IsADirectoryError: Errno 21 Is a directory (Fix)

IsADirectoryError: [Errno 21] Is a directory happens when your Python code tries to open, read, write, or otherwise use a folder path where Python expects a file path.

This is a common beginner error when working with open(), file paths, or folder loops. The fix is usually simple: check the path, confirm whether it points to a file or a directory, and use the right operation for that path.

Quick fix

path = "data.txt"  # use a file path, not a folder path

with open(path, "r") as file:
    content = file.read()
    print(content)

This error usually means the path points to a folder instead of a file.

What this error means

Python is telling you:

It expected a file path
The path you gave points to a directory (folder)
This often happens with open(), os.remove(), or similar file operations
A directory and a file are different things in Python and in your operating system

For example, this path is a directory:

path = "my_folder"

But this path is a file:

path = "my_folder/data.txt"

If you want to learn more about how file paths work, see working with file paths in Python.

Why this happens

This error usually appears for one of these reasons:

You passed a folder name to open()
You built the path incorrectly
You used a variable that contains a directory path instead of a file path
You looped through directory names and tried to open each one as a file
You expected a file like data.txt but only used the parent folder path

Common causes

Using open() with a folder path
Passing a directory to os.remove() or another file-only function
Forgetting to include the file name in the path
Incorrectly joining paths in loops
Reading items from os.listdir() without checking their type

Example that causes the error

Here is a simple example:

with open("my_folder", "r") as file:
    content = file.read()

If "my_folder" is a real folder, Python raises:

IsADirectoryError: [Errno 21] Is a directory: 'my_folder'

Python cannot read a folder as if it were a text file.

The fix is to open a real file inside that folder:

with open("my_folder/data.txt", "r") as file:
    content = file.read()
    print(content)

If you need a refresher on file reading, see how to read a file in Python or the Python open() function explained.

How to fix it

When you see this error, go through these checks:

Check the exact path stored in your variable
Make sure the path includes the file name and extension
Use os.path.isfile() before opening a path
Use os.path.isdir() if you want to work with folders
List files in the folder first, then open one file at a time

A safe pattern looks like this:

import os

path = "my_folder/data.txt"

if os.path.isfile(path):
    with open(path, "r") as file:
        print(file.read())
else:
    print("This path is not a file.")

This prevents your code from trying to open a directory as a file.

Fix by checking file vs directory

When a path might be either a file or a folder, test it first.

import os

path = "my_folder"

print(os.path.isfile(path))  # True if it is a file
print(os.path.isdir(path))   # True if it is a directory

This helps because:

os.path.isfile(path) confirms it is a file
os.path.isdir(path) confirms it is a directory
You can choose the correct operation
It is especially useful when paths come from user input or loops

Example:

import os

path = "my_folder"

if os.path.isfile(path):
    with open(path, "r") as file:
        print(file.read())
elif os.path.isdir(path):
    print("This is a folder, not a file.")
else:
    print("The path does not exist.")

You can also check whether a path exists first with os.path.exists().

Fix when looping through folder contents

This error often happens in loops.

A beginner might write code like this:

import os

folder = "my_folder"

for name in os.listdir(folder):
    with open(name, "r") as file:
        print(file.read())

There are two problems here:

name may be a directory, not a file
name is not joined to the folder path

A better version is:

import os

folder = "my_folder"

for name in os.listdir(folder):
    path = os.path.join(folder, name)

    if os.path.isfile(path):
        with open(path, "r") as file:
            print(file.read())
    else:
        print(f"Skipping directory: {path}")

This works better because:

Some folders contain subfolders as well as files
It does not assume every item from os.listdir() is a file
It joins the folder path and item name correctly
It skips directories or handles them separately

If you want to learn more, see os.listdir() explained.

Debugging steps

If you are not sure why the error is happening, check the path before the failing line.

Useful debugging commands:

print(path)

import os
print(os.path.exists(path))
print(os.path.isfile(path))
print(os.path.isdir(path))
print(os.getcwd())
print(os.listdir(path))

Here is what each one helps you find:

print(path) shows the exact value being used
os.path.exists(path) tells you whether the path exists
os.path.isfile(path) tells you whether it is a file
os.path.isdir(path) tells you whether it is a folder
os.getcwd() shows the current working directory
os.listdir(path) shows the contents of a folder

A simple debugging example:

import os

path = "my_folder"

print("Path:", path)
print("Exists:", os.path.exists(path))
print("Is file:", os.path.isfile(path))
print("Is directory:", os.path.isdir(path))
print("Current working directory:", os.getcwd())