You can easily generate an infinite stream:<\/p>\n
\nval x = Stream.from(1)\n\nx: scala.collection.immutable.Stream[Int] = Stream(1, ?)\n<\/pre>\nUsing this, you can pull values, however the “take” results will still be a lazy-loaded stream:<\/p>\n
\nx.take(10)\n\nres11: scala.collection.immutable.Stream[Int] = Stream(1, ?)\n<\/pre>\nYou have to manually force this to return what you want:<\/p>\n
\nx.take(10).toList\nres12: List[Int] = List(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)\n<\/pre>\nAlternately, you can use this as more complex constructions. For instance, shown here, you can find the first ten numbers divisible by five:<\/p>\n
\nx.filter(_ % 5 == 0).take(10).toList\n\nres16: List[Int] = \n List(5, 10, 15, 20, \n 25, 30, 35, 40, 45, 50)\n<\/pre>\n","protected":false},"excerpt":{"rendered":"You can easily generate an infinite stream: val x = Stream.from(1) x: scala.collection.immutable.Stream[Int] = Stream(1, ?) Using this, you can pull values, however the “take” results will still be a lazy-loaded stream: x.take(10) res11: scala.collection.immutable.Stream[Int] = Stream(1, ?) You have to manually force this to return what you want: x.take(10).toList res12: List[Int] = List(1, 2, … <\/p>\n