CODEWITHSUNDEEP
rframeapply
coding_heart
coding_heart

Reputation: 1295

Apply function to frame of characters in R

I've got a set of variables that are characters that I am trying to convert into a binary by creating the following function and using the apply() function:

a <- as.factor(c("n/a", "False", "False", "True"))
b <- as.factor(c("n/a", "True", "False", "True"))
y <- data.frame(a,b)


conv <- function(x){
    levels(x)[which(levels(x)=="n/a")] <- NA
    levels(x)[which(levels(x)=="False")] <- 0
    levels(x)[which(levels(x)=="True")] <- 1
    x <- as.numeric(levels(x))[x]
    return(x)
}

apply(y,2, conv)

However, when I do this, it outputs NAs. Alternatively, if you apply the function by column, it works:

conv(y[,1])
conv(y[,2])

The expected output should be:

y:
NA NA
0 1
0 0 
1 1

Any thoughts on why this is happening? Thanks.

Upvotes: 0

Views: 39

Answers (3)

akrun
akrun

Reputation: 887691

In R, logical values are TRUE/FALSE and not strings "True", "False". In addition, NA is the missing value

y[] <- NA^(is.na(replace(as.matrix(y), y=="n/a", NA)))*+(y=='True')
y
#   a  b
#1 NA NA
#2  0  1
#3  0  0
#4  1  1

Upvotes: 1

Sotos
Sotos

Reputation: 51592

A simple ifelse can take care of the NA requirement. grepl can then be used to convert to 0/1, i.e.

y[] <- lapply(y[], function(i) ifelse(i == 'n/a', NA, grepl('True', i)*1))
y
#   a  b
#1 NA NA
#2  0  1
#3  0  0
#4  1  1

Upvotes: 1

Chirayu Chamoli
Chirayu Chamoli

Reputation: 2076

Your function is fine you just need to use lapply.

conv <- function(x){
    levels(x)[which(levels(x)=="n/a")] <- NA
    levels(x)[which(levels(x)=="False")] <- 0
    levels(x)[which(levels(x)=="True")] <- 1
    x <- as.numeric(levels(x))[x]
    return(x)
}

lapply(y,conv)

Also if the order of levels is same for all the variables then you could just do this.

conv <- function(x){
    levels(x)=c(0,NA,1)
    return(x)
}

lapply(y, conv)

Upvotes: 1

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