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scala - 使用 scala 生成句子使用流

问题描述

我想使用 Stream 生成多个句子。我现在拥有的是我可以生成 1 个句子。

  def main(args: Array[String]): Unit = {
    println((generateSentence take 1).mkString)
  }

这是我到目前为止的结果

  zIYow5ZJn92TjbcKbTvCf vaRNqZs80Fi4LcU7 8izJggPbjz9brbMtWmvo bGK

现在如果我想取2个句子，流将继续写入第一个句子

现在我如何在该流中生成多个句子到一个数组（或一个列表）中。?

我想过将类型更改为，Stream[List[String]]但我不知道如何以正确的方式添加生成的（它给了我Exception in thread "main" java.lang.StackOverflowError）

使用 Stream[List[String]] 代码：

  /**
   * Generate one sentences between 2 and 25 words
   * @return
   */
  def generateSentence : Stream[List[String]] = {
    def sentences : List[String] = {
      sentences.::((generateWord take between(2, 25)).mkString) /* This line gave me the Exception StackOverflow */
    }
    Stream continually sentences
  }

我写的原始代码

  /**
   * Generate one word between 2 and 25 char
   * @return
   */
  def generateWord: Stream[String] = {
    def word : String = {
      (Random.alphanumeric take between(2, 25)).mkString.concat(" ")
    }
    Stream continually word
  }

  /**
   * Generate one sentences between 2 and 25 words
   * @return
   */
  def generateSentence : Stream[String] = {
    def sentences : String = {
      (generateWord take between(2, 25)).mkString 
    }
    Stream continually sentences
  }
  
  /* This one is from the Random library, as it was introduced with 2.13 (so I just backported it into my 2.12)*/

  def between(minInclusive: Int, maxExclusive: Int): Int = {
    require(minInclusive < maxExclusive, "Invalid bounds")

    val difference = maxExclusive - minInclusive
    if (difference >= 0) {
      Random.nextInt(difference) + minInclusive
    } else {
      /* The interval size here is greater than Int.MaxValue,
       * so the loop will exit with a probability of at least 1/2.
       */
      @tailrec
      def loop(): Int = {
        val n = Random.nextInt()
        if (n >= minInclusive && n < maxExclusive) n
        else loop()
      }
      loop()
    }
  }
}

标签： scalageneratescala-streams

解决方案

棘手的点:)

您的方法生成无限递归：

def sentences : List[String] = {
      sentences.::((generateWord take between(2, 25)).mkString)
}

是这样的：

def sentences : List[String] = {
      val result = sentences()
      result.::((generateWord take between(2, 25)).mkString)
}

它是方式，很明显它无限地调用自己。所以为了解决你的问题，你可以使用toList

def sentences : List[String] = {
  generateWord take between(2, 25) toList
}

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