The raster package can readily provide bilinear interpolation: library(raster) r <- rasterFromXY(df) ## due diligence, just a guess here you should check ## projection(r) <- "+proj=utm +zone=32 +datum=WGS84"
## coordinates to extract m <- matrix(c( 3454263, 5970687), ncol = 2) extract(r, m, method = "bilinear") [1] -1.686059 ## compare with extract(r, m, method = "simple") -1.6877 See ?extract - simplest usage is a query matrix of XY coordinates in the projection used by your raster, it will helpfully transform queries such as a "Spatial*DataFrame" if needed, as long as both raster x and query y have sufficient projection metadata (and it's up to you to make sure that's set right). (Generally building a raster from "XYZ" data is sub-optimal since there's so much redundancy in the XY coordinates, and so much room for things to go wrong in between. But sometimes there's no better option. ) Cheers, Mike. On Mon, Jul 28, 2014 at 11:07 PM, Kulupp <kul...@online.de> wrote: > Dear R-experts, > > I have a regular grid dataframe (here: the first 50 rows) : > > # data frame (regular grid) with x, y (UTM-coordinates) and z (depth) > # x=UTM coordinates (easting, zone 32) > # y=UTM coordinates (northing, zone 32) > # z=river-depth (meters) > df <- data.frame(x=c(3454240, 3454240, 3454240, 3454240, 3454240, 3454250, > 3454250, 3454250, 3454250, 3454250, 3454250, 3454250, 3454250, 3454250, > 3454250, 3454250, > 3454250, 3454250, 3454260, 3454260, 3454260, 3454260, > 3454260, 3454260, 3454260, 3454260, 3454260, 3454260, 3454260, 3454260, > 3454260, 3454260, > 3454260, 3454260, 3454260, 3454260, 3454260, 3454260, > 3454260, 3454260, 3454270, 3454270, 3454270, 3454270, 3454270, 3454270, > 3454270, 3454270, > 3454270, 3454270), > y=c(5970610, 5970620, 5970630, 5970640, 5970650, 5970610, > 5970620, 5970630, 5970640, 5970650, 5970660, 5970670, 5970680, 5970690, > 5970700, 5970710, > 5970720, 5970730, 5970610, 5970620, 5970630, 5970640, > 5970650, 5970660, 5970670, 5970680, 5970690, 5970700, 5970710, 5970720, > 5970730, 5970740, > 5970750, 5970760, 5970770, 5970780, 5970790, 5970800, > 5970810, 5970820, 5970610, 5970620, 5970630, 5970640, 5970650, 5970660, > 5970670, 5970680, > 5970690, 5970700), > z= c(-1.5621, -1.5758, -1.5911, -1.6079, -1.6247, -1.5704, > -1.5840, -1.5976, -1.6113, -1.6249, -1.6385, -1.6521, -1.6658, -1.6794, > -1.6930, -1.7067, > -1.7216, -1.7384, -1.5786, -1.5922, -1.6059, -1.6195, > -1.6331, -1.6468, -1.6604, -1.6740, -1.6877, -1.7013, -1.7149, -1.7285, > -1.7422, -1.7558, > -1.7694, -1.7831, -1.7967, -1.8103, -1.8239, -1.8376, > -1.8522, -1.8690, -1.5869, -1.6005, -1.6141, -1.6278, -1.6414, -1.6550, > -1.6686, -1.6823, > -1.6959, -1.7095)) > head(df) > plot(df[,1:2], las=3) # to show that it's a regular grid > > My question: is there a function to calculate the depth of any coordinate > pair (e.g. x=3454263, y=5970687) within the grid, e.g. by bilinear > interpolation or any other meaningful method? > > Thanks a lot for your help in anticipation > > Best wishes > > Thomas > > ______________________________________________ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. -- Michael Sumner Software and Database Engineer Australian Antarctic Division Hobart, Australia e-mail: mdsum...@gmail.com ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
