Expected str, bytes or os.PathLike object, not TextIOWrapper

# Table of Contents

1. Expected str, bytes or os.PathLike object, not TextIOWrapper
2. TypeError: expected str, bytes or os.PathLike object, not tuple

# Expected str, bytes or os.PathLike object, not TextIOWrapper

The Python "TypeError: expected str, bytes or os.PathLike object, not TextIOWrapper" occurs when we pass a file object instead of a string when opening a file.

To solve the error, pass the filename (as a string) to the open() function.

Here is an example of how the error occurs.

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with open('example.txt', 'r', encoding='utf-8') as file1:
    lines = file1.readlines()

    print(lines)

# ⛔️ TypeError: expected str, bytes or os.PathLike object, not TextIOWrapper
with open(file1, 'r', encoding='utf-8') as file2:
    lines = file2.readlines()
    print(lines)

The error is caused because we passed a file object instead of a string in the second call to the open() function.

# Pass the filename as a string to the open() function

The first argument the open() function expects is a string that represents the filename.

To solve the error, pass a string for the filename.

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with open('example.txt', 'r', encoding='utf-8') as file1:
    lines = file1.readlines()

    print(lines)

with open('example-2.txt', 'r', encoding='utf-8') as file2:
    lines = file2.readlines()
    print(lines)

The first argument the open() function takes is a string.

The second is the mode (e.g. open file for reading or writing). Here are the most common modes:

The third argument is the encoding - utf-8 in the example.

# Construct a filename by concatenating strings

If you need to construct the filename by concatenating strings, use a formatted string literal.

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name = 'example'
ext = 'txt'

filename = f'{name}.{ext}'

with open(filename, 'r', encoding='utf-8') as f:
    lines = f.readlines()

    for line in lines:
        print(line)

Make sure to wrap expressions in curly braces - {expression}.

Alternatively, you can use the addition (+) operator.

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name = 'example'
ext = 'txt'

filename = name + '.' + ext

with open(filename, 'r', encoding='utf-8') as f:
    lines = f.readlines()

    for line in lines:
        print(line)

Make sure the values of the left and right-hand sides of the addition (+) operator are strings.

We used the with statement in the example.

The with statement takes care of automatically closing the file after we finish reading or writing.

You can also use the open() function directly, but you have to close the file manually after you finish.

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file1 = open('example.txt', 'r', encoding='utf-8')
lines = file1.readlines()

print(lines)

file1.close() # 👈️ have to close file

On the other hand, the with statement closes the file automatically even if an error is raised.

# TypeError: expected str, bytes or os.PathLike object, not tuple

The Python "TypeError: expected str, bytes or os.PathLike object, not tuple" occurs when we pass a tuple instead of a string when opening a file.

To solve the error, use a for loop if you have to open multiple files or use the addition operator to get a filename from multiple strings.

Here is an example of how the error occurs.

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# 👇️ tuple
filenames = 'example-1.txt', 'example-2.txt'

# ⛔️ TypeError: expected str, bytes or os.PathLike object, not tuple
with open(filenames, 'r', encoding='utf-8') as f:
    lines = f.readlines()

    for line in lines:
        print(line)

The open() function expects a string for its first argument but we passed it a tuple.

# Use a for loop if you need to open multiple files

If you have to open multiple files, use a for loop.

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# 👇️ tuple
filenames = 'example-1.txt', 'example-2.txt'

for filename in filenames:
    with open(filename, 'r', encoding='utf-8') as f:
        lines = f.readlines()

        for line in lines:
            print(line)

We used a for loop to iterate over the tuple and passed each filename to the open() function.

Notice that we are passing a string for the filename and not a tuple.

# Another cause of the error

You might also get the error when trying to concatenate multiple strings to get the filename.

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# 👇️ tuple
filename = 'example', '.txt'

# ⛔️ TypeError: expected str, bytes or os.PathLike object, not tuple
with open(filename, 'r', encoding='utf-8') as f:
    lines = f.readlines()

    for line in lines:
        print(line)

The code sample creates a tuple by mistake.

One way to concatenate strings is to use the addition (+) operator.

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# 👇️ this is a string
filename = 'example' + '.txt'

with open(filename, 'r', encoding='utf-8') as f:
    lines = f.readlines()

    for line in lines:
        print(line)

The addition (+) operator can be used to concatenate strings.

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print('a' + 'b' + 'c')  # 👉️ 'abc'

Alternatively, you can use a formatted string literal.

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name = 'example'
ext = 'txt'

filename = f'{name}.{ext}'

with open(filename, 'r', encoding='utf-8') as f:
    lines = f.readlines()

    for line in lines:
        print(line)

Make sure to wrap expressions in curly braces - {expression}.

# How tuples are constructed in Python

In case you declared a tuple by mistake, tuples are constructed in multiple ways:

• Using a pair of parentheses () creates an empty tuple
• Using a trailing comma - a, or (a,)
• Separating items with commas - a, b or (a, b)
• Using the tuple() constructor

If you aren't sure what type a variable stores, use the built-in type() class.

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my_tuple = 'example', '.txt'
print(type(my_tuple))  # 👉️ <class 'tuple'>
print(isinstance(my_tuple, tuple))  # 👉️ True

my_str = 'example.txt'
print(type(my_str))  # 👉️ <class 'str'>
print(isinstance(my_str, str))  # 👉️ True

The type class returns the type of an object.

The isinstance() function returns True if the passed-in object is an instance or a subclass of the passed-in class.