Collection view vs withFilter
Both views and withFilter solve the problem of intermediate collection creations. What's the difference between them ?
List("a", "b", "c").withFilter(_ == "b").withFilter(_ == "c").map(x => x)
vs
List("a", "b", "c").view.filter(_ == "b").filter(_ == "c").map(x => x)
scala
asked Nov 12 '18 at 16:23
EugeneMiEugeneMi
1,36912243
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Both views and withFilter solve the problem of intermediate collection creations. What's the difference between them ?
List("a", "b", "c").withFilter(_ == "b").withFilter(_ == "c").map(x => x)
vs
List("a", "b", "c").view.filter(_ == "b").filter(_ == "c").map(x => x)
scala
asked Nov 12 '18 at 16:23
EugeneMiEugeneMi
1,36912243
add a comment |
Both views and withFilter solve the problem of intermediate collection creations. What's the difference between them ?
List("a", "b", "c").withFilter(_ == "b").withFilter(_ == "c").map(x => x)
vs
List("a", "b", "c").view.filter(_ == "b").filter(_ == "c").map(x => x)
scala
asked Nov 12 '18 at 16:23
EugeneMiEugeneMi
1,36912243
Both views and withFilter solve the problem of intermediate collection creations. What's the difference between them ?
List("a", "b", "c").withFilter(_ == "b").withFilter(_ == "c").map(x => x)
vs
List("a", "b", "c").view.filter(_ == "b").filter(_ == "c").map(x => x)
scala
scala
asked Nov 12 '18 at 16:23
EugeneMiEugeneMi
1,36912243
asked Nov 12 '18 at 16:23
EugeneMiEugeneMi
1,36912243
asked Nov 12 '18 at 16:23
EugeneMiEugeneMi
1,36912243
asked Nov 12 '18 at 16:23
EugeneMiEugeneMi
1,36912243
asked Nov 12 '18 at 16:23
EugeneMiEugeneMi
1,36912243
1,36912243
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
List("a", "b", "c").withFilter(_ == "b").withFilter(_ == "c").map(x => x)
Result:
List[String] = List()
Note: the result is no longer lazy.
List("a", "b", "c").view.filter(_ == "b").filter(_ == "c").map(x => x)
Result:
scala.collection.SeqView[String,Seq[_]] = SeqViewFFM(...)
The result has not been evaluated, it is still a view.
answered Nov 12 '18 at 17:00
Terry DactylTerry Dactyl
1,104412
add a comment |
The first is lazy until you call the map, while the second one is just lazy (not executed). For the second one it will finally execute when decide to call force - which you have not done in your example. So it'd look like:
List("a", "b", "c").view.filter(_ == "b").filter(_ == "c").map(x => x).force
this is equivalent to the first one.
See here and here about withFilter and view in Scala.
answered Nov 12 '18 at 22:54
TanjinTanjin
1,3951415
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
List("a", "b", "c").withFilter(_ == "b").withFilter(_ == "c").map(x => x)
Result:
List[String] = List()
Note: the result is no longer lazy.
List("a", "b", "c").view.filter(_ == "b").filter(_ == "c").map(x => x)
Result:
scala.collection.SeqView[String,Seq[_]] = SeqViewFFM(...)
The result has not been evaluated, it is still a view.
answered Nov 12 '18 at 17:00
Terry DactylTerry Dactyl
1,104412
add a comment |
List("a", "b", "c").withFilter(_ == "b").withFilter(_ == "c").map(x => x)
Result:
List[String] = List()
Note: the result is no longer lazy.
List("a", "b", "c").view.filter(_ == "b").filter(_ == "c").map(x => x)
Result:
scala.collection.SeqView[String,Seq[_]] = SeqViewFFM(...)
The result has not been evaluated, it is still a view.
answered Nov 12 '18 at 17:00
Terry DactylTerry Dactyl
1,104412
add a comment |
List("a", "b", "c").withFilter(_ == "b").withFilter(_ == "c").map(x => x)
Result:
List[String] = List()
Note: the result is no longer lazy.
List("a", "b", "c").view.filter(_ == "b").filter(_ == "c").map(x => x)
Result:
scala.collection.SeqView[String,Seq[_]] = SeqViewFFM(...)
The result has not been evaluated, it is still a view.
answered Nov 12 '18 at 17:00
Terry DactylTerry Dactyl
1,104412
List("a", "b", "c").withFilter(_ == "b").withFilter(_ == "c").map(x => x)
Result:
List[String] = List()
Note: the result is no longer lazy.
List("a", "b", "c").view.filter(_ == "b").filter(_ == "c").map(x => x)
Result:
scala.collection.SeqView[String,Seq[_]] = SeqViewFFM(...)
The result has not been evaluated, it is still a view.
answered Nov 12 '18 at 17:00
Terry DactylTerry Dactyl
1,104412
answered Nov 12 '18 at 17:00
Terry DactylTerry Dactyl
1,104412
answered Nov 12 '18 at 17:00
Terry DactylTerry Dactyl
1,104412
answered Nov 12 '18 at 17:00
Terry DactylTerry Dactyl
1,104412
1,104412
add a comment |
add a comment |
The first is lazy until you call the map, while the second one is just lazy (not executed). For the second one it will finally execute when decide to call force - which you have not done in your example. So it'd look like:
List("a", "b", "c").view.filter(_ == "b").filter(_ == "c").map(x => x).force
this is equivalent to the first one.
See here and here about withFilter and view in Scala.
answered Nov 12 '18 at 22:54
TanjinTanjin
1,3951415
add a comment |
The first is lazy until you call the map, while the second one is just lazy (not executed). For the second one it will finally execute when decide to call force - which you have not done in your example. So it'd look like:
List("a", "b", "c").view.filter(_ == "b").filter(_ == "c").map(x => x).force
this is equivalent to the first one.
See here and here about withFilter and view in Scala.
answered Nov 12 '18 at 22:54
TanjinTanjin
1,3951415
add a comment |
The first is lazy until you call the map, while the second one is just lazy (not executed). For the second one it will finally execute when decide to call force - which you have not done in your example. So it'd look like:
List("a", "b", "c").view.filter(_ == "b").filter(_ == "c").map(x => x).force
this is equivalent to the first one.
See here and here about withFilter and view in Scala.
answered Nov 12 '18 at 22:54
TanjinTanjin
1,3951415
The first is lazy until you call the map, while the second one is just lazy (not executed). For the second one it will finally execute when decide to call force - which you have not done in your example. So it'd look like:
List("a", "b", "c").view.filter(_ == "b").filter(_ == "c").map(x => x).force
this is equivalent to the first one.
See here and here about withFilter and view in Scala.
answered Nov 12 '18 at 22:54
TanjinTanjin
1,3951415
answered Nov 12 '18 at 22:54
TanjinTanjin
1,3951415
answered Nov 12 '18 at 22:54
TanjinTanjin
1,3951415
answered Nov 12 '18 at 22:54
TanjinTanjin
1,3951415
1,3951415
add a comment |
add a comment |
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