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 Jan 6 at 4:36

Nicholas K

6,59261132

asked Jan 5 at 17:31

Yassine Ben Hamida

370113

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 Jan 6 at 4:36

Nicholas K

6,59261132

asked Jan 5 at 17:31

Yassine Ben Hamida

370113

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 Jan 6 at 4:36

Nicholas K

6,59261132

asked Jan 5 at 17:31

Yassine Ben Hamida

370113

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 Jan 6 at 4:36

Nicholas K

6,59261132

asked Jan 5 at 17:31

Yassine Ben Hamida

370113

edited Jan 6 at 4:36

Nicholas K

6,59261132

asked Jan 5 at 17:31

Yassine Ben Hamida

370113

edited Jan 6 at 4:36

Nicholas K

6,59261132

edited Jan 6 at 4:36

Nicholas K

6,59261132

edited Jan 6 at 4:36

Nicholas K

6,59261132

asked Jan 5 at 17:31

Yassine Ben Hamida

370113

asked Jan 5 at 17:31

Yassine Ben Hamida

370113

asked Jan 5 at 17:31

Yassine Ben Hamida

370113

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 Jan 5 at 17:32

Aomine

41.9k74071

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 Jan 5 at 17:42

answered Jan 5 at 17:39

Hadi J

10.2k31744

1

@Aomine values() may involve a construction of an expensive intermediate object, thus simply ignoring k may be advantageous from a performance perspective.

– oakad
Jan 6 at 7:01

@oakad You're correct in the sense that values may involve creating a new object but note that the returned collection by values is lazily created once, thereafter, the same instance will be returned. So it's not so expensive after all. Further, I'd favour readability here which means map.values().forEach(list::addAll).

– Aomine
Jan 6 at 10:13

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 Jan 6 at 5:44

nullpointer

46.5k1199190

answered Jan 5 at 17:33

Ravindra Ranwala

9,06031634

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
Jan 5 at 17:46

9

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

– davidxxx
Jan 5 at 17:52

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 Jan 5 at 17:40

ETO

2,663526

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 Jan 6 at 4:39

answered Jan 5 at 17:38

nullpointer

46.5k1199190

add a comment |

Simply use :-

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

edited Jan 6 at 4:37

answered Jan 5 at 17:34

Nicholas K

6,59261132

add a comment |

Your Answer

StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});

}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f54054448%2fhow-to-get-a-liste-from-a-hashmapstring-liste%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

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 Jan 5 at 17:32

Aomine

41.9k74071

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 Jan 5 at 17:32

Aomine

41.9k74071

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 Jan 5 at 17:32

Aomine

41.9k74071

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 Jan 5 at 17:32

Aomine

41.9k74071

answered Jan 5 at 17:32

Aomine

41.9k74071

answered Jan 5 at 17:32

Aomine

41.9k74071

answered Jan 5 at 17:32

Aomine

41.9k74071

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 Jan 5 at 17:42

answered Jan 5 at 17:39

Hadi J

10.2k31744

1

@Aomine values() may involve a construction of an expensive intermediate object, thus simply ignoring k may be advantageous from a performance perspective.

– oakad
Jan 6 at 7:01

@oakad You're correct in the sense that values may involve creating a new object but note that the returned collection by values is lazily created once, thereafter, the same instance will be returned. So it's not so expensive after all. Further, I'd favour readability here which means map.values().forEach(list::addAll).

– Aomine
Jan 6 at 10:13

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 Jan 5 at 17:42

answered Jan 5 at 17:39

Hadi J

10.2k31744

1

@Aomine values() may involve a construction of an expensive intermediate object, thus simply ignoring k may be advantageous from a performance perspective.

– oakad
Jan 6 at 7:01

@oakad You're correct in the sense that values may involve creating a new object but note that the returned collection by values is lazily created once, thereafter, the same instance will be returned. So it's not so expensive after all. Further, I'd favour readability here which means map.values().forEach(list::addAll).

– Aomine
Jan 6 at 10:13

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 Jan 5 at 17:42

answered Jan 5 at 17:39

Hadi J

10.2k31744

Or use forEach

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

or as Aomine commented use this

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

edited Jan 5 at 17:42

answered Jan 5 at 17:39

Hadi J

10.2k31744

edited Jan 5 at 17:42

answered Jan 5 at 17:39

Hadi J

10.2k31744

answered Jan 5 at 17:39

Hadi J

10.2k31744

answered Jan 5 at 17:39

Hadi J

10.2k31744

1

@Aomine values() may involve a construction of an expensive intermediate object, thus simply ignoring k may be advantageous from a performance perspective.

– oakad
Jan 6 at 7:01

@oakad You're correct in the sense that values may involve creating a new object but note that the returned collection by values is lazily created once, thereafter, the same instance will be returned. So it's not so expensive after all. Further, I'd favour readability here which means map.values().forEach(list::addAll).

– Aomine
Jan 6 at 10:13

add a comment |

1

@Aomine values() may involve a construction of an expensive intermediate object, thus simply ignoring k may be advantageous from a performance perspective.

– oakad
Jan 6 at 7:01

@oakad You're correct in the sense that values may involve creating a new object but note that the returned collection by values is lazily created once, thereafter, the same instance will be returned. So it's not so expensive after all. Further, I'd favour readability here which means map.values().forEach(list::addAll).

– Aomine
Jan 6 at 10:13

@Aomine values() may involve a construction of an expensive intermediate object, thus simply ignoring k may be advantageous from a performance perspective.

– oakad
Jan 6 at 7:01

@oakad You're correct in the sense that values may involve creating a new object but note that the returned collection by values is lazily created once, thereafter, the same instance will be returned. So it's not so expensive after all. Further, I'd favour readability here which means map.values().forEach(list::addAll).

– Aomine
Jan 6 at 10:13

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 Jan 6 at 5:44

nullpointer

46.5k1199190

answered Jan 5 at 17:33

Ravindra Ranwala

9,06031634

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
Jan 5 at 17:46

9

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

– davidxxx
Jan 5 at 17:52

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 Jan 6 at 5:44

nullpointer

46.5k1199190

answered Jan 5 at 17:33

Ravindra Ranwala

9,06031634

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
Jan 5 at 17:46

9

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

– davidxxx
Jan 5 at 17:52

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 Jan 6 at 5:44

nullpointer

46.5k1199190

answered Jan 5 at 17:33

Ravindra Ranwala

9,06031634

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 Jan 6 at 5:44

nullpointer

46.5k1199190

answered Jan 5 at 17:33

Ravindra Ranwala

9,06031634

edited Jan 6 at 5:44

nullpointer

46.5k1199190

edited Jan 6 at 5:44

nullpointer

46.5k1199190

edited Jan 6 at 5:44

nullpointer

46.5k1199190

answered Jan 5 at 17:33

Ravindra Ranwala

9,06031634

answered Jan 5 at 17:33

Ravindra Ranwala

9,06031634

answered Jan 5 at 17:33

Ravindra Ranwala

9,06031634

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
Jan 5 at 17:46

9

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

– davidxxx
Jan 5 at 17:52

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
Jan 5 at 17:46

9

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

– davidxxx
Jan 5 at 17:52

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
Jan 5 at 17:46

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

– davidxxx
Jan 5 at 17:52

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 Jan 5 at 17:40

ETO

2,663526

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 Jan 5 at 17:40

ETO

2,663526

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 Jan 5 at 17:40

ETO

2,663526

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 Jan 5 at 17:40

ETO

2,663526

answered Jan 5 at 17:40

ETO

2,663526

answered Jan 5 at 17:40

ETO

2,663526

answered Jan 5 at 17:40

ETO

2,663526

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 Jan 6 at 4:39

answered Jan 5 at 17:38

nullpointer

46.5k1199190

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 Jan 6 at 4:39

answered Jan 5 at 17:38

nullpointer

46.5k1199190

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 Jan 6 at 4:39

answered Jan 5 at 17:38

nullpointer

46.5k1199190

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 Jan 6 at 4:39

answered Jan 5 at 17:38

nullpointer

46.5k1199190

edited Jan 6 at 4:39

answered Jan 5 at 17:38

nullpointer

46.5k1199190

answered Jan 5 at 17:38

nullpointer

46.5k1199190

answered Jan 5 at 17:38

nullpointer

46.5k1199190

add a comment |

Simply use :-

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

edited Jan 6 at 4:37

answered Jan 5 at 17:34

Nicholas K

6,59261132

add a comment |

Simply use :-

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

edited Jan 6 at 4:37

answered Jan 5 at 17:34

Nicholas K

6,59261132

add a comment |

Simply use :-

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

edited Jan 6 at 4:37

answered Jan 5 at 17:34

Nicholas K

6,59261132

Simply use :-

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

edited Jan 6 at 4:37

answered Jan 5 at 17:34

Nicholas K

6,59261132

edited Jan 6 at 4:37

answered Jan 5 at 17:34

Nicholas K

6,59261132

answered Jan 5 at 17:34

Nicholas K

6,59261132

answered Jan 5 at 17:34

Nicholas K

6,59261132

add a comment |

draft saved

draft discarded

Thanks for contributing an answer to Stack Overflow!

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

Search This Blog

Ufyukyu