By using Rule based layer styling for a polygon shapefile I want to display only those features whose centroid is separated by 1.000m from the feature centroids of another polygon layer ('layerB'). The following filter expression is invalid and I cannot understand why:

distance(centroid($geometry), centroid(geometry('layerB_011018_5154ef48_b1da_43ae_91b9_daeb20f6cab9'))) > 1000

Getting the feature geometries from 'layerB' is probably the problem here. The geometry operator doesn't seem to fetch all feature geometries from a 'layerB'. How can I solve this?

  • 3
    The functions geometry() has one argument: it expects a feature as input. However, you set a layer as input. that cannot work. Honestly I don't quite understand how your solution should look like: can you post a screenshot? Rule based style on layer A if distance to which centroid(s) of layer B? All of them? The closest one?
    – Babel
    Commented Aug 30, 2021 at 16:00
  • I want to style the features in layer A depending on whether the feature-centroid is within or beyond a 1.000m from any feature's centroid in layer B. Hope that's more concise :-) In any case, thanks for elaborating on geometry(), that wasn't clear to me yet. Commented Aug 31, 2021 at 7:09
  • See the help in the expression string builder - when you click on a function (like geometry()) in the middle, on the right side you see the help where you see exactly what kind of input (arguments) the functions expects. By the way: Make it clear if you still look for an answer or if the answer by @MrXsquared solved your problem (by accepting it). If not, give some information what you're missing - otherwise, it's difficult to guess. And: any substantial clarification about your problem should be included in the initial question by editing not - not in comments here.
    – Babel
    Commented Aug 31, 2021 at 7:30

1 Answer 1


How about:

distance(centroid($geometry), centroid(aggregate('layerB', 'collect', $geometry))) > 1000

The centroid(aggregate()) part does collect all geometries of 'layerB' as a Multi-Geometry and return the centroid this collection (so the whole layer).

The centroid() function expects a geometry as input, not a layer nor a feature. And the geometry() function would expect a feature, as already mentioned by @Babel.

To only take the nearest centroid into consideration, you can use overlay_nearest() since QGIS 3.16+ the following:

distance(centroid($geometry), centroid(overlay_nearest('layerB',$geometry)[0])) > 1000

You can also add a limit:=, filter:= or max_distance:= as optional arguments.

  • Thanks, this works in principle, but in my particular context it's not useful to consider layerB as a single multi-geometry. I don't think my idea is going anywhere and I'll abandon this approach. FYI I've used with get_feature() successfully in another project, to filter features from layer A laying within a single feature of layer B: within( $geometry , geometry(get_feature( 'layerB','fid', 213))). Commented Aug 31, 2021 at 7:26
  • 3
    @HansRoelofsen if you have QGIS 3.16+ available that would be possible by using overlay_nearest(), try my updated answer.
    – MrXsquared
    Commented Aug 31, 2021 at 8:01

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