Say I have a Python project that is structured as follows:
project /data test.csv /package __init__.py module.py main.py
from .module import test
import csv with open("..data/test.csv") as f: test = [line for line in csv.reader(f)]
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__")? How can deal with this, using relative paths only?
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]