Scala的List不仅可以指定循环区间，而且还能根据步长筛选元素。 List中的步长，by关键字：
scala> 1 to 100 by 3
res64: scala.collection.immutable.Range = Range(1, 4, 7, 10, 13, 16, 19, 22, 25,
28, 31, 34, 37, 40, 43, 46, 49, 52, 55, 58, 61, 64, 67, 70, 73, 76, 79, 82, 85,
88, 91, 94, 97, 100)
scala> 1 to 100 by 2
res65: scala.collection.immutable.Range = Range(1, 3, 5, 7, 9, 11, 13, 15, 17, 1
9, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 5
9, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 9
9)
scala> 1 to 100 by 1
res66: scala.collection.immutable.Range = Range(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1
1, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 3
1, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 5
1, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 7
1, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 9
1, 92, 93, 94, 95, 96, 97, 98, 99, 100)
可以看出，每个元素都相差了by后面的数字个步长
如果我们利用这个求奇数的平方和，11+33+55+ … + 9999，结果为166650
scala> List(1 to 100 by 2:_*) map (i=>i*i) sum
res59: Int = 166650
如果用filter也能做
scala
scala> (1 to 100) filter (_%2!=0) map (i=>i*i) sum
res60: Int = 166650
看一下源代码里怎么定义这个by的.
/** The `Range` class represents integer values in range
* ''[start;end)'' with non-zero step value `step`.
* It's a special case of an indexed sequence.
* For example:
*
*
* val r1 = 0 until 10
* val r2 = r1.start until r1.end by r1.step + 1
* println(r2.length) // = 5
*
*
* @param start the start of this range.
* @param end the exclusive end of the range.
* @param step the step for the range.
*
* @author Martin Odersky
* @author Paul Phillips
* @version 2.8
* @since 2.5
* @see [[http://docs.scala-lang.org/overviews/collections/concrete-immutable-collection-classes.html#ranges "Scala's Collection Library overview"]]
* section on `Ranges` for more information.
*
* @define Coll Range
* @define coll range
* @define mayNotTerminateInf
* @define willNotTerminateInf
* @define doesNotUseBuilders
* '''Note:''' this method does not use builders to construct a new range,
* and its complexity is O(1).
*/
@SerialVersionUID(7618862778670199309L)
class Range(val start: Int, val end: Int, val step: Int)
extends IndexedSeq[Int]
with collection.CustomParallelizable[Int, ParRange]
with Serializable
{
override def par = new ParRange(this)
// This member is designed to enforce conditions:
// (step != 0) && (length <= Int.MaxValue),
// but cannot be evaluated eagerly because we have a pattern where ranges
// are constructed like: "x to y by z"
// The "x to y" piece should not trigger an exception. So the calculation
// is delayed, which means it will not fail fast for those cases where failing
// was correct.
private lazy val numRangeElements: Int = Range.count(start, end, step, isInclusive)
protected def copy(start: Int, end: Int, step: Int): Range = new Range(start, end, step)
/** Create a new range with the `start` and `end` values of this range and
* a new `step`.
*
* @return a new range with a different step
*/
def by(step: Int): Range = copy(start, end, step)
def isInclusive = false
@inline final override def foreach[@specialized(Unit) U](f: Int => U) {
if (length > 0) {
val last = this.last
var i = start
while (i != last) {
f(i)
i += step
}
f(i)
}
}
这里重写了foreach方法，其实就是遍历每个元素时给i+=step。 其实返回的是一个range对象
scala> List(1 to 100 by 2)
res77: List[scala.collection.immutable.Range] = List(Range(1, 3, 5, 7, 9, 11, 13
, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53
, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93
, 95, 97, 99))
想要转换为int，则需要
scala> List(1 to 100 by 2:_*)
res78: List[Int] = List(1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 3
1, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 7
1, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99)