scala - string parsing without Regex

Question

I have various types of strings like the following:

sales_data_type
saledatatypes
sales_data.new.metric1
sales_data.type.other.metric2
sales_data.type3.metric3

I'm trying to parse them to get a substring with a word before and after the last dot. For example: new.metric1, other.metric2, type3.metric3. If a word doesn't contain dots, it has to be returned as is: sales_data_type, saledatatypes. With a Regex it may be done this way:

val infoStr = "sales_data.type.other.metric2"
val pattern = ".*?([^.]+\\.)?([^.]+)$"  
println(infoStr.replaceAll(pattern, "$1$2"))
// prints other.metric2
// for saledatatypes just prints nullsaledatatypes ??? but seems to work

I want to find a way to achieve this with Scala, without using Regex in order to expand my understanding of Scala features. Will be grateful for any ideas.

jwvh · Accepted Answer · 2018-08-01 08:05:09Z

3

One-liner:

dataStr.split('.').takeRight(2).mkString(".")

takeRight(2) will take the last 2 if there are 2 to take, else it will take the last, and only, 1. mkString(".") will re-insert the dot only if there are 2 elements for the dot to go between, else it will leave the string unaltered.

edited Aug 1, 2018 at 8:05

answered Aug 1, 2018 at 7:54

jwvh

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Comments

kag0 · Accepted Answer · 2018-07-31 23:32:12Z

2

Here's one with lots of scala features for you.

val string = "head.middle.last"

val split = string.split('.') // Array(head, middle, last)

val result = split.toSeq match {
  case Seq(word) ⇒ word
  case _ :+ before :+ after ⇒ s"$before.$after"
}

println(result) // middle.last

First we split the string on your . and get individual parts.
Then we pattern match those parts, first to check if there is only one (in which case we just return it), and second to grab the last two elements in the seq.
Finally we put a . back in between those last two using string interpolation.

answered Jul 31, 2018 at 23:32

kag0

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Comments

proximator · Accepted Answer · 2018-07-31 23:41:56Z

2

One way of doing it:

val value = "sales_data.type.other.metric2"
val elems = value.split("\\.").toList

elems match {
  case _:+beforeLast:+last => s"${beforeLast}.${last}"
  case _ => throw new NoSuchElementException
}

edited Jul 31, 2018 at 23:41

answered Jul 31, 2018 at 22:23

proximator

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Comments

RAGHHURAAMM · Accepted Answer · 2018-08-01 07:50:01Z

for(s<-strs) yield {val s1 = s.split('.');
     if(s1.size>=2)s1.takeRight(2).mkString(".") else s }

or

for(s<-strs) yield { val s1 = s.split('.');
    if(s1.size>=2)s1.init.last+'.'+s1.last else s }

In Scala REPL:

scala> val strs =
Vector("sales_data_type","saledatatypes","sales_data.new.metric1","sales_data.type.other.metric2","sales_d
ata.type3.metric3")
strs: scala.collection.immutable.Vector[String] = Vector(sales_data_type, saledatatypes, sales_data.new.metric1, sales_data.
type.other.metric2, sales_data.type3.metric3)


scala> for(s<-strs) yield { val s1 = s.split('.');if(s1.size>=2)s1.takeRight(2).mkString(".") else s }
res62: scala.collection.immutable.Vector[String] = Vector(sales_data_type, saledatatypes, new.metric1, other.metric2, type3.
metric3)

scala> for(s<-strs) yield { val s1 = s.split('.');if(s1.size>=2)s1.init.last+'.'+s1.last else s }
res60: scala.collection.immutable.Vector[String] = Vector(sales_data_type, saledatatypes, new.metric1, other.metric2, type3.
metric3)

Manoj Kumar Dhakad · Accepted Answer · 2018-08-01 09:57:57Z

1

Use scala match and do like this

def getFormattedStr(str:String):String={
  str.contains(".") match{
    case true=>{
      val arr=str.split("\\.")
      val len=arr.length
      len match{
        case 1=>str
        case _=>arr(len-2)+"."+arr(len-1)
      }
    }
    case _=>str
  }
}

answered Aug 1, 2018 at 9:57

Manoj Kumar Dhakad

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