1

I've created a Watershed raster data set (using the Watershed Tool) for a certain area. Within that watershed raster, there are 8 basins that appear

I'm trying add the total population to each of those 8 basins using census block data for the total watershed and then compare that data. I'll also need to sum all of the population for those block groups. Since that total population for each basin hasn't been calculated yet.

I have tried several tools. Clip, Join, Union, Intersect, etc. I've also tried turning the raster to polygon, and then adding the census data (but I end up with several small polygons instead of the 8 basins that I began with) so I know the population totals wouldnt be accurate.

I also turned the census block groups into a raster dataset, but really unsure if I did that correctly.

Anyone out there know how to accurately combine these two datasets into one?

  • You may want to search for previously asked questions, this is similar to this question: gis.stackexchange.com/questions/172843/… – Corey Pembleton Dec 7 '15 at 15:49
  • 1
    What software are you using? – Tom Dec 7 '15 at 16:54
  • There are a number of ways you could go about it. It's not quite like the other question, because your zones are rasters and your values are from polygons. Converting your polys to rasters in order to make this a zonal stats issue won't work, because you'd lose a lot of information. Instead, you want to convert your watershed raster to polygons. Then run an intersect or union (depending on your extents and what you want included). Then, you probably want to re-assign the population proportionate to size, and then dissolve on watershed id. Let me know about software and we can go from there. – Tom Dec 7 '15 at 17:24
  • Vector approach using proportion is accurate, however zonal statistics (SUM) tool WILL work as well. It is enough to convert census polygons into population 'density', i.e. person count per 1 cell. Smaller cell size produce more accurate population estimate – FelixIP Dec 7 '15 at 19:15

Your Answer

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

Browse other questions tagged or ask your own question.