Develop Reference

Glob doesn't return list of files from specified directory

I've found two ways of listing files from a specified directory from other posts here on Stack Overflow but I can't seem to get them working. The first one returns the path and second return the files I'm looking for but also the path. I have tried several ways like renaming the target directory and files but it doesn't seem to do the trick.
The code in question:
import glob
jpgFilenamesList = glob.glob(r"C:\Users\viodo\PycharmProjects\pythonProject")
print(jpgFilenamesList)
mydir = r"C:\Users\viodo\PycharmProjects\pythonProject"
file_list = glob.glob(mydir + "/*.jpg")
print(file_list)
what I get:
['C:\\Users\\viodo\\PycharmProjects\\pythonProject']
['C:\\Users\\viodo\\PycharmProjects\\pythonProject\\dngjknfjkg.jpg', 'C:\\Users\\viodo\\PycharmProjects\\pythonProject\\fjkdnfkl.jpg', 'C:\\Users\\viodo\\PycharmProjects\\pythonProject\\skdklenfkd.jpg']
Solution found in another thread: Python glob multiple filetypes
Some tweaking got it running smoth. Thanks for the help!

Glob returns a list of pathnames relative to the root directory. That root directory is assumed to be your current working directory unless the glob pattern specified is an absolute path. In short, because your pattern is an absolute path pattern, the returned files will not be relative, but absolute, including the entire path.
When not using an absolute path pattern, in some cases, you could get just a file name if a file name matches in the current working directory. That file name would of course be relative to the current working directory.
In Python 3.10, you should be able to change the assumed root directory without using an absolute pattern via a new root_dir parameter, but this is not currently available in 3.9 and below: https://docs.python.org/3.10/library/glob.html.
In your case, as mentioned in the comments by othes, os.path.basename should be able to get just the file name if that is what you are after. Alternatively, you could change the current working directory via os.chdir and provide a glob pattern of simply *.jpg and get just the file names relative to the that current working directory, both are reasonable solutions.
Extracting the base name:
mydir = r"C:\Users\viodo\PycharmProjects\pythonProject"
file_list = [os.path.basename(f) for f in glob.glob(mydir + "/*.jpg")]
or returning the files relative to an arbitrary "current working directory":
os.chdir(r"C:\Users\viodo\PycharmProjects\pythonProject")
file_list = glob.glob("*.jpg")
Depending on your requirements, one solution may be better than the other.

Can i shorten a file location to the .py files location

first post here so sorry if it's hard to understand. Is it possible to shorten the directory in python to the location of the .py file?. For example, if I wanted to grab something from the directory "C:\Users\Person\Desktop\Code\Data\test.txt", and if the .py was located in the Code folder, could I shorten it to "\data\test.txt". I'm new to python so sorry if this is something really basic and I just didn't understand it correctly.
I forgot to add i plan to use this with multiple files, for example: "\data\test.txt" and \data\test2.txt

import os
CUR_FILE = os.path.abspath(__file__)
TARGET_FILE = "./data/test.txt"
print(os.path.join(CUR_FILE, TARGET_FILE))
With this, you can move around your Code directory anywhere and not have to worry about getting the full path to the file.
Also, you can run the script from anywhere and it will work (you don't have to move to Code's location to run the script.

You can import os and get current working directory ,this will give you the location of python file and then you can add the location of folder data and the file stored in that ,code is given below
import os
path=os.getcwd()
full_path1=path+"\data\test.txt"
full_path2=path+"\data\test2.txt"
print(full_path1)
print(full_path2)
I think this will work for your case and if it doesn't work then add a comment

Os.path gives unexpected output

lately I started working with the Os module in python . And I finally arrived to this Os.path method . So here is my question . I ran this method in one of my kivy project just for testing and it actually didn't returned the correct output.The method consisted of finding if any directory exist and return a list of folders in the directory . otherwise print Invalid Path and return -1 . I passed in an existing directory and it returned -1 but the weird path is that when I run similar program out of my kivy project using the same path present in thesame folder as my python file it return the desired output .here is the image with the python file and the directory name image I have tested which returns invalid path.
and here is my code snippet
def get_imgs(self, img_path):
if not os.path.exists(img_path):
print("Invalid Path...")
return -1
else:
all_files = os.listdir(img_path)
imgs = []
for f in all_files:
if (
f.endswith(".png")
or f.endswith(".PNG")
or f.endswith(".jpg")
or f.endswith(".JPG")
or f.endswith(".jpeg")
or f.endswith(".JPEG")
):
imgs.append("/".join([img_path, f]))
return imgs

It's tough to tell without seeing the code with your function call. Whatever argument you're passing must not be a valid path. I use the os module regularly and have slowly learned a lot of useful methods. I always print out paths that I'm reading or where I'm writing before doing it in case anything unexpected happens, I can see that img_path variable, for example. Copy and paste the path in file explorer up to the directory and make sure that's all good.
Some other useful os.path methods you will find useful, based on your code:
os.join(<directory>, <file_name.ext>) is much more intuitive than imgs.append("/".join([img_path, f]))
os.getcwd() gets your working directory (which I print at the start of scripts in dev to quickly address issues before debugging). I typically use full paths to play it safe because Python pathing can cause differences/issues when running from cmd vs. PyCharm
os.path.basename(f) gives you the file, while os.path.dirname(f) gives you the directory.
It seems like a better approach to this is to use pathlib and glob. You can iterate over directories and use wild cards.
Look at these:
iterating over directories: How can I iterate over files in a given directory?
different file types: Python glob multiple filetypes
Then you don't even need to check whether os.path.exists(img_path) because this will read the files directly from your file system. There's also more wild cards in the glob library such as * for anything/any length, ? for any character, [0-9] for any number, found here: https://docs.python.org/3/library/glob.html

How to get the proper absolute path of a file using Python 3?

I am trying to get the absolute path of a file, but it's not working. My Code is only showing the path to the directory of my running code. This are the two way I have tried:
1)
import os
print(os.path.abspath("More_Types.py"))
2)
import os
print(os.path.realpath("More_Types.py"))
But I continue to get the full path to my running program. How can I get the correct path to file, that is located some where else in my Computer?
PS: I am sorry I can't provide the output because it will reveal all the folders to my running program.

More_Types.py, subdir/More_Types.py and ../More_Types.py are relative paths.
Outcome
If you provide a relative path, realpath and abspath will return an absolute path relative to the current working directory. So essentially, they behave like os.path.join with the current working directory as first argument:
>>> import os.path
>>> os.path.join(os.getcwd(), 'More_Types.py') == os.path.abspath('More_Types.py')
True
This explains, why you get the result you explained.
Explanation
The purpose of abspath is to convert a relative path into an absolute path:
>>> os.path.abspath('More_Types.py')
'/home/meisterluk/More_Types.py'
Unlike abspath, relpath also follows symbolic links. Technically, this is dangerous and can get you caught up in infinite loops (if a link points to any of its parent directories):
>>> os.path.abspath('More_Types.py')
'/net/disk-user/m/meisterluk/More_Types.py'
Proposed solution
However, if you want to retrieve the absolute path relative to some other directory, you can use os.path.join directly:
>>> directory = '/home/foo'
>>> os.path.join(directory, 'More_Types.py')
'/home/foo/More_Types.py'
>>>
I think the best answer to your question is: You need to specify the directory the file is in. Then you can use os.path.join to create an absolute filepath.

You could use this if you want:
import os
a='MoreTypes.py'
for root , dirs , files in os.walk('.') :
for file in files:
if file==a:
print(os.path.join(root,file))

Python: import a file from another directory

I am working in a set of directories with the following structure:
Master/
Subfolder_1/file_1_1.py
file_1_2.txt
Subfolder_2/file_2_1.py
I import file_1_1 in file_2_1 as follows:
import sys
sys.path.append('../file_1_1')
file_1_1 is reading file_1_2.txt which is in the same directory. However, when I call the function that reads file_1_2.txt from file_2_1.py, it says no such file and directory and it gives me the path of file_1_2.txt as:
Master/Subfolder_2/file_1_2.txt
which is a wrong path. It seems like python in this case is using the working directory as a reference. How can I solve this error given that I don't want to include the absolute path for each file I read.

Don't mess with sys.path, and don't think of imports as working against files and directories. Ultimately everything has to live in a file somewhere, but the module hierarchy is a little more subtle than "replace dot with slash and stick a .py at the end".
You almost certainly want to be in Master and run python -m Subfolder_1.file_1_1. You can use pkg_resources to get the text file:
pkg_resources.resource_string('Subfolder_1', 'file_1_1.txt')

info=[]
with open(os.path.realpath('yourtextfile.txt','r') as myfile:
for line in myfile:
info.append(line)