I have the following file which I would like to read, as you can see its incomplete:
file = 'dir2/file.hdf5'
However, I would like to get the full path of file (*):
'/home/user/git_hub_repo/dir1/dir2/file.hdf5'
However, when I do:
from pathlib import Path
filename = Path('dir2/file.hdf5').resolve()
print(filename)
I get:
'/home/user/git_hub_repo/dir2/file.hdf5'
Which is wrong because a dir1 is missing in the retrieved path, how can I get (*) path
Note, that in my terminal i am in:
/home/user/git_hub_repo/
If your current directory is
/home/user/git_hub_repo/
and your file is in
/home/user/git_hub_repo/dir1/dir2/file.hdf5
You should change this
file = 'dir2/file.hdf5'
to
file = 'dir1/dir2/file.hdf5'
Related
So here's my code :
climate_data = np.genfromtxt('climate.txt', delimiter = ',', skip_header = 1)
I put this python file in the same directory as the "climate.txt" file and I also already import numpy to my python file
And I get this error :
OSError: climate.txt not found.
What can I do to fix this?
Putting this python script file in the same directory doesn't guarantee your code will be run in the same directory. To make sure of that, use absolute path to climate.txt, or you can get the absolute path to the script directory by writing:
import os
DIR_PATH = os.path.dirname(os.path.abspath(__file__))
climate_data = np.genfromtxt(os.path.join(DIR_PATH, 'climate.txt'), delimiter = ',', skip_header = 1)
I'm trying to get an image to show on the screen. I have the image file and code file in the same folder. I tried using the image file name and it said that there is no such file or directory. How do I do that?
I think you should try to write absolute path. you can use following code
import os
path = os.getcwd()
abs_path = path + img_name
Say I have a Python project that is structured as follows:
project
/data
test.csv
/package
__init__.py
module.py
main.py
__init__.py:
from .module import test
module.py:
import csv
with open("..data/test.csv") as f:
test = [line for line in csv.reader(f)]
main.py:
import package
print(package.test)
When I run main.py I get the following error:
C:\Users\Patrick\Desktop\project>python main.py
Traceback (most recent call last):
File "main.py", line 1, in <module>
import package
File "C:\Users\Patrick\Desktop\project\package\__init__.py", line 1, in <module>
from .module import test
File "C:\Users\Patrick\Desktop\project\package\module.py", line 3, in <module>
with open("../data/test.csv") as f:
FileNotFoundError: [Errno 2] No such file or directory: '../data/test.csv'
However, if I run module.py from the package directory, I don’t get any errors. So it seems that the relative path used in open(...) is only relative to where the originating file is being run from (i.e __name__ == "__main__")? I don't want to use absolute paths. What are some ways to deal with this?
Relative paths are relative to current working directory.
If you do not want your path to be relative, it must be absolute.
But there is an often used trick to build an absolute path from current script: use its __file__ special attribute:
from pathlib import Path
path = Path(__file__).parent / "../data/test.csv"
with path.open() as f:
test = list(csv.reader(f))
This requires python 3.4+ (for the pathlib module).
If you still need to support older versions, you can get the same result with:
import csv
import os.path
my_path = os.path.abspath(os.path.dirname(__file__))
path = os.path.join(my_path, "../data/test.csv")
with open(path) as f:
test = list(csv.reader(f))
[2020 edit: python3.4+ should now be the norm, so I moved the pathlib version inspired by jpyams' comment first]
For Python 3.4+:
import csv
from pathlib import Path
base_path = Path(__file__).parent
file_path = (base_path / "../data/test.csv").resolve()
with open(file_path) as f:
test = [line for line in csv.reader(f)]
This worked for me.
with open('data/test.csv') as f:
My Python version is Python 3.5.2 and the solution proposed in the accepted answer didn't work for me. I've still were given an error
FileNotFoundError: [Errno 2] No such file or directory
when I was running my_script.py from the terminal. Although it worked fine when I run it through Run/Debug Configurations from the PyCharm IDE (PyCharm 2018.3.2 (Community Edition)).
Solution:
instead of using:
my_path = os.path.abspath(os.path.dirname(__file__)) + some_rel_dir_path
as suggested in the accepted answer, I used:
my_path = os.path.abspath(os.path.dirname(os.path.abspath(__file__))) + some_rel_dir_path
Explanation:
Changing os.path.dirname(__file__) to os.path.dirname(os.path.abspath(__file__))
solves the following problem:
When we run our script like that: python3 my_script.py
the __file__ variable has a just a string value of "my_script.py" without path leading to that particular script. That is why method dirname(__file__) returns an empty string "". That is also the reason why my_path = os.path.abspath(os.path.dirname(__file__)) + some_rel_dir_path is actually the same thing as my_path = some_rel_dir_path. Consequently FileNotFoundError: [Errno 2] No such file or directory is given when trying to use open method because there is no directory like "some_rel_dir_path".
Running script from PyCharm IDE Running/Debug Configurations worked because it runs a command python3 /full/path/to/my_script.py (where "/full/path/to" is specified by us in "Working directory" variable in Run/Debug Configurations) instead of justpython3 my_script.py like it is done when we run it from the terminal.
Try
with open(f"{os.path.dirname(sys.argv[0])}/data/test.csv", newline='') as f:
I was surprised when the following code worked.
import os
for file in os.listdir("../FutureBookList"):
if file.endswith(".adoc"):
filename, file_extension = os.path.splitext(file)
print(filename)
print(file_extension)
continue
else:
continue
So, I checked the documentation and it says:
Changed in version 3.6: Accepts a path-like object.
path-like object:
An object representing a file system path. A path-like object is
either a str or...
I did a little more digging and the following also works:
with open("../FutureBookList/file.txt") as file:
data = file.read()
I'm trying to rename an audio file but I keep getting OSError: [Errno 2] No such file or directory.
In my program, each user has a directory that holds the users files. I obtain the path for each user by doing the following:
current_user_path = os.path.join(current_app.config['UPLOAD_FOLDER'], user.username)
/Users/johnsmith/Documents/pythonprojects/project/files/john
Now I want to obtain the path for the existing file I want to rename:
current_file_path = os.path.join(current_user_path,form.currentTitle.data)
/Users/johnsmith/Documents/pythonprojects/project/files/john/File1
The path for the rename file will be:
new_file_path = os.path.join(current_user_path, form.newTitle.data)
/Users/johnsmith/Documents/pythonprojects/project/files/john/One
Now I'm just running the rename command:
os.rename(current_file_path, new_file_path)
you can use os.rename for rename single file.
to avoid
OSError: [Errno 2] No such file or directory.
check if file exist or not.
here is the working example:
import os
src = "D:/test/Untitled003.wav"
dst = "D:/test/Audio001.wav"
if os.path.isfile(src):
os.rename(src, dst)
If the OS says there's no such file or directory, that's the gospel truth. You're making a lot of assumptions about where the file is, constructing a path to it, and renaming it. It's a safe bet there's no such file as the one named by current_file_path, or no directory to new_file_path.
Try os.stat(current_file_path), and similarly double-check the new file path with os.stat(os.posixpath.dirname(new_file_path)). Once you've got them right, os.rename will work if you've got permissions.
Try changing the current working directory to the one you want to work with. This code below should give you a simple walk through of how you should go about it:
import os
print (os.getcwd())
os.chdir(r"D:\Python\Example")
print (os.getcwd())
print ("start")
def rename_files():
file_list= os.listdir(r"D:\Python\Example")
print(file_list)
for file_name in file_list:
os.rename(file_name,file_name.translate(None,"0123456789")) rename_files()
print("stop")
print (os.getcwd())
I have a small text (XML) file that I want a Python function to load. The location of the text file is always in a fixed relative position to the Python function code.
For example, on my local computer, the files text.xml and mycode.py could reside in:
/a/b/text.xml
/a/c/mycode.py
Later at run time, the files could reside in:
/mnt/x/b/text.xml
/mnt/x/c/mycode.py
How do I ensure I can load in the file? Do I need the absolute path? I see that I can use os.path.isfile, but that presumes I have a path.
you can do a call as follows:
import os
BASE_DIR = os.path.dirname(os.path.realpath(__file__))
This will get you the directory of the python file you're calling from mycode.py
then accessing the xml files is as simple as:
xml_file = "{}/../text.xml".format(BASE_DIR)
fin = open(xml_file, 'r+')
If the parent directory of the two directories are always the same this should work:
import os
path_to_script = os.path.realpath(__file__)
parent_directory = os.path.dirname(path_to_script)
for root, dirs, files in os.walk(parent_directory):
for file in files:
if file == 'text.xml':
path_to_xml = os.path.join(root, file)
You can use the special variable __file__ which gives you the current file name (see http://docs.python.org/2/reference/datamodel.html).
So in your first example, you can reference text.xml this way in mycode.py:
xml_path = os.path.join(__file__, '..', '..', 'text.xml')