How to get first 100 prime numbers in scala as i g

2019-08-16 12:45发布

问题:

output : prime numbers 2 3 () 5 () 7 () ()

i want as 2 3 5 7

def primeNumber(range: Int): Unit ={

    val primeNumbers: immutable.IndexedSeq[AnyVal] =

      for (number :Int <- 2 to range) yield{

        var isPrime = true

        for(checker : Int <- 2 to Math.sqrt(number).toInt if number%checker==0 if isPrime) isPrime = false

        if(isPrime)  number
      }

    println("prime numbers")
    for(prime <- primeNumbers)
       println(prime)
 }

回答1:

You can use collect on your output

primeNumbers.collect{
  case i : Int => i
}
res2: IndexedSeq[Int] = Vector(2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97)


回答2:

so the underlying problem here is that your yield block effectively will return an Int or a Unit depending on isPrime this leads your collection to be of type AnyVal because that's pretty much the least upper bound that can represent both types. Unit is a type only inhabited by one value which is represented as an empty set of round brackets in scala () so that's what you see in your list.

As Puneeth Reddy V said you can use collect to filter out all the non-Int values but I think that is a suboptimal approach (partial functions are often considered a code-smell depending on what type of scala-style you do). More idiomatic would be to rethink your loop (such for loops are scarcely used in scala) and this could be definitely be done using a foldLeft operation maybe even something else.



回答3:

The reason is if else returning two types of value one is prime value and another is empty. You could return 0 in else case and filter it before printing it.

scala> def primeNumber(range: Int): Unit ={
     | 
     |     val primeNumbers: IndexedSeq[Int] =
     | 
     |       for (number :Int <- 2 to range) yield{
     | 
     |         var isPrime = true
     | 
     |         for(checker : Int <- 2 to Math.sqrt(number).toInt if number%checker==0 if isPrime) isPrime = false
     | 
     |         if(isPrime)  number
     |         else 
     |         0
     |       }
     | 
     |     println("prime numbers" + primeNumbers)
     |     for(prime <- primeNumbers.filter(_ > 0))
     |        println(prime)
     |  }
primeNumber: (range: Int)Unit

scala> primeNumber(10)
prime numbersVector(2, 3, 0, 5, 0, 7, 0, 0, 0)
2
3
5
7

But we should not write the code the way you have written it. You are using mutable variables. Try to write code in an immutable way. For example

scala> def isPrime(number: Int) =
     | number > 1 && !(2 to number - 1).exists(e => e % number == 0)
isPrime: (number: Int)Boolean

scala> def generatePrimes(starting: Int): Stream[Int] = {
     | if(isPrime(starting))
     | starting #:: generatePrimes(starting + 1)
     |  else
     | generatePrimes(starting + 1)
     | }
generatePrimes: (starting: Int)Stream[Int]

scala> generatePrimes(2).take(100).toList
res12: List[Int] = List(2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101)


标签: scala