Change the level of a factor based on the level of another factor

Question

I have a data set with many variables, two of which called "animal" and "plant". Both variable are factors, and both are binary, ie they are either a text value, or NA.

For example:

animal <- c(NA, NA, "cat", "cat", NA)
plant  <- c("ivy", NA, "ivy", NA, NA)
value  <- c(1:5)
df     <- data.frame(animal, plant, value)

> df
  animal plant value
1   <NA>   ivy     1
2   <NA>  <NA>     2
3    cat   ivy     3
4    cat  <NA>     4
5   <NA>  <NA>     5

When the value of plant is "ivy" and the value of animal is "cat", I want to change the value of plant to NA (i,e, the two things can not be true and the animal value takes priority. I don't any changes in my other variables

I've tried the following but get an error message:

df <- df %>% if (isTRUE(animal == "cat")) {plant==NA}

Error in if (.) isTRUE(animal == "cat") else { : 
  argument is not interpretable as logical
In addition: Warning message:
In if (.) isTRUE(animal == "cat") else { :
  the condition has length > 1 and only the first element will be used

My goal output is:

> df
  animal plant value
1   <NA>   ivy     1
2   <NA>  <NA>     2
3    cat  <NA>     3
4    cat  <NA>     4
5   <NA>  <NA>     5

I would really appreciate any help. I'm sure there is a really simple way of doing this, maybe I can't see the wood for the trees.

Answer 1

library(dplyr)    

df %>% 
      mutate(plant = case_when(animal == 'cat' & plant == 'ivy' ~ NA_character_,
                               TRUE ~ plant))

This gives us:

  animal plant value
1   <NA>   ivy     1
2   <NA>  <NA>     2
3    cat  <NA>     3
4    cat  <NA>     4
5   <NA>  <NA>     5

Answer 2

You could also do:

df[!(is.na(df$animal)|is.na(df$plant)),'plant'] <- NA
df
  animal plant value
1   <NA>   ivy     1
2   <NA>  <NA>     2
3    cat  <NA>     3
4    cat  <NA>     4
5   <NA>  <NA>     5

This can also be expressed as:

df[!is.na(df$animal) & !is.na(df$plant),'plant'] <- NA

Answer 3

Your problem seems to be simpler than you think. You can achieve the same result simply by turning all plants, where animal is not NA , to NA :

df$plant[!is.na(df$animal)] <- NA

Or a bit fancier:

is.na(df$plant) <- !is.na(df$animal)

Answer 4

The problem here is that == does not work intuitively with the NA values in R.

> df[df$animal=="cat",]
     animal plant value
NA     <NA>  <NA>    NA
NA.1   <NA>  <NA>    NA
3       cat   ivy     3
4       cat  <NA>     4
NA.2   <NA>  <NA>    NA

Here for example all lines are returned because NA == "ANYTHING" returns NA .

You could define this function which returns TRUE if both x and y are equal and not NA , or if both are NA .

is.equal.force <- `%===%` <- function(x,y, vect=T) {
  res <- ifelse(is.na(x),is.na(y),ifelse(!is.na(y)&!is.na(x),x==y, NA))
  if(!vect){
    res <- all(res)
  }
  return(res)
}

Then the solution to your problem becomes simply:

df[df$animal%===%"cat"&df$plant%===%"ivy","plant"] <- NA
df
  animal plant value
1   <NA>   ivy     1
2   <NA>  <NA>     2
3    cat  <NA>     3
4    cat  <NA>     4
5   <NA>  <NA>     5

Note that the correct syntax was used here.