I would like to filter a layer with the maximum value of field_a, grouped by field_b. I tried this expression: "field_a"= maximum ( "field_a","field_b") But it doesn't work (there are no more features in the layer). When I use this expression for Select by Expression, the features are selected. Is there a way to filter them (without selecting)?

[Edit] I want to filter a existing layer (right mouse click -> filter). I don't want to select features or to create a new layer.


2 Answers 2


The filter query (layer properties/source/provider feature filter) allows you to specify an SQL where clause. If you enter id=1 it will translate it to select * from mylayer where id=1.

From there, you can add a sub-query in this where clause to identify the IDs (or else) of interest:

field_a IN (SELECT MAX(field_a) FROM myLayer GROUP BY field_b)
  • Thank you very much, that is what I looked for!
    – Nane
    Oct 20, 2019 at 13:38
  • @JGH sorry to ask but i don't see how to input SQL syntax in that part of the interface ... how do u do that ? where exactly are you ?
    – Snaileater
    Nov 4, 2019 at 19:04
  • @snaileater layer properties, source, query builder
    – JGH
    Nov 4, 2019 at 19:09
  • ok i was at the bottom of the attribute table ... but ... following your advice i don't succeed to enter full sql syntax (i mean using the FROM clause) ... there must be something i don't get ...
    – Snaileater
    Nov 4, 2019 at 19:25
  • @snaileater I tested it with a Postgres layer and a Shapefile. It failed with an in-memory layer though, so this solution seems to be data provider-dependent
    – JGH
    Nov 4, 2019 at 19:40

If this is for visualization purpose, I suggest Rule-based symbology.

"field_a"  =  maximum("field_a", group_by:= "field_b" )

enter image description here

[EDIT] To actually reduce the features by the expression, try Extract by expression tool (in Processing toolbox > Vector selection).

The expression is the same as the above.

  • No, it is not for visualization purpose. I want to reduces the number of features in the attribute table with a filter. And unfortunatly the expression does not work with the filter.
    – Nane
    Oct 20, 2019 at 9:48
  • @Nane Got that, thanks for clarification.
    – Kazuhito
    Oct 20, 2019 at 9:55
  • @ Kazuhita. Thank you for your help. Somehow I describe my problem not clear, sorry. I don't want to get a new layer, I just want to filter an existing layer (right mouse click on the layer, -> filter). But there this expression doesn't work...
    – Nane
    Oct 20, 2019 at 10:27
  • @Nane Aah, sorry I misunderstood the question. I now understand this expression does not work,
    – Kazuhito
    Oct 20, 2019 at 10:38

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