在r中，如何使函数将特定字母识别为我要解决的未知值？

Question

I am new to R and I am just trying to write a simple function where the user will input 2 known sides of a right triangle and then the output will be the length of the third side. I have it working with a "fix" but it's a duct tape solution. The idea is that the user will be able to call on my function, named pythag. I want them to be able to enter in values by typing pythag(3,b,5). This would tell the program that one leg of the triangle is length 3, the hypotenuse is length 5, solve for the other leg, leg b. But the only way I've gotten it to work is if the user types in pythag(3, "b", 5). Here is the code:

pythag <- function(a, b, c) {
    if (a %in% "a") {
            answer <- sqrt(c^2 - b^2)
    } else if ( b %in% "b") {
            answer <- sqrt(c^2 - a^2)
    } else if (c %in% "c") {
            answer <- sqrt(a^2 + b^2)
    }
    answer }

If I take away the "" in the code so that it is stated as (a %in% a), it doesn't work. Any help would be appreciated. I've also tried (a==a) and (a=a).

Answer 1

I am not sure why you would want to design your function this way, since the labels of triangle sides are largely arbitrary. But if you insist, I would approach it this way:

pythag <- function(A=NA, B=NA, C=NA) {
  # want to check inputs
  inputs <- c(A, B, C)
  if (sum(is.na(inputs)) != 1)
    stop("Exactly two inputs are required!")
  if (is.na(A)) return(sqrt(C^2 - B^2))
  if (is.na(B)) return(sqrt(C^2 - A^2))
  if (is.na(C)) return(sqrt(A^2 + B^2))
}
pythag(A=3, B=4)
[1] 5
pythag(A=3, C=5)
[1] 4
pythag(B=4, C=5)
[1] 3

Of course, this function is still not great, since it is not "immune" from bad inputs. But it should generally work for you.

Answer 2

This answer relies on your users naming the known inputs. It then spits out the name and value of the unknown input.

# Create the Pythageorean Theorem function
# A squared + B squared = C squared
PythagoreanTheorem <- function( a = NULL, b = NULL, c = NULL ){
  # Takes in two inputs and calculates 
  # the missing parameter's value

  # Args: two numeric values

  # Stop messages
  if( !is.null( x = a ) &&
      !is.null( x = b ) &&
      !is.null( x = c ) ){
    stop( "All three arguments cannot be used. Use two.")
  }

  if( !is.numeric( x = a ) && !is.null( x = a ) | 
      !is.numeric( x = b ) && !is.null( x = b ) |
      !is.numeric( x = c ) && !is.null( x = c ) ){
    stop( "Ensure both inputs are numeric.")
  }

  if( !is.null( a ) && !is.null( b ) ){

    return( 
      setNames( 
        object = sqrt( x = c( a^2 + b^2 ) )
        , nm = "c"
      )
    )

  } else if( !is.null( a ) && !is.null( c ) ){

    return( 
      setNames( 
        object = sqrt( x = c( a^2 + c^2 ) )
        , nm = "b"
      )
    )

  } else if( !is.null( b ) && !is.null( c ) ){

    return( 
      setNames( 
        object = sqrt( x = c( b^2 + c^2 ) )
        , nm = "a"
      )
    )
  }

} # end of PythagoreanTheorem() function

# test PythagoreanTheorem() function
PythagoreanTheorem( c = 3, a = 4 )
# b 
# 5 
class( x = PythagoreanTheorem( c = 3, a = 4 ) )
# [1] "numeric"
PythagoreanTheorem( a = 2, b = 5, c = 6)
# Error in PythagoreanTheorem(a = 2, b = 5, c = 6) : 
#   All three arguments cannot be used. Use two.

# end of script #

Answer 3

Try this with a slightly different user interface:

pythag <- function(a = 0, b = 0, c = 0) {
  stopifnot(is.numeric(a), is.numeric(b), is.numeric(c), (a!=0) + (b!=0) + (c!=0) == 2)
  if (c == 0) sqrt(a^2 + b^2) else sqrt(c^2 - a^2 - b^2)
}

# tests

pythag(a = 3, c = 5)
## [1] 4
pythag(3,,5) # same
## [1] 4

pythag(a = 4, b = 3)
## [1] 5
pythag(4, 3) # same
## [1] 5

pythag(3, "X")
## Error: is.numeric(b) is not TRUE

pythag(1)
## Error: (a != 0) + (b != 0) + (c != 0) == 2 is not TRUE

pythag(1, 2, 3)
## Error: (a != 0) + (b != 0) + (c != 0) == 2 is not TRUE