2

I am trying to extract values of raster cells on points using extract from raster package in R (similar to 'Extract Values to Points' in ArcGIS-10.2). After doing so in order to check integrity, I computed the total value of all the raster cells (i.e. sum) ~66,000 and total extracted values on points ~49,000. We have tried both techniques simple and bilinear interpolation techniques. Data loss gets only slightly reduced with bilinear technique, i.e. ~50,000 points are extracted. There is a huge data loss, if anyone can provide a way forward or has experienced the same.

  • If you are only sampling 49,000 points then obviously they are going to sum to less than 66,000 cells. What did you expect, and can you give us a reproducible example maybe using random data on a much smaller grid? – Spacedman Nov 20 '14 at 12:00
  • @Spacedman, he means the sum of point values, as a proxy for how different they are, I guess. – user21313 Nov 20 '14 at 13:49
  • @CincoSauces yes. But how can the sum of the values of a small number of points sampled over a grid ever be anything but less than the sum of the grid cells? We really need to see some code and some clarification. I don't understand why this Q has been upvoted. – Spacedman Nov 20 '14 at 15:01
  • @Spacedman : I did not mean 66K cells or 49K cells, To brief you more: 66 K is the total road length(adding cell by cell) in the raster file and 49 K is the total road length (adding extracted raster value on points) after the raster file is extracted on Points. – Tisha Nov 25 '14 at 4:35
1

You might want to give a try to the GMT tool grdtrack. You can start by reconverting to your raster to a NetCDF file using gdal_translate:

gdal_translate -of NetCDF myraster.tif myraster.grd

and then put your x,y locations in a text file:

x1, y1
x2, y2
.., ..

then call grdtrack on that NetCDF grid based on your table of locations:

grdtrack mylocations.xy -Gmyraster.grd > myvalues.txt

You will get a new table with the sampled points. This procedure should work well and it is relatively fast. Good luck!

  • How this answers the question? – user32309 Nov 21 '14 at 5:40
  • Thanks for your inquiry @Pascal, this answer addresses the question by providing a simple yet effective method of solving the problem. – user21313 Nov 21 '14 at 14:01
1

You said:

I am trying to extract values of raster cells on points using extract from raster package in R (similar to 'Extract Values to Points' in ArcGIS-10.2).

Lets set up a test raster:

require(raster)
r = raster(ncol=100,nrow=100,xmn=0,xmx=1,ymn=0,ymx=1)
r[]=runif(100*100)

You said:

After doing so in order to check integrity, I computed the total value of all the raster cells (i.e. sum) ~66,000 and total extracted values on points ~49,000.

Okay, lets do that:

pts = cbind(runif(100),runif(100))
vr = extract(r,pts)
sum(vr)
[1] 55.41762
sum(values(r))
[1] 5003.098

Obviously the sum of 100 points sampled at random across that raster is going to be approximately 0.5*100 = 50. The sum of the whole raster is 0.5*100*100 because there's 100*100 cells with an average value of 0.5.

You said:

We have tried both techniques simple and bilinear interpolation techniques. Data loss gets only slightly reduced with bilinear technique, i.e. ~50,000 points are extracted. There is a huge data loss, if anyone can provide a way forward or has experienced the same.

I do not understand why you think this is a good idea for some kind of "integrity test" unless I misunderstand what you are doing. If this answer doesn't help please clarify your question with some sample code like mine. Otherwise this question is unclear.

  • I appreciate your inputs but the problem is little more deep. I have a raster which consists of road length i.e. each cell represents a value of total road length present per sq Km on earth in kilometers and I need to extract this raster on Points (~ at 30 arc seconds which is 928 m). So after the extraction, i try to compute the Total Road Length on raster cells and points and found a whooping difference of 17K in kilometers. I suspect there will be some loss due to sampling but to this degree was unaware. – Tisha Nov 25 '14 at 4:42
  • Edit your question to set up an example like I've done that illustrates the problem. Is it just a case that your new sample cell areas are different, and you've forgotten that the original raster is km/**per unit area** and you need to scale up by the change in cell area? I don't know, because you haven't given us any code. – Spacedman Nov 25 '14 at 8:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.