In the following example, I am trying to replace the value 10 in raster r with the respective pixel value from raster r2:
library(raster)
r <- raster(nrows=25, ncols=25, vals=round(rnorm(625, 3), 0)) #land-use/cover raster
r[ r > 2 ] <- NA
r[ r < 1 ] <- 10
plot(r)
r2 <- raster(nrows=25, ncols=25, vals=round(rnorm(625, 3), 0)) #land-use/cover raster
plot(r2)
I was hoping the following code would work:
r3 <- overlay(r, r2, fun = function(x, y) { x[ x == 10 ] <- y ; return(x) })
... but it returns an error:
Error in (function (x, fun, filename = "", recycle = TRUE, forcefun = FALSE, : cannot use this formula, probably because it is not vectorized
I am sure part of the problem is with "<- y" in the overlay function. I tried using "<- y[ x == 10 ]" instead but I still run into the same problem... Any ideas? Thanks in advance.
If you're using random values for your example, I would suggest to also use a random seed so it's reproducible and comparable.
Since your two rasters have the same dimensions, you can just use logical indexing.
Here's a small example:
library(raster)
set.seed(42)
r <-setValues(raster(ncol=10,nrow=10),round(runif(100,1,10)))
r2 <-setValues(raster(ncol=10,nrow=10),round(runif(100,1,10)))
# create logical index (for later comparison)
ix <- r == 10
# returns FALSE, so r2 doesn't have any 10 where r has one
any(r[ix] == r2[ix])
r[ix] <- r2[ix]
# prints TRUE, so all values have been correctly replaced
all(r[ix] == r2[ix])
You can do the replacement also in a simple one liner:
r[r == 10] <- r2[r == 10]
(Edit: Thanks @Nemesi)