Question

假设x是一个holdem手牌（棋盘+底牌），其中J：A = 11:14且A可能= 1.套装并不重要。你只需要检查一下。

x <- c(2,5,6,7,8,9,14)

这是holdem包的straight1功能。除了功能结束时的for循环，我理解了一切。有人可以解释这部分是如何工作的。我输了。

function(x){
a1 = sort(unique(x))
if (length(a1)<4.5) return(0)
a3 = 0
n = length(a1)
if(a1[n] == 14) a1 = c(1,a1) ## count ace as both 1 and 14
a2 = length(a1)
for(j in c(5:a2)){ ## j will be the potential highest card of straight
if( sum(15^c(1:5) * a1[(j-4):j]) == sum(15^c(1:5) * ((a1[j]-4):a1[j]))) a3 = a1[j]
}
a3
}   ## end of straight1

Answer 1

我认为这个问题需要澄清一下需要什么：

Qstraight <-function(x){
     a1 = sort(unique(x))
     if (length(a1)<4.5) return(0)
     a3 = 0
     n = length(a1)
     if(a1[n] == 14) a1 = c(1,a1) ## count ace as both 1 and 14
     a2 = length(a1)
     for(j in c(5:a2)){ 
         ## j will be the potential highest card of straight
         if( sum(15^c(1:5) * a1[(j-4):j]) == 
             sum(15^c(1:5) * ((a1[j]-4):a1[j]))) a3 = a1[j]
                        }
    a3
    }   ## end

所以结果是......

 Qstraight(x)
#[1] 9  # i.e a "nine-high straight
x2 <- c(2,5,6,7,8,10,14)
 Qstraight(x2)
#[1] 0   # i.e not a straight at all.

我可能会对唯一值进行排序，然后取rle的最大长度（diff（unique（sort（x）））$ values == 1 ..或者更少依赖于模运算。

扑克：Holdem包直1功能？

1 个答案: