I have a variable which has the following data structure, I want to replace all "foo" with all "goo" for the String part. Is there any one line neat code to do that? Want to see if any smart solutions to skip to write a loop. :)
var result = List[List[(List[String], Double)]]
regards, Lin
I am not sure if I got it right but maybe this is what you are looking for?
scala> val a: List[List[(List[String], Double)]] = List(List((List("foo asd", "asd foo"), 2.6)))
scala> a map (_ map { case (k, v) => (k map (_.replaceAll("foo", "goo")), v) })
res1: List[List[(List[String], Double)]] = List(List((List(goo asd, asd goo),2.6)))
to answer the comment let me first remove spaces and use dots
scala> a.map(_.map { case (k, v) => (k.map(_.replaceAll("foo", "goo")), v) })
and now, expand _.method(param) to x => x.method(param)
scala> a.map(b => b.map { case (k, v) => (k.map(c => c.replaceAll("foo", "goo")), v) })
You have 3 levels of nested lists, and one is inside a tuple, it won't be pretty, you need to map over each of them and extract last one from tuple.
scala> val l = List(List((List("foo","afoo"),3.4),(List("gfoo","cfoo"),5.6)))
l: List[List[(List[String], Double)]] = List(List((List(foo, afoo),3.4), (List(gfoo, cfoo),5.6)))
scala> def replaceFoo(y:List[String]) = y.map(s => s.replace("foo","goo"))
replaceFoo: (y: List[String])List[String]
scala> l.map(x => x.map(y => (replaceFoo(y._1),y._2)))
res0: List[List[(List[String], Double)]] = List(List((List(goo, agoo),3.4), (List(ggoo, cgoo),5.6)))
l.map(x => x.map(y => (y._1.map(s => s.replace("foo","goo")),y._2)))? Still seems a bit advanced to me. :)