Logical operators and strings: function error

Question

Here is a minimal reproducible example that generates the error:

 comb3 <- function(x) {
      if (x == "Unable to do") {
        x = 0
      } 
    }

Here is my original function:

 comb <- function(x) {
      if (x == "Unable to do") {
        x = 0
      } else if (x == "Very difficult to do") {
        x = 1
      } else if (x == "Somewhat difficult to do") {
        x = 2
      } else if (x == "Not difficult") {
        x = 3
      } 
    }

I am trying to use this function on a column sampled below. I get this error:

Warning messages:
1: In if (x == "Unable to do") { :
  the condition has length > 1 and only the first element will be used
2: In if (x == "Very difficult to do") { :
  the condition has length > 1 and only the first element will be used

Here is a sample of what the data in one column looks like:
sample <- c("Unable to do", "Somewhat difficult to do", "Very difficult to do", "Unable to do", "Not difficult","Unable to do","Very difficult to do", "Not difficult", "Unable to do", "Not difficult")

It looks like you want `case_when` instead if you are using `mutate_at`. But you should show your `dplyr` code as well and include some sample data to make a proper [reproducible example](https://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example) — MrFlick, Oct 22 '19 at 15:26
I am getting the error when performing function(column) on a single column without using the dplyr function that is why I didn't include it. I haven't attempted to use this function with dplyr yet. @MrFlick — Krishna, Oct 22 '19 at 15:48
@slava-kohut Actually the sample I provided is one column. Let me clarify that. — Krishna, Oct 22 '19 at 15:49
@Krishna `if()` is not a vectorized control flow statement. If i have `x<-c(3,10)` and I do `if(x>5) print("ok")`, this will not work because there are two values of `x` and some of the values are less than 5 and some greater. One `if` statement doesn't work in that case. You need `ifelse()` or `if_else` or `case_when` since functions operate on all rows of a data.frame at a time. — MrFlick, Oct 22 '19 at 15:51
@MrFlick Ok that makes a lot of sense. I will investigate those three options. Thank you so much for your guidance! — Krishna, Oct 22 '19 at 16:00
@MrFlick Using case_when in my function worked perfectly! Thanks again. — Krishna, Oct 22 '19 at 16:06

score 0 · Answer 1 · answered Oct 22 '19 at 16:05

The warning message describes the issue with your code fairly well. if is a function which expects a length one logical vector as input. Thus to use conditionals over a vector, you should instead use something like ifelse, or as MrFlick said, to use case_when or mutate_at.

An equivalent version of your function using ifelse would be something like this:

comb1 <- function(x) {
  ifelse(x == "Unable to do", 
    0,
    ifelse (x == "Very difficult to do",
      1,
      ifelse(x == "Somewhat difficult to do",
        2,
        ifelse(x == "Not difficult",
          3,
          ## If not match then NA
          NA
        )
      )
    )
  )
}

Note that this is very difficult to read, as the ifelse calls are chained together. You could therefore avoid this by using a slightly modified version of your function in a call to sapply to accomplish the same thing

comb2 <- function(x) {
  sapply(x, function(y) {
    if (y == "Unable to do") {
      0
    } else if (y == "Very difficult to do") {
      1
    } else if (y == "Somewhat difficult to do") {
      2
    } else if (y == "Not difficult") {
       3
    }
  ## USE.NAMES = FALSE means that the output is not named, and has no other effect
  }, USE.NAMES = FALSE)
}

You could also use factors, which are internally coded as integers starting at 1, and (ab)use this to convert from strings to numbers:

comb3 <- function(x) {
  fac <- factor(x, 
    levels = c(
      "Unable to do",
      "Very difficult to do",
      "Somewhat difficult to do",
      "Not difficult"
    )
  )
  as.numeric(fac) - 1
}

The output of these 3 versions is identical, and is a great example of how there can be many ways to accomplish things in R. This can sometimes be a curse rather than a gift.

sample <- c("Unable to do", "Somewhat difficult to do", "Very difficult to do", "Unable to do", "Not difficult","Unable to do","Very difficult to do", "Not difficult", "Unable to do", "Not difficult")
comb1(sample)
# [1] 0 2 1 0 3 0 1 3 0 3
comb2(sample)
# [1] 0 2 1 0 3 0 1 3 0 3
comb3(sample)
# [1] 0 2 1 0 3 0 1 3 0 3

Thank you for the detailed answer aocall and demonstrating the results of each option. I will look into their implementation! — Krishna, Nov 05 '19 at 16:00

Logical operators and strings: function error

1 Answers1