4

I am currently trying to do a grouped reduction, but I am having trouble manipulating the outputs. The code calculates areas per land cover type and outputs a list of objects with two properties, 'code' i.e. land cover type and 'area', the area of the land cover in question. The issue is that I am having trouble filtering the output list, as I would like to access the area of each specific land cover without knowing the index.

The code is as follows:

// Load land cover
var lcStudy = ee.Image("COPERNICUS/Landcover/100m/Proba-V-C3/Global/2019")
                   .select('discrete_classification').clip(neighborhood);

// Add pixel area as a band to the Image
var lcStudyAreas = ee.Image.pixelArea().rename('area').addBands(lcStudy);
print(lcStudyAreas);

// Do a grouped reduction
var output = lcStudyAreas.reduceRegion({
  reducer: ee.Reducer.sum().group({
    groupField: 1,
    groupName: 'code',
  }),
  geometry: neighborhood,
  scale: 1000,
  maxPixels: 1e13
}).get('groups');

print(output);

The issue is I would like to know, for example, the area of the land cover number 20 (shrubland). But when I haven't found the correct way to filter the list. Writing:

var shrubArea = ee.List(output).filter(ee.Filter.eq('code',20));

Produces an error:

List.filter: Filters used in List.filter should refer to the property named 'item'.

But replacing 'code' for 'item' produces an empty list.

Here is a link to the code: https://code.earthengine.google.com/b144b1c5645b69748576e7d426a3bb3b

1
  • Edited because ee.Image.pixelArea() should be used without multiplying
    – M. Nicolas
    Feb 17, 2021 at 23:37

1 Answer 1

4

ee.Filters are best when working with features and their properties, not dictionaries. Therefore, I recommend converting the dictionaries to features and the list to a feature collection:

var groupsList = lcStudyAreas.reduceRegion({
  reducer: ee.Reducer.sum().group({
    groupField: 1,
    groupName: 'code',
  }),
  geometry: neighborhood,
  scale: 1000,
  maxPixels: 1e13
}).get('groups');

var groupFeatures = ee.FeatureCollection(
  ee.List(groupsList).map(function (dict) {
    // Convert the dictionary to properties of a feature (with no geometry)
    return ee.Feature(null, dict);
  })
);

// All the features (for demonstration)
print(groupFeatures);

// Filtered to one matching feature
print(groupFeatures.filter(ee.Filter.eq('code', 20)));

It's also possible to filter a list using ee.List.map with the dropNulls option, but that is not particularly advantageous in this case. Converting to a FeatureCollection also makes it easy to export your data as a table later.

2
  • I would like to know more about "ee.Filters are best when working with features and their properties, not dictionaries"? There is a clue in the docs for ee.List.filter. "Filters a list to only the elements that match the given filter. To filter list items that aren't images or features, test a property named 'item', e.g.: ee.Filter.gt('item', 3)". This doesn't help for a list of dictionaries.
    – intotecho
    Sep 14, 2021 at 11:24
  • @intotecho If you have a question, please ask it — with more concrete details of what exactly you want to know — on the Ask Question page. You can include a link to this question if it is useful context.
    – Kevin Reid
    Sep 14, 2021 at 14:39

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