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Scala使用随机值填充地图

[英]Scala populate map with random values

I need to create a test for various collections based on Map and HashMap. 我需要为基于Map和HashMap的各种集合创建测试。

I have two functions that create test data, eg: 我有两个创建测试数据的函数,例如:

def f1: String = { ... )
def f2: String = { ... }

these functions create random data every time they are called. 这些函数在每次调用时都会创建随机数据。

My map is: 我的地图是:

val m:Map[String,String] = ...

what I try to accomplish is construct an immutable map with 10000 random items that were generated by calling f1/f2. 我试图完成的工作是构造一个具有10000个随机项的不可变映射,这些随机项是通过调用f1 / f2生成的。 so protocode would be: 所以protocode将是:

for 1 to 10000
   add-key-value-to-map (key = f1(), value = f2() )
end for

how can I accomplish this in scala, without destroying and rebuilding the list 10000 times? 如何在scala中完成此操作,而又不破坏并重建列表10000次?

EDIT: Since it wasn't clear in the original post above, I am trying to run this with various types of maps (Map, HashMap, TreeMap). 编辑:由于上面的原始帖子中不清楚,我试图用各种类型的地图(Map,HashMap,TreeMap)运行它。

You can use List.fill to create a List of couple (String, String) and then call .toMap on it: 您可以使用List.fill来创建一对夫妇(String, String)的列表,然后在其上调用.toMap

scala> def f1 = util.Random.alphanumeric take 5 mkString
f1: String

scala> def f2 = util.Random.alphanumeric take 5 mkString
f2: String

scala> val m = List.fill(5)(f1 -> f2).toMap
m: scala.collection.immutable.Map[String,String] =
     Map(T7hD8 -> BpAa1, uVpno -> 6sMjc, wdaRP -> XSC1V, ZGlC0 -> aTwBo, SjfOr -> hdzIN)

Alternatively you could use Map / HashMap / TreeMap 's .apply function: 或者,您可以使用Map / HashMap / TreeMap.apply函数:

scala> val m = collection.immutable.TreeMap(List.fill(5)(f1 -> f2) : _*)
m: scala.collection.immutable.TreeMap[String,String] = 
     Map(3cieU -> iy0KV, 8oUb1 -> YY6NC, 95ol4 -> Sf9qp, GhXWX -> 8U8wt, ZD8Px -> STMOC)
List.fill(10000)((f1, f2)).toMap
val m = (1 to 10000).foldLeft(Map.empty[String,String]) { (m, _) => m + (f1 -> f2) }

Using tabulate as follows, 使用以下tabulate

Seq.tabulate(10000)(_ => f1 -> f2).toMap

It proves unclear whether the random key generator function may duplicate some keys, in which case 10000 iterations would not suffice to produce a map of such size. 尚不清楚随机密钥生成器功能是否可以复制某些密钥,在这种情况下, 10000次迭代不足以生成这种大小的映射。

An intuitive approach, 直观的方法

(1 to 10000).map(_ => f1 -> f2).toMap

Using a recursive function instead of generating a range to iterate over, (although numerous intermediate Maps are created), 使用递归函数而不是生成要迭代的范围(尽管创建了大量中间图),

def g(i: Int): Map[String,String] = {
  if (i<=0) 
    Map() 
  else 
    Map(f1 -> f2) ++ g(i-1)
}

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