How to generate a filmstrip image in python from a folder of images?

I would like to do the equivalent off this (ruby code) in python for a Django project I am working on. I want to make a filmstrip image of X number of images in a folder.

From stackoverflow

• Do you mnean something like this? Use PIL to make a "contact sheet" of images?

  Perhaps there are others here that are closer to what you want: http://code.activestate.com/recipes/tags/graphics/

  Espen Christensen : It's something like this, but this is over complicated, i want a singlecolumn "list" of images, and need the images to be loaded from a folder. I guess it is possible to modify that script to do that.

  S.Lott : My thought is that the row number is always 0. Based on that, you can simplify this considerably.

• Here is a function that wraps the contact sheet function S.Lott mentioned.

  #!/usr/bin/env python
  
  import os, os.path
  from contactsheet import make_contact_sheet
  
  def make_film_strip(fnames,
                     (photow,photoh),
                     (marl,mart,marr,marb),
                     padding):
      return make_contact_sheet(fnames,
                                (1, len(fnames)),
                                (photow,photoh),
                                (marl,mart,marr,marb),
                                padding)
  

  It is assuming the recipe is saved as contactsheet.py. Usage is:

  fstrip = filmstrip.make_film_strip(filmstrip.fnames, (120, 120), (0,0,0,0), 0)
  fstrip.save('/path/to/file.format')
  

  Tested.

  Espen Christensen : Thanks, the wrapping is good, thanks but i think i will try to make å decoupled solution, just modify the script and see if i can make it work...