Develop Reference

How to get the full path of a file in python including the subdirectory?

I tried every method, but it simply doesn't work. I have a file that is within a subdirectory and that subdirectory lies within a directory. Calling something like
os.path.abspath(__file__))
only yields
directory\\file
but I need
directory\\subdirectory\\file
There should be an easy way to do this, right? I have no idea why abspath doesn't recognize the subdirectory.

from pathlib import Path
my_path = Path("my_file.txt")
full_path = Path.resolve()
Then you can convert full_path to a string if you need it as such.

You should look at the pathlib module

Is there a way to be able to use a variable path using os

The goal is to run through a half stable and half variable path.
I am trying to run through a path (go to lowest folder which is called Archive) and fill a list with files that have a certain ending. This works quite well for a stable path such as this.
fileInPath='\\server123456789\provider\COUNTRY\CATEGORY\Archive
My code runs through the path (recursive) and lists all files that have a certain ending. This works well. For simplicity I will just print the file name in the following code.
import csv
import os
fileInPath='\\\\server123456789\\provider\\COUNTRY\\CATEGORY\\Archive
fileOutPath=some path
csvSeparator=';'
fileList = []
for subdir, dirs, files in os.walk(fileInPath):
for file in files:
if file[-3:].upper()=='PAR':
print (file)
The problem is that I can manage to have country and category to be variable e.g. by using *

The standard library module pathlib provides a simple way to do this.
Your file list can be obtained with
from pathlib import Path
list(Path("//server123456789/provider/".glob("*/*/Archive/*.PAR"))
Note I'm using / instead of \\ pathlib handles the conversion for you on windows.

How to normalize a relative path using pathlib

I'm trying to use relative paths in Python, and I want to put my csv files in a separate folder from my python code.
My python program is in the following folder:
G:\projects\code
I want to read this file which is one level up:
G:\projects\data\sales.csv
How do I specify a path using pathlib that is one level up from my current working folder? I don't want to change the current working folder.
I tried this:
from pathlib import Path
file = Path.cwd() /'..'/'data'/'sales.csv'
But now the 'file' variable equals this:
'G:/projects/code/../data/sales.csv'
I read through the docs and either it isn't explained there or I'm just missing it.

Although it's not a problem that your path includes '..' (you can still use this path to open files, etc. in Python), you can normalize the path using resolve():
from pathlib import Path
path = Path.cwd() / '..' / 'data' / 'sales.csv'
print(path) # WindowsPath('G:/projects/code/../data/sales.csv')
print(path.resolve()) # WindowsPath('G:/projects/data/sales.csv')
NB: I personally would name a variable that contains a path path, not file. So you could later on do file = open(path).

print(
Path(__file__).parent, # the folder
Path(__file__).parent.parent, # the folder's parent
sep='\n'
)
print(
Path(
Path(__file__).parent.parent, 'hello.py'
)
)
results in
C:\Users\isik\Desktop\Python\MessAround\project\module
C:\Users\isik\Desktop\Python\MessAround\project
C:\Users\isik\Desktop\Python\MessAround\project\hello.py
with this file structure
-project
-module
-__init__.py
-hello.py
-__init__.py
while the code is located inside project.module.__init__.py

Do you mean "read my csv files"?
The import keyword has a different meaning in Python (you import only other Python modules).
In any case, in order to read a file located one folder above your Python file, you can use this:
import os
filePath = os.path.dirname(__file__)+'/../'+fileName
fileDesc = open(filePath)
fileData = fileDesc.read()
fileDesc.close()
...

here is an example I used:
import json
from pathlib import Path
def read_files(folder_name, file_name):
base_path = Path.cwd().joinpath('configs','resources')
path = base_path.joinpath(folder_name,file_name)
open_file = open(path,'r')
return json.load(open_file.read())

This is pretty old, but I happened on it looking for something else.
Oddly you never got a direct, obvious, answer -- you want the parent property:
from pathlib import Path
file = Path.cwd().parent / 'data' / 'sales.csv'
Note that some of the answers that say you want __file__ rather than the current working directory may be correct (depending on your use case), in which case it's:
from pathlib import Path
file = Path(__file__).parent.parent / 'data' / 'sales.csv'
(parent of the python file is the code dir, parent of that is the projects dir.
However, It's not great practice to refer to your data by its relative path to your code -- I think using the cwd is a better option -- though what you should do is pass the path to the data in to the script via sys.argv.

python: cd into file

I have a script that creates a folder foo-12345. Issue is, the numbers in the folder change name anytime I run a per script that creates it. I'm trying to find a way to change directory into the folder so I can do a search. I tried using a variable:
import os
output = /var/foo-*
os.chdir(ouput)
This does not seem to work. Is there a way to capture that folder name in a variable and use that variable instead?

You can use the glob module to do it.
import glob
dirs = glob.glob('/var/foo-*')
The resulting dirs is a list, so you will need to process it as such. Have a look at the documentation

How to import using a path that is a variable in Python

I am trying to make a program that will go through and visit an array of directories and run a program and create a file inside.
I have everything working except that I need to figure out a way to import from a new path each time to get to a new directory.
For example:
L =["directory1", "directory2", "directory3"]
for i in range(len(L)):
#I know this is wrong, but just to give an idea
myPath = "parent."+L[i]
from myPath import file
#make file... etc.
Obviously when I use myPath as a variable for the path to import, I get an error. I have tried several different ways by searching online through Stack Overflow and reading OS and Sys documentation, but have come to no working result.

You can use 'imp' module to load source code of python scrips
import imp
root_dir = '/root/'
dirs =["directory1", "directory2", "directory3"]
for _dir in dirs:
module_path = os.path.join(root_dir,_dir,'module.py')
mod = imp.load_source("module_name", module_path)
# now you can call function in regular way, like mod.some_func()

I want to create a text file inside each directory. To do this I must
cycle through my array and take each directory name so I can visit it.
import is for loading external modules, not creating new files, if creating new files is what you want to do, use the open statement, and open the not yet existing file with 'w' mode. Note: the directory must exist.
from os.path import join
L =["directory1", "directory2", "directory3"]
for d in L: # loop through the directories
with open(join(d,"filename.txt"), "w") as file:
pass # or do stuff with the newly created file