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The histograms are different. head(table(one)) ##one ##31 32 33 34 35 36 ## 1 2 1 2 2 2 head(table(two)) ##two ##18 19 22 23 24 25 ## 2 1 3 4 2 5 stem(one, 0.5) # The decimal point …

You can do something along these lines: Example data: library(raster) r <- raster(ncol=10,nrow=10, ext=extent(c(0,10,0,10)), crs='+proj=utm +zone=1') r[] <- 1 r[15:42] <- 2 set.seed(0) r[sample(ncell …

The general approach you are taking works for me with terra and with raster. … dem_rast <- raster(floor(extent(dem_points_clip)), res=5) …

'sp' is a Spatial* object library(raster) e <- extent(c(0,10,0,10)) intersect(e, extent(sp)) But it may indeed be, as whuber points out, that the polygon is in fact not inside extent e. …

Also, as you want to fill in missing values, not change existing values, I would use NAonly = TRUE, together with pad=TRUE (to pad virtual rows and columns with NAs outside of the raster). … Here is an example: library(raster) # example data logo <- raster(system.file("external/rlogo.grd", package="raster")) set.seed(0) i <- sample(ncell(logo), 200) logo[i] <- NA plot(logo) m <- matrix …

You can use the sampleRandom function: library(raster) r <- raster(ncol=30,nrow=20) r[] <- 1:ncell(r) x <- sampleRandom(r, ncell(r)*.3, asRaster=TRUE) …

Here are three alternative approaches to terra::app that Jeffrey Evans proposed; instead using ifel, selectRange and lapp. I am not saying these are better in any way, but they may be easier to use. J …

To load all values in RasterBrick into RAM you can use readAll library(raster) b <-brick("tmmn_2020.nc") x <- readAll(b) But this will not help you if you cannot have that much data in RAM, as you say …

The NA's here refer to the fact that the min and max value are unknown (this is no longer the case in the current version of 'raster'). …

Impossible to know from your description. But you can try things like the below to see what is going on: # get values of a few cells cells <- 100000:100010 x <- data.frame(w1[cells]) x predict(s2.rf …

x <- calc(st1, function(x) movingFun(x, 3, mean)) y <- st1 - x

<- function(x,y) { z <- rep(NA, length(x)) i <- which(y == 1) z[i] <- x[i] * 3 i <- which(y == 2) z[i] <- x[i] * 33 i <- which(y == 3) z[i] <- x[i] * 333 z } library(raster … ) r1 <- r2 <- raster(nrow=5, ncol=5) r1[] <- runif(ncell(r1)) * 10 r2[] <- round(runif(ncell(r2),1,3)) x <- overlay(r1, r2, fun=f) …

What you are seeing is a map of one raster cell (~ 9.5 - 10.5; 46.3 - 47.3) that has a single value (36). Had you provided more information, e.g. show(rMerge), it would have been very easy to spot. … Or use the raster(u, nrow=10, ncol=10) u <- union(union(extent(ro),extent(ra)),extent(co)) rast <- raster(u, ncol=10, nrow=10) r1 <- rasterize(ro, rast) r2 <- rasterize(ra, rast) r3 <- rasterize(co, rast …

whuber showed the way, but here's another way to get there (using the functions that match the operations he suggests) d <- c(.25,.50,.75) m <- cbind(1:3, d) r3 <- reclassify(r2, m) out <- r1 * r3

If so, yo can do the following # Your code (simplified) library(raster) files= list.files(pattern='\\.grd$', full.names=TRUE) s <- stack(files) pt<- read.csv("latandlong") point <- SpatialPoints(pt) df …