Why does it appear a window in a flash when I use "call" in execjs, python?

Question

I'm sure the problem happens here:

self.js_fun.call('myfunc')

and I have initialized js_fun:

self.js_fun = execjs.compile(f.read())

just as the title, there will always be a sudden window as it executes,
and disappeared as fast as a HK reporter

Lukily, I catched it anyway:

QAQ...Failed to upload image; an error occured on the server
OK then I paste the link here: The image

Thanks in advance.