Scala sort list based on second attribute and then first

Question

I wish to sort a list containing (word, word.length) first based on length and then words alphabetically. So given: "I am a girl" the output should be a:1, I:1, am:2, girl:4 I have the following piece of code which works but not for all examples

val lengths = words.map(x => x.length)
val wordPairs = words.zip(lengths).toList
val mapwords = wordPairs.sort (_._2 < _._2).sortBy(_._1)

Sergey Passichenko · Accepted Answer · 2013-01-23 06:39:40Z

12

You can sort by tuple:

scala>  val words = "I am a girl".split(" ")
words: Array[java.lang.String] = Array(I, am, a, girl)

scala>  words.sortBy(w => w.length -> w)
res0: Array[java.lang.String] = Array(I, a, am, girl)

scala>  words.sortBy(w => w.length -> w.toLowerCase)
res1: Array[java.lang.String] = Array(a, I, am, girl)

answered Jan 23, 2013 at 6:39

Sergey Passichenko

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2 Comments

korefn Over a year ago

to get the ouput from the above you add a map function val output:Array[String] = words.sortBy(w => w.length -> w.toLowerCase) map {x => x + ":" + x.length }

Régis Jean-Gilles Over a year ago

+1, you learn something everyday. After looking it up in the docs, it works because the Ordering object provides lexicographic ordering for tuples (see implicits Ordering.Tuple*)

Ghashange · Accepted Answer · 2013-01-23 10:26:24Z

-2

U can do that in one line:

 "I am a girl".toLowerCase.split(" ").map(x => (x,x.length)).sortWith { (x: (String,Int), y: (String,Int)) => x._1 < y._1 }

or in two lines:

 val wordPairs = "I am a girl".split(" ").map(x => (x,x.length))
 val result = wordPairs.toLowerCase.sortWith { (x: (String,Int), y: (String,Int)) => x._1 < y._1 }

edited Jan 23, 2013 at 10:26

answered Jan 23, 2013 at 10:05

Ghashange

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1 Comment

Régis Jean-Gilles Over a year ago

You are only sorting by length, not by length + alphabetic order

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