r - 尝试在 R 中随机化石头、剪刀、布游戏

Question

我创建了一个 RPS 函数，可以使计算机的移动随机化一回合。例如，如果您输入 rps("rock")，该函数将随机吐出石头、布、剪刀以及结果(赢、输、平)。

rps = function(move){
  options = c("rock", "paper", "scissors")
  comp.move = sample(options, size = 1)
  if(move == "rock" & comp.move == "rock"){
    names(comp.move) = "tie"
}else
if(move == "rock" & comp.move == "scissors"){
    names(comp.move) = "loss"
}else
if(move == "rock" & comp.move == "paper"){
    names(comp.move) = "win"
}else
if(move == "paper" & comp.move == "paper"){
    names(comp.move) = "tie"
}else
if(move == "paper" & comp.move == "scissors"){
    names(comp.move) = "win"
}else
if(move == "paper" & comp.move == "rock"){
    names(comp.move) = "loss"
}else
if(move == "scissors" & comp.move == "scissors"){
    names(comp.move) = "tie"
}else
if(move == "scissors" & comp.move == "rock"){
    names(comp.move) = "win"
}else
if(move == "scissors" & comp.move == "paper"){
    names(comp.move) = "loss"
}
  return(comp.move)
}

现在，我想玩一个发生了 50 次这种情况的游戏版本。我已经创建了一个长度为 50 的向量，其中包含预先记录的人类 Action 。

human.move = c("rock", "paper", "scissors", "paper", "paper", "rock", "scissors", "rock", "rock", "paper", "paper", "scissors", "rock", "rock", "paper", "paper", "paper", "scissors", "paper", "rock", "paper", "rock", "rock", "scissors", "scissors", "paper", "rock", "paper", "scissors", "rock", "paper", "paper", "scissors", "rock", "paper", "rock", "paper", "paper", "scissors", "scissors", "paper", "rock", "rock", "scissors", "scissors", "rock", "paper", "scissors", "scissors", "rock")

如何让计算机对该向量中的每个元素运行 rps 函数，并给出结果？目前，我的函数仅采用人类移动向量的第一个元素(即岩石)，并且仅将其随机移动与其进行比较。

rps.random=function(move,n=50){
comp.moves = vector("character")

for(i in human.move){
comp.move=rps(move[i])
comp.moves = append(comp.moves,comp.move, after = length(comp.moves))
}
return(comp.moves)
}

Answer 1

这里有一些建议。首先，更新您的功能。您可以通过使用一些带有 | 运算符的“或”语句来减少所有这些 if() 语句，以表示获胜或失败，并且使用 identical( ) 平局函数。另外，据我了解，像这样的随机游戏应该一次玩一轮，并重复一轮以上。因此，该函数检查 move 中给出的单长度向量。

rps <- function (move) {
    opts <- c("rock", "paper", "scissors")
    stopifnot(length(move) == 1L, move %in% opts)
    cpu <- sample(opts, 1)
    names(cpu) <- if (identical(move, cpu)) {
        "tie"
    }
    else if (move == "rock" & cpu == "scissors" | move == "paper" & 
        cpu == "rock" | move == "scissors" & cpu == "paper") {
        "win"
    }
    else {
        "loss"
    }
    cpu
}

这里有几个运行示例 -

rps("paper")
#    win 
# "rock" 
rps("rock")
#    loss 
# "paper" 

现在您无需手动创建包含 50 个项目的向量。我们可以通过使用 move 参数中的示例来重复该过程 50 次。

replicate(50, rps(sample(c("rock", "paper", "scissors"), 1)))
#        win        win       loss        win       loss        tie        tie 
# "scissors"    "paper"    "paper"    "paper"     "rock"    "paper" "scissors" 
#        win       loss       loss        tie       loss        win        win 
#     "rock"    "paper"    "paper"    "paper"    "paper" "scissors" "scissors" 
#        tie       loss        tie        win        win        tie       loss 
# "scissors"     "rock" "scissors"    "paper"    "paper" "scissors"     "rock" 
#        tie        tie       loss        tie        win        win        win 
#     "rock" "scissors" "scissors" "scissors"     "rock" "scissors" "scissors" 
#        win        win       loss        win        tie        tie        win 
# "scissors" "scissors"    "paper"    "paper"    "paper"    "paper"    "paper" 
#        win        tie        win        tie        tie        win        win 
#    "paper"     "rock"     "rock"    "paper" "scissors"    "paper" "scissors" 
#       loss        tie        tie       loss        tie        win        win 
#     "rock"     "rock"    "paper" "scissors" "scissors"     "rock"     "rock" 
#        win 
#    "paper" 

当然，如果您确实有输入向量，我们可以通过 sapply() 运行它，以查看与上面相同的结果(名称中包含或不包含输入向量)。

sapply(c("rock", "paper", "scissors"), rps)
#   rock.loss    paper.win scissors.tie 
#     "paper"       "rock"   "scissors" 
sapply(c("rock", "paper", "scissors"), rps, USE.NAMES = FALSE)
#       tie       loss        win 
#    "rock" "scissors"    "paper" 

更新:我突然意识到，在上述情况下，我们从 replicate() 获得的输入是不知道的，所以这里有一个使固定。如果我们将函数的最后一行更改为矩阵(或列表)，我们还可以包含输入内容。我们可以在最后一行使用类似的内容 =

matrix(c(move, cpu), 1, dimnames = list(wlt, c("move", "cpu")))

然后我们会得到以下结果 -

rps("paper")
#      move    cpu       
# loss "paper" "scissors"
rps("paper")
#     move    cpu    
# tie "paper" "paper"

现在我们知道我们输入了什么作为输入，而无需做任何额外的工作。我们还可以与我们的老 friend do.call(rbind, ...) 一起将复制放入一个矩阵中。