scala 如何从范围创建列表
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How to create List from Range
提问by Ula Krukar
I am new to Scala, just started learning, so this is basic beginner question.
我是 Scala 的新手，刚开始学习，所以这是基本的初学者问题。
I try to implement Sieve of Eratosthenesalgorithm. Here is what I got so far:
我尝试实现Eratosthenes算法的筛分。这是我到目前为止所得到的：
def sieve_core(cross: Int, lst: Seq[Int]): List[Int] = {
val crossed = lst.filter(_ % cross != 0)
crossed match {
case a :: rest => cross :: sieve_core(a, crossed)
case _ => cross :: Nil
}
}
def sieve(max: Int): List[Int] = {
sieve_core(2, (2 to max))
}
println(sieve(100))
The result is:
结果是：
List(2)
As far as I understand, case _ => cross :: Nilis matched in first iteration of sieve_core, which means that crossedis not an instance of a List.
据我了解，case _ => cross :: Nil在 的第一次迭代中匹配sieve_core，这意味着它crossed不是 List 的实例。
I changed lstparameters type to List[Int]and now the code won't compile with an error:
我将lst参数类型更改为List[Int]，现在代码不会编译并出现错误：
(fragment of Problem3.scala):24: error: type mismatch;
found : Range.Inclusive
required: List[Int]
sieve_core(2, (2 to max))
^
Apparently Rangeis not a List.
显然Range不是List.
Question: how can I turn Range into a List? Or is it some bigger problem with my code, I have made some bad assumption somewhere along the way?
问题：如何将 Range 变成 List？还是我的代码有更大的问题，我在此过程中的某个地方做了一些错误的假设？
Any help appreciated.
任何帮助表示赞赏。
回答by DigitalRoss
There's an applymethod on the Listcompanion object which takes a range and returns a List:
伴随对象apply上有一个方法，List它接受一个范围并返回一个List：
scala> List.range(2, 11)
res0: List[Int] = List(2, 3, 4, 5, 6, 7, 8, 9, 10)
There are lots of useful Listfactory methods in the Listcollection documentation.
有很多有用的List工厂方法中List采集文件。
回答by Jonathan Graehl
To turn any sequence sinto a list, use s.toList
要将任何序列s转换为列表，请使用s.toList
I'm sure digitalross' is more efficient in this case, though.
不过，我确信在这种情况下，digitalross 的效率更高。
回答by Sal Borrelli
(2 to max)is not a scala.collection.immutable.Listindeed but a scala.collection.immutable.Range, more precisely an instance of scala.collection.immutable.Range.Inclusive, as mentioned in your error message. Just in passing note that Inclusiveand Exclusiveare themselves members of Range, with a fairly auto-explanatory meaning.
(2 to max)确实不是scala.collection.immutable.List而是scala.collection.immutable.Range，更准确地说是 的实例scala.collection.immutable.Range.Inclusive，如您的错误消息中所述。顺便提一下，Inclusive和Exclusive本身就是 的成员Range，具有相当不言自明的含义。
Luckily the Range class offers the handy method toList, which you can leverage for converting the range into a list and resolve the problem, like in the following code snippet:
幸运的是 Range 类提供了方便的方法toList，您可以利用该方法将范围转换为列表并解决问题，如下面的代码片段所示：
scala> val max = 10
max: Int = 10
scala> val r = (2 to max)
r: scala.collection.immutable.Range.Inclusive = Range 2 to 10
scala> val l = r.toList
l: List[Int] = List(2, 3, 4, 5, 6, 7, 8, 9, 10)
