Converting a text file to Map<String, List> using lambda
12
I am trying to convert the following text input file:
A=groupA1
A=groupA2
A=groupA3
B=groupB1
B=groupB2
into Map<String, List<String>> by splitting each line on "="
So far I manged to get this sort of output:
KEY: A
VALUE: A=groupA1
VALUE: A=groupA2
VALUE: A=groupA3
KEY: B
VALUE: B=groupB1
VALUE: B=groupB2
using such code:
File reqFile = new File("test.config");
try (Stream<String> stream = Files.lines(reqFile.toPath())) {
Map<String, List<String>> conf = stream.collect(Collectors.groupingBy(s -> s.split("=")[0]));
for (Map.Entry<String, List<String>> entry: conf.entrySet()) {
System.out.println("KEY: " + entry.getKey());
for (String value : entry.getValue()) {
System.out.println("VALUE: " + value);
}
}
} catch (Exception e) {
e.printStackTrace();
}
How to tweak the above lambda to get something like this:
KEY: A
VALUE: groupA1
VALUE: groupA2
VALUE: groupA3
KEY: B
VALUE: groupB1
VALUE: groupB2
java java-8 java-stream
edited Jan 26 at 23:06
Deadpool
5,7632528
asked Jan 26 at 22:43
BartDBartD
635
add a comment |
12
I am trying to convert the following text input file:
A=groupA1
A=groupA2
A=groupA3
B=groupB1
B=groupB2
into Map<String, List<String>> by splitting each line on "="
So far I manged to get this sort of output:
KEY: A
VALUE: A=groupA1
VALUE: A=groupA2
VALUE: A=groupA3
KEY: B
VALUE: B=groupB1
VALUE: B=groupB2
using such code:
File reqFile = new File("test.config");
try (Stream<String> stream = Files.lines(reqFile.toPath())) {
Map<String, List<String>> conf = stream.collect(Collectors.groupingBy(s -> s.split("=")[0]));
for (Map.Entry<String, List<String>> entry: conf.entrySet()) {
System.out.println("KEY: " + entry.getKey());
for (String value : entry.getValue()) {
System.out.println("VALUE: " + value);
}
}
} catch (Exception e) {
e.printStackTrace();
}
How to tweak the above lambda to get something like this:
KEY: A
VALUE: groupA1
VALUE: groupA2
VALUE: groupA3
KEY: B
VALUE: groupB1
VALUE: groupB2
java java-8 java-stream
edited Jan 26 at 23:06
Deadpool
5,7632528
asked Jan 26 at 22:43
BartDBartD
635
2
much like regular expressions, once you decide to solve a problem with lambdas now you have two problems ( if you do not 100% comprehend what you are doing and why you are doing it that way ). Regular non-lambda solution would be far more efficient and more importantly readable and maintainable.
– Jarrod Roberson
Jan 26 at 22:47
add a comment |
12
12
12
3
I am trying to convert the following text input file:
A=groupA1
A=groupA2
A=groupA3
B=groupB1
B=groupB2
into Map<String, List<String>> by splitting each line on "="
So far I manged to get this sort of output:
KEY: A
VALUE: A=groupA1
VALUE: A=groupA2
VALUE: A=groupA3
KEY: B
VALUE: B=groupB1
VALUE: B=groupB2
using such code:
File reqFile = new File("test.config");
try (Stream<String> stream = Files.lines(reqFile.toPath())) {
Map<String, List<String>> conf = stream.collect(Collectors.groupingBy(s -> s.split("=")[0]));
for (Map.Entry<String, List<String>> entry: conf.entrySet()) {
System.out.println("KEY: " + entry.getKey());
for (String value : entry.getValue()) {
System.out.println("VALUE: " + value);
}
}
} catch (Exception e) {
e.printStackTrace();
}
How to tweak the above lambda to get something like this:
KEY: A
VALUE: groupA1
VALUE: groupA2
VALUE: groupA3
KEY: B
VALUE: groupB1
VALUE: groupB2
java java-8 java-stream
edited Jan 26 at 23:06
Deadpool
5,7632528
asked Jan 26 at 22:43
BartDBartD
635
I am trying to convert the following text input file:
A=groupA1
A=groupA2
A=groupA3
B=groupB1
B=groupB2
into Map<String, List<String>> by splitting each line on "="
So far I manged to get this sort of output:
KEY: A
VALUE: A=groupA1
VALUE: A=groupA2
VALUE: A=groupA3
KEY: B
VALUE: B=groupB1
VALUE: B=groupB2
using such code:
File reqFile = new File("test.config");
try (Stream<String> stream = Files.lines(reqFile.toPath())) {
Map<String, List<String>> conf = stream.collect(Collectors.groupingBy(s -> s.split("=")[0]));
for (Map.Entry<String, List<String>> entry: conf.entrySet()) {
System.out.println("KEY: " + entry.getKey());
for (String value : entry.getValue()) {
System.out.println("VALUE: " + value);
}
}
} catch (Exception e) {
e.printStackTrace();
}
How to tweak the above lambda to get something like this:
KEY: A
VALUE: groupA1
VALUE: groupA2
VALUE: groupA3
KEY: B
VALUE: groupB1
VALUE: groupB2
java java-8 java-stream
java java-8 java-stream
edited Jan 26 at 23:06
Deadpool
5,7632528
asked Jan 26 at 22:43
BartDBartD
635
edited Jan 26 at 23:06
Deadpool
5,7632528
asked Jan 26 at 22:43
BartDBartD
635
edited Jan 26 at 23:06
Deadpool
5,7632528
edited Jan 26 at 23:06
Deadpool
5,7632528
edited Jan 26 at 23:06
Deadpool
5,7632528
5,7632528
asked Jan 26 at 22:43
BartDBartD
635
asked Jan 26 at 22:43
BartDBartD
635
asked Jan 26 at 22:43
BartDBartD
635
635
2
much like regular expressions, once you decide to solve a problem with lambdas now you have two problems ( if you do not 100% comprehend what you are doing and why you are doing it that way ). Regular non-lambda solution would be far more efficient and more importantly readable and maintainable.
– Jarrod Roberson
Jan 26 at 22:47
add a comment |
2
much like regular expressions, once you decide to solve a problem with lambdas now you have two problems ( if you do not 100% comprehend what you are doing and why you are doing it that way ). Regular non-lambda solution would be far more efficient and more importantly readable and maintainable.
– Jarrod Roberson
Jan 26 at 22:47
2
2
much like regular expressions, once you decide to solve a problem with lambdas now you have two problems ( if you do not 100% comprehend what you are doing and why you are doing it that way ). Regular non-lambda solution would be far more efficient and more importantly readable and maintainable.
– Jarrod Roberson
Jan 26 at 22:47
much like regular expressions, once you decide to solve a problem with lambdas now you have two problems ( if you do not 100% comprehend what you are doing and why you are doing it that way ). Regular non-lambda solution would be far more efficient and more importantly readable and maintainable.
– Jarrod Roberson
Jan 26 at 22:47
add a comment |
3 Answers
3
active
oldest
votes
9
Map and collect:
Map<String, List<String>> res = lines.stream()
.map(s -> Arrays.asList(s.split("=")))
.collect(HashMap::new,
(map, item) -> map.computeIfAbsent(item.get(0), k -> new ArrayList<>()).add(item.get(1)),
HashMap::putAll);
Or map and group by:
Map<String, List<String>> res = lines.stream()
.map(s -> Arrays.asList(s.split("=")))
.collect(Collectors.groupingBy(s -> s.get(0), Collectors.mapping(v->v.get(1), Collectors.toList())));
Stream.collectdocumentation
answered Jan 26 at 23:15
Michał ZioberMichał Ziober
13.1k967102
No problem. Mapping solves problem with splitting items two times. I have also added a version with grouping by collectors. If you think it is helpful for you, please, upvote it.
– Michał Ziober
Jan 26 at 23:35
add a comment |
5
Use Collectors.mapping while groupingBy, for more information look at this doc-with-example
Map<String, List<String>> conf = stream.
collect(Collectors.groupingBy(s -> s.split("=")[0], Collectors.mapping(v->v.split("=")[1], Collectors.toList())));
System.out.println(conf); //{A=[groupA1, groupA2, groupA3], B=[groupB1, groupB2]}
answered Jan 26 at 23:04
DeadpoolDeadpool
5,7632528
1
That is exactly what I was looking for!Map<String, List<String>> conf = stream.collect(Collectors.groupingBy(s -> s.split("=")[0], Collectors.mapping(v->v.split("=")[1], Collectors.toList())));and the output:KEY: A VALUE: groupA1 VALUE: groupA2 VALUE: groupA3 KEY: B VALUE: groupB1 VALUE: groupB2
– BartD
Jan 26 at 23:12
add a comment |
2
If you are open to using a third-party library, the following will work using Eclipse Collections.
ListMultimap<String, String> strings = stream
.map(s -> s.split("="))
.collect(Collectors2.toListMultimap(a -> a[0], a -> a[1]));
Collectors2.toListMultimap takes a Function to calculate the key and a separate Function to calculate the value. The ListMultimap<K, V> type is equivalent to Map<K, List<V>>.
Note: I am a committer for Eclipse Collections.
answered Jan 27 at 2:17
Donald RaabDonald Raab
4,30112030
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');
});
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%2f54383490%2fconverting-a-text-file-to-mapstring-liststring-using-lambda%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
9
Map and collect:
Map<String, List<String>> res = lines.stream()
.map(s -> Arrays.asList(s.split("=")))
.collect(HashMap::new,
(map, item) -> map.computeIfAbsent(item.get(0), k -> new ArrayList<>()).add(item.get(1)),
HashMap::putAll);
Or map and group by:
Map<String, List<String>> res = lines.stream()
.map(s -> Arrays.asList(s.split("=")))
.collect(Collectors.groupingBy(s -> s.get(0), Collectors.mapping(v->v.get(1), Collectors.toList())));
Stream.collectdocumentation
answered Jan 26 at 23:15
Michał ZioberMichał Ziober
13.1k967102
No problem. Mapping solves problem with splitting items two times. I have also added a version with grouping by collectors. If you think it is helpful for you, please, upvote it.
– Michał Ziober
Jan 26 at 23:35
add a comment |
9
Map and collect:
Map<String, List<String>> res = lines.stream()
.map(s -> Arrays.asList(s.split("=")))
.collect(HashMap::new,
(map, item) -> map.computeIfAbsent(item.get(0), k -> new ArrayList<>()).add(item.get(1)),
HashMap::putAll);
Or map and group by:
Map<String, List<String>> res = lines.stream()
.map(s -> Arrays.asList(s.split("=")))
.collect(Collectors.groupingBy(s -> s.get(0), Collectors.mapping(v->v.get(1), Collectors.toList())));
Stream.collectdocumentation
answered Jan 26 at 23:15
Michał ZioberMichał Ziober
13.1k967102
No problem. Mapping solves problem with splitting items two times. I have also added a version with grouping by collectors. If you think it is helpful for you, please, upvote it.
– Michał Ziober
Jan 26 at 23:35
add a comment |
9
9
9
Map and collect:
Map<String, List<String>> res = lines.stream()
.map(s -> Arrays.asList(s.split("=")))
.collect(HashMap::new,
(map, item) -> map.computeIfAbsent(item.get(0), k -> new ArrayList<>()).add(item.get(1)),
HashMap::putAll);
Or map and group by:
Map<String, List<String>> res = lines.stream()
.map(s -> Arrays.asList(s.split("=")))
.collect(Collectors.groupingBy(s -> s.get(0), Collectors.mapping(v->v.get(1), Collectors.toList())));
Stream.collectdocumentation
answered Jan 26 at 23:15
Michał ZioberMichał Ziober
13.1k967102
Map and collect:
Map<String, List<String>> res = lines.stream()
.map(s -> Arrays.asList(s.split("=")))
.collect(HashMap::new,
(map, item) -> map.computeIfAbsent(item.get(0), k -> new ArrayList<>()).add(item.get(1)),
HashMap::putAll);
Or map and group by:
Map<String, List<String>> res = lines.stream()
.map(s -> Arrays.asList(s.split("=")))
.collect(Collectors.groupingBy(s -> s.get(0), Collectors.mapping(v->v.get(1), Collectors.toList())));
Stream.collectdocumentation
answered Jan 26 at 23:15
Michał ZioberMichał Ziober
13.1k967102
edited Jan 26 at 23:34
answered Jan 26 at 23:15
Michał ZioberMichał Ziober
13.1k967102
answered Jan 26 at 23:15
Michał ZioberMichał Ziober
13.1k967102
answered Jan 26 at 23:15
Michał ZioberMichał Ziober
13.1k967102
13.1k967102
No problem. Mapping solves problem with splitting items two times. I have also added a version with grouping by collectors. If you think it is helpful for you, please, upvote it.
– Michał Ziober
Jan 26 at 23:35
add a comment |
No problem. Mapping solves problem with splitting items two times. I have also added a version with grouping by collectors. If you think it is helpful for you, please, upvote it.
– Michał Ziober
Jan 26 at 23:35
No problem. Mapping solves problem with splitting items two times. I have also added a version with grouping by collectors. If you think it is helpful for you, please, upvote it.
– Michał Ziober
Jan 26 at 23:35
No problem. Mapping solves problem with splitting items two times. I have also added a version with grouping by collectors. If you think it is helpful for you, please, upvote it.
– Michał Ziober
Jan 26 at 23:35
add a comment |
5
Use Collectors.mapping while groupingBy, for more information look at this doc-with-example
Map<String, List<String>> conf = stream.
collect(Collectors.groupingBy(s -> s.split("=")[0], Collectors.mapping(v->v.split("=")[1], Collectors.toList())));
System.out.println(conf); //{A=[groupA1, groupA2, groupA3], B=[groupB1, groupB2]}
answered Jan 26 at 23:04
DeadpoolDeadpool
5,7632528
1
That is exactly what I was looking for!Map<String, List<String>> conf = stream.collect(Collectors.groupingBy(s -> s.split("=")[0], Collectors.mapping(v->v.split("=")[1], Collectors.toList())));and the output:KEY: A VALUE: groupA1 VALUE: groupA2 VALUE: groupA3 KEY: B VALUE: groupB1 VALUE: groupB2
– BartD
Jan 26 at 23:12
add a comment |
5
Use Collectors.mapping while groupingBy, for more information look at this doc-with-example
Map<String, List<String>> conf = stream.
collect(Collectors.groupingBy(s -> s.split("=")[0], Collectors.mapping(v->v.split("=")[1], Collectors.toList())));
System.out.println(conf); //{A=[groupA1, groupA2, groupA3], B=[groupB1, groupB2]}
answered Jan 26 at 23:04
DeadpoolDeadpool
5,7632528
1
That is exactly what I was looking for!Map<String, List<String>> conf = stream.collect(Collectors.groupingBy(s -> s.split("=")[0], Collectors.mapping(v->v.split("=")[1], Collectors.toList())));and the output:KEY: A VALUE: groupA1 VALUE: groupA2 VALUE: groupA3 KEY: B VALUE: groupB1 VALUE: groupB2
– BartD
Jan 26 at 23:12
add a comment |
5
5
5
Use Collectors.mapping while groupingBy, for more information look at this doc-with-example
Map<String, List<String>> conf = stream.
collect(Collectors.groupingBy(s -> s.split("=")[0], Collectors.mapping(v->v.split("=")[1], Collectors.toList())));
System.out.println(conf); //{A=[groupA1, groupA2, groupA3], B=[groupB1, groupB2]}
answered Jan 26 at 23:04
DeadpoolDeadpool
5,7632528
Use Collectors.mapping while groupingBy, for more information look at this doc-with-example
Map<String, List<String>> conf = stream.
collect(Collectors.groupingBy(s -> s.split("=")[0], Collectors.mapping(v->v.split("=")[1], Collectors.toList())));
System.out.println(conf); //{A=[groupA1, groupA2, groupA3], B=[groupB1, groupB2]}
answered Jan 26 at 23:04
DeadpoolDeadpool
5,7632528
edited Jan 26 at 23:12
answered Jan 26 at 23:04
DeadpoolDeadpool
5,7632528
answered Jan 26 at 23:04
DeadpoolDeadpool
5,7632528
answered Jan 26 at 23:04
DeadpoolDeadpool
5,7632528
5,7632528
1
That is exactly what I was looking for!Map<String, List<String>> conf = stream.collect(Collectors.groupingBy(s -> s.split("=")[0], Collectors.mapping(v->v.split("=")[1], Collectors.toList())));and the output:KEY: A VALUE: groupA1 VALUE: groupA2 VALUE: groupA3 KEY: B VALUE: groupB1 VALUE: groupB2
– BartD
Jan 26 at 23:12
add a comment |
1
That is exactly what I was looking for!Map<String, List<String>> conf = stream.collect(Collectors.groupingBy(s -> s.split("=")[0], Collectors.mapping(v->v.split("=")[1], Collectors.toList())));and the output:KEY: A VALUE: groupA1 VALUE: groupA2 VALUE: groupA3 KEY: B VALUE: groupB1 VALUE: groupB2
– BartD
Jan 26 at 23:12
1
1
That is exactly what I was looking for!
Map<String, List<String>> conf = stream.collect(Collectors.groupingBy(s -> s.split("=")[0], Collectors.mapping(v->v.split("=")[1], Collectors.toList()))); and the output: KEY: A VALUE: groupA1 VALUE: groupA2 VALUE: groupA3 KEY: B VALUE: groupB1 VALUE: groupB2– BartD
Jan 26 at 23:12
That is exactly what I was looking for!
Map<String, List<String>> conf = stream.collect(Collectors.groupingBy(s -> s.split("=")[0], Collectors.mapping(v->v.split("=")[1], Collectors.toList()))); and the output: KEY: A VALUE: groupA1 VALUE: groupA2 VALUE: groupA3 KEY: B VALUE: groupB1 VALUE: groupB2– BartD
Jan 26 at 23:12
add a comment |
2
If you are open to using a third-party library, the following will work using Eclipse Collections.
ListMultimap<String, String> strings = stream
.map(s -> s.split("="))
.collect(Collectors2.toListMultimap(a -> a[0], a -> a[1]));
Collectors2.toListMultimap takes a Function to calculate the key and a separate Function to calculate the value. The ListMultimap<K, V> type is equivalent to Map<K, List<V>>.
Note: I am a committer for Eclipse Collections.
answered Jan 27 at 2:17
Donald RaabDonald Raab
4,30112030
add a comment |
2
If you are open to using a third-party library, the following will work using Eclipse Collections.
ListMultimap<String, String> strings = stream
.map(s -> s.split("="))
.collect(Collectors2.toListMultimap(a -> a[0], a -> a[1]));
Collectors2.toListMultimap takes a Function to calculate the key and a separate Function to calculate the value. The ListMultimap<K, V> type is equivalent to Map<K, List<V>>.
Note: I am a committer for Eclipse Collections.
answered Jan 27 at 2:17
Donald RaabDonald Raab
4,30112030
add a comment |
2
2
2
If you are open to using a third-party library, the following will work using Eclipse Collections.
ListMultimap<String, String> strings = stream
.map(s -> s.split("="))
.collect(Collectors2.toListMultimap(a -> a[0], a -> a[1]));
Collectors2.toListMultimap takes a Function to calculate the key and a separate Function to calculate the value. The ListMultimap<K, V> type is equivalent to Map<K, List<V>>.
Note: I am a committer for Eclipse Collections.
answered Jan 27 at 2:17
Donald RaabDonald Raab
4,30112030
If you are open to using a third-party library, the following will work using Eclipse Collections.
ListMultimap<String, String> strings = stream
.map(s -> s.split("="))
.collect(Collectors2.toListMultimap(a -> a[0], a -> a[1]));
Collectors2.toListMultimap takes a Function to calculate the key and a separate Function to calculate the value. The ListMultimap<K, V> type is equivalent to Map<K, List<V>>.
Note: I am a committer for Eclipse Collections.
answered Jan 27 at 2:17
Donald RaabDonald Raab
4,30112030
answered Jan 27 at 2:17
Donald RaabDonald Raab
4,30112030
answered Jan 27 at 2:17
Donald RaabDonald Raab
4,30112030
answered Jan 27 at 2:17
Donald RaabDonald Raab
4,30112030
4,30112030
add a comment |
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');
});
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%2f54383490%2fconverting-a-text-file-to-mapstring-liststring-using-lambda%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
2
much like regular expressions, once you decide to solve a problem with lambdas now you have two problems ( if you do not 100% comprehend what you are doing and why you are doing it that way ). Regular non-lambda solution would be far more efficient and more importantly readable and maintainable.
– Jarrod Roberson
Jan 26 at 22:47