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

edited 1 hour ago

Nicholas K

5,81951031

asked 12 hours ago

Yassine Ben Hamida

14111

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

edited 1 hour ago

Nicholas K

5,81951031

asked 12 hours ago

Yassine Ben Hamida

14111

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

edited 1 hour ago

Nicholas K

5,81951031

asked 12 hours ago

Yassine Ben Hamida

14111

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

edited 1 hour ago

Nicholas K

5,81951031

asked 12 hours ago

Yassine Ben Hamida

14111

edited 1 hour ago

Nicholas K

5,81951031

asked 12 hours ago

Yassine Ben Hamida

14111

edited 1 hour ago

Nicholas K

5,81951031

edited 1 hour ago

Nicholas K

5,81951031

edited 1 hour ago

Nicholas K

5,81951031

asked 12 hours ago

Yassine Ben Hamida

14111

asked 12 hours ago

Yassine Ben Hamida

14111

asked 12 hours ago

Yassine Ben Hamida

14111

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());

answered 12 hours ago

Aomine

40.7k73870

add a comment |

Here's an alternate way to do it with Java-9 and above:

List<E> result = map.values()

                    .stream()

                    .collect(Collectors.flatMapping(List::stream, Collectors.toList()));

edited 28 mins ago

nullpointer

43.4k1093178

answered 12 hours ago

Ravindra Ranwala

8,53231634

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
12 hours ago

3

It decreases indeed the readability here. The Java 8 way is clearly more relevant here.
– davidxxx
12 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
12 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.

answered 12 hours ago

ETO

1,886422

add a comment |

Or use forEach

 map.forEach((k,v)->list.addAll(v));

or as Aomine commented use this

map.values().forEach(list::addAll);

edited 12 hours ago

answered 12 hours ago

Hadi J

9,87231742

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
12 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)

edited 1 hour ago

answered 12 hours ago

nullpointer

43.4k1093178

add a comment |

Simply use :-

map.values().stream().flatMap(List::stream).collect(Collectors.toList());

edited 1 hour ago

answered 12 hours ago

Nicholas K

5,81951031

add a comment |

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6 Answers
6

active

oldest

votes

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());

answered 12 hours ago

Aomine

40.7k73870

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());

answered 12 hours ago

Aomine

40.7k73870

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());

answered 12 hours ago

Aomine

40.7k73870

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 12 hours ago

Aomine

40.7k73870

answered 12 hours ago

Aomine

40.7k73870

answered 12 hours ago

Aomine

40.7k73870

answered 12 hours ago

Aomine

40.7k73870

add a comment |

Here's an alternate way to do it with Java-9 and above:

List<E> result = map.values()

                    .stream()

                    .collect(Collectors.flatMapping(List::stream, Collectors.toList()));

edited 28 mins ago

nullpointer

43.4k1093178

answered 12 hours ago

Ravindra Ranwala

8,53231634

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
12 hours ago

3

It decreases indeed the readability here. The Java 8 way is clearly more relevant here.
– davidxxx
12 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
12 hours ago

add a comment |

Here's an alternate way to do it with Java-9 and above:

List<E> result = map.values()

                    .stream()

                    .collect(Collectors.flatMapping(List::stream, Collectors.toList()));

edited 28 mins ago

nullpointer

43.4k1093178

answered 12 hours ago

Ravindra Ranwala

8,53231634

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
12 hours ago

3

It decreases indeed the readability here. The Java 8 way is clearly more relevant here.
– davidxxx
12 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
12 hours ago

add a comment |

Here's an alternate way to do it with Java-9 and above:

List<E> result = map.values()

                    .stream()

                    .collect(Collectors.flatMapping(List::stream, Collectors.toList()));

edited 28 mins ago

nullpointer

43.4k1093178

answered 12 hours ago

Ravindra Ranwala

8,53231634

Here's an alternate way to do it with Java-9 and above:

List<E> result = map.values()

                    .stream()

                    .collect(Collectors.flatMapping(List::stream, Collectors.toList()));

edited 28 mins ago

nullpointer

43.4k1093178

answered 12 hours ago

Ravindra Ranwala

8,53231634

edited 28 mins ago

nullpointer

43.4k1093178

edited 28 mins ago

nullpointer

43.4k1093178

edited 28 mins ago

nullpointer

43.4k1093178

answered 12 hours ago

Ravindra Ranwala

8,53231634

answered 12 hours ago

Ravindra Ranwala

8,53231634

answered 12 hours ago

Ravindra Ranwala

8,53231634

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
12 hours ago

3

It decreases indeed the readability here. The Java 8 way is clearly more relevant here.
– davidxxx
12 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
12 hours ago

add a comment |

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
12 hours ago

3

It decreases indeed the readability here. The Java 8 way is clearly more relevant here.
– davidxxx
12 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
12 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
12 hours ago

It decreases indeed the readability here. The Java 8 way is clearly more relevant here.
– davidxxx
12 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
12 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.

answered 12 hours ago

ETO

1,886422

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.

answered 12 hours ago

ETO

1,886422

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.

answered 12 hours ago

ETO

1,886422

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 12 hours ago

ETO

1,886422

answered 12 hours ago

ETO

1,886422

answered 12 hours ago

ETO

1,886422

answered 12 hours ago

ETO

1,886422

add a comment |

Or use forEach

 map.forEach((k,v)->list.addAll(v));

or as Aomine commented use this

map.values().forEach(list::addAll);

edited 12 hours ago

answered 12 hours ago

Hadi J

9,87231742

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
12 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);

edited 12 hours ago

answered 12 hours ago

Hadi J

9,87231742

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
12 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);

edited 12 hours ago

answered 12 hours ago

Hadi J

9,87231742

Or use forEach

 map.forEach((k,v)->list.addAll(v));

or as Aomine commented use this

map.values().forEach(list::addAll);

edited 12 hours ago

answered 12 hours ago

Hadi J

9,87231742

edited 12 hours ago

answered 12 hours ago

Hadi J

9,87231742

answered 12 hours ago

Hadi J

9,87231742

answered 12 hours ago

Hadi J

9,87231742

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
12 hours ago

add a comment |

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
12 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
12 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)

edited 1 hour ago

answered 12 hours ago

nullpointer

43.4k1093178

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)

edited 1 hour ago

answered 12 hours ago

nullpointer

43.4k1093178

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)

edited 1 hour ago

answered 12 hours ago

nullpointer

43.4k1093178

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 1 hour ago

answered 12 hours ago

nullpointer

43.4k1093178

edited 1 hour ago

answered 12 hours ago

nullpointer

43.4k1093178

answered 12 hours ago

nullpointer

43.4k1093178

answered 12 hours ago

nullpointer

43.4k1093178

add a comment |

Simply use :-

map.values().stream().flatMap(List::stream).collect(Collectors.toList());

edited 1 hour ago

answered 12 hours ago

Nicholas K

5,81951031

add a comment |

Simply use :-

map.values().stream().flatMap(List::stream).collect(Collectors.toList());

edited 1 hour ago

answered 12 hours ago

Nicholas K

5,81951031

add a comment |

Simply use :-

map.values().stream().flatMap(List::stream).collect(Collectors.toList());

edited 1 hour ago

answered 12 hours ago

Nicholas K

5,81951031

Simply use :-

map.values().stream().flatMap(List::stream).collect(Collectors.toList());

edited 1 hour ago

answered 12 hours ago

Nicholas K

5,81951031

edited 1 hour ago

answered 12 hours ago

Nicholas K

5,81951031

answered 12 hours ago

Nicholas K

5,81951031

answered 12 hours ago

Nicholas K

5,81951031

add a comment |

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