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" )
# [31] 0.00000000 -0.67994832 0.50287454 0.63979527 1.48410571 -2.90402836
The NaN value was showing to be zero but when I typed in the name of the data frame and tried to review it, the value was still remaining NaN.
data123$contri_us
# [31] NaN -0.67994832 0.50287454 0.63979527 1.48410571 -2.90402836
I am not sure whether the rapply
command was actually applying the adjustment in the data frame, or just replaced the value as per shown.
Any idea how to actually change the NaN
value to zero?
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
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
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)