CODEWITHSUNDEEP
scalaenumerate
Abhinav Sharma
Abhinav Sharma

Reputation: 1255

Is there a Scala equivalent for the python enumerate?

I'd like the convienience of 

for i, line in enumerate(open(sys.argv[1])):
  print i, line

when doing the following in Scala

for (line <- Source.fromFile(args(0)).getLines()) {
  println(line)
}

Upvotes: 46

Views: 13853

Answers (2)

janluke
janluke

Reputation: 1667

As already answered by others, if you want your index to start from 0, you can use zipWithIndex:

for ((elem, i) <- collection.zipWithIndex) {
    println(i, elem)
}

Because zipWithIndex creates a copy of the collection if called on the collection itself, you may want to call it to a view of the colleciton instead: collection.view.zipWithIndex.

Nonetheless, Python's enumerate has an optional parameter to set the start value of your index. In scala, you can do:

for ((elem, i) <- collection.zip(Stream from 1) {
    println(i, elem)
}

For a longer discussion, read https://alvinalexander.com/scala/how-to-use-zipwithindex-create-for-loop-counters-scala-cookbook.

Upvotes: 11

rafalotufo
rafalotufo

Reputation: 3932

You can use the zipWithIndex from Iterable trait:

for ((line, i) <- Source.fromFile(args(0)).getLines().zipWithIndex) {
   println(i, line)
}

Upvotes: 57

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