I have a data frame that I construct as such:

> yyz <- data.frame(a = c("1","2","n/a"), b = c(1,2,"n/a"))
> apply(yyz, 2, class) a b
"character" "character"

I am attempting to convert the last column to numeric while still maintaining the first column as a character. I tried this:

> yyz$b <- as.numeric(as.character(yyz$b))
> yyz a b 1 1 2 2 n/a NA

But when I run the apply class it is showing me that they are both character classes.

> apply(yyz, 2, class) a b
"character" "character"

Am I setting up the data frame wrong? Or is it the way R is interpreting the data frame?

1

1 Answer

If we need only one column to be numeric

yyz$b <- as.numeric(as.character(yyz$b))

But, if all the columns needs to changed to numeric, use lapply to loop over the columns and convert to numeric by first converting it to character class as the columns were factor.

yyz[] <- lapply(yyz, function(x) as.numeric(as.character(x)))

Both the columns in the OP's post are factor because of the string "n/a". This could be easily avoided while reading the file using na.strings = "n/a" in the or if we are using data.frame, we can have character columns with stringsAsFactors=FALSE (the default is stringsAsFactors=TRUE)

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Regarding the usage of apply, it converts the dataset to matrix and matrix can hold only a single class. To check the class, we need

lapply(yyz, class)

Or

sapply(yyz, class)

Or check

str(yyz)

5

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