File path handling in Python - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this code using pathlib?

Consider the following Python code that uses pathlib to manipulate file paths. What will be printed?

Python

from pathlib import Path
p = Path('/home/user/docs')
new_path = p / 'projects' / '..' / 'notes.txt'
print(new_path.resolve())

A /home/user/docs/projects/../notes.txt

B /home/user/projects/notes.txt

C /home/user/docs/notes.txt

D /home/user/docs/projects/notes.txt

💡 Hint

Remember that '..' means to go up one directory, and resolve() simplifies the path.

The resolve() method simplifies the path by resolving '..' to go up one directory. So, '/home/user/docs/projects/../notes.txt' becomes '/home/user/docs/notes.txt'.

❓ Predict Output

intermediate

What does this os.path code output?

Given this Python code using os.path, what will be the output?

Python

import os
path = '/var/log/syslog'
print(os.path.basename(os.path.dirname(path)))

A syslog

B log

C /var/log

D var

💡 Hint

os.path.dirname() gets the folder path, and os.path.basename() gets the last part of a path.

First, os.path.dirname('/var/log/syslog') returns '/var/log'. Then, os.path.basename('/var/log') returns 'log'.

❓ Predict Output

advanced

What is the output of this code using pathlib's parts?

Look at this code that uses pathlib.Path.parts. What will it print?

Python

from pathlib import Path
p = Path('/usr/local/bin/python3')
print(p.parts[2])

A python3

B usr

C bin

D local

💡 Hint

The parts attribute splits the path into a tuple starting with root '/' as the first part.

The parts tuple is ('/', 'usr', 'local', 'bin', 'python3'). Index 2 is 'local'.

❓ Predict Output

advanced

What error does this code raise?

What error will this Python code raise when run?

Python

import os
path = '/tmp/testfile.txt'
print(os.path.join(path, 'subdir'))

A No error, output: '/tmp/testfile.txt/subdir'

B ValueError

C TypeError

D No error, output: '/tmp/subdir'

💡 Hint

Check how os.path.join works when the first argument is an absolute path.

os.path.join appends the second argument to the first path. Since the first is absolute, it joins as expected.

🧠 Conceptual

expert

Which option correctly normalizes a Windows path?

You have a Windows path string with mixed slashes: "C:/Users\\Admin/../Documents\\file.txt". Which option correctly normalizes it to a clean absolute path using Python?

A Use <code>os.path.normpath('C:/Users\\Admin/../Documents\\file.txt')</code>

B Use <code>Path('C:/Users\\Admin/../Documents\\file.txt').resolve()</code>

C Use <code>os.path.abspath('C:/Users\\Admin/../Documents\\file.txt')</code>

D Use <code>Path('C:/Users\\Admin/../Documents\\file.txt').absolute()</code>

💡 Hint

Normalization fixes slashes and removes '..' parts without requiring the file to exist.

os.path.normpath cleans the path string by fixing slashes and resolving '..' without checking the file system. Path.resolve() requires the path to exist and may raise errors.

Practice

1. Which Python module is recommended for safe and easy file path handling?

easy

A. pathlib

B. os.path

C. sys

D. math

Solution

1. Step 1: Identify the module for file paths

   The pathlib module provides an easy and modern way to handle file paths safely.

2. Step 2: Compare with other modules

   While os.path also handles paths, pathlib is recommended for its simplicity and object-oriented approach.

3. Final Answer:

   pathlib -> Option A

4. Quick Check:

   File path handling = pathlib [OK]

Hint: Remember: pathlib is the modern way to handle paths [OK]

Common Mistakes:

• Confusing pathlib with os.path
• Using sys for paths
• Choosing unrelated modules like math

2. Which of the following is the correct way to join paths using pathlib in Python?

easy

A. Path('folder') + 'file.txt'

B. Path('folder') / 'file.txt'

C. Path('folder').join('file.txt')

D. Path('folder').append('file.txt')

Solution

1. Step 1: Understand pathlib path joining

   In pathlib, the slash operator / is overloaded to join paths safely.

2. Step 2: Check each option

   Path('folder') / 'file.txt' uses / correctly. Path('folder') + 'file.txt' uses + which is invalid. The .append() and .join() methods do not exist on Path objects.

3. Final Answer:

   Path('folder') / 'file.txt' -> Option B

4. Quick Check:

   Use slash (/) to join paths [OK]

Hint: Use / operator to join pathlib paths [OK]

Common Mistakes:

• Using + to join paths
• Calling non-existent join or append methods
• Forgetting to import pathlib

3. What will be the output of this code?

from pathlib import Path
p = Path('folder') / 'subfolder' / 'file.txt'
print(p.parts)

medium

A. ('folder/subfolder/file.txt',)

B. ['folder', 'subfolder', 'file.txt']

C. ['folder/subfolder/file.txt']

D. ('folder', 'subfolder', 'file.txt')

Solution

1. Step 1: Understand Path.parts attribute

   The parts attribute returns a tuple of each part of the path as separate strings.

2. Step 2: Analyze the given path

   The path is 'folder/subfolder/file.txt', so parts will be ('folder', 'subfolder', 'file.txt').

3. Final Answer:

   ('folder', 'subfolder', 'file.txt') -> Option D

4. Quick Check:

   Path.parts returns tuple of path parts [OK]

Hint: Path.parts returns a tuple of path components [OK]

Common Mistakes:

• Expecting a list instead of tuple
• Getting full path as one string
• Confusing parts with name or stem

4. What is wrong with this code snippet?

from pathlib import Path
p = Path('folder') + 'file.txt'
print(p)

medium

A. Using + operator to join paths causes TypeError

B. Missing import statement for os module

C. Path object cannot be printed directly

D. The path string should use backslashes instead of forward slashes

Solution

1. Step 1: Check path joining method

   The code uses + operator to join a Path object and a string, which is not supported and raises a TypeError.

2. Step 2: Verify other options

   Import is correct, Path objects can be printed, and forward slashes are valid on most systems.

3. Final Answer:

   Using + operator to join paths causes TypeError -> Option A

4. Quick Check:

   Use /, not +, to join pathlib paths [OK]

Hint: Never use + to join pathlib paths; use / instead [OK]

Common Mistakes:

• Using + operator for path joining
• Thinking Path can't be printed
• Confusing path separators

5. You want to check if a file named data.csv exists inside a folder reports before reading it. Which code correctly does this using pathlib?

hard

A. p = Path('reports') + 'data.csv'
if p.is_file():
print('File found')

B. p = 'reports/data.csv'
if os.path.exists(p):
print('File found')

C. p = Path('reports') / 'data.csv'
if p.exists():
print('File found')

D. p = Path('reports/data.csv')
if p.is_dir():
print('File found')

Solution

1. Step 1: Create path using pathlib and join correctly

   p = Path('reports') / 'data.csv' if p.exists(): print('File found') uses Path('reports') / 'data.csv' which correctly joins folder and file.

2. Step 2: Check if file exists

   p = Path('reports') / 'data.csv' if p.exists(): print('File found') uses p.exists() to check if the file exists before reading, which is correct.

3. Final Answer:

   p = Path('reports') / 'data.csv'\nif p.exists():\n print('File found') -> Option C

4. Quick Check:

   Use pathlib with / and exists() to check files [OK]

Hint: Use Path(...) / filename and exists() to check files [OK]

Common Mistakes:

• Using + to join paths
• Using os.path without import
• Checking is_dir() instead of exists() or is_file()