使用 while 循环重复一个函数，直到在 r 中满足条件

Question

这可能不是最有效的方法，但我在书的帮助下为水果机创建了一个代码（有效）（不确定我是否可以分享标题，但它是非常适合新手）。不过这段代码很简单，它运行一个函数，依次调用另外两个函数：

1. pull() 通过从预先确定的向量中随机选择字符来模拟轮子机制。
2. prize() 分析来自 pull() 的响应，以使用查找表为符号添加值。

代码有效，但我想对其进行修改，以便代码运行，直到达到三颗钻石的头奖； "DD"。我还希望它量化所有其他响应，直到它达到这个目标，所以我尝试了一个 while 循环，它似乎没有产生任何结果：

 play2 <- function(){
  response <- pull()

  # I put the while loop straight after calling the first function,
  # hoping that if the condition is not met, then the first function
  # runs again, creating a new "response" until conditions are met.

  while (sum(response == "DD") != 3) {

  # I set the condition so that if the jackpot is not achieved, the 
  # while loop causes the other responses to be investigated. It 
  # it determines this by counting booleans. 

    # The following assignments are designed to begin the count for how
    # many responses that are considered prizes occur, and how often no
    # prize occurs.

    cherry_prize <- 0
    B_prize <- 0
    BB_prize <- 0
    BBB_prize <- 0
    seven_prize <- 0
    no_prize <- 0

    if ("C" %in% response){
      cherry_prize <- cherry_prize + 1



      # A cherry prize occurs if any "C" is returned. The following
      # statements will asses the other non-jackpot three of a kinds.

    }
    else if (sum(response == "B") != 3) {
      B_prize <- B_prize + 1
    }
    else if (sum(response == "BB") != 3) {
      BB_prize <- BB_prize + 1
    }
    else if (sum(response == "BBB") != 3) {
      BBB_prize <- BBB_prize + 1
    }
    else if (sum(response == "7") != 3) {
      seven_prize <- seven_prize + 1
    }
    else {
      no_prize <- no_prize + 1
    }
  }

  # After the jackpot is achieved, the number of other prizes (or lack
  # thereof) that were returned are quantified, prior to the jackpot 
  # were won. The jackpot triplicate would then also be returned and
  # quantified.

  print(cherry_prize)
  print(B_prize)
  print(BB_prize)
  print(BBB_prize)
  print(seven_prize)
  print(no_prize)
  print(response)

  # The last function quantifies the jackpot, once it has been reached.
  prize(response)
}

希望我提供了足够的信息。非常感谢所有帮助，希望它可以帮助其他人。

Answer 1

感谢以上建议。因此，我按照说明进行操作，遇到了阻止该函数运行的警告。

Error in symbols == "DD" : 
comparison (1) is possible only for atomic and list types

所以我再次认为我使用的编码方法不是最有效的方法，但我用as.list()纠正了这个问题。

play2 <- function(){

  runs <- 0
  cherry_prize <- 0
  B_prize <- 0
  BB_prize <- 0
  BBB_prize <- 0
  seven_prize <- 0
  no_prize <- 0

  while (sum(as.list(response) == "DD") != 3) {

    runs <- runs + 1

    response <- pull()

    if ("C" %in% response){
      cherry_prize <- cherry_prize + 1
    }
    else if (sum(as.list(response) == "B") != 3) {
      B_prize <- B_prize + 1
    }
    else if (sum(as.list(response) == "BB") != 3) {
      BB_prize <- BB_prize + 1
    }
    else if (sum(as.list(response) == "BBB") != 3) {
      BBB_prize <- BBB_prize + 1
    }
    else if (sum(as.list(response) == "7") != 3) {
      seven_prize <- seven_prize + 1
    }
    else {
      no_prize <- no_prize + 1
    }
  }
  print(runs)
  print(cherry_prize)
  print(B_prize)
  print(BB_prize)
  print(BBB_prize)
  print(seven_prize)
  print(no_prize)
  print(response)
  score(response)
}

这给出了输出：

> play2()
[1] 430118
[1] 12526
[1] 411199
[1] 6393
[1] 0
[1] 0
[1] 0
[1] "DD" "DD" "DD"
[1] 800

我未显示的部分代码影响了我的 BBB，7 并且没有奖品。