Better way to find absolute paths during os.walk()?

Question

I am practicing with the os module and more specifically os.walk(). I am wondering if there is an easier/more efficient way to find the actual path to a file considering this produces a path that suggests the file is in the original folder when os.walk() is first ran:

import os

threshold_size = 500

for folder, subfolders, files in os.walk(os.getcwd()):
    for file in files:
        filePath = os.path.abspath(file)
        if os.path.getsize(filePath) >= threshold_size:
            print filePath, str(os.path.getsize(filePath))+"kB"

This is my current workaround:

import os

threshold_size = 500

for folder, subfolders, files in os.walk(os.getcwd()):
    path = os.path.abspath(folder)
    for file in files:
        filePath = path + "\\" + file
        if os.path.getsize(filePath) >= threshold_size:
            print filePath, str(os.path.getsize(filePath))+"kB"

For shaktimaan, this:

for folder, subfolders, files in os.walk(os.getcwd()):
    for file in files:
        filePath = os.path.abspath(file)
        print filePath

produces this(most of these files are in a subfolder of projects, not projects itself):

C:\Python27\projects\ps4.py
C:\Python27\projects\ps4_encryption_sol.py
C:\Python27\projects\ps4_recursion_sol.py
C:\Python27\projects\words.txt
C:\Python27\projects\feedparser.py
C:\Python27\projects\feedparser.pyc
C:\Python27\projects\news_gui.py
C:\Python27\projects\news_gui.pyc
C:\Python27\projects\project_util.py
C:\Python27\projects\project_util.pyc
C:\Python27\projects\ps5.py
C:\Python27\projects\ps5.pyc
C:\Python27\projects\ps5_test.py
C:\Python27\projects\test.py
C:\Python27\projects\triggers.txt
C:\Python27\projects\ps6.py
C:\Python27\projects\ps6_pkgtest.py
C:\Python27\projects\ps6_solution.py
C:\Python27\projects\ps6_visualize.py
C:\Python27\projects\ps6_visualize.pyc
C:\Python27\projects\capitalsquiz1.txt
C:\Python27\projects\capitalsquiz2.txt
C:\Python27\projects\capitalsquiz3.txt
C:\Python27\projects\capitalsquiz4.txt
C:\Python27\projects\capitalsquiz5.txt
C:\Python27\projects\capitalsquiz_answers1.txt
C:\Python27\projects\capitalsquiz_answers2.txt
C:\Python27\projects\capitalsquiz_answers3.txt
C:\Python27\projects\capitalsquiz_answers4.txt
C:\Python27\projects\capitalsquiz_answers5.txt
C:\Python27\projects\quiz.py
C:\Python27\projects\file2.txt
C:\Python27\projects\regexes.txt
C:\Python27\projects\regexsearch.py
C:\Python27\projects\testfile.txt
C:\Python27\projects\renamedates.py

Answer 1

I think there you mistook what abspath does. abspath just convert a relative path to a complete absolute filename.

For e.g.

os.path.abspath(os.path.join(r"c:\users\anonymous\", ".."))
#produces this output : c:\users

Without any other information, abspath can only form an absolute path from the only directory it can know about, for your case the current working directory. So currently what it is doing is it joins os.getcwd() and your file

So what you would have to do is:

for folder, subfolders, files in os.walk(os.getcwd()):
    for file in files:
        filePath = os.path.join(os.path.abspath(folder), file)