4

An interpolator is considered exact if it returns the value of an observation when you are exactly at the same location (if your interpolation output is a raster, then the center of the pixel must be exactly at the same location than the observed values). With this definition in mind : Natural neighbors, nearest neighbor, inverse distance weigth and TIN are ...


3

Yes, gstat::idw can predict at any x,y location, but if you want to show that as a continuous map then you need a raster. The example does: # Interpolate the grid cells using a power value of 2 (idp=2.0) P.idw <- gstat::idw(Precip_in ~ 1, P, newdata=grd, idp=2.0) to predict at a regular grid of x,y coordinates generated by spsample It then converts the ...


3

Yes, you can call this from a Python script. But you don't directly call the low-level C API. First, take a look at the GDAL Grid Tutorial for background info. From the Python library, the relevant function is gdal.Grid(destName, srcDS, **kwargs). You can see some examples of how it's used in test_gdal_grid_lib.py (from the test suite). Or a made-up example:...


3

When you refer to the output being "gridded", I'm assuming you mean that the output is a raster. The various IDW Interpolation tools all create rasters as their outputs. You can turn a raster into contours by an additional step, using any of these tools/algorithms: GDAL Contour r.contour SAGA Contour lines To find these tools, search for "contour" in the ...


2

I think there's a difference in the variogram between the model fitted here with gstat and the one in the linked question. In the linked question, the model is expressed in terms of the covariance as C: C(h) = σ² = 4, if |h| = 0 C(h) = (σ² - a)exp(-3|h|/r) = (4 - 1)exp(-3|h|/6), if |h| > 0 In R, C = function(h){3*exp(-3*h/6)} And the variogram from ...


2

Take a look at the spatial::surf.ls function although, I would seriously question the use of a 12th order polynomial. At that point you should be using a more robust interpolator and as far as de-trending, with a 12th order fit you have no idea as to what spatial process you are actually de-trending. Plausibly, you could end up with an exact adherence to the ...


2

Assuming you have access to the Spatial Analyst Extension, you could try the following: In the Spatial Analyst Toolbar, choose Interpolation and Topo To Raster. Choose the contours and the height fields as the inputs. You can also define the cell size, try the spacing that you want your X and Y coordinate table. Once you have the raster, use the Conversion ...


2

I would like to suggest Multileve B-spline Interpolation tool in Processing Toolbox > SAGA > Raster creation tools.


2

It is relatively rare, but sometimes we encounter a situation GdalTools plugin is not activated. In an extreme case like below, we miss lots of raster tools. In your case, the Plugin Manager will have below tools activated: Heatmap Interpolotaion plugin Raster Terrain Analysis Zonal statistics But probably the GdalTools plugin is unchecked. Ticking it on ...


2

You can replace the values using where(). // Replace masked pixels by the mean of the previous and next months // (otherwise, how to deal with the first images??) var replacedVals = composites.map(function(image){ var currentDate = ee.Date(image.get('system:time_start')); var meanImage = composites.filterDate( currentDate.advance(-2, '...


2

It is a bad idea to measure distances in lat-long, as a degree of longitude is not the same ground length (in meters) as a degree of latitude. The linestring has 3 points that are not aligned. The first (top) half of the linestring vary a bit in longitude and a lot in latitude. The second (bottom) half of the linestring vary much more in longitude than in ...


2

The final problem were different CRS. The points were in 4326 and my Project layer in 25832. I changed everything to the same CRS and it worked. Thanks for the help.


1

At this point dat isn't a spatial object, so when predict refers back to it it can't get any spatial data: idw_ff = gstat(formula = ff ~ 1, data = dat, nmax = neighbors, set = list(idp = beta)) make it spatial: coordinates(dat)=~dist+rel_alt and it works. idw_ff = gstat(formula = ff ~ 1, ...


1

Not sure why you called it tin_interpolator when you want to use QgsIDWInterpolator :) But you need to replace the following lines: layer_data.SourceType =SourcePoints to layer_data.sourceType = QgsInterpolator.SourcePoints layer_data.distanceCoefficient=2 to tin_interpolator.setDistanceCoefficient(2) cos= canvas.extent() to cos = iface.mapCanvas()....


1

if you need the class proportions, you could use those steps: 1) reclassify your DEM into categories (using the reclassify tool) 2) use tabulate area to have the count of pixels of each category inside each polygon Warning: in case of large area, the default size of the pixel in the analysis (there is an internal conversion of the feature class into ...


1

I figured it out. You can see the attribute table below. Before, I was adding 3 different shape files to add 3 locations. After I did the same thing with only one shape file (and three different locations [points]) it worked. I also had to remove the Type (text) column, otherwise it didn't work again.


1

I continue with going through the code and it is not a bug but change in the API (and mainly my fault) which leads to the problems with triangulation. In python there is zCoordInterpolation = False property but in the new API there is an enumeration so instead of ValueZ (which I wrongly set up) you need set up the different value from enumeration (of ...


1

First check in the layer properties and check the fields tab. Make sure that the field that you want to interpolate is not an integer64 one. The interpolation field can be a decimal number (real) or an integer with a width less than ten characters. It seems that the interpolation plugin can not understand integer64 fields. If interpolation field is an ...


1

As commented, the pits/holes in the final intensity image is due to missing data. Using OP's example, a resolution of 0.5 units yields 9244 NA cells, while doubling it (squaring in area) yields only 28 NA cells. When gridding LiDAR data, it is a best practice starting the cell size with at least the average point spacing from the point cloud (which in this ...


1

You can use Spatial Analyst extension Interpolate Tools (IDW) to create a Raster from the input point features. Then convert back to points if that is your needed format. Make sure to specify environment setting or specify cellSize to desired resolution (ie 1 = 1m) if you are in projected coordinate system import arcpy from arcpy.sa import * arcpy....


1

You can use Add Polygon Attributes to points tool under SAGA which will transfer the land cover field from the polygon attribute table to the points attribute table. The tool is located under Processing toolbox -> SAGA -> Vector Points tools -> Add Polygon Attributes to points: Make sure that both vector files have the same CRS. Another option is ...


1

Select by attribute all urban areas from the landcover layer and export them to a new layer. Then select by position all buffers which intersect with the newly exported layer. Export the selected features to a new layer.


1

You can do it in raster calculator, but instead of conditional use the result of relational operator e.g. river@1 = 9999 gives 0 (false) or 1 (true). Try this: ((river@1 = 9999) * river@1) + ((river@1 != 9999) * dem@1) The first part or the second part will be zero, so the value of river raster or the value of the dem raster will be the result.


1

It is possible that the black lines are pixels with missing data (no data). In this case, some options are: i) start over by accessing the raw point cloud data and generate a DEM with a method that process tiles altogether so to avoid having edge's artifacts. Take a look at Merging multiple LiDAR (.las) files?. ii) interpolate (fill in no data values) ...


1

if you have demographic data at hand, co-kriging with population density computed on the same radius would be very powerful. Otherwise, I would consider bilinear interpolation between the points, because your density is not a punctual measure.


1

Here's a start: (untested) Supposing you have a spatial points data.frame 'points_spdf' of points with attribute 'z' to be interpolated, and the result grid as 'grd' library(gstat) pwr = 3.0 # Put you inverse dist power here interp <- idw(z~1, locations=points_spdf, newdata=grd, idp=pwr) Check help(idw) after you have loaded gstat.


1

http://pro.arcgis.com/en/pro-app/tool-reference/data-management/resample.htm "If the center of the pixel in output space falls exactly the same as one of the >pixels in the input cells, that particular cell value gets all the weights, >thereby causing the output pixel to be the same as the cell center. This will >affect the result of bilinear ...


1

I have been simulating your problem here and unfortunately I have no advanced license, but a normal one. Anyways, I got to the following workflow, which I believe will give you what you want. Create and empty raster covering your study area with the tool Create Raster Dataset. Choose your cell size wisely (I would go for the minimum distance between the ...


1

I regret that I didn't see your post back when you wrote it. But I've made a picture that I think shows the essential difference between the two. Both interpolations produce continuous surfaces and both will preserve the original values (so that if you take a query point at the position of an input sample point, you get back the original value associated ...


1

As indiciated here https://issues.qgis.org/issues/20105 the problem is most likely a lack of space. I had the same problem and was able to solve it by reducing the radius and thus also the overall number of rows and columns. The solution provided by Lu76 probably also reduced the necessary space. I tried it too and in one case it worked for me and in the ...


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