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I am using the formula:

($perimeter -sqrt($perimeter^2 - 16 * $area)) /4

to calculate the width of irregular polygons in a layer. However, I get an error

"cannot convert 'nan' to double"

I have looked at other stackoverflow reports on this error but cannot make it run correctly. I have tried changing the output type and the field length with no avail.

enter image description here

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  • 1
    somewhere you are dividing by 0 or taking the sqrt of a negative number or some such issue - try checking the inputs to your equation
    – Ian Turton ♦
    Oct 21, 2022 at 17:09
  • @MrXsquared it is a polygon layer and the $perimeter and $area should be inherently calculated from qgis, so they are not calling upon attribute columns (to my understanding). All of the features have valid geometries, so I don't understand how this could be an issue. I don't know how sharing any data would help.
    – vermicellion
    Oct 21, 2022 at 17:54
  • What other “stackoverflow reports” have you looked at? Please provide links to them within the body of your question.
    – PolyGeo ♦
    Oct 21, 2022 at 20:18
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    just try to calculate this $perimeter^2 - 16 * $area and see if there are any negative values
    – pigreco
    Oct 21, 2022 at 20:58
  • 1
    I managed to reproduce error: if there are polygons similar to circles, the expression $perimeter ^2 - 16 * $area is negative !!!
    – pigreco
    Oct 21, 2022 at 21:07

1 Answer 1

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8

The expression returns the error if there were circles or polygons very similar to circles among the polygons; in fact assuming the presence of a circle:

(2*pi*r)^2 < 16*pi*r^2

pi(4*pi*r^2) < 4(4*pi*r^2)

that is pi < 4

enter image description here

EDIT1:

To prevent the error from appearing you can use the expression:

try(
   ($perimeter -sqrt($perimeter^2 - 16 * $area)) /4
   ,0)

which will calculate the expression in all polygons except the ones that generate error ((in these polygons, it will always return 0).

try function: https://docs.qgis.org/3.22/en/docs/user_manual/expressions/functions_list.html#try

EDIT2:

you also need to pay attention to the number of digits after the comma (for the perimeter and area) because the difference $perimeter^2-16*$area could be negative, so I also suggest this solution:

with_variable('perim', round($perimeter,1),
try(
    (@perim-sqrt((@perim)^2 - 16*round($area,1)))/4
 ,0))
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  • This is great, thank you. I know you said "assuming" presence of a circle. My polygons are definitely not circles (but QGIS thinks they are). Do you know how I could Identify which polygons it thinks are circles and then exclude them from this calculation?
    – vermicellion
    Oct 22, 2022 at 14:17
  • 1
    @hcf, I added a possible solution to the answer to prevent the calculation from freezing.
    – pigreco
    Oct 22, 2022 at 16:14

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