just the apiNote :- TheflatMapping()
collectors are most useful when used in a multi-level reduction, such as downstream of agroupingBy
orpartitioningBy
.
– nullpointer
11 hours ago
3
It decreases indeed the readability here. The Java 8 way is clearly more relevant here.
– davidxxx
11 hours ago
2
In fact it is not the Java 9 solution. It is a way possible with Java 9. But the Java 9 way to favor is of course the same as with Java 8 : stackoverflow.com/a/54054456/270371
– davidxxx
10 hours ago
add a comment |
Here's the Java9 solution,
List<E> result = map.values().stream()
.collect(Collectors.flatMapping(List::stream, Collectors.toList()));
Here's the Java9 solution,
List<E> result = map.values().stream()
.collect(Collectors.flatMapping(List::stream, Collectors.toList()));
answered 11 hours ago
Ravindra Ranwala
8,52231634
8,52231634
6
Just the apiNote :- TheflatMapping()
collectors are most useful when used in a multi-level reduction, such as downstream of agroupingBy
orpartitioningBy
.
– nullpointer
11 hours ago
3
It decreases indeed the readability here. The Java 8 way is clearly more relevant here.
– davidxxx
11 hours ago
2
In fact it is not the Java 9 solution. It is a way possible with Java 9. But the Java 9 way to favor is of course the same as with Java 8 : stackoverflow.com/a/54054456/270371
– davidxxx
10 hours ago
add a comment |
6
Just the apiNote :- TheflatMapping()
collectors are most useful when used in a multi-level reduction, such as downstream of agroupingBy
orpartitioningBy
.
– nullpointer
11 hours ago
3
It decreases indeed the readability here. The Java 8 way is clearly more relevant here.
– davidxxx
11 hours ago
2
In fact it is not the Java 9 solution. It is a way possible with Java 9. But the Java 9 way to favor is of course the same as with Java 8 : stackoverflow.com/a/54054456/270371
– davidxxx
10 hours ago
6
6
Just the apiNote :- The
flatMapping()
collectors are most useful when used in a multi-level reduction, such as downstream of a groupingBy
or partitioningBy
.– nullpointer
11 hours ago
Just the apiNote :- The
flatMapping()
collectors are most useful when used in a multi-level reduction, such as downstream of a groupingBy
or partitioningBy
.– nullpointer
11 hours ago
3
3
It decreases indeed the readability here. The Java 8 way is clearly more relevant here.
– davidxxx
11 hours ago
It decreases indeed the readability here. The Java 8 way is clearly more relevant here.
– davidxxx
11 hours ago
2
2
In fact it is not the Java 9 solution. It is a way possible with Java 9. But the Java 9 way to favor is of course the same as with Java 8 : stackoverflow.com/a/54054456/270371
– davidxxx
10 hours ago
In fact it is not the Java 9 solution. It is a way possible with Java 9. But the Java 9 way to favor is of course the same as with Java 8 : stackoverflow.com/a/54054456/270371
– davidxxx
10 hours ago
add a comment |
In addition to other answers:
List<E> result = map.values()
.stream()
.collect(ArrayList::new, List::addAll, List::addAll);
This could also do the trick.
add a comment |
In addition to other answers:
List<E> result = map.values()
.stream()
.collect(ArrayList::new, List::addAll, List::addAll);
This could also do the trick.
add a comment |
In addition to other answers:
List<E> result = map.values()
.stream()
.collect(ArrayList::new, List::addAll, List::addAll);
This could also do the trick.
In addition to other answers:
List<E> result = map.values()
.stream()
.collect(ArrayList::new, List::addAll, List::addAll);
This could also do the trick.
answered 11 hours ago
ETO
1,886422
1,886422
add a comment |
add a comment |
Or use forEach
map.forEach((k,v)->list.addAll(v));
or as Aomine commented use this
map.values().forEach(list::addAll);
3
good idea in showing a non-stream version. btw it would be better to iterate over thevalues
since you're not doing anything with thek
i.e. you can doList<String> result = new ArrayList<>(); map.values().forEach(result::addAll);
– Aomine
11 hours ago
add a comment |
Or use forEach
map.forEach((k,v)->list.addAll(v));
or as Aomine commented use this
map.values().forEach(list::addAll);
3
good idea in showing a non-stream version. btw it would be better to iterate over thevalues
since you're not doing anything with thek
i.e. you can doList<String> result = new ArrayList<>(); map.values().forEach(result::addAll);
– Aomine
11 hours ago
add a comment |
Or use forEach
map.forEach((k,v)->list.addAll(v));
or as Aomine commented use this
map.values().forEach(list::addAll);
Or use forEach
map.forEach((k,v)->list.addAll(v));
or as Aomine commented use this
map.values().forEach(list::addAll);
edited 11 hours ago
answered 11 hours ago
Hadi J
9,86231742
9,86231742
3
good idea in showing a non-stream version. btw it would be better to iterate over thevalues
since you're not doing anything with thek
i.e. you can doList<String> result = new ArrayList<>(); map.values().forEach(result::addAll);
– Aomine
11 hours ago
add a comment |
3
good idea in showing a non-stream version. btw it would be better to iterate over thevalues
since you're not doing anything with thek
i.e. you can doList<String> result = new ArrayList<>(); map.values().forEach(result::addAll);
– Aomine
11 hours ago
3
3
good idea in showing a non-stream version. btw it would be better to iterate over the
values
since you're not doing anything with the k
i.e. you can do List<String> result = new ArrayList<>(); map.values().forEach(result::addAll);
– Aomine
11 hours ago
good idea in showing a non-stream version. btw it would be better to iterate over the
values
since you're not doing anything with the k
i.e. you can do List<String> result = new ArrayList<>(); map.values().forEach(result::addAll);
– Aomine
11 hours ago
add a comment |
You can use Collection.stream
with flatMap
as:
Map<String, List<E>> map = new HashMap<>(); // program to interface
List<E> list = map.values()
.stream()
.flatMap(Collection::stream)
.collect(Collectors.toList());
or use a non-stream version as:
List<E> list = new ArrayList<>();
map.values().forEach(list::addAll)
add a comment |
You can use Collection.stream
with flatMap
as:
Map<String, List<E>> map = new HashMap<>(); // program to interface
List<E> list = map.values()
.stream()
.flatMap(Collection::stream)
.collect(Collectors.toList());
or use a non-stream version as:
List<E> list = new ArrayList<>();
map.values().forEach(list::addAll)
add a comment |
You can use Collection.stream
with flatMap
as:
Map<String, List<E>> map = new HashMap<>(); // program to interface
List<E> list = map.values()
.stream()
.flatMap(Collection::stream)
.collect(Collectors.toList());
or use a non-stream version as:
List<E> list = new ArrayList<>();
map.values().forEach(list::addAll)
You can use Collection.stream
with flatMap
as:
Map<String, List<E>> map = new HashMap<>(); // program to interface
List<E> list = map.values()
.stream()
.flatMap(Collection::stream)
.collect(Collectors.toList());
or use a non-stream version as:
List<E> list = new ArrayList<>();
map.values().forEach(list::addAll)
edited 16 mins ago
answered 11 hours ago
nullpointer
43.4k1093178
43.4k1093178
add a comment |
add a comment |
Simply use :-
map.values().stream().flatMap(List::stream).collect(Collectors.toList());
1
This wont' work, rather it will give youList<List<E>>
– Ravindra Ranwala
11 hours ago
add a comment |
Simply use :-
map.values().stream().flatMap(List::stream).collect(Collectors.toList());
1
This wont' work, rather it will give youList<List<E>>
– Ravindra Ranwala
11 hours ago
add a comment |
Simply use :-
map.values().stream().flatMap(List::stream).collect(Collectors.toList());
Simply use :-
map.values().stream().flatMap(List::stream).collect(Collectors.toList());
edited 18 mins ago
answered 11 hours ago
Nicholas K
5,81951031
5,81951031
1
This wont' work, rather it will give youList<List<E>>
– Ravindra Ranwala
11 hours ago
add a comment |
1
This wont' work, rather it will give youList<List<E>>
– Ravindra Ranwala
11 hours ago
1
1
This wont' work, rather it will give you
List<List<E>>
– Ravindra Ranwala
11 hours ago
This wont' work, rather it will give you
List<List<E>>
– Ravindra Ranwala
11 hours ago
add a comment |
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