Help with small Python Script

cmfarley19 · 05-01-2005, 08:01 AM

Gretting all.

I have a need to convert PNG files to EPS format. I did some searching and found som python commands I can execute from the python interpreter and they seem to work just fine.

Code:

 >>> import Image
 >>> im=Image.open("logo_nosonis.png")
 >>> im.save("logo_nosonis.eps")

I have no experience with python.

How would I turn this into a script that could take the base file name as an argument and spit out the eps file?

Thanks,

Chris

nhs · 05-01-2005, 10:20 AM

Try something along the lines of:

Code:

#!/usr/bin/env python
import sys
import Image

def main():
	if len(sys.argv) < 2:
		print """Usage:
	%s input_file [output_file]""" % (sys.argv[0],)
		return
	if len(sys.argv) == 2:
		sys.argv.append(sys.argv[1].replace(".png",".eps"))
	img = Image.open(sys.argv[1])
	img.save(sys.argv[2])

if __name__ == '__main__':
	main()

And if you "chmod u+x" the file then executing it (like a normal executable) will run it fine (although I've never actually tested the code)

cmfarley19 · 05-01-2005, 12:13 PM

Thanks for the response.
I do get an error when trying to execute.
I named the script png2eps.py.
Here's what I get.

Code:

gcnpd:/home/user/temp# ./png2eps.py NPMTV_FP_EX.PNG 
  File "./png2eps.py", line 9
    return
    ^
SyntaxError: invalid syntax

Any advice?

nhs · 05-01-2005, 12:19 PM

Sorry, my error. Use

Code:

return None

instead.

APB_4 · 05-01-2005, 03:18 PM

I've been using python for a while and seen "if __name__ == '__main__'" but what is the point of it? You could just call the function main without that line.

nhs · 05-01-2005, 03:49 PM

The advantage isn't so great in this example but the idea is that it allows making an executable script which is also usable as a module. In this case another program could "import png2eps" (where __name__ != '__main__') and it won't attempt to execute main() at load. The script could then set up sys.argv and call png2eps.main(). This is fairly pointless here but it is not uncommon to have an executable script structured so that it can be imported. The usage is varied (think code + test case) or a program+lib in one file where the other functions are useful in a modular context. In any case it's never the decision of the programmer how the code will be used and it's generally polite to make sure that the right thing happens if the module is imported regardless of whether it makes sense or not.

APB_4 · 05-01-2005, 04:06 PM

Ah ok cheers nhs. Thanks for clearing that up for me.

cmfarley19 · 05-02-2005, 07:02 AM

nhs,
Here's my script. I culled the bits trying to make it a module.

Code:

#!/usr/bin/python
import sys
import Image
if len(sys.argv) < 2:
        print """Usage:
%s input_file [output_file]""" % (sys.argv[0],)
        return None
if len(sys.argv) == 2:
        sys.argv.append(sys.argv[1].replace(".png",".eps"))
img = Image.open(sys.argv[1])
img.save(sys.argv[2])

I am still getting the same error:

Code:

user@gcnpd:~/temp$ ./png2eps.py NPMTV_FP_EX.PNG 
  File "./png2eps.py", line 7
    return None
SyntaxError: 'return' outside function

Should that be a quit or exit command? Does python have such a construct?

APB_4 · 05-02-2005, 08:10 AM

That's because you haven't made it as a function. It needs to start "def func_name():" otherwise return doesn't work. If you don't want it as a function you could use sys.exit() to close the script.