2

I'm trying to extract the raster value of the nearest pixel for a group of points.

I want the value for each point rather than the mean for example. Also, some of the points are in masked areas, hence the nearest pixel value.

//import Pekel data
var gsw = ee.Image('JRC/GSW1_0/GlobalSurfaceWater');
var change = gsw.select("change_abs");


//points 
var p1 = ee.Geometry.Point([89.74319458007812,22.076065830651526])
var p2 = ee.Geometry.Point([89.87743377685547,22.0827468707419])
var p3 = ee.Geometry.Point([89.95536804199219,21.943622724707772])
var pts = ee.FeatureCollection(ee.List([ee.Feature(p1),ee.Feature(p2),ee.Feature(p3)]))


//////////////////////////////////////////////////////////////
// Constants
//////////////////////////////////////////////////////////////

var VIS_CHANGE = {
    min:-100,
    max:100,
    palette: ['red', 'black', 'limegreen']
};




//////////////////////////////////////////////////////////////
// Map Layers
//////////////////////////////////////////////////////////////

//Change in water intensity
Map.addLayer({
  eeObject: change,
  visParams: VIS_CHANGE,
  name: 'occurrence change intensity', 
  shown: true
});


Map.addLayer(pts, {'color': 'blue'}, 'points'); 
1

You can compute the distance to the nearest unmasked value, buffer the points by that much, and then reduce within each buffer. Something like this:

var distance = change.fastDistanceTransform().sqrt().multiply(ee.Image.pixelArea().sqrt()).rename("distance")
pts = distance.reduceRegions(pts, ee.Reducer.first().setOutputs(["distance"]))
  .map(function(f) {
    var distance = ee.Number(f.get('distance'))
    f = ee.Algorithms.If(distance, 
        f.buffer(distance.add(30), 1),
        f)
    f = ee.Feature(f)
    return f.set(change.reduceRegion(ee.Reducer.mean().unweighted(), f.geometry(), 30))
  })
print(pts)

You buffer by 1 extra pixel to make sure the buffer hits something; for the pixels that have a distance of 0, you skip the buffer step. Unweighted mean reducer will take any fraction of a pixel it encounters without normalizing by the fractional portion.

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.