Skip to content

2022-10-13 Open that file

How to be explicit with filepaths while still being portable

Maybe you've been here before; you are coding up a small script that needs to open a file, read the contents, and take some actions. You put your program (reader.py) in the same directory as the file (source.txt) and code up the following:

reader.py
with open("source.txt", "r"): as infile:
    file_contents = infile.read()

print(file_contents)

Here you tell Python to open the file source.txt and you don't give any type of filepath to the file because it's right there, in the same place as your program reader.py. So you hit Run in your editor of choice and...

console output
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
FileNotFoundError: [Errno 2] No such file or directory: 'source.txt'

What?

Now we're sure the reader.py and the source.txt are together in a folder named project. It looks like this:

directory tree
C
└── Users
    └── preocts
        └── Documents
            └── project
                ├── reader.py
                └── source.txt

So, reader.py should totally be able to see source.txt and open it. Right? Well, if we add a few lines of code we see the cause of our issue.

reader.py
import pathlib

print(pathlib.Path().cwd())
print(pathlib.Path(__file__))

with open("source.txt", "r"): as infile:
    file_contents = infile.read()

print(contents)
Console output
C:\Users\preocts
C:\Users\preocts\Documents\project\reader.py

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
FileNotFoundError: [Errno 2] No such file or directory: 'source.txt'

We import pathlib which is a great standard library for working with filepaths. The first print is showing us what is called the "current working directory". This is where the Python interpreter was run.

The second print is telling us where the reader.py is. __file__ is a special value set by Python which holds the location of the file.

Notice our paths here do not match! Python isn't "working" where our file is located. We can think of that current working directory as a "You are here" arrow on a map of our file system. That's where the interpreter is and, unless told otherwise, where the interpreter starts to look for files.

This is a common issue for anyone that uses the built-in terminal of their editors to run the code. Unless told otherwise, many editors default to the user's home directory as the current working directory.

So when we used open("source.txt")... we assumed the interpreter would look in the same path as the reader.py (at least, I did at first). Instead, python is looking for the file at C:\Users\preocts\source.txt.

Key Point

The Python interpreter's current working directory is set where the interpreter is run. This will affect where Python looks for files with the open statement (and imports too).

So that's what is wrong with our program. We're just looking for our file in the wrong place. Now, we just need a solution.

A fast and easy solution would be to use an absolute path to open our file. Using C:\Users\preocts\Documents\project\source.txt would get the job done, but it's not very portable. The code wouldn't work on anyone else's computer or if we moved the files. Great for local-only projects.

Another solution would be to configure the editor to use the .\project directory as the working directory. This can be a good habit for development but if you shared the project, the next dev might have the same issue. Again, portability is lost.

Instead, we can leverage that special __file__ value and open source.txt relative to the location of reader.py. Meaning that so long as we know where source.txt is in relation to reader.py we will be able to find and open it. To do this, we'll, again, use pathlib.

filepath = pathlib.Path(__file__).parent / "source.txt"

This looks complex if you aren't familiar with pathlib but it is actually very straight-forward.

  • pathlib.Path(__file__) gives us the full path to our file: C:\Users\preocts\Documents\project\reader.py
  • .parent removes the right-most piece of the path, leaving us with just a location: C:\Users\preocts\Documents\project
  • / "source.txt is how we add to a Path object.
  • Together we get C:\Users\preocts\Documents\project\source.txt which is the absolute path of our target file.

The neat thing is that if we move reader.py the code will still work so long as we move source.txt with it.

Our end code looks like this:

reader.py
import pathlib

filepath = pathlib.Path(__file__).parent / "source.txt"

with open(filepath, "r"): as infile:
    file_contents = infile.read()

print(file_contents)

With an expected output of our file without needing to worry about where the Python interpreter is running (working directory).

Console output
Eggs are amazing. # (1)!
  1. In fact, they are eggcellent. that's it, that's the yolk. Now go have a shell of a good time programming. Don't let little things like this scramble you.