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Whenever I try to run a program in VisualStudio in python that involves the simple procedure of opening a file and reading it, or creating a DataFrame with pd.pandas I come across the following error:

FileNotFoundError: [Errno 2] No such file or directory: 'file-name.csv'

I have already checked several different paths, like trying to connect VS with the conda virtual environment (I use python through the Anaconda package) through different paths - either launching the application from within anaconda navigator or through command lines in the internal terminal, and in all cases encounter the same error.

A very simple example of code that I tried to run was the following, which opens a file with a .json extension and converts it to a .csv file using pandas:

import pandas as pd

dataframe = pd.read_json('original-file-name.json', lines = True)

dataframe.to_csv('converted-file.csv', index=False, encoding='utf-8')

I believe that it is some foolishness of mine regarding the functioning of VS Code, since when running the same code mentioned above in another IDE, such as Spyder, the same works.

I appreciate any kind of help or guidance! thanks in advance

Ruben Esteche
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  • If you don't provide the full path to the file, Python treats the name as being in what it thinks is the current working directory. This location is often not what novices expect. For example, it is usually not the folder where the running `.py` file is. – BoarGules Jun 08 '22 at 18:32
  • @BoarGules thanks mate! So I must put full path to the file in the terminal of VS whenever I run some code that needs to open a file -and not just hit compile and run-. Do you know if there is any extension to automate this in VS? I imagine it's not that hard to write it anyway – Ruben Esteche Jun 08 '22 at 18:38
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    There are ways to find the current working directory (see the `os` module). Your code can also discover the location of the running Python program. See https://stackoverflow.com/questions/50499/how-do-i-get-the-path-and-name-of-the-file-that-is-currently-executing. – BoarGules Jun 08 '22 at 18:44
  • That will probably do it. Thanks for your attention! – Ruben Esteche Jun 08 '22 at 18:52
  • I am not a python expert, but it's calls, like all languages that work with an operating system, pass arguments to operating systems that typically understand not only absolute paths, but also relative paths. So for example if the target file is in a directory at the same level as the project directory, then `..\neighbor_dir\file-name.csv` should work. ( or if Linux, `../neighbor_dir/file-name.csv` ) – ryyker Jun 08 '22 at 19:07

1 Answers1

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Answering just to close the question, in case anyone falls into the same rookie mistake I did, @BoarGules comments on the topic above will probably solve it for you too :P

For path problems while trying to open files in your current IDE you should check as recommended the os module, a working code example to get the current working directory (CWD) is

import os

cwd = os.getcwd()
print("Current working directory:", cwd)

OUTPUT:

Current working directory: /your-current-working-directory

In case you want to change your current working directory you can use the os.chdir() (it only takes a single argument as a new directory path) and then you can automate your code so that it runs in the directory where your .py code is saved (obviously this is the same directory where your .json or .csv file is saved, since that's the one you're actually wanting to open)

Example:

import os
   
def current_path():
    print("Current working directory before")
    print(os.getcwd())
    print()
   
current_path()
   
os.chdir('../')
   
current_path()

OUTPUT

Current working directory before C:\Users\User\Desktop\file

Current working directory after C:\Users\User\Desktop

That should do (at least it works for my original problem!)

Ruben Esteche
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