In python I have variables base_dir and filename. I would like to concatenate them to obtain fullpath. But under windows I should use \ and for POSIX / .
fullpath = “%s/%s” % ( base_dir, filename ) # for Linux
How can I make this platform independent?
….
I have a situation very much like the one at Error “ImportError: DLL load failed: %1 is not a valid Win32 application”, but the answer there isn’t working for me.
My Python code says:
import cv2
But that line throws the error shown in the title of this question.
I have OpenCV installed in C:\lib\opencv on this 64-bit machine. I’m using 64-bit Pyt….
I am getting the following error:
WindowsError: [Error 2] The system cannot find the file specified
My code is:
subprocess.call([“<
Windows 7, 64 bit. Python 3.x latest, stable.
Any ideas?
Thanks,
….
Where is Python’s sys.path initialized from?
UPD: Python is adding some paths before refering to PYTHONPATH:
>>> import sys
>>> from pprint import pprint as p
>>> p(sys.path)
[”,
‘C:\\Python25\\lib\\site-packages\\setuptools-0.6c9-py2.5.egg’,
‘C:\\Python25\\lib\\site-packages\\orbited-0.7.8-py2.5.egg’,
‘C:\\Python25….
Ok…I dont know where module x is, but I know that I need to get the path to the directory two levels up.
So, is there a more elegant way to do:
import os
two_up = os.path.dirname(os.path.dirname(__file__))
Solutions for both Python 2 and 3 are welcome!
….
I have just joined a project with a rather large existing code base. We develop in linux and do not use and IDE. We run through the command line. I’m trying to figure out how to get python to search for the right path when I run project modules. For instance, when I run something like:
python someprojectfile.py
I get
ImportError: no module named ….
this code is get the templates/blog1/page.html in b.py:
path = os.path.join(os.path.dirname(__file__), os.path.join(‘templates’, ‘blog1/page.html’))
but i want to get the parent dir location:
aParent
|–a
| |—b.py
| |—templates
| |——–blog1
| |——-page.html
|–templates
….
I would like to get just the folder path from the full path to a file.
For example T:\Data\DBDesign\DBDesign_93_v141b.mdb and I would like to get just T:\Data\DBDesign (excluding the \DBDesign_93_v141b.mdb).
I have tried something like this:
existGDBPath = r’T:\Data\DBDesign\DBDesign_93_v141b.mdb’
wkspFldr = str(existGDBPath.split(‘\\’)[0:-1])
pri….
UNIX absolute path starts with ‘/’, whereas Windows starts with alphabet ‘C:’ or ‘\’.
Does python have a standard function to check if a path is absolute or relative?
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I’m trying to make a script to list all directory, subdirectory, and files in a given directory.
I tried this:
import sys, os
root = “/home/patate/directory/”
path = os.path.join(root, “targetdirectory”)
for r, d, f in os.walk(path):
for file in f:
print(os.path.join(root, file))
Unfortunatly it doesn’t work properly.
I get all the ….