write R codes to find median of 3 numbers with if-CodePudding

I still get errors with the following code. Could anyone help me please? I need to find the median number of a set of three numbers.

find_median = function(a, b, c){
    if(a > b){
      if(a > c){
      if(c > b){
        return(b)}
      else{return(c)}
   }
   else {return(a)}
  }
  else if(b > a){
    if(b > c){
      if(c > a){
        return(c)}
      else {return(a)}
    }
    else {return(b)}
  }
  else if(c > a){
      if(c > b){
        if(a > b){
          return(a)}
        else {return(b)}
     }
      else {return(c)}
  }
}

CodePudding user response：

Just expanding on my comment, but the hardest part of truth tables (or any function really) is accounting for all the variations of inputs that might present. Often this is handled by error checking, essentially saying, not designed to do that or work with that. Here, the question is what's the number in the middle, so taking the elegant permutations by @Museful :

permutations <- function(n){
    if(n==1){
        return(matrix(1))
    } else {
        sp <- permutations(n-1)
        p <- nrow(sp)
        A <- matrix(nrow=n*p,ncol=n)
        for(i in 1:n){
            A[(i-1)*p 1:p,] <- cbind(i,sp (sp>=i))
        }
        return(A)
    }
}

my3<-matrix(c(1:3)[permutations(3)], ncol = 3)
> my3
     [,1] [,2] [,3]
[1,]    1    2    3
[2,]    1    3    2
[3,]    2    1    3
[4,]    2    3    1
[5,]    3    1    2
[6,]    3    2    1

# all possibles now known

if (my3[6,1] >= my3[6,2] & my3[6,2] >= my3[6,3]) print(my3[6,2])
[1] 2
> if (my3[5,1] >= my3[5,2] & my3[5,2] <= my3[5,3]) print(my3[5,3])
[1] 2
> if (my3[4,1] >= my3[4,3] & my3[4,1] <= my3[4,2]) print(my3[4,1])
[1] 2
> if (my3[3,1] >= my3[3,2] & my3[3,2] <= my3[3,3]) print(my3[3,1])
[1] 2
> if (my3[2,1] <= my3[2,2] & my3[2,2] >= my3[2,3]) print(my3[2,3])
[1] 2
> if (my3[1,1] <= my3[1,2] & my3[1,2] <= my3[1,3]) print(my3[1,2])
[1] 2

Now substitute the [,1], [,2], [,3], to a,b,c; print to return, and else between...

CodePudding user response：

Here is another approach that uses if statements to sort the values like the code in the original question but more compact. Getting the nested "{}" is the real challenge. We can order the values by making three comparisons a<b, a<c, and b<c. That makes it a bit easier to follow the code:

find_median = function(a, b, c) {
ab <- a < b
ac <- a < c
bc <- b < c

if (ab) { 
    if (ac) { 
        if (bc) {return(b) 
        } else return(c)
    } else return(a)
} else if (ac) { return(a)
    } else if (bc) { return(c)
    } else return(b)
}

Now test it:

a <- sample.int(50, 1)
b <- sample.int(50, 1)
c <- sample.int(50, 1) 
a; b; c
# [1] 25
# [1] 10
# [1] 36
find_median(a, b, c)
# [1] 25