If you’re new to Scala, the idea of immutable data structures can be a pain at times, especially when you need to replace “for (int x=0; x < arr1.length; x++)\".\n\nOne easy way to handle this is zipWithIndex:\n\n\n\n
\nval arr1 = List(“a”, “b”, “c”)\n\narr1.zipWithIndex\n<\/pre>\nNote that you call this WITHOUT the parens:<\/p>\n
\nres14: List[(String, Int)] = List((a,0), (b,1), (c,2))\n<\/pre>\nOtherwise, you’ll get an awful type error, which indicates you probably weren’t doing what you wanted:<\/p>\n
\n:9: error: not enough arguments \nfor method zipWithIndex: \n(implicit bf: scala.collection.generic.CanBuildFrom\n[List[String],(A1, Int),That])That.\nUnspecified value parameter bf.\n<\/pre>\nUsing the above technique, you get a new list, with tuples of object + index. If you need something that doesn’t pre-compute everything, like your old “for” loop, you can do something like this:<\/p>\n
\narr1\n .view\n .zipWithIndex\n .flatMap(\n x => { \n System.out.println(x); \n None \n }).toList\n<\/pre>\n\n(a,0)\n(b,1)\n(c,2)\nres49: List[Nothing] = List()\n<\/pre>\n","protected":false},"excerpt":{"rendered":"If you’re new to Scala, the idea of immutable data structures can be a pain at times, especially when you need to replace “for (int x=0; x < arr1.length; x++)\". One easy way to handle this is zipWithIndex: val arr1 = List(“a”, “b”, “c”) arr1.zipWithIndex Note that you call this WITHOUT the parens: res14: List[(String, … \n\n