Flatten directory with incrementing names

ok, i understand that shell scripting is a really power tool but it is real mess:

i have tried with thefourtheye answer but the sort command messed everything up because of the order was like this:

Sequence 1.wav
Sequence 10.wav
Sequence 11.wav
Sequence 12.wav
Sequence 2.wav
Sequence 3.wav
Sequence 4.wav
...

and the spaces was a problem so i had to add " but there was some mess too i don't remember exactly where, then i tried to put some debug print and it didn't work so... In the end I tested it on a Mac OS/X terminal and it has another shell i never heard about (zsh) that uses different commands.. a big mess.

I came up with a small python script. Maybe it is dirty but it works:

import os, sys

count = 1
for dirpath, diname, filenames in os.walk('.'):

    if dirpath=='.' or dirpath=='./out':
        continue

    filenames = [f for f in filenames if f.split('.')[1]=='aif' and f[0]!='.']
    filenames = sorted(filenames, key=lambda f: int(f.split('.')[0].split(' ')[1]))

    for f in filenames:
        filepath = dirpath+'/'+f
        ffmpeg_command = 'ffmpeg -i "'+filepath+'" out/track'+"%04d" % (count,)+'.mp3'
        count += 1
        print ffmpeg_command
        os.system(ffmpeg_command)

Here's roughly how I would do it. Iterate the directories in the right order, then the files. It assumes there's no dir higher than in9/ and no file higher than file99.wav . If there are, extend the loops accordingly. E.g. for dir in in[0-9]/ in[1-9][0-9]/; do

#!/bin/bash
i=0
for dir in in[0-9]/; do
    for file in "$dir"/file[0-9].wav "$dir"/file[1-9][0-9].wav; do
        printf -v dest 'out/file%04d.mp3' "$((++i))"
        ffmpeg -i "$file" "$dest"
    done
done