sort map groupingBy counting
Is there way to sort map by value inside this collector without creation a new stream?
Now it prints: AAS : 5 ABA : 195 ABC : 12 ABE : 52.
Desired: ABA : 195 ABE : 52 ...
getTrigraphStream(path) returns: HOW THE TRA GRI ONC GRI ONC INS WHE INS WHE
public Map<String, Long> getTrigraphStatisticsMapAlphabet(String path) {
return getTrigraphStream(path)
.collect(groupingByConcurrent(Function.identity(),
ConcurrentSkipListMap::new, counting()));
}
sorting java-8 java-stream counting collectors
add a comment |
Is there way to sort map by value inside this collector without creation a new stream?
Now it prints: AAS : 5 ABA : 195 ABC : 12 ABE : 52.
Desired: ABA : 195 ABE : 52 ...
getTrigraphStream(path) returns: HOW THE TRA GRI ONC GRI ONC INS WHE INS WHE
public Map<String, Long> getTrigraphStatisticsMapAlphabet(String path) {
return getTrigraphStream(path)
.collect(groupingByConcurrent(Function.identity(),
ConcurrentSkipListMap::new, counting()));
}
sorting java-8 java-stream counting collectors
add a comment |
Is there way to sort map by value inside this collector without creation a new stream?
Now it prints: AAS : 5 ABA : 195 ABC : 12 ABE : 52.
Desired: ABA : 195 ABE : 52 ...
getTrigraphStream(path) returns: HOW THE TRA GRI ONC GRI ONC INS WHE INS WHE
public Map<String, Long> getTrigraphStatisticsMapAlphabet(String path) {
return getTrigraphStream(path)
.collect(groupingByConcurrent(Function.identity(),
ConcurrentSkipListMap::new, counting()));
}
sorting java-8 java-stream counting collectors
Is there way to sort map by value inside this collector without creation a new stream?
Now it prints: AAS : 5 ABA : 195 ABC : 12 ABE : 52.
Desired: ABA : 195 ABE : 52 ...
getTrigraphStream(path) returns: HOW THE TRA GRI ONC GRI ONC INS WHE INS WHE
public Map<String, Long> getTrigraphStatisticsMapAlphabet(String path) {
return getTrigraphStream(path)
.collect(groupingByConcurrent(Function.identity(),
ConcurrentSkipListMap::new, counting()));
}
sorting java-8 java-stream counting collectors
sorting java-8 java-stream counting collectors
edited Nov 20 '18 at 19:58
Stefan Zobel
2,44031828
2,44031828
asked Nov 20 '18 at 19:12
a_chubenkoa_chubenko
4615
4615
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Is there way to sort map by value inside this collector without
creation a new stream?
The answer is No. You can only sort by the value of the grouping after you've finished grouping.
Options:
- You'll need to invoke the
stream()
method after grouping and sort by
value then collect to a map. - use
collectingAndThen
with a finishing function to do the sorting
and then collect to map.
i.e.
return getTrigraphStream(path)
.collect(collectingAndThen(groupingByConcurrent(Function.identity(), counting()),
map -> map.entrySet().stream().sorted(....).collect(...)));
By the way, what's the problem with creating a new stream? it's a cheap operation.
or use aSortedMap
maybe?
– nullpointer
Nov 21 '18 at 0:37
1
In this operation, there is no need to use aConcurrentSkipListMap
.
– Holger
Nov 21 '18 at 8:03
@nullpointer that sorts the map by keys as opposed to values I believe.
– Aomine
Nov 21 '18 at 9:06
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53399973%2fsort-map-groupingby-counting%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Is there way to sort map by value inside this collector without
creation a new stream?
The answer is No. You can only sort by the value of the grouping after you've finished grouping.
Options:
- You'll need to invoke the
stream()
method after grouping and sort by
value then collect to a map. - use
collectingAndThen
with a finishing function to do the sorting
and then collect to map.
i.e.
return getTrigraphStream(path)
.collect(collectingAndThen(groupingByConcurrent(Function.identity(), counting()),
map -> map.entrySet().stream().sorted(....).collect(...)));
By the way, what's the problem with creating a new stream? it's a cheap operation.
or use aSortedMap
maybe?
– nullpointer
Nov 21 '18 at 0:37
1
In this operation, there is no need to use aConcurrentSkipListMap
.
– Holger
Nov 21 '18 at 8:03
@nullpointer that sorts the map by keys as opposed to values I believe.
– Aomine
Nov 21 '18 at 9:06
add a comment |
Is there way to sort map by value inside this collector without
creation a new stream?
The answer is No. You can only sort by the value of the grouping after you've finished grouping.
Options:
- You'll need to invoke the
stream()
method after grouping and sort by
value then collect to a map. - use
collectingAndThen
with a finishing function to do the sorting
and then collect to map.
i.e.
return getTrigraphStream(path)
.collect(collectingAndThen(groupingByConcurrent(Function.identity(), counting()),
map -> map.entrySet().stream().sorted(....).collect(...)));
By the way, what's the problem with creating a new stream? it's a cheap operation.
or use aSortedMap
maybe?
– nullpointer
Nov 21 '18 at 0:37
1
In this operation, there is no need to use aConcurrentSkipListMap
.
– Holger
Nov 21 '18 at 8:03
@nullpointer that sorts the map by keys as opposed to values I believe.
– Aomine
Nov 21 '18 at 9:06
add a comment |
Is there way to sort map by value inside this collector without
creation a new stream?
The answer is No. You can only sort by the value of the grouping after you've finished grouping.
Options:
- You'll need to invoke the
stream()
method after grouping and sort by
value then collect to a map. - use
collectingAndThen
with a finishing function to do the sorting
and then collect to map.
i.e.
return getTrigraphStream(path)
.collect(collectingAndThen(groupingByConcurrent(Function.identity(), counting()),
map -> map.entrySet().stream().sorted(....).collect(...)));
By the way, what's the problem with creating a new stream? it's a cheap operation.
Is there way to sort map by value inside this collector without
creation a new stream?
The answer is No. You can only sort by the value of the grouping after you've finished grouping.
Options:
- You'll need to invoke the
stream()
method after grouping and sort by
value then collect to a map. - use
collectingAndThen
with a finishing function to do the sorting
and then collect to map.
i.e.
return getTrigraphStream(path)
.collect(collectingAndThen(groupingByConcurrent(Function.identity(), counting()),
map -> map.entrySet().stream().sorted(....).collect(...)));
By the way, what's the problem with creating a new stream? it's a cheap operation.
edited Nov 21 '18 at 8:59
answered Nov 20 '18 at 20:35


AomineAomine
42k74071
42k74071
or use aSortedMap
maybe?
– nullpointer
Nov 21 '18 at 0:37
1
In this operation, there is no need to use aConcurrentSkipListMap
.
– Holger
Nov 21 '18 at 8:03
@nullpointer that sorts the map by keys as opposed to values I believe.
– Aomine
Nov 21 '18 at 9:06
add a comment |
or use aSortedMap
maybe?
– nullpointer
Nov 21 '18 at 0:37
1
In this operation, there is no need to use aConcurrentSkipListMap
.
– Holger
Nov 21 '18 at 8:03
@nullpointer that sorts the map by keys as opposed to values I believe.
– Aomine
Nov 21 '18 at 9:06
or use a
SortedMap
maybe?– nullpointer
Nov 21 '18 at 0:37
or use a
SortedMap
maybe?– nullpointer
Nov 21 '18 at 0:37
1
1
In this operation, there is no need to use a
ConcurrentSkipListMap
.– Holger
Nov 21 '18 at 8:03
In this operation, there is no need to use a
ConcurrentSkipListMap
.– Holger
Nov 21 '18 at 8:03
@nullpointer that sorts the map by keys as opposed to values I believe.
– Aomine
Nov 21 '18 at 9:06
@nullpointer that sorts the map by keys as opposed to values I believe.
– Aomine
Nov 21 '18 at 9:06
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53399973%2fsort-map-groupingby-counting%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown