How map stream to immutable java class?
Consider a piece of code:
imnport reactor.util.context.Context
public Context addAll (Context ctx, Map.Entry<String, Object> hashMap) {
Context ctxVar = ctx;
for (Map.Entry<String, Object> e : hashMap.entrySet()) {
if (e.getValue() != null) {
ctxVar = ctxVar.put(e.getKey(), e.getValue());
}
}
return ctxVar;
}
reactor.util.context.Context is immutable class. So put merges old context with new added value and returns new
context.
The question is - is there more compact way to "combine" HashMap into immutable object using java 8 streams? (Not for Context class only)
Note: I have read about java stream collect and it seems that does not work because I have to supply initial Context
and combine several contexts after map but recreate entire context for combine operations I think is too much.
java java-8 java-stream
add a comment |
Consider a piece of code:
imnport reactor.util.context.Context
public Context addAll (Context ctx, Map.Entry<String, Object> hashMap) {
Context ctxVar = ctx;
for (Map.Entry<String, Object> e : hashMap.entrySet()) {
if (e.getValue() != null) {
ctxVar = ctxVar.put(e.getKey(), e.getValue());
}
}
return ctxVar;
}
reactor.util.context.Context is immutable class. So put merges old context with new added value and returns new
context.
The question is - is there more compact way to "combine" HashMap into immutable object using java 8 streams? (Not for Context class only)
Note: I have read about java stream collect and it seems that does not work because I have to supply initial Context
and combine several contexts after map but recreate entire context for combine operations I think is too much.
java java-8 java-stream
Shouldn'tMap.Entry<String, Object> hashMapbeMap<String, Object> hashMap? You use that variable as if it's aMap.
– Eran
Nov 22 '18 at 7:46
add a comment |
Consider a piece of code:
imnport reactor.util.context.Context
public Context addAll (Context ctx, Map.Entry<String, Object> hashMap) {
Context ctxVar = ctx;
for (Map.Entry<String, Object> e : hashMap.entrySet()) {
if (e.getValue() != null) {
ctxVar = ctxVar.put(e.getKey(), e.getValue());
}
}
return ctxVar;
}
reactor.util.context.Context is immutable class. So put merges old context with new added value and returns new
context.
The question is - is there more compact way to "combine" HashMap into immutable object using java 8 streams? (Not for Context class only)
Note: I have read about java stream collect and it seems that does not work because I have to supply initial Context
and combine several contexts after map but recreate entire context for combine operations I think is too much.
java java-8 java-stream
Consider a piece of code:
imnport reactor.util.context.Context
public Context addAll (Context ctx, Map.Entry<String, Object> hashMap) {
Context ctxVar = ctx;
for (Map.Entry<String, Object> e : hashMap.entrySet()) {
if (e.getValue() != null) {
ctxVar = ctxVar.put(e.getKey(), e.getValue());
}
}
return ctxVar;
}
reactor.util.context.Context is immutable class. So put merges old context with new added value and returns new
context.
The question is - is there more compact way to "combine" HashMap into immutable object using java 8 streams? (Not for Context class only)
Note: I have read about java stream collect and it seems that does not work because I have to supply initial Context
and combine several contexts after map but recreate entire context for combine operations I think is too much.
java java-8 java-stream
java java-8 java-stream
asked Nov 22 '18 at 7:27
CherryCherry
9,4292597178
9,4292597178
Shouldn'tMap.Entry<String, Object> hashMapbeMap<String, Object> hashMap? You use that variable as if it's aMap.
– Eran
Nov 22 '18 at 7:46
add a comment |
Shouldn'tMap.Entry<String, Object> hashMapbeMap<String, Object> hashMap? You use that variable as if it's aMap.
– Eran
Nov 22 '18 at 7:46
Shouldn't
Map.Entry<String, Object> hashMap be Map<String, Object> hashMap? You use that variable as if it's a Map.– Eran
Nov 22 '18 at 7:46
Shouldn't
Map.Entry<String, Object> hashMap be Map<String, Object> hashMap? You use that variable as if it's a Map.– Eran
Nov 22 '18 at 7:46
add a comment |
2 Answers
2
active
oldest
votes
You can use reduce:
Context ctxVar = hashMap.entrySet()
.stream()
.filter(e -> e.getValue() != null)
.reduce(ctx,
(c, e) -> c.put(e.getKey(), e.getValue()),
(c1, c2) -> c1.putAll(c2));
It does seem wasteful, though (in the same way your original loop is wasteful), since it creates multiple Context instances when only the last one is needed.
It would make more sense if you write a static method of the Context class (or a constructor) that accepts a Map and only creates a single Context instance for the entries of the Map. However, I now noticed that you didn't write this Context class, so you can't change it.
It's a bit long winded but... you could implement a mutable version ofContextwhich accumulates (key, value) pairs, build it up while streaming the map and then hand it over to the provided context withcontext.addAll( myMutableContext)
– David Soroko
Nov 22 '18 at 15:52
@DavidSoroko Your approach will iteratively create a newContextfor each entry fromContext(containing previous entries) provided as a parameter toContext.putAll.
– ETO
Nov 23 '18 at 1:58
add a comment |
Your question interested me. I researched the subject.
Here is what I came up with:
package reactor.util.context;
import java.util.Map;
import java.util.Map.Entry;
import static java.util.stream.Collectors.toMap;
public class ContextUtils {
public static Context putAll(Context context, Map map) {
if (map.isEmpty()) {
return context;
} else {
Map contextMap = context.stream()
.filter(e -> e.getValue() != null)
.collect(toMap(Entry::getKey, Entry::getValue));
return new ContextN(contextMap, map);
}
}
}
This utility method creates a new Context containing the entries from both initial context and provided map.
This solution may still look not optimal, as it creates a new HashMap with initiaCapacity = 0 and loadFactor = 0.75f. In fact, this is not a problem, since ContextN itself is a subclass of HashMap with precise capacity and loadFactor = 1. The data from intermediary map will be copied to the context and later it will be collected by GC.
Note: the utility class has to be in package reactor.util.context. ContextN is package-private so it is not accessible from other packages.
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%2f53425826%2fhow-map-stream-to-immutable-java-class%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can use reduce:
Context ctxVar = hashMap.entrySet()
.stream()
.filter(e -> e.getValue() != null)
.reduce(ctx,
(c, e) -> c.put(e.getKey(), e.getValue()),
(c1, c2) -> c1.putAll(c2));
It does seem wasteful, though (in the same way your original loop is wasteful), since it creates multiple Context instances when only the last one is needed.
It would make more sense if you write a static method of the Context class (or a constructor) that accepts a Map and only creates a single Context instance for the entries of the Map. However, I now noticed that you didn't write this Context class, so you can't change it.
It's a bit long winded but... you could implement a mutable version ofContextwhich accumulates (key, value) pairs, build it up while streaming the map and then hand it over to the provided context withcontext.addAll( myMutableContext)
– David Soroko
Nov 22 '18 at 15:52
@DavidSoroko Your approach will iteratively create a newContextfor each entry fromContext(containing previous entries) provided as a parameter toContext.putAll.
– ETO
Nov 23 '18 at 1:58
add a comment |
You can use reduce:
Context ctxVar = hashMap.entrySet()
.stream()
.filter(e -> e.getValue() != null)
.reduce(ctx,
(c, e) -> c.put(e.getKey(), e.getValue()),
(c1, c2) -> c1.putAll(c2));
It does seem wasteful, though (in the same way your original loop is wasteful), since it creates multiple Context instances when only the last one is needed.
It would make more sense if you write a static method of the Context class (or a constructor) that accepts a Map and only creates a single Context instance for the entries of the Map. However, I now noticed that you didn't write this Context class, so you can't change it.
It's a bit long winded but... you could implement a mutable version ofContextwhich accumulates (key, value) pairs, build it up while streaming the map and then hand it over to the provided context withcontext.addAll( myMutableContext)
– David Soroko
Nov 22 '18 at 15:52
@DavidSoroko Your approach will iteratively create a newContextfor each entry fromContext(containing previous entries) provided as a parameter toContext.putAll.
– ETO
Nov 23 '18 at 1:58
add a comment |
You can use reduce:
Context ctxVar = hashMap.entrySet()
.stream()
.filter(e -> e.getValue() != null)
.reduce(ctx,
(c, e) -> c.put(e.getKey(), e.getValue()),
(c1, c2) -> c1.putAll(c2));
It does seem wasteful, though (in the same way your original loop is wasteful), since it creates multiple Context instances when only the last one is needed.
It would make more sense if you write a static method of the Context class (or a constructor) that accepts a Map and only creates a single Context instance for the entries of the Map. However, I now noticed that you didn't write this Context class, so you can't change it.
You can use reduce:
Context ctxVar = hashMap.entrySet()
.stream()
.filter(e -> e.getValue() != null)
.reduce(ctx,
(c, e) -> c.put(e.getKey(), e.getValue()),
(c1, c2) -> c1.putAll(c2));
It does seem wasteful, though (in the same way your original loop is wasteful), since it creates multiple Context instances when only the last one is needed.
It would make more sense if you write a static method of the Context class (or a constructor) that accepts a Map and only creates a single Context instance for the entries of the Map. However, I now noticed that you didn't write this Context class, so you can't change it.
edited Nov 22 '18 at 7:41
answered Nov 22 '18 at 7:35
EranEran
289k37473561
289k37473561
It's a bit long winded but... you could implement a mutable version ofContextwhich accumulates (key, value) pairs, build it up while streaming the map and then hand it over to the provided context withcontext.addAll( myMutableContext)
– David Soroko
Nov 22 '18 at 15:52
@DavidSoroko Your approach will iteratively create a newContextfor each entry fromContext(containing previous entries) provided as a parameter toContext.putAll.
– ETO
Nov 23 '18 at 1:58
add a comment |
It's a bit long winded but... you could implement a mutable version ofContextwhich accumulates (key, value) pairs, build it up while streaming the map and then hand it over to the provided context withcontext.addAll( myMutableContext)
– David Soroko
Nov 22 '18 at 15:52
@DavidSoroko Your approach will iteratively create a newContextfor each entry fromContext(containing previous entries) provided as a parameter toContext.putAll.
– ETO
Nov 23 '18 at 1:58
It's a bit long winded but... you could implement a mutable version of
Context which accumulates (key, value) pairs, build it up while streaming the map and then hand it over to the provided context with context.addAll( myMutableContext)– David Soroko
Nov 22 '18 at 15:52
It's a bit long winded but... you could implement a mutable version of
Context which accumulates (key, value) pairs, build it up while streaming the map and then hand it over to the provided context with context.addAll( myMutableContext)– David Soroko
Nov 22 '18 at 15:52
@DavidSoroko Your approach will iteratively create a new
Context for each entry from Context(containing previous entries) provided as a parameter to Context.putAll.– ETO
Nov 23 '18 at 1:58
@DavidSoroko Your approach will iteratively create a new
Context for each entry from Context(containing previous entries) provided as a parameter to Context.putAll.– ETO
Nov 23 '18 at 1:58
add a comment |
Your question interested me. I researched the subject.
Here is what I came up with:
package reactor.util.context;
import java.util.Map;
import java.util.Map.Entry;
import static java.util.stream.Collectors.toMap;
public class ContextUtils {
public static Context putAll(Context context, Map map) {
if (map.isEmpty()) {
return context;
} else {
Map contextMap = context.stream()
.filter(e -> e.getValue() != null)
.collect(toMap(Entry::getKey, Entry::getValue));
return new ContextN(contextMap, map);
}
}
}
This utility method creates a new Context containing the entries from both initial context and provided map.
This solution may still look not optimal, as it creates a new HashMap with initiaCapacity = 0 and loadFactor = 0.75f. In fact, this is not a problem, since ContextN itself is a subclass of HashMap with precise capacity and loadFactor = 1. The data from intermediary map will be copied to the context and later it will be collected by GC.
Note: the utility class has to be in package reactor.util.context. ContextN is package-private so it is not accessible from other packages.
add a comment |
Your question interested me. I researched the subject.
Here is what I came up with:
package reactor.util.context;
import java.util.Map;
import java.util.Map.Entry;
import static java.util.stream.Collectors.toMap;
public class ContextUtils {
public static Context putAll(Context context, Map map) {
if (map.isEmpty()) {
return context;
} else {
Map contextMap = context.stream()
.filter(e -> e.getValue() != null)
.collect(toMap(Entry::getKey, Entry::getValue));
return new ContextN(contextMap, map);
}
}
}
This utility method creates a new Context containing the entries from both initial context and provided map.
This solution may still look not optimal, as it creates a new HashMap with initiaCapacity = 0 and loadFactor = 0.75f. In fact, this is not a problem, since ContextN itself is a subclass of HashMap with precise capacity and loadFactor = 1. The data from intermediary map will be copied to the context and later it will be collected by GC.
Note: the utility class has to be in package reactor.util.context. ContextN is package-private so it is not accessible from other packages.
add a comment |
Your question interested me. I researched the subject.
Here is what I came up with:
package reactor.util.context;
import java.util.Map;
import java.util.Map.Entry;
import static java.util.stream.Collectors.toMap;
public class ContextUtils {
public static Context putAll(Context context, Map map) {
if (map.isEmpty()) {
return context;
} else {
Map contextMap = context.stream()
.filter(e -> e.getValue() != null)
.collect(toMap(Entry::getKey, Entry::getValue));
return new ContextN(contextMap, map);
}
}
}
This utility method creates a new Context containing the entries from both initial context and provided map.
This solution may still look not optimal, as it creates a new HashMap with initiaCapacity = 0 and loadFactor = 0.75f. In fact, this is not a problem, since ContextN itself is a subclass of HashMap with precise capacity and loadFactor = 1. The data from intermediary map will be copied to the context and later it will be collected by GC.
Note: the utility class has to be in package reactor.util.context. ContextN is package-private so it is not accessible from other packages.
Your question interested me. I researched the subject.
Here is what I came up with:
package reactor.util.context;
import java.util.Map;
import java.util.Map.Entry;
import static java.util.stream.Collectors.toMap;
public class ContextUtils {
public static Context putAll(Context context, Map map) {
if (map.isEmpty()) {
return context;
} else {
Map contextMap = context.stream()
.filter(e -> e.getValue() != null)
.collect(toMap(Entry::getKey, Entry::getValue));
return new ContextN(contextMap, map);
}
}
}
This utility method creates a new Context containing the entries from both initial context and provided map.
This solution may still look not optimal, as it creates a new HashMap with initiaCapacity = 0 and loadFactor = 0.75f. In fact, this is not a problem, since ContextN itself is a subclass of HashMap with precise capacity and loadFactor = 1. The data from intermediary map will be copied to the context and later it will be collected by GC.
Note: the utility class has to be in package reactor.util.context. ContextN is package-private so it is not accessible from other packages.
edited Nov 23 '18 at 2:01
answered Nov 22 '18 at 22:49
ETOETO
2,7161629
2,7161629
add a comment |
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%2f53425826%2fhow-map-stream-to-immutable-java-class%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
Shouldn't
Map.Entry<String, Object> hashMapbeMap<String, Object> hashMap? You use that variable as if it's aMap.– Eran
Nov 22 '18 at 7:46