How to convert java.util.List[java.lang.Double] to Scala's List[Double]?
I'd like to convert a Java list of Java doubles (java.util.List[java.lang.Double]) to a Scala list of Scala doubles (List[Double]) in an efficient way.
Currently I'm mapping over the list converting each Double value to a Scala Double. I'd prefer not to map over every value and am looking for a more efficient way of doing this.
import collection.JavaConversions._
import collection.mutable.Buffer
val j: java.util.List[java.lang.Double] = Buffer(new java.lang.Double(1.0), new java.lang.Double(2.0))
val s: List[Double] = ...
I've seen documentation to go from Scala --> Java, but not much going the other way.
java scala implicit-conversion
edited Dec 1 '15 at 9:56
Jacek Laskowski
43.2k17126256
asked May 23 '14 at 5:30
Erik Shilts
2,75321940
add a comment |
I'd like to convert a Java list of Java doubles (java.util.List[java.lang.Double]) to a Scala list of Scala doubles (List[Double]) in an efficient way.
Currently I'm mapping over the list converting each Double value to a Scala Double. I'd prefer not to map over every value and am looking for a more efficient way of doing this.
import collection.JavaConversions._
import collection.mutable.Buffer
val j: java.util.List[java.lang.Double] = Buffer(new java.lang.Double(1.0), new java.lang.Double(2.0))
val s: List[Double] = ...
I've seen documentation to go from Scala --> Java, but not much going the other way.
java scala implicit-conversion
edited Dec 1 '15 at 9:56
Jacek Laskowski
43.2k17126256
asked May 23 '14 at 5:30
Erik Shilts
2,75321940
"I've seen documentation to go from Scala --> Java, but not much going the other way."... did you now? Care to share that information? ;-)
– Make42
Oct 19 '17 at 12:12
I found out the answer: stackoverflow.com/q/46830217/4533188
– Make42
Oct 19 '17 at 12:48
add a comment |
I'd like to convert a Java list of Java doubles (java.util.List[java.lang.Double]) to a Scala list of Scala doubles (List[Double]) in an efficient way.
Currently I'm mapping over the list converting each Double value to a Scala Double. I'd prefer not to map over every value and am looking for a more efficient way of doing this.
import collection.JavaConversions._
import collection.mutable.Buffer
val j: java.util.List[java.lang.Double] = Buffer(new java.lang.Double(1.0), new java.lang.Double(2.0))
val s: List[Double] = ...
I've seen documentation to go from Scala --> Java, but not much going the other way.
java scala implicit-conversion
edited Dec 1 '15 at 9:56
Jacek Laskowski
43.2k17126256
asked May 23 '14 at 5:30
Erik Shilts
2,75321940
I'd like to convert a Java list of Java doubles (java.util.List[java.lang.Double]) to a Scala list of Scala doubles (List[Double]) in an efficient way.
Currently I'm mapping over the list converting each Double value to a Scala Double. I'd prefer not to map over every value and am looking for a more efficient way of doing this.
import collection.JavaConversions._
import collection.mutable.Buffer
val j: java.util.List[java.lang.Double] = Buffer(new java.lang.Double(1.0), new java.lang.Double(2.0))
val s: List[Double] = ...
I've seen documentation to go from Scala --> Java, but not much going the other way.
java scala implicit-conversion
java scala implicit-conversion
edited Dec 1 '15 at 9:56
Jacek Laskowski
43.2k17126256
asked May 23 '14 at 5:30
Erik Shilts
2,75321940
edited Dec 1 '15 at 9:56
Jacek Laskowski
43.2k17126256
asked May 23 '14 at 5:30
Erik Shilts
2,75321940
edited Dec 1 '15 at 9:56
Jacek Laskowski
43.2k17126256
edited Dec 1 '15 at 9:56
Jacek Laskowski
43.2k17126256
edited Dec 1 '15 at 9:56
Jacek Laskowski
43.2k17126256
43.2k17126256
asked May 23 '14 at 5:30
Erik Shilts
2,75321940
asked May 23 '14 at 5:30
Erik Shilts
2,75321940
asked May 23 '14 at 5:30
Erik Shilts
2,75321940
2,75321940
"I've seen documentation to go from Scala --> Java, but not much going the other way."... did you now? Care to share that information? ;-)
– Make42
Oct 19 '17 at 12:12
I found out the answer: stackoverflow.com/q/46830217/4533188
– Make42
Oct 19 '17 at 12:48
add a comment |
"I've seen documentation to go from Scala --> Java, but not much going the other way."... did you now? Care to share that information? ;-)
– Make42
Oct 19 '17 at 12:12
I found out the answer: stackoverflow.com/q/46830217/4533188
– Make42
Oct 19 '17 at 12:48
"I've seen documentation to go from Scala --> Java, but not much going the other way."... did you now? Care to share that information? ;-)
– Make42
Oct 19 '17 at 12:12
"I've seen documentation to go from Scala --> Java, but not much going the other way."... did you now? Care to share that information? ;-)
– Make42
Oct 19 '17 at 12:12
I found out the answer: stackoverflow.com/q/46830217/4533188
– Make42
Oct 19 '17 at 12:48
I found out the answer: stackoverflow.com/q/46830217/4533188
– Make42
Oct 19 '17 at 12:48
add a comment |
2 Answers
2
active
oldest
votes
The recommendation is to use JavaConverters, not JavaConversions:
import collection.JavaConverters._
import scala.collection.breakOut
val l: List[Double] = j.asScala.map(_.doubleValue)(breakOut)
doubleValue will convert it from a java.lang.Double into a scala.Double. breakOut tells it to produce a List as the result of the map, without having to traverse it another time to convert to a List. You could also just do toList instead of breakOut if you didn't care about the extra traversal.
The Java classes are entirely different objects from the Scala ones; it's not just a name-change. Thus, it would be impossible to convert without traversing. You have to create an entirely new collection and then look at each element in order to (convert it and) move it over.
The reason the types are different is that java.lang.Double is a "boxed primitive" whereas scala.Double is equivalent to Java's double primitive. The conversion here is essentially "unboxing" the primitive.
edited Dec 1 '15 at 9:57
Jacek Laskowski
43.2k17126256
answered May 23 '14 at 5:36
dhg
45.2k698131
Is there a way to do it without any traversals? Perhaps that's not possible, but I naively imagine that casting won't involve any traversals.
– Erik Shilts
May 23 '14 at 5:47
Note that if you are willing to accept aSeqrather than aListprecisely, you can avoid the traversal by converting the Scala-converted JavaListinto a view by calling theviewmethod before callingmap. That will give you a lazily evaluatedSeqViewthat will apply the conversions to the elements on an as-needed basis.
– reggert
May 24 '14 at 17:59
add a comment |
import collection.JavaConverters._
val javaList : java.util.List[Double] = ...
val scalaList : List[Double] = j.asScala.toList
answered May 23 '14 at 5:36
ggreiner
9,04213749
This is able to cast a java.util.list[Double] to a List[Double], but the tricky part is also converting from java.lang.Double to Scala Double.
– Erik Shilts
May 23 '14 at 5:48
1
@Erik, It's not a cast:asScalatraverses the java-based collection and builds a brand new scala-based collection. It has to since the classes are completely different.
– dhg
May 23 '14 at 5:57
@dhg, you're in fact wrong.asScaladoes not traverse a collection;asScalacreates a view into the original collection. It'stoListwhich performs actual traversal. Seescala.collection.convert.Wrapperstrait: it contains implementations of these views, and it is obvious that they only forward collection operations to the original Java collections.
– Vladimir Matveev
May 23 '14 at 7:34
@dhg, you're right, however, thatjavaList.asScala.map(_.doubleValue).toListwill perform two traversals, the first one happening inmap, and the second intoList.
– Vladimir Matveev
May 23 '14 at 7:41
add a comment |
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2 Answers
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2 Answers
2
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The recommendation is to use JavaConverters, not JavaConversions:
import collection.JavaConverters._
import scala.collection.breakOut
val l: List[Double] = j.asScala.map(_.doubleValue)(breakOut)
doubleValue will convert it from a java.lang.Double into a scala.Double. breakOut tells it to produce a List as the result of the map, without having to traverse it another time to convert to a List. You could also just do toList instead of breakOut if you didn't care about the extra traversal.
The Java classes are entirely different objects from the Scala ones; it's not just a name-change. Thus, it would be impossible to convert without traversing. You have to create an entirely new collection and then look at each element in order to (convert it and) move it over.
The reason the types are different is that java.lang.Double is a "boxed primitive" whereas scala.Double is equivalent to Java's double primitive. The conversion here is essentially "unboxing" the primitive.
edited Dec 1 '15 at 9:57
Jacek Laskowski
43.2k17126256
answered May 23 '14 at 5:36
dhg
45.2k698131
Is there a way to do it without any traversals? Perhaps that's not possible, but I naively imagine that casting won't involve any traversals.
– Erik Shilts
May 23 '14 at 5:47
Note that if you are willing to accept aSeqrather than aListprecisely, you can avoid the traversal by converting the Scala-converted JavaListinto a view by calling theviewmethod before callingmap. That will give you a lazily evaluatedSeqViewthat will apply the conversions to the elements on an as-needed basis.
– reggert
May 24 '14 at 17:59
add a comment |
The recommendation is to use JavaConverters, not JavaConversions:
import collection.JavaConverters._
import scala.collection.breakOut
val l: List[Double] = j.asScala.map(_.doubleValue)(breakOut)
doubleValue will convert it from a java.lang.Double into a scala.Double. breakOut tells it to produce a List as the result of the map, without having to traverse it another time to convert to a List. You could also just do toList instead of breakOut if you didn't care about the extra traversal.
The Java classes are entirely different objects from the Scala ones; it's not just a name-change. Thus, it would be impossible to convert without traversing. You have to create an entirely new collection and then look at each element in order to (convert it and) move it over.
The reason the types are different is that java.lang.Double is a "boxed primitive" whereas scala.Double is equivalent to Java's double primitive. The conversion here is essentially "unboxing" the primitive.
edited Dec 1 '15 at 9:57
Jacek Laskowski
43.2k17126256
answered May 23 '14 at 5:36
dhg
45.2k698131
Is there a way to do it without any traversals? Perhaps that's not possible, but I naively imagine that casting won't involve any traversals.
– Erik Shilts
May 23 '14 at 5:47
Note that if you are willing to accept aSeqrather than aListprecisely, you can avoid the traversal by converting the Scala-converted JavaListinto a view by calling theviewmethod before callingmap. That will give you a lazily evaluatedSeqViewthat will apply the conversions to the elements on an as-needed basis.
– reggert
May 24 '14 at 17:59
add a comment |
The recommendation is to use JavaConverters, not JavaConversions:
import collection.JavaConverters._
import scala.collection.breakOut
val l: List[Double] = j.asScala.map(_.doubleValue)(breakOut)
doubleValue will convert it from a java.lang.Double into a scala.Double. breakOut tells it to produce a List as the result of the map, without having to traverse it another time to convert to a List. You could also just do toList instead of breakOut if you didn't care about the extra traversal.
The Java classes are entirely different objects from the Scala ones; it's not just a name-change. Thus, it would be impossible to convert without traversing. You have to create an entirely new collection and then look at each element in order to (convert it and) move it over.
The reason the types are different is that java.lang.Double is a "boxed primitive" whereas scala.Double is equivalent to Java's double primitive. The conversion here is essentially "unboxing" the primitive.
edited Dec 1 '15 at 9:57
Jacek Laskowski
43.2k17126256
answered May 23 '14 at 5:36
dhg
45.2k698131
The recommendation is to use JavaConverters, not JavaConversions:
import collection.JavaConverters._
import scala.collection.breakOut
val l: List[Double] = j.asScala.map(_.doubleValue)(breakOut)
doubleValue will convert it from a java.lang.Double into a scala.Double. breakOut tells it to produce a List as the result of the map, without having to traverse it another time to convert to a List. You could also just do toList instead of breakOut if you didn't care about the extra traversal.
The Java classes are entirely different objects from the Scala ones; it's not just a name-change. Thus, it would be impossible to convert without traversing. You have to create an entirely new collection and then look at each element in order to (convert it and) move it over.
The reason the types are different is that java.lang.Double is a "boxed primitive" whereas scala.Double is equivalent to Java's double primitive. The conversion here is essentially "unboxing" the primitive.
edited Dec 1 '15 at 9:57
Jacek Laskowski
43.2k17126256
answered May 23 '14 at 5:36
dhg
45.2k698131
edited Dec 1 '15 at 9:57
Jacek Laskowski
43.2k17126256
edited Dec 1 '15 at 9:57
Jacek Laskowski
43.2k17126256
edited Dec 1 '15 at 9:57
Jacek Laskowski
43.2k17126256
43.2k17126256
answered May 23 '14 at 5:36
dhg
45.2k698131
answered May 23 '14 at 5:36
dhg
45.2k698131
answered May 23 '14 at 5:36
dhg
45.2k698131
45.2k698131
Is there a way to do it without any traversals? Perhaps that's not possible, but I naively imagine that casting won't involve any traversals.
– Erik Shilts
May 23 '14 at 5:47
Note that if you are willing to accept aSeqrather than aListprecisely, you can avoid the traversal by converting the Scala-converted JavaListinto a view by calling theviewmethod before callingmap. That will give you a lazily evaluatedSeqViewthat will apply the conversions to the elements on an as-needed basis.
– reggert
May 24 '14 at 17:59
add a comment |
Is there a way to do it without any traversals? Perhaps that's not possible, but I naively imagine that casting won't involve any traversals.
– Erik Shilts
May 23 '14 at 5:47
Note that if you are willing to accept aSeqrather than aListprecisely, you can avoid the traversal by converting the Scala-converted JavaListinto a view by calling theviewmethod before callingmap. That will give you a lazily evaluatedSeqViewthat will apply the conversions to the elements on an as-needed basis.
– reggert
May 24 '14 at 17:59
Is there a way to do it without any traversals? Perhaps that's not possible, but I naively imagine that casting won't involve any traversals.
– Erik Shilts
May 23 '14 at 5:47
Is there a way to do it without any traversals? Perhaps that's not possible, but I naively imagine that casting won't involve any traversals.
– Erik Shilts
May 23 '14 at 5:47
Note that if you are willing to accept a
Seq rather than a List precisely, you can avoid the traversal by converting the Scala-converted Java List into a view by calling the view method before calling map. That will give you a lazily evaluated SeqView that will apply the conversions to the elements on an as-needed basis.– reggert
May 24 '14 at 17:59
Note that if you are willing to accept a
Seq rather than a List precisely, you can avoid the traversal by converting the Scala-converted Java List into a view by calling the view method before calling map. That will give you a lazily evaluated SeqView that will apply the conversions to the elements on an as-needed basis.– reggert
May 24 '14 at 17:59
add a comment |
import collection.JavaConverters._
val javaList : java.util.List[Double] = ...
val scalaList : List[Double] = j.asScala.toList
answered May 23 '14 at 5:36
ggreiner
9,04213749
This is able to cast a java.util.list[Double] to a List[Double], but the tricky part is also converting from java.lang.Double to Scala Double.
– Erik Shilts
May 23 '14 at 5:48
1
@Erik, It's not a cast:asScalatraverses the java-based collection and builds a brand new scala-based collection. It has to since the classes are completely different.
– dhg
May 23 '14 at 5:57
@dhg, you're in fact wrong.asScaladoes not traverse a collection;asScalacreates a view into the original collection. It'stoListwhich performs actual traversal. Seescala.collection.convert.Wrapperstrait: it contains implementations of these views, and it is obvious that they only forward collection operations to the original Java collections.
– Vladimir Matveev
May 23 '14 at 7:34
@dhg, you're right, however, thatjavaList.asScala.map(_.doubleValue).toListwill perform two traversals, the first one happening inmap, and the second intoList.
– Vladimir Matveev
May 23 '14 at 7:41
add a comment |
import collection.JavaConverters._
val javaList : java.util.List[Double] = ...
val scalaList : List[Double] = j.asScala.toList
answered May 23 '14 at 5:36
ggreiner
9,04213749
This is able to cast a java.util.list[Double] to a List[Double], but the tricky part is also converting from java.lang.Double to Scala Double.
– Erik Shilts
May 23 '14 at 5:48
1
@Erik, It's not a cast:asScalatraverses the java-based collection and builds a brand new scala-based collection. It has to since the classes are completely different.
– dhg
May 23 '14 at 5:57
@dhg, you're in fact wrong.asScaladoes not traverse a collection;asScalacreates a view into the original collection. It'stoListwhich performs actual traversal. Seescala.collection.convert.Wrapperstrait: it contains implementations of these views, and it is obvious that they only forward collection operations to the original Java collections.
– Vladimir Matveev
May 23 '14 at 7:34
@dhg, you're right, however, thatjavaList.asScala.map(_.doubleValue).toListwill perform two traversals, the first one happening inmap, and the second intoList.
– Vladimir Matveev
May 23 '14 at 7:41
add a comment |
import collection.JavaConverters._
val javaList : java.util.List[Double] = ...
val scalaList : List[Double] = j.asScala.toList
answered May 23 '14 at 5:36
ggreiner
9,04213749
import collection.JavaConverters._
val javaList : java.util.List[Double] = ...
val scalaList : List[Double] = j.asScala.toList
answered May 23 '14 at 5:36
ggreiner
9,04213749
answered May 23 '14 at 5:36
ggreiner
9,04213749
answered May 23 '14 at 5:36
ggreiner
9,04213749
answered May 23 '14 at 5:36
ggreiner
9,04213749
9,04213749
This is able to cast a java.util.list[Double] to a List[Double], but the tricky part is also converting from java.lang.Double to Scala Double.
– Erik Shilts
May 23 '14 at 5:48
1
@Erik, It's not a cast:asScalatraverses the java-based collection and builds a brand new scala-based collection. It has to since the classes are completely different.
– dhg
May 23 '14 at 5:57
@dhg, you're in fact wrong.asScaladoes not traverse a collection;asScalacreates a view into the original collection. It'stoListwhich performs actual traversal. Seescala.collection.convert.Wrapperstrait: it contains implementations of these views, and it is obvious that they only forward collection operations to the original Java collections.
– Vladimir Matveev
May 23 '14 at 7:34
@dhg, you're right, however, thatjavaList.asScala.map(_.doubleValue).toListwill perform two traversals, the first one happening inmap, and the second intoList.
– Vladimir Matveev
May 23 '14 at 7:41
add a comment |
This is able to cast a java.util.list[Double] to a List[Double], but the tricky part is also converting from java.lang.Double to Scala Double.
– Erik Shilts
May 23 '14 at 5:48
1
@Erik, It's not a cast:asScalatraverses the java-based collection and builds a brand new scala-based collection. It has to since the classes are completely different.
– dhg
May 23 '14 at 5:57
@dhg, you're in fact wrong.asScaladoes not traverse a collection;asScalacreates a view into the original collection. It'stoListwhich performs actual traversal. Seescala.collection.convert.Wrapperstrait: it contains implementations of these views, and it is obvious that they only forward collection operations to the original Java collections.
– Vladimir Matveev
May 23 '14 at 7:34
@dhg, you're right, however, thatjavaList.asScala.map(_.doubleValue).toListwill perform two traversals, the first one happening inmap, and the second intoList.
– Vladimir Matveev
May 23 '14 at 7:41
This is able to cast a java.util.list[Double] to a List[Double], but the tricky part is also converting from java.lang.Double to Scala Double.
– Erik Shilts
May 23 '14 at 5:48
This is able to cast a java.util.list[Double] to a List[Double], but the tricky part is also converting from java.lang.Double to Scala Double.
– Erik Shilts
May 23 '14 at 5:48
1
1
@Erik, It's not a cast:
asScala traverses the java-based collection and builds a brand new scala-based collection. It has to since the classes are completely different.– dhg
May 23 '14 at 5:57
@Erik, It's not a cast:
asScala traverses the java-based collection and builds a brand new scala-based collection. It has to since the classes are completely different.– dhg
May 23 '14 at 5:57
@dhg, you're in fact wrong.
asScala does not traverse a collection; asScala creates a view into the original collection. It's toList which performs actual traversal. See scala.collection.convert.Wrappers trait: it contains implementations of these views, and it is obvious that they only forward collection operations to the original Java collections.– Vladimir Matveev
May 23 '14 at 7:34
@dhg, you're in fact wrong.
asScala does not traverse a collection; asScala creates a view into the original collection. It's toList which performs actual traversal. See scala.collection.convert.Wrappers trait: it contains implementations of these views, and it is obvious that they only forward collection operations to the original Java collections.– Vladimir Matveev
May 23 '14 at 7:34
@dhg, you're right, however, that
javaList.asScala.map(_.doubleValue).toList will perform two traversals, the first one happening in map, and the second in toList.– Vladimir Matveev
May 23 '14 at 7:41
@dhg, you're right, however, that
javaList.asScala.map(_.doubleValue).toList will perform two traversals, the first one happening in map, and the second in toList.– Vladimir Matveev
May 23 '14 at 7:41
add a comment |
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"I've seen documentation to go from Scala --> Java, but not much going the other way."... did you now? Care to share that information? ;-)
– Make42
Oct 19 '17 at 12:12
I found out the answer: stackoverflow.com/q/46830217/4533188
– Make42
Oct 19 '17 at 12:48