Dealing with TRUE, FALSE, NA and NaN

Question

Here is a vector

a <- c(TRUE, FALSE, FALSE, NA, FALSE, TRUE, NA, FALSE, TRUE)

I\'d like a simple function that returns TRUE

Answer 1

I like the is.element-function:

is.element(a, T)

Answer 2

So you want TRUE to remain TRUE and FALSE to remain FALSE, the only real change is that NA needs to become FALSE, so just do this change like:

a[ is.na(a) ] <- FALSE

Or you could rephrase to say it is only TRUE if it is TRUE and not missing:

a <- a & !is.na(a)

Answer 3

To answer your questions in order:

1) The == operator does indeed not treat NA's as you would expect it to. A very useful function is this compareNA function from r-cookbook.com:

  compareNA <- function(v1,v2) {
    # This function returns TRUE wherever elements are the same, including NA's,
    # and false everywhere else.
    same <- (v1 == v2)  |  (is.na(v1) & is.na(v2))
    same[is.na(same)] <- FALSE
    return(same)
   }

2) NA stands for "Not available", and is not the same as the general NaN ("not a number"). NA is generally used for a default value for a number to stand in for missing data; NaN's are normally generated because a numerical issue (taking log of -1 or similar).

3) I'm not really sure what you mean by "logical things"--many different data types, including numeric vectors, can be used as input to logical operators. You might want to try reading the R logical operators page: http://stat.ethz.ch/R-manual/R-patched/library/base/html/Logic.html.

Hope this helps!

Answer 4

You don't need to wrap anything in a function - the following works

a = c(T,F,NA)

a %in% TRUE

[1]  TRUE FALSE FALSE

Answer 5

Taking Ben Bolker's suggestion above you could set your own function following the is.na() syntax

is.true <- function(x) {
  !is.na(x) & x
}

a = c(T,F,F,NA,F,T,NA,F,T)

is.true(a)
[1]  TRUE FALSE FALSE FALSE FALSE  TRUE FALSE FALSE  TRUE

This also works for subsetting data.

b = c(1:9)
df <- as.data.frame(cbind(a,b))

df[is.true(df$a),]

  a b
1 1 1
6 1 6
9 1 9

And helps avoid accidentally incorporating empty rows where NA do exist in the data.

df[df$a == TRUE,]

      a  b
1     1  1
NA   NA NA
6     1  6
NA.1 NA NA
9     1  9