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.