The data

I have a polygon layer with polygons (download the data here) of different sizes as can be seen on the screenshot. Each polygon contains a value for the attribute called count. The range of these values is from 1 to 997.

What I want to do

Using QGIS 3.24 or later, I want to merge adjacent polygons in a way to create contiguous polygons (of any shape, not necessarily rectangles - but still as compact as possible) with the aggregated sum of the merged polygons being at least of value 500, but in no case be larger then 1000.

The question

How do I merge polygons in a way to get an accumalted sum of between 500 and 1000 for each resulting polygon?

What I tried

This is a follow up question. See here for context.


For example: on the upper right (see screenshot below), there is a polygon with value 51; it could be merged with the polygon to its upper left (12): 51+12=63. It can't be, however, merged with the polygon 983, as the sum would be too large (51+983 > 1000). However, to met the condition that the sum should be > 500, polygon 51 could be merged with two other ones at the left: 146 and 387, thus: 51+12+146+387 = 596 : this is larger than 500 and smaller than 1000.


Like this, adjacent polygons should be merged to come as close to the conditions I defined as possible. I am aware that an exact solution probably does not exist and I do not really need a solution to meet 100% of the conditions - but as close as possible with reasonable effort. The conditions in decreasing importance:

  • aggregeate sum of merged polygons being min. 500
  • aggregeate sum of merged polygons being max. 1000
  • shapes should be as compact as possible, square shapes prefered (avoiding "gerrymandering" style polygons as much as possible)

Screenshot: polygons are in graduated style, for lower values in white, higher values in red: enter image description here


1 Answer 1


Algorithm on this set will work for larger totals, e.g.:

enter image description here

940 polygons not enough to create approx. 370 groups with even totals.

enter image description here

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.