How to read a file in other directory in python
Question:
I have a file named 5_1.txt in a directory named direct, how can I read that file using read?
I verified the path using :
import os
os.getcwd()
os.path.exists(direct)
the result was
True
x_file=open(direct,'r')
and I got this error :
Traceback (most recent call last):
File "<pyshell#17>", line 1, in <module>
x_file=open(direct,'r')
IOError: [Errno 13] Permission denied
I don’t know why I can’t read the file ? Any suggestions?
thanks .
Answers:
As error message said your application has no permissions to read from the directory. It can be the case when you created the directory as one user and run script as another user.
In case you’re not in the specified directory (i.e. direct), you should use (in linux):
x_file = open('path/to/direct/filename.txt')
Note the quotes and the relative path to the directory.
This may be your problem, but you also don’t have permission to access that file.
Maybe you’re trying to open it as another user.
You can’t “open” a directory using the open function. This function is meant to be used to open files.
Here, what you want to do is open the file that’s in the directory. The first thing you must do is compute this file’s path. The os.path.join function will let you do that by joining parts of the path (the directory and the file name):
fpath = os.path.join(direct, "5_1.txt")
You can then open the file:
f = open(fpath)
And read its content:
content = f.read()
Additionally, I believe that on Windows, using open on a directory does return a PermissionDenied exception, although that’s not really the case.
Looks like you are trying to open a directory for reading as if it’s a regular file. Many OSs won’t let you do that. You don’t need to anyway, because what you want (judging from your description) is
x_file = open(os.path.join(direct, "5_1.txt"), "r")
or simply
x_file = open(direct+"/5_1.txt", "r")
i found this way useful also.
import tkinter.filedialog
from_filename = tkinter.filedialog.askopenfilename()
here a window will appear so you can browse till you find the file , you click on it then you can continue using open , and read .
from_file = open(from_filename, 'r')
contents = from_file.read()
contents
For folks like me looking at the accepted answer, and not understanding why it’s not working, you need to add quotes around your sub directory, in the green checked example,
x_file = open(os.path.join(direct, "5_1.txt"), "r")
should actually be
x_file = open(os.path.join('direct', "5_1.txt"), "r")
For windows you can either use the full path with ‘\’ (‘/’ for Linux and Mac) as separator of you can use os.getcwd to get the current working directory and give path in reference to the current working directory
data_dir = os.getcwd()+'\child_directory'
file = open(data_dir+'\filename.txt', 'r')
When I tried to give the path of child_diectory entirely it resulted in error. For e.g. in this case:
file = open('child_directory\filename.txt', 'r')
Resulted in error. But I think it must work or I am doing it somewhat wrong way but it doesn’t work for me. The about way always works.
I have a file named 5_1.txt in a directory named direct, how can I read that file using read?
I verified the path using :
import os
os.getcwd()
os.path.exists(direct)
the result was
True
x_file=open(direct,'r')
and I got this error :
Traceback (most recent call last):
File "<pyshell#17>", line 1, in <module>
x_file=open(direct,'r')
IOError: [Errno 13] Permission denied
I don’t know why I can’t read the file ? Any suggestions?
thanks .
As error message said your application has no permissions to read from the directory. It can be the case when you created the directory as one user and run script as another user.
In case you’re not in the specified directory (i.e. direct), you should use (in linux):
x_file = open('path/to/direct/filename.txt')
Note the quotes and the relative path to the directory.
This may be your problem, but you also don’t have permission to access that file.
Maybe you’re trying to open it as another user.
You can’t “open” a directory using the open function. This function is meant to be used to open files.
Here, what you want to do is open the file that’s in the directory. The first thing you must do is compute this file’s path. The os.path.join function will let you do that by joining parts of the path (the directory and the file name):
fpath = os.path.join(direct, "5_1.txt")
You can then open the file:
f = open(fpath)
And read its content:
content = f.read()
Additionally, I believe that on Windows, using open on a directory does return a PermissionDenied exception, although that’s not really the case.
Looks like you are trying to open a directory for reading as if it’s a regular file. Many OSs won’t let you do that. You don’t need to anyway, because what you want (judging from your description) is
x_file = open(os.path.join(direct, "5_1.txt"), "r")
or simply
x_file = open(direct+"/5_1.txt", "r")
i found this way useful also.
import tkinter.filedialog
from_filename = tkinter.filedialog.askopenfilename()
here a window will appear so you can browse till you find the file , you click on it then you can continue using open , and read .
from_file = open(from_filename, 'r')
contents = from_file.read()
contents
For folks like me looking at the accepted answer, and not understanding why it’s not working, you need to add quotes around your sub directory, in the green checked example,
x_file = open(os.path.join(direct, "5_1.txt"), "r")
should actually be
x_file = open(os.path.join('direct', "5_1.txt"), "r")
For windows you can either use the full path with ‘\’ (‘/’ for Linux and Mac) as separator of you can use os.getcwd to get the current working directory and give path in reference to the current working directory
data_dir = os.getcwd()+'\child_directory'
file = open(data_dir+'\filename.txt', 'r')
When I tried to give the path of child_diectory entirely it resulted in error. For e.g. in this case:
file = open('child_directory\filename.txt', 'r')
Resulted in error. But I think it must work or I am doing it somewhat wrong way but it doesn’t work for me. The about way always works.