Problem with the "stackApply" function in R - r-raster

Question

I have a problem with the "stackApply" function from the raster-package. First I want to stack three raster layers (each layer has one band) - that works. And then I want to create a raster-object that shows in which of the three bands/layers the minimum value occurs (each pixel in the raster layers has a different value). But I get various error messages. Does anyone have an idea how I can solve the problem?
Thank you
stacktest<-stack(test,test1,test2)
min_which <- stackApply(stacktest, indices=1, fun=function(x, na.rm=NULL)which.min(x))
Error in setValues(out, v) : values must be a vector
Error in is.infinite(v) : not implemented standard method for type 'list'

Answer 1

Here is a minimal, self-contained, reproducible example:
Example data from ?stackApply
library(raster)
r <- raster(ncol=10, nrow=10)
values(r) <- 1:ncell(r)
s <- stack(r,r,r,r,r,r)
s <- s * 1:6
Now use these data with your function (I removed the na.rm=NULL as it is not used)
w <- stackApply(s, indices=1, fun=function(x, ...) which.min(x) )
w
#class : RasterLayer
#dimensions : 10, 10, 100 (nrow, ncol, ncell)
#resolution : 36, 18 (x, y)
#extent : -180, 180, -90, 90 (xmin, xmax, ymin, ymax)
#crs : +proj=longlat +datum=WGS84 +no_defs
#source : memory
#names : index_1
#values : 1, 1 (min, max)
Same for which.max
w <- stackApply(s, indices=1, fun=function(x, na.rm=NULL) which.max(x) )
w
# (...)
#values : 6, 6 (min, max)
This suggest it works fine. In most cases that means that you probably have cells that are NA
s[1:10] <- NA
w <- stackApply(s, indices=1, fun=function(x, ...) which.min(x) )
# Error in setValues(out, v) : values must be numeric, logical or factor
It is easy to see why this error occurs
which.min(3:1)
#[1] 3
which.min(c(3:1, NA))
#[1] 3
which.min(c(NA, NA, NA))
#integer(0)
If all values are NA, which.min does not return NA as expected. Instead it returns an empty vector. That can be fixed like this
which.min(c(NA, NA, NA))[1]
#[1] NA
And you can do
w <- stackApply(s, indices=1, fun=function(x, ...) which.min(x)[1] )
However, using stackApply with indices=1 is not a good approach. You should generally use calc to compute cell values across all layers.
y <- calc(s, function(x) which.min(x)[1])
But in this case you can use the more straightforward
z <- which.min(s)