Convert factor into logical datatype

Question

I have a two levels factor in my data that I want to convert to logical

a <- str(df$y)
a
Factor w/ 2 levels "0","1": 1 1 1 1 1 1 1 1 1 1 ...

I use as.logical(df$y) to convert them into logical. However, the data turn into NA

summary(a)

      Mode    NA's 
    logical  500000

At which point do I fail to convert the data?

Poisonous

Since you have a factor variable, I suggest you convert to character, then to integer and finally to logical: as.logical(as.integer(levels(a))[a]). You can read about it in the "Value" section of ?as.logical – Awlwort

There is a short cut: y <- factor(0:1); str(y); as.logical(as.integer(y) - 1L) returns [1] FALSE TRUE – Quilt

Just test for equality with whatever level you want to be "true": df$y == "1". The result will be your logical vector. – Leer

Answer 1

At which point do I fail to convert the data?

I'd argue that you at no point fail to convert the data, it's the function that is a bit odd and fails to understand the nature of your data.

If you read ?as.logical you'll see that when input is factor the levels (which are character) are used in the conversion. The only valid character strings are all variations of "true" and "false", everything else, including "0" and "1", returns NA. 0 and 1 are however interpreted as FALSE and TRUE, respectively, when they are given as numeric, hence all the following works:

y <- factor(c(0, 1, 1, 0))

as.logical(as.integer(levels(y)[y]))
as.logical(as.integer(y) - 1L)
as.logical(as.integer(as.character(y)))

A bit cumbersome, I know, but that's how it is.

Jeanelle

Also y == 1, much less cumbersome. – Leer

Hah, true. Doesn't tell you much about why the straight as.logical() didn't work though. – Jeanelle

it's working very well. I do think the ?as.logical have to be read carefully to understand the function. Thanks – Poisonous

Answer 2

Indeed, there is a strightforward method.

As you have 2 levels factor, identify whats true and false

df <- data.frame(y=factor(sample(c("0","1"),10,replace = TRUE)))

str(df$y)
#  Factor w/ 2 levels "0","1": 2 2 2 1 1 2 2 2 2 2

levels(df$y) <- c(FALSE,TRUE)
df$y <- as.logical(df$y)

str(df$y)
# logi [1:10] TRUE TRUE TRUE FALSE FALSE TRUE ...

Intendant

Answer 3

This is probably a little too late to be helpful, but I ran into a similar problem and found a fix:

as.logical(as.integer(data.frame$column))

should do the trick.

Expressway

Should probably note that this works only if the factor representation for TRUE is 1 and FALSE is 0. Any other value for the factors gets coerced into TRUE – Lowrey

Answer 4

You can use == to create TRUE and FALSE values:

y = factor(c(0, 1, NA))
y == "1"
# [1] FALSE  TRUE    NA

Leer

Answer 5

You can use the == operator in as.logical() to identify which factor value is TRUE.

as.logical(factor_vector == 'True value here')

Gravitate

#40956997 – Lowrey