I assume you want your irregular point data on a regular raster. In that case, rasterize should work, and the examples in ?rasterize show how. Here is something based on your data
s100 <- matrix(c(267573.9, 2633781, 213.29545, 262224.4, 2633781, 69.78261, 263742.7, 2633781, 51.21951, 259328.4, 2633781, 301.98413, 264109.8, 2633781, 141.72414, 255094.8, 2633781, 88.90244), ncol=3, byrow=TRUE)
colnames(s100) <- c('X', 'Y', 'Z')
library(raster)
# set up an 'empty' raster, here via an extent object derived from your data
e <- extent(s100[,1:2])
e <- e + 1000 # add this as all y's are the same
r <- raster(e, ncol=10, nrow=2)
# or r <- raster(xmn=, xmx=, ...
# you need to provide a function 'fun' for when there are multiple points per cell
x <- rasterize(s100[, 1:2], r, s100[,3], fun=mean)
plot(x)
Best Answer
This is likely not entirely an issue with the interpolation model. Bathymetric data can exhibit considerable noise. Because of an equal weight associated with each TIN facet and outlier effect, A TIN base interpolation can extenuate this noise and is not recommended. I would apply a Topogrid (Topo to raster tool) Spline interpolation and then apply a smoothing filter to the result. I commonly use a Gaussian weighted filter with a sigma of 2, but in ArcGIS you could just use a focal mean. The size of the window will depend on the resolution of the interpolated surface and an error criteria. You do not want to oversmooth the data so, assessing the Root Mean Squared Error (RMSE) of the observed vs. predicted is essential. Find a window size that exhibits an acceptable balance between smoothness and error.