How to convert a Stream[IO, List[A]] to Stream[IO, A]

Question

I want to parse a json file which output a collection of A. The signature of the Output is IO[List[A]]

How can I convert this value to a Stream: Stream[IO, A] ? I can convert to a Stream[IO, List[A]] but it is not what I want

fs2.Stream.eval(input).flatMap(x => fs2.Stream.apply(x)) Thanks

score 12 · Answer 1 · answered Jun 19 '18 at 19:55

12

You can also use Stream.emits, which accepts a Seq, so fs2.Stream.eval(output).flatMap(fs2.Stream.emits(_)).

This is more efficient than using varargs with apply because it avoids wrapping and unwrapping the sequence structure - this can save a lot in the case of specialized primitive collections.

answered Jun 19 '18 at 19:55

Daenyth

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Thank you so much! I was getting annoyed with the amount of vararg conversions I was dealing with. This is much cleaner – Henry Parker Jul 12 '21 at 16:00

Dmytro Mitin · Accepted Answer · 2018-04-10T12:18:57.697

4

Try

fs2.Stream.eval(output).flatMap(x => fs2.Stream.apply(x: _*))

What does `:_*` (colon underscore star) do in Scala?

edited Apr 10 '18 at 12:18

answered Apr 10 '18 at 12:11

Dmytro Mitin

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How to convert a Stream[IO, List[A]] to Stream[IO, A]

2 Answers2