I tried writing some commenly used function in a seperate file and import the same into mainApp file, but not able to use import.
I did find many questions regarding the this same question but, the solution was to keep the files in the same folder
I tried without .py as well, but the same error:
Can you please help me how can i fix this issue ?
No '.py'. Just import seperate
Try using this in mainApp.py:
from seperate import *
a()
where seperate.py looks like this:
def a():
print('hi')
Well, sorry, those two files need to be in the same folder. This is not a solution to your problem.
The syntax of a relative import depends on the current location as well as the location of the module, package, or object to be imported. Here are a few examples of relative imports:
from .some_module import some_class
from ..some_package import some_function
from . import some_class
Read more about Absolute vs Relative Imports in Python
In your case it should be:
from .seperate import a
Also check this question:
Importing from a relative path in Python
add your project directory into your path variable so that python know from where you want to import file
When I run the predict.py file alone, it finds and reads the data.csv file. but it fails by running the predict.py file from the asd.py file in another file path;
My Files
-sbin
-master
+asd.py
-scripts
-bash
-dataset
+data.csv
+predict.py
asd.py
import os
import sys
runPath = os.path.dirname(os.path.realpath(__file__))
sys.path.append(os.path.join(runPath, "../../scripts/bash"))
from predict import pred
pred().main()
predict.py
import pandas as pd
class pred:
def main(self):
data = pd.read_csv('dataset/data.csv', sep=';')
Could the reason for this error be caused by changing the path of operation? Or I didn't get it because of another mistake.
Error:
FileNotFoundError: File b'dataset/data.csv' does not exist
Longer answer on the comment above:
This happens because although you appended the scripts folder to your sys path in order to import stuff from predict.py, whenever you call that code inside asd.py, it will run from the calling script's (asd.py) current working directory.
What that means for you is that the relative path dataset/dataset.csv does not exist in the current working directory of asd.py (sbin/master) and consequently the code will raise a FileNotFound exception.
The way to fix this and be able to run your code from anywhere, would be to give predict.py the absolute path of your dataset file.
To do that in a way that is not hardcoded, I would do what you did to get your runPath, namely get the absolute path of your predict.py file inside a varible and use os.path.join to join that to the dataset file. That way you can always be sure the dataset file will be found by whatever calling script uses the code in predict.py.
Example below:
predict.py
import pandas as pd
import os
current_dir = os.path.dirname(os.path.realpath(__file__))
class pred:
def main(self):
data_fullpath = os.path.join(current_dir, 'dataset/dataset.csv')
data = pd.read_csv(data_fullpath, sep=';')
I think you should use absolute path, instead of relative path
sys.path documentations says:
A list of strings that specifies the search path for modules
So I would not expect that relative paths starts from the path you set in asd.py
The script is trying to open dataset/data.csv starting from current folder where the script was started.
In your case I would try to pass that path somehow to the second script.
Preliminary:
I have Anaconda 3 on Windows 10, and a folder, folder_default, that I have put on the Python path. I'm not actually sure whether that's the right terminology, so to be clear: regardless to where my Python script is, if I have a line of code that says import myFile, that line of code will succeed if myFile.py is in folder_default.
My issue:
In folder_default, I have:
A subfolder called useful_files which contains a text file called useful_file_1.txt.
A python script called my_misc.py.
my_misc.py has a line similar to: np.loadtxt('useful_files/useful_file_1.txt'). This line does not work if I use import my_script in a python file in a location other than folder_default, since useful_files/useful_file_1.txt is not the folder path relative to the python file that imports my_misc.py. I don't want to start using global file paths if I can avoid it.
How can I access files using file paths relative to the imported python module, rather than relative to the python script that imports that module?
Please let me know if the question is unclear - I tried to write a fake, minimal version of the setup that's actually on my computer in the hopes that that would simplify things, but I can change it if that actually makes things more confusing.
Thanks.
You can get the path to current module using the getfile method of inspect module as inspect.getfile(inspect.currentframe()).
For example:
# File : my_misc.py
import os, inspect
module_dir_path = os.path.dirname(os.path.abspath(inspect.getfile(inspect.currentframe()))) # get path to the directory of current module
useful_file_path = os.path.join(module_dir_path,'useful_files','useful_file_1.txt')
# path stored in useful_file_path. do whatever you want!
np.loadtxt(useful_file_path)
Is there a universal approach in Python, to find out the path to the file that is currently executing?
Failing approaches
path = os.path.abspath(os.path.dirname(sys.argv[0]))
This does not work if you are running from another Python script in another directory, for example by using execfile in 2.x.
path = os.path.abspath(os.path.dirname(__file__))
I found that this doesn't work in the following cases:
py2exe doesn't have a __file__ attribute, although there is a workaround
When the code is run from IDLE using execute(), in which case there is no __file__ attribute
On Mac OS X v10.6 (Snow Leopard), I get NameError: global name '__file__' is not defined
Test case
Directory tree
C:.
| a.py
\---subdir
b.py
Content of a.py
#! /usr/bin/env python
import os, sys
print "a.py: sys.argv[0]=", sys.argv[0]
print "a.py: __file__=", __file__
print "a.py: os.getcwd()=", os.getcwd()
print
execfile("subdir/b.py")
Content of subdir/b.py
#! /usr/bin/env python
import os, sys
print "b.py: sys.argv[0]=", sys.argv[0]
print "b.py: __file__=", __file__
print "b.py: os.getcwd()=", os.getcwd()
print
Output of python a.py (on Windows)
a.py: __file__= a.py
a.py: os.getcwd()= C:\zzz
b.py: sys.argv[0]= a.py
b.py: __file__= a.py
b.py: os.getcwd()= C:\zzz
Related (but these answers are incomplete)
Find path to currently running file
Path to current file depends on how I execute the program
How can I know the path of the running script in Python?
Change directory to the directory of a Python script
First, you need to import from inspect and os
from inspect import getsourcefile
from os.path import abspath
Next, wherever you want to find the source file from you just use
abspath(getsourcefile(lambda:0))
You can't directly determine the location of the main script being executed. After all, sometimes the script didn't come from a file at all. For example, it could come from the interactive interpreter or dynamically generated code stored only in memory.
However, you can reliably determine the location of a module, since modules are always loaded from a file. If you create a module with the following code and put it in the same directory as your main script, then the main script can import the module and use that to locate itself.
some_path/module_locator.py:
def we_are_frozen():
# All of the modules are built-in to the interpreter, e.g., by py2exe
return hasattr(sys, "frozen")
def module_path():
encoding = sys.getfilesystemencoding()
if we_are_frozen():
return os.path.dirname(unicode(sys.executable, encoding))
return os.path.dirname(unicode(__file__, encoding))
some_path/main.py:
import module_locator
my_path = module_locator.module_path()
If you have several main scripts in different directories, you may need more than one copy of module_locator.
Of course, if your main script is loaded by some other tool that doesn't let you import modules that are co-located with your script, then you're out of luck. In cases like that, the information you're after simply doesn't exist anywhere in your program. Your best bet would be to file a bug with the authors of the tool.
This solution is robust even in executables:
import inspect, os.path
filename = inspect.getframeinfo(inspect.currentframe()).filename
path = os.path.dirname(os.path.abspath(filename))
I was running into a similar problem, and I think this might solve the problem:
def module_path(local_function):
''' returns the module path without the use of __file__. Requires a function defined
locally in the module.
from http://stackoverflow.com/questions/729583/getting-file-path-of-imported-module'''
return os.path.abspath(inspect.getsourcefile(local_function))
It works for regular scripts and in IDLE. All I can say is try it out for others!
My typical usage:
from toolbox import module_path
def main():
pass # Do stuff
global __modpath__
__modpath__ = module_path(main)
Now I use _modpath_ instead of _file_.
You have simply called:
path = os.path.abspath(os.path.dirname(sys.argv[0]))
instead of:
path = os.path.dirname(os.path.abspath(sys.argv[0]))
abspath() gives you the absolute path of sys.argv[0] (the filename your code is in) and dirname() returns the directory path without the filename.
The short answer is that there is no guaranteed way to get the information you want, however there are heuristics that work almost always in practice. You might look at How do I find the location of the executable in C?. It discusses the problem from a C point of view, but the proposed solutions are easily transcribed into Python.
See my answer to the question Importing modules from parent folder for related information, including why my answer doesn't use the unreliable __file__ variable. This simple solution should be cross-compatible with different operating systems as the modules os and inspect come as part of Python.
First, you need to import parts of the inspect and os modules.
from inspect import getsourcefile
from os.path import abspath
Next, use the following line anywhere else it's needed in your Python code:
abspath(getsourcefile(lambda:0))
How it works:
From the built-in module os (description below), the abspath tool is imported.
OS routines for Mac, NT, or Posix depending on what system we're on.
Then getsourcefile (description below) is imported from the built-in module inspect.
Get useful information from live Python objects.
abspath(path) returns the absolute/full version of a file path
getsourcefile(lambda:0) somehow gets the internal source file of the lambda function object, so returns '<pyshell#nn>' in the Python shell or returns the file path of the Python code currently being executed.
Using abspath on the result of getsourcefile(lambda:0) should make sure that the file path generated is the full file path of the Python file.
This explained solution was originally based on code from the answer at How do I get the path of the current executed file in Python?.
This should do the trick in a cross-platform way (so long as you're not using the interpreter or something):
import os, sys
non_symbolic=os.path.realpath(sys.argv[0])
program_filepath=os.path.join(sys.path[0], os.path.basename(non_symbolic))
sys.path[0] is the directory that your calling script is in (the first place it looks for modules to be used by that script). We can take the name of the file itself off the end of sys.argv[0] (which is what I did with os.path.basename). os.path.join just sticks them together in a cross-platform way. os.path.realpath just makes sure if we get any symbolic links with different names than the script itself that we still get the real name of the script.
I don't have a Mac; so, I haven't tested this on one. Please let me know if it works, as it seems it should. I tested this in Linux (Xubuntu) with Python 3.4. Note that many solutions for this problem don't work on Macs (since I've heard that __file__ is not present on Macs).
Note that if your script is a symbolic link, it will give you the path of the file it links to (and not the path of the symbolic link).
You can use Path from the pathlib module:
from pathlib import Path
# ...
Path(__file__)
You can use call to parent to go further in the path:
Path(__file__).parent
Simply add the following:
from sys import *
path_to_current_file = sys.argv[0]
print(path_to_current_file)
Or:
from sys import *
print(sys.argv[0])
If the code is coming from a file, you can get its full name
sys._getframe().f_code.co_filename
You can also retrieve the function name as f_code.co_name
The main idea is, somebody will run your python code, but you need to get the folder nearest the python file.
My solution is:
import os
print(os.path.dirname(os.path.abspath(__file__)))
With
os.path.dirname(os.path.abspath(__file__))
You can use it with to save photos, output files, ...etc
import os
current_file_path=os.path.dirname(os.path.realpath('__file__'))
I have a Python module and I'd like to get that modules directory from inside itself. I want to do this because I have some files that I'd like to reference relative to the module.
First you need to get a reference to the module inside itself.
mod = sys.__modules__[__name__]
Then you can use __file__ to get to the module file.
mod.__file__
Its directory is a dirname of that.
As you are inside the module all you need is this:
import os
path_to_this_module = os.path.dirname(__file__)
However, if the module in question is actually your programs entry point, then __file__ will only be the name of the file and you'll need to expand the path:
import os
path_to_this_module = os.path.dirname(os.path.abspath(__file__))
I think this is what you are looking for:
import <module>
import os
print os.path.dirname(<module>.__file__)
You should be using pkg_resources for this, the resource* family of functions do just about everything you need without having to muck about with the filesystem.
import pkg_resources
data = pkg_resources.resource_string(__name__, "some_file")