Working with operating system paths in Python - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of joining paths?

Consider the following Python code using os.path.join. What will be printed?

Python

import os
path = os.path.join('folder', 'subfolder', 'file.txt')
print(path)

A "folder.subfolder.file.txt"

B "folder/subfolder/file.txt"

C "folder-subfolder-file.txt"

D "folder\\subfolder\\file.txt"

💡 Hint

Think about how paths are joined on Unix-like systems.

The os.path.join function joins parts of a path using the system's path separator. On Unix-like systems, it uses forward slashes (/). So the output will show forward slashes between parts.

❓ Predict Output

intermediate

What does os.path.basename return?

What will the following code print?

Python

import os
filename = os.path.basename('/home/user/docs/report.pdf')
print(filename)

A "report.pdf"

B "/home/user/docs/report.pdf"

C "/home/user/docs"

D "docs/report.pdf"

💡 Hint

Think about what the basename function extracts from a path.

os.path.basename returns the last part of the path, which is the file name with extension.

❓ Predict Output

advanced

What is the output of normpath with redundant separators?

What will this code print?

Python

import os
path = os.path.normpath('folder//subfolder/../file.txt')
print(path)

A "folder//file.txt"

B "folder/../file.txt"

C "folder/file.txt"

D "folder/subfolder/file.txt"

💡 Hint

Think about how normpath simplifies paths by removing redundant parts.

os.path.normpath removes redundant separators and resolves '..' to go up one directory, so 'folder//subfolder/../file.txt' becomes 'folder/file.txt'.

❓ Predict Output

advanced

What error does this code raise?

What error will this code raise?

Python

import os
path = os.path.join('folder', None, 'file.txt')
print(path)

A TypeError

B ValueError

C AttributeError

D No error, prints 'folder/None/file.txt'

💡 Hint

Consider what happens when you pass None to os.path.join.

Passing None to os.path.join causes a TypeError because it expects strings, bytes, or os.PathLike objects, not NoneType.

🧠 Conceptual

expert

How many items are in the list after splitting a path?

Given this code, how many items will the list parts contain?

Python

import os
path = '/usr/local/bin/python3'
parts = path.split(os.sep)
print(len(parts))

A 4

B 3

C 6

D 5

💡 Hint

Remember that splitting by '/' on an absolute path creates an empty string at the start.

Splitting '/usr/local/bin/python3' by '/' results in ['', 'usr', 'local', 'bin', 'python3'], which has 5 items.

Practice

1. Which function is used to safely combine folder and file names into a full path in Python?

easy

A. os.path.join()

B. os.path.exists()

C. os.path.basename()

D. os.path.dirname()

Solution

1. Step 1: Understand purpose of os.path.join()

   It combines parts of a path into one full path, handling separators correctly.

2. Step 2: Compare with other functions

   os.path.exists() checks if a path exists, basename() gets file name, dirname() gets folder name.

3. Final Answer:

   os.path.join() -> Option A

4. Quick Check:

   Combine paths = os.path.join() [OK]

Hint: Use os.path.join() to build paths safely [OK]

Common Mistakes:

• Confusing join() with exists()
• Using basename() to join paths
• Using dirname() to combine paths

2. Which of the following is the correct syntax to get the folder name from a path stored in variable path?

easy

A. os.path.basename(path)

B. os.path.dirname(path)

C. os.path.join(path)

D. os.path.exists(path)

Solution

1. Step 1: Identify function to get folder name

   os.path.dirname(path) returns the directory part of the path.

2. Step 2: Check other options

   basename() returns file name, join() combines paths, exists() checks path existence.

3. Final Answer:

   os.path.dirname(path) -> Option B

4. Quick Check:

   Folder name = os.path.dirname(path) [OK]

Hint: Use os.path.dirname() to get folder name from path [OK]

Common Mistakes:

• Using basename() to get folder
• Calling join() with one argument
• Confusing exists() with dirname()

3. What will be the output of this code?

import os
path = os.path.join('folder', 'subfolder', 'file.txt')
print(os.path.basename(path))

medium

A. folder

B. subfolder

C. file.txt

D. folder/subfolder/file.txt

Solution

1. Step 1: Understand os.path.join()

   It creates 'folder/subfolder/file.txt' (with correct separator).

2. Step 2: Understand os.path.basename()

   It returns the last part of the path, which is the file name 'file.txt'.

3. Final Answer:

   file.txt -> Option C

4. Quick Check:

   basename() returns file name [OK]

Hint: basename() returns last part (file) of path [OK]

Common Mistakes:

• Expecting folder name instead of file
• Confusing join() output with basename()
• Printing full path instead of basename

4. What is wrong with this code snippet?

import os
path = os.path.join('folder', 'file.txt')
if os.path.exists:
    print('Path exists')

medium

A. os.path.exists is used without parentheses

B. os.path.join() cannot join two parts

C. print statement is missing parentheses

D. Variable path is not defined

Solution

1. Step 1: Check usage of os.path.exists

   It is a function and must be called with parentheses and argument: os.path.exists(path).

2. Step 2: Verify other parts

   join() usage is correct, print() has parentheses, path is defined.

3. Final Answer:

   os.path.exists is used without parentheses -> Option A

4. Quick Check:

   Call exists() with parentheses [OK]

Hint: Call os.path.exists() with path argument [OK]

Common Mistakes:

• Forgetting () after exists
• Passing no argument to exists()
• Misusing join() function

5. You want to check if a file named data.csv exists inside a folder reports located in the user's home directory. Which code correctly builds the path and checks existence?

hard

A. os.path.exists(os.path.join(os.getcwd(), 'reports', 'data.csv'))

B. os.path.exists(os.path.join('reports', 'data.csv'))

C. os.path.exists('~/reports/data.csv')

D. os.path.exists(os.path.join(os.path.expanduser('~'), 'reports', 'data.csv'))

Solution

1. Step 1: Get user's home directory

   Use os.path.expanduser('~') to get the home folder path.

2. Step 2: Join home, reports, and file name

   Use os.path.join() to combine home path, 'reports', and 'data.csv'.

3. Step 3: Check if the full path exists

   Pass the full joined path to os.path.exists() to check existence.

4. Final Answer:

   os.path.exists(os.path.join(os.path.expanduser('~'), 'reports', 'data.csv')) -> Option D

5. Quick Check:

   Expand user + join + exists = os.path.exists(os.path.join(os.path.expanduser('~'), 'reports', 'data.csv')) [OK]

Hint: Use expanduser('~') to get home folder before joining [OK]

Common Mistakes:

• Using relative path without home folder
• Using literal '~/...' without expanduser
• Using current working directory instead of home