Question

我在课程上做scala课程。我将完成第6周的任务。我创建了一个函数，用于查找单词中char的出现。测试方法相对于预期的结果序列失败。

以下是问题描述：

/** A word is simply a `String`. */
type Word = String
/** Converts the word into its character occurrence list.
*
*  Note: the uppercase and lowercase version of the character are treated as the
*  same character, and are represented as a lowercase character in the occurrence list.
*
*  Note: you must use `groupBy` to implement this method!
*/

def wordOccurrences(w: Word): Occurrences = {

   def breakWord(s: List[Char], xs: List[(Char, Int)]): List[(Char, Int)] = s match {
   case Nil => xs
   case char :: rest =>     breakWord(s.tail, matchTuple(xs, char))
   }

   breakWord(w.toList, List[(Char, Int)]())
}

 def matchTuple(tupleList: List[(Char, Int)], char: Char): List[(Char, Int)] = tupleList match {
 case Nil => tupleList:+ (char.toLower, 1)
 case pair :: restOfList => {
   if(pair._1.toLower == char.toLower)
      restOfList :+(pair._1, pair._2+1)
   else
      tupleList:+ (char.toLower, 1)
  }
}

有人可以指出我做错了什么。我不需要直接回答，只是对我的序列有什么问题的逻辑提示。这是测试及其结果：

 assert(wordOccurrences("Robert") === List(('b', 1), ('e', 1), ('o', 1), ('r', 2), ('t', 1)))

这是输出：

[info] - wordOccurrences: Robert *** FAILED ***
[info]   List((o,1), (b,1), (e,1), (r,2), (t,1)) did not equal List((b,1), (e,1), (o,1), (r,2), (t,1)) (AnagramsSuite.scala:20)

更新：重构了我的功能：

def wordOccurrences(w: Word): Occurrences = {

    def breakWord(s: List[Char], xs: List[(Char, Int)]): List[(Char, Int)] = s match {
      case Nil => xs
      case char :: rest =>     breakWord(s.tail, xs:+ (char, 1))
    }

    breakWord(w.toList, List[(Char, Int)]()).groupBy( pair => pair._1.toLower)
      .map(entry => (entry._1.toLower, (entry._1.toLower, entry._2.size)) )
      .values.toList.sorted
  }

看起来太糟糕了，但我会继续努力改进这种方法。

Answer 1

你得到的结果很好，它只是按照不同的顺序，~~但是任务似乎没有说明需要特定订单~~。因此，您需要在之后对其进行排序（在修改解决方案之前或之后使用groupBy）。您可以调用sorted方法来执行此操作（或查找sortBy / sortWith）。

Answer 2

您可以通过此方法计算出现次数 -

REM Used to keep variable for duration of script
setlocal enabledelayedexpansion

cls
@echo off 

echo Math Experts Commands
set /p loop=Loop how many times: 

for /L %%A IN (1,1,!loop!) DO (
    REM Request the first number
    echo Enter your first digit you wish to add

    REM Set first user input to "num1"
    set /p num1=

    REM Request the second number
    echo Enter the second digit you wish to add the first by

    REM Set second usr input to "num2"
    set /p num2=

    REM Add the 2 values then set the outcome to "total"
    set /a total=num1+num2

    REM Clear the screen
    cls

    if !total! leq 4 (
    echo ####################################################################################
    echo #    Hosts Req:      #      Mask:      #      Netmask:      #      Max Supported:  #                                                  
    echo #    !total!                      /30           255.255.255.252                4         #
    echo ####################################################################################
    echo.
    ) else if !total! gtr 4 and leq 8 (   REM <----WHERE THE PROBLEM IS. 
    echo ####################################################################################
    echo #    Hosts Req:      #      Mask:      #      Netmask:      #      Max Supported:  #                                                  
    echo #    !total!                      /29            255.255.255.248               8        #
    echo ####################################################################################
    echo. 
    )
    )

    REM Pause the console
    pause
    echo.
)

此方法在第4周分配中使用：）

scala coursera分配方法失败测试

2 个答案: