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Can't replace "\" with "/"

I am trying to replace a string here in Python.
This is my Input -
link = (r"C:\dell\Documents\Ms\Realm")
I want my output to be like this:
C:/dell/Documents/Ms/Realm
I tried the replace method it didn't work.
the code tried:
link = (r"C:\dell\Documents\Ms\Realm")
link.replace("\","/")

In Python strings, the backslash "\" is a special character, also called the "escape" character. You need to add second backslash to "escape" it and use in string search.
link.replace("\\", "/")

There's nothing wrong with that Windows path. There's no reason to replace anything. A real improvement would be to use pathlib instead of raw strings:
from pathlib import Path
link = Path(r"C:\dell\Documents\Ms\Realm")
That would allow you to construct paths from parts using, eg joinpath, get the parts of the path with parts, the name with name, directory with parent etc.
var filePath=link.joinpath("some_file.txt")
print(filePath)
-------------------
C:\dell\Documents\Ms\Realm\some_file.txt
and more
>>> print(link.parts)
('C:\\', 'dell', 'Documents', 'Ms', 'Realm')
>>> print(link.parent)
C:\dell\Documents\Ms
Or search for files in a folder recursively:
var files=(for file in link.rglob("*.txt") if file.is_file())

Printing File Names

I am very new to python and just installed Eric6 I am wanting to search a folder (and all sub dirs) to print the filename of any file that has the extension of .pdf I have this as my syntax, but it errors saying
The debugged program raised the exception unhandled FileNotFoundError
"[WinError 3] The system can not find the path specified 'C:'"
File: C:\Users\pcuser\EricDocs\Test.py, Line: 6
And this is the syntax I want to execute:
import os
results = []
testdir = "C:\Test"
for folder in testdir:
for f in os.listdir(folder):
if f.endswith('.pdf'):
results.append(f)
print (results)

Use the glob module.
The glob module finds all the pathnames matching a specified pattern
import glob, os
parent_dir = 'path/to/dir'
for pdf_file in glob.glob(os.path.join(parent_dir, '*.pdf')):
print (pdf_file)
This will work on Windows and *nix platforms.
Just make sure that your path is fully escaped on windows, could be useful to use a raw string.
In your case, that would be:
import glob, os
parent_dir = r"C:\Test"
for pdf_file in glob.glob(os.path.join(parent_dir, '*.pdf')):
print (pdf_file)
For only a list of filenames (not full paths, as per your comment) you can do this one-liner:
results = [os.path.basename(f) for f in glob.glob(os.path.join(parent_dir, '*.pdf')]

Right now, you search each character string inside of testdir's variable.
so it's searching the folder for values "C", ":", "\", "T" etc. You'll want to also escape your escape character like "C:\...\...\"
You probably was to use os.listdir(testdir) instead.

Try running your Python script from C:. From the Command Prompt, you might wanna do this:
> cd C:\
> python C:\Users\pcuser\EricDocs\Test.py
As pointed out by Tony Babarino, use r"C:\Test" instead of "C:\Test" in your code.

There are a few problems in your code, take a look at how I've modified it below:
import os
results = []
testdir = "C:\\Test"
for f in os.listdir(testdir):
if f.endswith('.pdf'):
results.append(f)
print (results)
Note that I have escaped your path name, and removed your first if folder.... That wasn't getting the folders as you expected, but rather selecting a character of the path string one at a time.
You will need to modify the code to get it to look through all folders, this currently doesn't. Take a look at the glob module.

You will need to escape the backslash on windows and you can use os.walk to get all the pdf files.
for root,dirs,files in os.walk(testdir):
for f in files:
if f.endswith('.pdf'):
results.append(f)
print (results)

You are basically iterating through the string testdir with the first for loop then passing each character to os.listdir(folder) does not make any sense then, just remove that first for loop and use fnmatch method from fnmatch module:
import os
from fnmatch import fnmatch
ext = '*.pdf'
results = []
testdir = "C:\Test"
for f in os.listdir(testdir):
if fnmatch(f, ext):
results.append(f)
print (results)

Try testdir = r"C:\Test" instead of testdir = "C:\Test". In python You have to escape special characters like for example \. You can escape them also with symbol '\' so it would be "C:\\Test". By using r"C:\Test", You are telling python to use raw string.
Also for folder in testdir: line doesn't make sense because testdir is a string so You are basically trying to iterate over a string.

I had to mention the names of training images for my Yolo model,
Here's what i did to print names of all images which i kept for training YoloV3 Model
import os
for root, dirs, files in os.walk("."):
for filename in files:
print(filename)
It prints out all the file names from the current directory

How to process files from one subfolder to another in each directory using Python?

I have a basic file/folder structure on the Desktop where the "Test" folder contains "Folder 1", which in turn contains 2 subfolders:
An "Original files" subfolder which contains shapefiles (.shp).
A "Processed files" subfolder which is empty.
I am attempting to write a script which looks into each parent folder (Folder 1, Folder 2 etc) and if it finds an Original Files subfolder, it will run a function and output the results into the Processed files subfolder.
I made a simple diagram to showcase this where if Folder 1 contains the relevant subfolders then the function will run; if Folder 2 does not contain the subfolders then it's simply ignored:
I looked into the following posts but having some trouble:
python glob issues with directory with [] in name
Getting a list of all subdirectories in the current directory
How to list all files of a directory?
The following is the script which seems to run happily, annoying thing is that it doesn't produce an error so this real noob can't see where the problem is:
import os, sys
from os.path import expanduser
home = expanduser("~")
for subFolders, files in os.walk(home + "\Test\\" + "\*Original\\"):
if filename.endswith('.shp'):
output = home + "\Test\\" + "\*Processed\\" + filename
# do_some_function, output

I guess you mixed something up in your os.walk()-loop.
I just created a simple structure as shown in your question and used this code to get what you're looking for:
root_dir = '/path/to/your/test_dir'
original_dir = 'Original files'
processed_dir = 'Processed files'
for path, subdirs, files in os.walk(root_dir):
if original_dir in path:
for file in files:
if file.endswith('shp'):
print('original dir: \t' + path)
print('original file: \t' + path + os.path.sep + file)
print('processed dir: \t' + os.path.sep.join(path.split(os.path.sep)[:-1]) + os.path.sep + processed_dir)
print('processed file: ' + os.path.sep.join(path.split(os.path.sep)[:-1]) + os.path.sep + processed_dir + os.path.sep + file)
print('')
I'd suggest to only use wildcards in a directory-crawling script if you are REALLY sure what your directory tree looks like. I'd rather use the full names of the folders to search for, as in my script.
Update: Paths
Whenever you use paths, take care of your path separators - the slashes.
On windows systems, the backslash is used for that:
C:\any\path\you\name
Most other systems use a normal, forward slash:
/the/path/you/want
In python, a forward slash could be used directly, without any problem:
path_var = '/the/path/you/want'
...as opposed to backslashes. A backslash is a special character in python strings. For example, it's used for the newline-command: \n
To clarify that you don't want to use it as a special character, but as a backslash itself, you either have to "escape" it, using another backslash: '\\'. That makes a windows path look like this:
path_var = 'C:\\any\\path\\you\\name'
...or you could mark the string as a "raw" string (or "literal string") with a proceeding r. Note that by doing that, you can't use special characters in that string anymore.
path_var = r'C:\any\path\you\name'
In your comment, you used the example root_dir = home + "\Test\\". The backslash in this string is used as a special character there, so python tries to make sense out of the backslash and the following character: \T. I'm not sure if that has any meaning in python, but \t would be converted to a tab-stop. Either way - that will not resolve to the path you want to use.
I'm wondering why your other example works. In "C:\Users\me\Test\\", the \U and \m should lead to similar errors. And you also mixed single and double backslashes.
That said...
When you take care of your OS path separators and trying around with new paths now, also note that python does a lot of path-concerning things for you. For example, if your script reads a directory, as os.walk() does, on my windows system the separators are already processed as double backslashes. There's no need for me to check that - it's usually just hardcoded strings, where you'll have to take care.
And finally: The Python os.path module provides a lot of methods to handle paths, seperators and so on. For example, os.path.sep (and os.sep, too) wil be converted in the correct seperator for the system python is running on. You can also build paths using os.path.join().
And finally: The home-directory
You use expanduser("~") to get the home-path of the current user. That should work fine, but if you're using an old python version, there could be a bug - see: expanduser("~") on Windows looks for HOME first
So check if that home-path is resolved correct, and then build your paths using the power of the os-module :-)
Hope that helps!

Python filename change

I have a number of videos in a directory on my Mac that all have a specific string in the file name that I want to remove, but I want to keep the rest of the file name as it is. I'm running this python script from terminal.
I have this syntax but it doesn't seem to work. Is it practical to use the following? It seems to simple to be the best way to do this sort of thing which is why I don't think it works.
from os import rename, listdir
text = "Some text I want to remove from file name"
files = listdir("/Users/Admin/Desktop/Dir_of_videos/")
for x in files:
if text in files:
os.rename(files, files.replace(text, ""))

the problem is that you get incomplete paths when you are using listdir, basically, it returns only the files in the directory without the prepending path to the directory
this should do the job:
import os
in_dir = './test'
remove = 'hello'
paths = [os.path.join(in_dir,file) for file in os.listdir(in_dir) if remove in file]
for file in paths:
os.rename(file, file.replace(remove, ""))

Manipulating paths in python

I am writing a python script 2.5 in Windows whose CurrentDir = C:\users\spring\projects\sw\demo\753\ver1.1\011\rev120\source my file is test.py. From this path I would like to access files in this path: C:\users\spring\projects\sw\demo\753\ver1.1\011\rev120\Common\
I tried using os.path.join but it does not work and I from the docs I understand why.
So what could be the best pythonic solution for this?
currentdir = os.getcwd()
config_file_path = os.path.join(currentdir,"\\..\\Common")

Your problem can be solved by using os.path.join, but you're not using it properly.
currentdir = os.getcwd()
config_file_path = os.path.join(currentdir,"\\..\\Common")
"\\..\\Common" is not a relative path, as it starts with \.
You need to join with ..\\Common, which is a relative path.
Please note that os.path.join is not a simple string concatenation function, you don't need to insert the in-between antislashes.
So fixed code would be :
config_file_path = os.path.join(currentdir,"..\\Common")
or, alternatively :
config_file_path = os.path.join(currentdir, "..", "Common")

from os.path import dirname, join
join(dirname(dirname(__file__)), 'Common')
should work.

Try this:
joined = os.path.join('C:\\users\\spring\\projects\\sw\\demo\\753\\ver1.1\\011\\rev120\\source', '..\\Common\\')
# 'C:\\users\\spring\\projects\\sw\\demo\\753\\ver1.1\\011\\rev120\\source\\..\\Common\\'
canonical = os.path.realpath(joined)
# 'C:\\users\\spring\\projects\\sw\\demo\\753\\ver1.1\\011\\rev120\\Common'