Thanks for sharing the form!
So looking at some of your calculates, for example:
(countSelected(data(‘mysc’)) > 0 ? 1 : 0) + (countSelected(data(‘mysp2’)) > 0 ? 1 : 0) + (countSelected(data(‘pehy6’)) > 0 ? 1 : 0) + (countSelected(data(‘pope10’)) > 0 ? 1 : 0) + (countSelected(data(‘phar3’)) > 0 ? 1 : 0) + (countSelected(data(‘phau7’)) > 0 ? 1 : 0) + (countSelected(data(‘poco’)) > 0 ? 1 : 0) + (countSelected(data(‘potr2’)) > 0 ? 1 : 0) + (countSelected(data(‘prav’)) > 0 ? 1 : 0) + (countSelected(data(‘rafi’)) > 0 ? 1 : 0) + (countSelected(data(‘reja2’)) > 0 ? 1 : 0) + (countSelected(data(‘rhca3’)) > 0 ? 1 : 0) + (countSelected(data(‘romu’)) > 0 ? 1 : 0) + (countSelected(data(‘ruph’)) > 0 ? 1 : 0) + (countSelected(data(‘sodu’)) > 0 ? 1 : 0) + (countSelected(data(‘trna’)) > 0 ? 1 : 0) + (countSelected(data(‘trre3’)) > 0 ? 1 : 0) + (countSelected(data(‘tufa’)) > 0 ? 1 : 0) + (countSelected(data(‘tygl’)) > 0 ? 1 : 0) + (countSelected(data(‘veth’)) > 0 ? 1 : 0) + (countSelected(data(‘veof2’)) > 0 ? 1 : 0) + (countSelected(data(‘adeltsug’)) > 0 ? 1 : 0) + (countSelected(data(‘agriplan’)) > 0 ? 1 : 0) + (countSelected(data(‘anapglab’)) > 0 ? 1 : 0) + (countSelected(data(‘cipangup’)) > 0 ? 1 : 0) + (countSelected(data(‘dendfront’)) > 0 ? 1 : 0) + (countSelected(data(‘halyhaly’)) > 0 ? 1 : 0) + (countSelected(data(‘orcorust’)) > 0 ? 1 : 0) + (countSelected(data(‘lymadisp’)) > 0 ? 1 : 0)
I can see how this might slow things down. I feel like there should be a more efficient way to do this, either some other way to count selected across multiple prompts so there is not so much processing, or a different way to structure your question. Can you do something like a select multiple for, for example, ALL the trees? Then count the number of responses in the select multiple? I think that will be one calculation rather than the many embedded here. Alas, not sure exactly how to do this myself, but would try that general direction.