Given a list of filenames files = ['a.txt', 'b.txt', 'c.txt'] and a folder path folder = '/data', which code creates a list of full paths correctly?

hard📝 Application Q9 of 15

Python - Standard Library Usage

Given a list of filenames files = ['a.txt', 'b.txt', 'c.txt'] and a folder path folder = '/data', which code creates a list of full paths correctly?

A[os.path.join(folder + f) for f in files]

B[folder + '/' + f for f in files]

C[os.path.join(f, folder) for f in files]

D[os.path.join(folder, f) for f in files]

Step-by-Step Solution

Solution:

Step 1: Use list comprehension with os.path.join
Join folder and each filename as separate arguments inside comprehension.
Step 2: Avoid string concatenation or wrong argument order
Concatenation may cause wrong separators; order matters for correct paths.
Final Answer:
[os.path.join(folder, f) for f in files] -> Option D
Quick Check:
List comprehension with join(folder, file) [OK]

Quick Trick: Use join(folder, filename) inside list comprehension [OK]

Common Mistakes:

Concatenating strings manually
Swapping join arguments
Passing single string to join

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Given a list of filenames files = ['a.txt', 'b.txt', 'c.txt'] and a folder path folder = '/data', which code creates a list of full paths correctly?

Step 1: Use list comprehension with os.path.join

Step 2: Avoid string concatenation or wrong argument order

Final Answer:

Quick Check:

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