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

r replace

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

r replace

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)] <- 0a

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)] <- 0b

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

r replace

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)

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