I just experienced a weird behaviour in Python.
I created a copy of a script.py file in a sub-folder within the folder that contains the initial script.
The script at the end exports some data into a .txt file by using:
with open('clayList.2203.txt', 'w',encoding='utf-8') as f:
for item in claysUniqueList:
f.write("%s\n" % item)
The problem is that Python writes the new file on the parent directory instead of the current one.
I checked the path with:
print(sys.path[0])
and it prints the current path correctly.
By default, relative paths are relative to the working directory, that is the directory from which is run the command that run the script.
If you want the path to be relative from the script directory, you will have to explicitly code this behaviour:
import os
filepath = os.path.join(os.path.dirname(__file__), 'clayList.2203.txt')
with open(filepath, 'w',encoding='utf-8') as f:
# ...
When you run code in Visual Studio, there are debugging options.
One of these it the directory to run from, called "Working directory".
(Right click your project and go to settings).
To run from the sub-directory you need to change this.
If you want to start in a sub directory, type that in instead, in the "working directory" shown here:
path for creating the file should be relative to the directory of execution
e.g. your pwd is parent and your script is in parent/child1/child2/script.py then path of the file to be created should be ./child1/child2/clayList.2203.txt
Related
I know there are a lot of answers on this subject, but no one works once you compile a script in an executable.
In my python script, I create a file within the same directory of the script.
to get the path of the current dir I use pathlib
basepath = Path(__file__).parent
filename='myfile'
filepath=os.path.join(basepath, filename)
if I print the directory I get the file wrote in the good directory and everything works fine within python
(i.e desktop/myname/myscriptdir/myfile)
but once I "compile" with pyinstaller with --onefile, if I launch the executable, the directory will be
like
/var/folders/nr/w0698dl96j39_fq33lqd8pk80000gn/T/_MEIP12KxC/myfile
believed me, I tried a lot of various method (abspath, os.realpath..)to get the current dir, no one worked fine once in an executable file.
When you compile an app using pyinstaller with the --onefile or -F flag, the file that it creates is actually an archive file, like a .zip file.
When you execute that file it launches a process that extracts itself into a temporary folder somewhere in your OS filesystem. This is the path that is reported when you use the __file__ variable in the compiled application.
It then continues to launch your application from there and the temporary directory becomes the runtime durectory for the duration of the apps life. When the app is finally closed it deletes the temporary runtime directory on it's way out.
Since this is the case there are alternatives.
To get the current working directory during runtime use:
path = '.'
#or
path = os.getcwd()
To get the path to the compiled executable file during runtime:
path = sys.executable
My directory structure like below:
./outputsetting.json
./myapp/app.py
I load outputsetting.json file in app.py like below:
with open("..\outputpath.json","r") as f:
j=json.load(f)
And run in myapp directory it's ok:
python app.py
But if I run app.py in the parent directory:
python .\myapp\app.py
It raise error
FileNotFoundError: [Errno 2] No such file or directory: '..\\outputpath.json'
How can I load file from disk by the relative directory to the app.py? So I can run it from any place and needn't modify code every time.
Thanks!
When you start the script from the parent directory, the working directory for the script changes. If you want to access a file that has a specific location relative to the script itself, you can do this:
from pathlib import Path
location = Path(__file__).parent
with open(location / "../outputsetting.json","r") as f:
j=json.load(f)
Path(__file__) gets the location of the script it is executed in. The .parent thus gives you the folder the script is in, still as a Path object. And you can then navigate to the parent directory from that with location / "../etc"
Or of course in one go:
with open(Path(__file__).parent / "../outputsetting.json","r") as f:
j=json.load(f)
(Note: your code says outputpath.json, but I assume that's the same outputsetting.json)
Another solution would be to just change the working directory to be the folder the script is in - but that of course only works if all your scripts and modules are OK with that change:
from pathlib import Path
from os import chdir
chdir(Path(__file__).parent)
with open("../outputsetting.json","r") as f:
j=json.load(f)
I'd prefer constructing the absolute path as in the previous example though.
so I recently tried using playsound and the .mp3 is in the same folder as the .py, and it says it could not find the file.
I printed out the current directory and it says it's at C:\Users\me
Shouldn't this be at the same directory as where the .py script is at? This has been happening with my other python scripts where I have to explicitly give it the directory, whereas before I didn't have to and it was just where the python script was at.
Is there a setting for this?
The current directory is defined by where you execute the python file not the location of it. If you want to change the directory you can use the os module:
import os
os.chdir(PATH)
If you're looking for a file in the same directory as the executing python program:
basename = os.path.basename(__file__)
my_file_name = basename + "/myfile"
with open(my_file_name) as my_file:
...
The variable __file__ contains the full pathname of the currently executing Python program.
I am trying to print the directory of the .py file however I want to run the file from another directory by adding the file's directory to the system Path file. I ran in to some problems when I turned to file to an executable with pyinstaller.
The code in the file:
import pathlib
path= pathlib.Path(__file__).parent.resolve()
print(path)
Let's say the directory of the file is: C:\Users\User\Documents\file.py
And let's say the current working directory is: C:\Users\User
When I run file.py I get:
C:\Users\User\Documents
which is the output I want
When I run file.exe I get:
C:\Users\User
which is not what I wan't
I don't understand why the .py and .exe files are working differently.
What can I do to fix the problem?
I tried a bunch of stuff from here but none of them seem to work:
How do I get the full path of the current file's directory?
My tests are at:
src/com/xyz/tests/api_test/<Test File>
My test file calls my libraries at:
src/com/xyz/libs/api_libs/<Library File>
My library file has to open a JSON file at:
src/com/xyz/libs/api_libs/configs/<Config File>
In my library file, since its at the same parent directory as the JSON configs, I have used the following code to open the JSON.
with open('configs/sample_wlan_json'):
<Do Some action>
I tried various paths like:
.../libs/api_libs/configs/<ConfigFileName>
src/com/mist/libs/api_libs/configs/<ConfigFileName>.json
The whole path from /Users/...... but nothing seems to work.
A relative path is relative to the current working directory. Current working directory depends on how an application is started, and not where it is.
So, if you want to have a path relative to your source code, you should not rely on the current working directory, but construct the absolute path instead.
You can construct a path which is relative to your source code by using the __file__ variable, which is the path to the current py file.
Something like this should work:
configs_dir = os.path.join(__file__, '..', 'configs')
with open(os.path.join(configs_dir, 'sample_wlan.json'), 'rt') as f:
...