How to Resolve Python Error "TypeError: expected str, bytes or os.PathLike object, not ..." (when using open())

When working with file I/O in Python, the built-in open() function is fundamental. However, a common error encountered is TypeError: expected str, bytes or os.PathLike object, not X (where X is often _io.TextIOWrapper, tuple, or another type). This error clearly indicates that you passed an object of an unexpected type as the first argument (the file path) to the open() function.

This guide explains why this happens and shows how to correctly provide the file path to open().

Understanding the Error: What open() Expects

The open() function's primary purpose is to establish a connection (a "file handle" or "file object") to a file on your filesystem. Its first and most crucial argument is the path to that file. Python expects this path to be provided as one of the following types:

• str: A standard string representing the relative or absolute path (e.g., 'my_file.txt', r'C:\data\report.csv').
• bytes: A bytes object representing the path (less common for open(), more often used in lower-level OS interactions).
• os.PathLike object: An object implementing the os.PathLike protocol (like pathlib.Path objects), which can represent file paths.

The TypeError occurs when you provide something else, like an already opened file object or a tuple containing filenames, where open() expects the path string itself.

Cause 1: Passing a File Object (TextIOWrapper) to open()

This often happens when you accidentally reuse the variable holding an already opened file handle in a subsequent open() call.

Error Scenario

filename1 = 'config.ini'
filename2 = 'settings.ini'

try:
    # Correctly open the first file
    with open(filename1, 'r', encoding='utf-8') as file_handle1:
        print(f"Successfully opened {filename1}. Handle type: {type(file_handle1)}")
        # Output: Handle type: <class '_io.TextIOWrapper'>
        content1 = file_handle1.read()

    # Incorrectly try to open the file handle instead of the second filename
    with open(file_handle1, 'r', encoding='utf-8') as file_handle2:
        # ⛔️ TypeError: expected str, bytes or os.PathLike object, not _io.TextIOWrapper
        content2 = file_handle2.read()

except TypeError as e:
    print(f"Error on second open(): {e}")
except FileNotFoundError as e:
    print(f"Error finding file: {e}")
except Exception as e:
    print(f"An unexpected error occurred: {e}")

The second with open(...) block incorrectly passes file_handle1 (which is the already opened file object, a _io.TextIOWrapper) instead of the string filename2.

Solution: Pass the Filename String

Always pass the filename (as a string or PathLike object) as the first argument to open().

filename1 = 'config.ini'
filename2 = 'settings.ini' # Assume this file exists for the example

try:
    # Open the first file
    with open(filename1, 'r', encoding='utf-8') as file_handle1:
        print(f"Reading {filename1}...")
        content1 = file_handle1.read()
        print(f"Read {len(content1)} bytes from {filename1}")

    # ✅ Correctly open the second file using its filename string
    with open(filename2, 'r', encoding='utf-8') as file_handle2:
        print(f"Reading {filename2}...")
        content2 = file_handle2.read()
        print(f"Read {len(content2)} bytes from {filename2}")

except FileNotFoundError as e:
    print(f"Error finding file: {e}")
except Exception as e:
    print(f"An unexpected error occurred: {e}")

Cause 2: Passing a Tuple to open()

This can happen if you group multiple filenames together in a tuple and pass the tuple directly, or if you accidentally create a tuple when constructing a filename.

Error Scenario (Multiple Filenames)

You might have a collection of filenames and mistakenly pass the whole collection to open().

files_to_process = ('log_a.txt', 'log_b.txt', 'log_c.txt') # This is a tuple

print(f"Type of files_to_process: {type(files_to_process)}") # Output: <class 'tuple'>

try:
    # ⛔️ TypeError: expected str, bytes or os.PathLike object, not tuple
    # Passing the entire tuple to open()
    with open(files_to_process, 'r', encoding='utf-8') as f:
        content = f.read()
except TypeError as e:
    print(e)

Solution: Loop Through Filenames

If you need to process multiple files, iterate through the tuple (or list) and call open() for each filename individually within the loop.

files_to_process = ('log_a.txt', 'log_b.txt', 'log_c.txt') # Assume these exist

print("Processing files:")
for filename in files_to_process:
    try:
        # ✅ Open each filename string inside the loop
        with open(filename, 'r', encoding='utf-8') as f:
            print(f"  Reading {filename}...")
            content = f.read(50) # Read first 50 chars for demo
            print(f"    Content start: '{content[:50]}...'")
    except FileNotFoundError:
        print(f"  Error: File '{filename}' not found.")
    except Exception as e:
        print(f"  Error processing {filename}: {e}")

Error Scenario (Accidental Tuple Creation)

Sometimes, while trying to build a filename from parts, you might inadvertently create a tuple due to stray commas.

base_name = 'report'
extension = '.csv'

# Accidental tuple creation due to comma before string concatenation attempt
filename_tuple = base_name, extension # Creates ('report', '.csv')

print(f"Type of filename_tuple: {type(filename_tuple)}") # Output: <class 'tuple'>

try:
    # ⛔️ TypeError: expected str, bytes or os.PathLike object, not tuple
    with open(filename_tuple, 'w', encoding='utf-8') as f:
        f.write("data")
except TypeError as e:
    print(e)

Solution: Correctly Construct Filename String

Ensure you are actually concatenating the parts into a single string before passing it to open().

base_name = 'report'
extension = '.csv'

# ✅ Correct: Concatenate strings using '+'
filename_str_plus = base_name + extension
print(f"Filename (+): '{filename_str_plus}', Type: {type(filename_str_plus)}")

# ✅ Correct: Using f-string (often preferred)
filename_str_f = f"{base_name}{extension}"
print(f"Filename (f): '{filename_str_f}', Type: {type(filename_str_f)}")

# Now open using the correctly formed string
try:
    with open(filename_str_f, 'w', encoding='utf-8') as f:
        f.write("Report data.\n")
    print(f"Successfully wrote to {filename_str_f}")
except Exception as e:
    print(f"Error writing to file: {e}")

Best Practices for Filename Construction

• Using F-Strings: Clear and concise for embedding variables.

  name = "results"
  date_str = "2023-10-27"
  filename = f"{name}_{date_str}.log" # 'results_2023-10-27.log'
• Using os.path.join(): Creates paths reliably across different operating systems (handles / vs \ correctly).

  import os
  dir_path = "data"
  filename_base = "report.txt"
  full_path = os.path.join(dir_path, filename_base) # 'data/report.txt' or 'data\report.txt'

Debugging the Type Error (type(), isinstance())

When you get this TypeError:

1. Examine the line where open() is called.
2. Look at the first argument being passed to it.
3. Print its type: print(type(argument_passed_to_open)).
4. This will reveal the incorrect type (e.g., _io.TextIOWrapper, tuple) instead of the expected str, bytes, or os.PathLike.
5. Use isinstance(argument, str) to confirm if it's a string.

Conclusion

The TypeError: expected str, bytes or os.PathLike object, not ... when using open() is a direct consequence of passing the wrong type of object as the file path argument.

The solutions involve:

1. Ensuring you pass a string (or bytes/os.PathLike) representing the filename/path to open().
2. Do not pass an already opened file object (_io.TextIOWrapper) back into open().
3. If you have multiple filenames (e.g., in a tuple or list), loop through them and call open() for each string individually.
4. If constructing a filename from parts, ensure you correctly concatenate them into a single string (+ operator, f-string, os.path.join) before passing it to open().

By providing open() with the correct path type, you can avoid this common TypeError.