Suppose my python code is executed a directory called main and the application needs to access main/2091/data.txt.
how should I use open(location)? what should the parameter location be?
I found that below simple code will work.. does it have any disadvantages?
file = "\2091\sample.txt"
path = os.getcwd()+file
fp = open(path, 'r+');
With this type of thing you need to be careful what your actual working directory is. For example, you may not run the script from the directory the file is in. In this case, you can't just use a relative path by itself.
If you are sure the file you want is in a subdirectory beneath where the script is actually located, you can use __file__ to help you out here. __file__ is the full path to where the script you are running is located.
So you can fiddle with something like this:
import os
script_dir = os.path.dirname(__file__) #<-- absolute dir the script is in
rel_path = "2091/data.txt"
abs_file_path = os.path.join(script_dir, rel_path)
This code works fine:
import os
def read_file(file_name):
file_handle = open(file_name)
print file_handle.read()
file_handle.close()
file_dir = os.path.dirname(os.path.realpath('__file__'))
print file_dir
#For accessing the file in the same folder
file_name = "same.txt"
read_file(file_name)
#For accessing the file in a folder contained in the current folder
file_name = os.path.join(file_dir, 'Folder1.1/same.txt')
read_file(file_name)
#For accessing the file in the parent folder of the current folder
file_name = os.path.join(file_dir, '../same.txt')
read_file(file_name)
#For accessing the file inside a sibling folder.
file_name = os.path.join(file_dir, '../Folder2/same.txt')
file_name = os.path.abspath(os.path.realpath(file_name))
print file_name
read_file(file_name)
I created an account just so I could clarify a discrepancy I think I found in Russ's original response.
For reference, his original answer was:
import os
script_dir = os.path.dirname(__file__)
rel_path = "2091/data.txt"
abs_file_path = os.path.join(script_dir, rel_path)
This is a great answer because it is trying to dynamically creates an absolute system path to the desired file.
Cory Mawhorter noticed that __file__ is a relative path (it is as well on my system) and suggested using os.path.abspath(__file__). os.path.abspath, however, returns the absolute path of your current script (i.e. /path/to/dir/foobar.py)
To use this method (and how I eventually got it working) you have to remove the script name from the end of the path:
import os
script_path = os.path.abspath(__file__) # i.e. /path/to/dir/foobar.py
script_dir = os.path.split(script_path)[0] #i.e. /path/to/dir/
rel_path = "2091/data.txt"
abs_file_path = os.path.join(script_dir, rel_path)
The resulting abs_file_path (in this example) becomes: /path/to/dir/2091/data.txt
It depends on what operating system you're using. If you want a solution that is compatible with both Windows and *nix something like:
from os import path
file_path = path.relpath("2091/data.txt")
with open(file_path) as f:
<do stuff>
should work fine.
The path module is able to format a path for whatever operating system it's running on. Also, python handles relative paths just fine, so long as you have correct permissions.
Edit:
As mentioned by kindall in the comments, python can convert between unix-style and windows-style paths anyway, so even simpler code will work:
with open("2091/data/txt") as f:
<do stuff>
That being said, the path module still has some useful functions.
I spend a lot time to discover why my code could not find my file running Python 3 on the Windows system. So I added . before / and everything worked fine:
import os
script_dir = os.path.dirname(__file__)
file_path = os.path.join(script_dir, './output03.txt')
print(file_path)
fptr = open(file_path, 'w')
Try this:
from pathlib import Path
data_folder = Path("/relative/path")
file_to_open = data_folder / "file.pdf"
f = open(file_to_open)
print(f.read())
Python 3.4 introduced a new standard library for dealing with files and paths called pathlib. It works for me!
Code:
import os
script_path = os.path.abspath(__file__)
path_list = script_path.split(os.sep)
script_directory = path_list[0:len(path_list)-1]
rel_path = "main/2091/data.txt"
path = "/".join(script_directory) + "/" + rel_path
Explanation:
Import library:
import os
Use __file__ to attain the current script's path:
script_path = os.path.abspath(__file__)
Separates the script path into multiple items:
path_list = script_path.split(os.sep)
Remove the last item in the list (the actual script file):
script_directory = path_list[0:len(path_list)-1]
Add the relative file's path:
rel_path = "main/2091/data.txt
Join the list items, and addition the relative path's file:
path = "/".join(script_directory) + "/" + rel_path
Now you are set to do whatever you want with the file, such as, for example:
file = open(path)
import os
def file_path(relative_path):
dir = os.path.dirname(os.path.abspath(__file__))
split_path = relative_path.split("/")
new_path = os.path.join(dir, *split_path)
return new_path
with open(file_path("2091/data.txt"), "w") as f:
f.write("Powerful you have become.")
If the file is in your parent folder, eg. follower.txt, you can simply use open('../follower.txt', 'r').read()
Get the path of the parent folder, then os.join your relative files to the end.
# get parent folder with `os.path`
import os.path
BASE_DIR = os.path.dirname(os.path.abspath(__file__))
# now use BASE_DIR to get a file relative to the current script
os.path.join(BASE_DIR, "config.yaml")
The same thing with pathlib:
# get parent folder with `pathlib`'s Path
from pathlib import Path
BASE_DIR = Path(__file__).absolute().parent
# now use BASE_DIR to get a file relative to the current script
BASE_DIR / "config.yaml"
Python just passes the filename you give it to the operating system, which opens it. If your operating system supports relative paths like main/2091/data.txt (hint: it does), then that will work fine.
You may find that the easiest way to answer a question like this is to try it and see what happens.
Not sure if this work everywhere.
I'm using ipython in ubuntu.
If you want to read file in current folder's sub-directory:
/current-folder/sub-directory/data.csv
your script is in current-folder
simply try this:
import pandas as pd
path = './sub-directory/data.csv'
pd.read_csv(path)
When I was a beginner I found these descriptions a bit intimidating. As at first I would try
For Windows
f= open('C:\Users\chidu\Desktop\Skipper New\Special_Note.txt','w+')
print(f)
and this would raise an syntax error. I used get confused alot. Then after some surfing across google. found why the error occurred. Writing this for beginners
It's because for path to be read in Unicode you simple add a \ when starting file path
f= open('C:\\Users\chidu\Desktop\Skipper New\Special_Note.txt','w+')
print(f)
And now it works just add \ before starting the directory.
In Python 3.4 (PEP 428) the pathlib was introduced, allowing you to work with files in an object oriented fashion:
from pathlib import Path
working_directory = Path(os.getcwd())
path = working_directory / "2091" / "sample.txt"
with path.open('r+') as fp:
# do magic
The with keyword will also ensure that your resources get closed properly, even if you get something goes wrong (like an unhandled Exception, sigint or similar)
Related
My script is under the directory:
'C:\\Users\\rikesh.kayastha\\project1\\daas\\src'
My script code is :
file_name_csv = "sample.csv"
os.chdir('C:\\Users\\rikesh.kayastha\\project1\\data')
df.to_csv(file_name_csv,index=False,encoding="utf-8")
This code saves the csv file in my desired directory. But this is just for local machine. How to adjust this without mentioning the local path. The base path is just project1 I want to remove the C:\\Users\\rikesh.kayastha\\ part so that this code will work on every machine.
import os
cudir = os.getcwd()
additional_dir = "\\project1\\data"
newdir = os.path.join(cudir, additional_dir)
print(newdir)
You can use sys.argv[0] to get the location of the script on the local machine, which should allow you to locate the data directory aswell
You can get the user's home directory using pathlib with Path.home().
from pathlib import Path
file_name_csv = "sample.csv"
user_home_directory = Path.home()
project_directory = user_home_directory / "project1"
output_filepath = project_directory / file_name_csv
df.to_csv(output_filepath, index=False, encoding="utf-8")
From which directory are you opening your file?
If it is
C:\roboczy\PythonScripts\test\project1\daas\src>python main.py
Then you can use:
filename = "sample.csv"
df.to_csv(f'../../data/{filename}',index=False,encoding="utf-8")
If you are ruinning scripts from project1 folder:
C:\roboczy\PythonScripts\test\project1>python daas\src\myscript.py
Then you can simply use:
filename = "sample.csv"
df.to_csv(f'data/{filename}',index=False,encoding="utf-8")
Use pathlib library.
from pathlib import Path
file_name_csv = "sample.csv"
current_dir = Path(__file__).parent # returns the folder where your python file is
df.to_csv(current_dir.joinpath(file_name_csv), index=False,encoding="utf-8")
I am working on a program that will edit all local files ending in a csv extension. When I call the location of the directory and then change directory I get an error. The error is due to extra \'s being added to the path. How can I call the path without these extra \'s?
I've looked around and there are similar issues but every example I see is for a hard written location and not a movable one.
import os
import glob
import sys
path = os.path.abspath(__file__)
extension = '.csv'
os.chdir(os.path.abspath(__file__))
result = glob.glob('*'.format(extension))
print(path)
print(result)
os.chdir() needs a directory not a file which is what you are giving it. try changing os.chdir(os.path.abspath(__file__)) to os.chdir(os.path.dirname(path))
import os
import glob
import sys
__file__ = 'test.txt'
path = os.path.abspath(__file__)
print(path)
extension = '.csv'
os.chdir(os.path.dirname(path))
result = glob.glob('*'.format(extension))
print(path)
print(result)
Preferably without the entire directory address. Can I define the folder relative to my executable script's position?
dataframe.to_csv('findorb_data.txt',header=False, index=False)
You can use the __file__ variable that contains the full path to the script. So, something like this:
import os
file_dir = os.path.dirname(os.path.abspath(__file__))
csv_folder = 'csv files'
file_path = os.path.join(file_dir, csv_folder, 'findorb_data.txt')
dataframe.to_csv(file_path, header=False, index=False)
You can always use relative path syntax but you can also use something like this to find the parent directory of your script.
import os
import sys
script_dir = os.path.abspath(os.path.dirname(sys.argv[0]) or '.')
So if your csv is in the folder `../csv_dir/csv.csv' you can use
csv_path = os.path.join(script_dir, '../csv_dir/csv.csv')
I've something like this in my program:
A main script main.py inside a folder named 'OpenFileinaModule'. There's a folder called 'sub' inside it with a script called subScript.py and a file xlFile.xlsx, which is opened by subScript.py.
OpenFileinaModule/
main.py
sub/
__init__.py (empty)
subScript.py
xlFile.xlsx
Here is the code:
sub.Script.py:
import os, openpyxl
class Oop:
def __init__(self):
__file__='xlFile.xlsx'
__location__ = os.path.realpath(
os.path.join(os.getcwd(), os.path.dirname(__file__)))
print os.path.join(__location__, __file__)
self.wrkb = openpyxl.load_workbook(os.path.join(__location__,
__file__),read_only=True)
main.py:
import sub.subScript
objt=sub.subScript.Oop()
When I execute main.py, I get the error:
IOError: [Errno 2] No such file or directory: 'C:\\Users\\...\\OpenFileInaModule\\xlFile.xlsx'
It jumps the sub folder...
I've tried
__file__='sub/xlFile.xlsx'
But then the "sub" folder is duplicated:
IOError: [Errno 2] No such file or directory: 'C:\\Users\\...\\OpenFileInaModule\\sub\\sub/xlFile.xlsx'
How to open xlFile.xlsx with subScript.py from main.py?
you're overriding __file__ with __file='xlFile.xlsx', do you mean to do this?
I think you want something like
import os
fname = 'xlFile.xlsx'
this_file = os.path.abspath(__file__)
this_dir = os.path.dirname(this_file)
wanted_file = os.path.join(this_dir, fname)
I'd suggest always using the absolute path for a file, especially if you're on windows when a relative path might not make sense if the file is on a different drive (I actually have no idea what it would do if you asked it for a relative path between devices).
Please avoid using __file__ and __location__ to name your variables, these are more like builtin variables which might cause a confusion.
Note something here:
__location__ = os.path.realpath(
os.path.join(os.getcwd(), os.path.dirname(__file__)))
You have not included sub directory and the above joins only the CWD + os.path.dirname(__file__). This doesn't get you to the file. Please read the documentation of os.path.dirname: os.path.dirname(__file__) returns an empty string here.
def __init__(self):
file = 'xlFile.xlsx'
location = os.path.join('sub', file)
location = os.path.abspath(location) # absolute path to file
location = os.path.realpath(location) # rm symbolic links in path
self.wrkb = openpyxl.load_workbook(location)
I want to get the path of the current directory under which a .py file is executed.
For example a simple file D:\test.py with code:
import os
print os.getcwd()
print os.path.basename(__file__)
print os.path.abspath(__file__)
print os.path.dirname(__file__)
It is weird that the output is:
D:\
test.py
D:\test.py
EMPTY
I am expecting the same results from the getcwd() and path.dirname().
Given os.path.abspath = os.path.dirname + os.path.basename, why
os.path.dirname(__file__)
returns empty?
Because os.path.abspath = os.path.dirname + os.path.basename does not hold. we rather have
os.path.dirname(filename) + os.path.basename(filename) == filename
Both dirname() and basename() only split the passed filename into components without taking into account the current directory. If you want to also consider the current directory, you have to do so explicitly.
To get the dirname of the absolute path, use
os.path.dirname(os.path.abspath(__file__))
import os.path
dirname = os.path.dirname(__file__) or '.'
os.path.split(os.path.realpath(__file__))[0]
os.path.realpath(__file__)return the abspath of the current script; os.path.split(abspath)[0] return the current dir
can be used also like that:
dirname(dirname(abspath(__file__)))
print(os.path.join(os.path.dirname(__file__)))
You can also use this way
Since Python 3.4, you can use pathlib to get the current directory:
from pathlib import Path
# get parent directory
curr_dir = Path(__file__).parent
file_path = curr_dir.joinpath('otherfile.txt')
None of the above answers is correct. OP wants to get the path of the current directory under which a .py file is executed, not stored.
Thus, if the path of this file is /opt/script.py...
#! /usr/bin/env python3
from pathlib import Path
# -- file's directory -- where the file is stored
fd = Path(__file__).parent
# -- current directory -- where the file is executed
# (i.e. the directory of the process)
cwd = Path.cwd()
print(f'{fd=} {cwd=}')
Only if we run this script from /opt, fd and cwd will be the same.
$ cd /
$ /opt/script.py
cwd=PosixPath('/') fd=PosixPath('/opt')
$ cd opt
$ ./script.py
cwd=PosixPath('/opt') fd=PosixPath('/opt')
$ cd child
$ ../script.py
cwd=PosixPath('/opt/child') fd=PosixPath('/opt/child/..')
I guess this is a straight forward code without the os module..
__file__.split(__file__.split("/")[-1])[0]