calling xll function python

Question

Hi I am trying to call a function of an xll addin through python. I have loaded the addin and everything runs fine, up to the point where I need to call a function where I an argument is optional and thus I can leave it blank, and then I get an error. Namely: in excel, function1 works fine if i type in a cell

=function1('a','b',,'d')

but if in python I write:

import os
import numpy as np
from win32com.client import Dispatch
Path  = 'myWeirdApp.xll'
xlApp = Dispatch("Excel.Application")
xlApp.RegisterXLL(Path)

the following line doesn't work:

xllApp.run('function1', 'a','b', , 'd')

does anybody have a solution for this type of situation?

Hi. I also have some xll add-in that I want to be able to load and use in python. Can you please share your solution? Thanks — Confounded
– Confounded, Commented Nov 8, 2019 at 14:02
xllApp is a typo, this should read xlApp.run('function1', ...) — Avidadollars
– Avidadollars, Commented Jan 26, 2022 at 14:09

NickD1 · Accepted Answer · 2015-02-13 17:22:08Z

2

Looks like that if you substitute the gaps with None it works fine!

answered Feb 13, 2015 at 17:22

NickD1

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