Python 3: Open file in read mode without raising an exception?

Question

I am trying to write a block of code which opens a new file every time a Python3 script is run.

I am constructing the filename using an incrementing number.

For example, the following are some examples of valid filenames which should be produced:

output_0.csv
output_1.csv
output_2.csv
output_3.csv

On the next run of the script, the next filename to be used should be output_4.csv.

In C/C++ I would do this in the following way:

Enter an infinite loop
Try to open the first filename, in "read" mode
If the file is open, increment the filename number and repeat
If the file is not open, break out of the loop and re-open the file in "write" mode

This doesn't seem to work in Python 3, as opening a non-existing file in read mode causes an exception to be raised.

One possible solution might be to move the open file code block inside a try-catch block. But this doesn't seem like a particularly elegant solution.

Here is what I tried so far in code

# open a file to store output data
filename_base = "output"
filename_ext = "csv"
filename_number = 0

while True:
    filename_full = f"{filename_base}_{filename_number}.{filename_ext}"
    with open(filename_full, "r") as f:
        if f.closed:
            print(f"Writing data to {filename_full}")
            break
        else:
            print(f"File {filename_full} exists")
            filename_number += 1

with open(filename_full, "w") as f:
    pass

As explained above this code crashes when trying to open a file which does not exist in "read" mode.

The exception is there to inform you "If the file is not open". Catching the exception is how you check it. There is nothing inelegant about this. — MisterMiyagi, Jul 09 '21 at 12:27
This article may help you, https://dev.to/zhiyueyi/how-to-open-a-file-in-python-like-a-pro-3oe0 — Akshay Saambram, Jul 09 '21 at 12:28

Alex · Answer 1 · 2021-07-09T13:08:21.403

4

Using pathlib you can check with Path.is_file() which returns True when it encounters a file or a symbolic link to a file.

from pathlib import Path

filename_base = "output"
filename_ext = "csv"
filename_number = 0

filename_full = f"{filename_base}_{filename_number}.{filename_ext}"
p = Path(filename_full)

while p.is_file() or p.is_dir():
    filename_number += 1
    p = Path(f"{filename_base}_{filename_number}.{filename_ext}")

This loop should exit when the file isn’t there so you can open it for writing.

edited Jul 09 '21 at 13:08

answered Jul 09 '21 at 12:28

Alex

6,610
3
20
38

A nice solution. I see only one potential issue, which is where something, like a directory, exists, and is not a file. Is there a version of this which uses something in place of `is_file()` to detect the presence of any path, rather than just files. (Simlinks, dirs, etc) – FreelanceConsultant Jul 09 '21 at 12:52
`is_file()` will return True for symbolic links, I've updated the answer a little to account for directories – Alex Jul 09 '21 at 13:09

score 2 · Answer 2 · answered Jul 09 '21 at 12:27

2

you can check if a file exists prior using

os.path.exists(filename)

answered Jul 09 '21 at 12:27

Ramtin Nouri

310
1
8

score 1 · Answer 3 · answered Jul 09 '21 at 12:26

1

You could use the OS module to check if the file path is a file, and then open it:

import os

file_path = './file.csv'

if(os.path.isfile(file_path)):
    with open(file_path, "r") as f:

answered Jul 09 '21 at 12:26

andrepz

443
6
16

score 1 · Answer 4 · answered Jul 09 '21 at 12:29

This should work:

filename_base = "output"
filename_ext = "csv"
filename_number = 0

while True:
    filename_full = f"{filename_base}_{filename_number}.{filename_ext}"
    try:
      with open(filename_full, "r") as f:
          print(f"File {filename_full} exists")
          filename_number += 1
    except FileNotFoundError:
      print("Creating new file")
      open(filename_full, 'w');
      break;

score 1 · Answer 5 · answered Jul 09 '21 at 12:29

You might os.path.exists to check if file already exists for example

import os
print(os.path.exists("output_0.csv"))

or harness fact that your names

output_0.csv
output_1.csv
output_2.csv
output_3.csv

are so regular, exploit glob.glob like so

import glob
existing = glob.glob("output_*.csv")
print(existing)  # list of existing files

Python 3: Open file in read mode without raising an exception?

5 Answers5