How to replace NaN value with zero in a huge data frame?

Question

I tried to replace NaN values with zeros using the following script:

rapply( data123, f=function(x) ifelse(is.nan(x),0,x), how=\"replace\" )
#          


        

Answer 1

The following should do what you want:

x <- data.frame(X1=sample(c(1:3,NaN), 200, replace=TRUE), X2=sample(c(4:6,NaN), 200, replace=TRUE))
head(x)
x <- replace(x, is.na(x), 0)
head(x)

Answer 2

In fact, in R, this operation is very easy:

If the matrix 'a' contains some NaN, you just need to use the following code to replace it by 0:

a <- matrix(c(1, NaN, 2, NaN), ncol=2, nrow=2)
a[is.nan(a)] <- 0
a

If the data frame 'b' contains some NaN, you just need to use the following code to replace it by 0:

#for a data.frame: 
b <- data.frame(c1=c(1, NaN, 2), c2=c(NaN, 2, 7))
b[is.na(b)] <- 0
b

Note the difference is.nan when it's a matrix vs. is.na when it's a data frame.

Doing

#...
b[is.nan(b)] <- 0
#...

yields: Error in is.nan(b) : default method not implemented for type 'list' because b is a data frame.

Note: Edited for small but confusing typos

Answer 3

It would seem that is.nan doesn't actually have a method for data frames, unlike is.na. So, let's fix that!

is.nan.data.frame <- function(x)
do.call(cbind, lapply(x, is.nan))

data123[is.nan(data123)] <- 0