Import from file with same name but other directory - python

I have two files /somewhere/main.py and /somewhere/repeated_name.py and then I have other file /random/path/repeated_name.py. How do I import a function from /random/path/repeated_name.py into /somewhere/main.py?

I think you can use shutil for that. All you need is the absolute path and then use the shutil.move(src, dst) method. And then you just have to use the from file import method without bothering about the folder.
But if you want to import a function to another file, I think this article should answer your question: https://www.geeksforgeeks.org/python-import-module-from-different-directory/

You can use:
from somewhere.repeated_name import func_name
Just make sure folder also contains an init.py, this allows it to be included as a package.

Related

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

How to display a variable without a specific part?

So, i have a variable, for example dir = "Crypter.aes". I need to variable like dir, but without .aes. What gotta I do for that? I use directory parser, that make many dir with file name in that directory, and for each file I need to remove a certain part at the end - .aes
This is a task for the os.path module in the standard library.
import os.path
dir, _ = os.path.splitext("Crypter.aes")
If you're working a lot with file paths, you also might want to take a look at the pathlib module.
from pathlib import Path
dir = Path("Crypter.aes").stem

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 access file in parent directory using python?

I am trying to access a text file in the parent directory,
Eg : python script is in codeSrc & the text file is in mainFolder.
script path:
G:\mainFolder\codeSrc\fun.py
desired file path:
G:\mainFolder\foo.txt
I am currently using this syntax with python 2.7x,
import os
filename = os.path.dirname(os.getcwd())+"\\foo.txt"
Although this works fine, is there a better (prettier :P) way to do this?
While your example works, it is maybe not the nicest, neither is mine, probably. Anyhow, os.path.dirname() is probably meant for strings where the final part is already a filename. It uses os.path.split(), which provides an empty string if the path end with a slash. So this potentially can go wrong. Moreover, as you are already using os.path, I'd also use it to join paths, which then becomes even platform independent. I'd write
os.path.join( os.getcwd(), '..', 'foo.txt' )
...and concerning the readability of the code, here (as in the post using the environ module) it becomes evident immediately that you go one level up.
To get a path to a file in the parent directory of the current script you can do:
import os
file_path = os.path.join(os.path.dirname(os.path.dirname(__file__)), 'foo.txt')
You can try this
import environ
environ.Path() - 1 + 'foo.txt'
to get the parent dir the below code will help you:
import os
os.path.abspath(os.path.join('..', os.getcwd()))

Fast way to read filename from directory?

Given a local directory structure of /foo/bar, and assuming that a given path contains exactly one file (filename and content does not matter), what is a reasonably fast way to get the filename of that single file (NOT the file content)?
1st element of os.listdir()
import os
os.listdir('/foo/bar')[0]
Well I know this code works...
for file in os.listdir('.'):
#do something
you can also use glob
import glob
print glob.glob("/path/*")[0]
os.path.basename will return the file name for you
so you can use it for the exact one file by adding your file path :
os.path.basename("/foo/bar/file.file")
or you can run through the files in the folder and read all names
file_src = "/foo/bar/"
for x in os.listdir(file_src):
print(os.path.basename(x))

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