What is the expected output of the following code? data = [ [1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16] ] for i in range(0, 4): print(data[i].pop(), end='')

The correct answer is D. 481216. D is correct because Python's .pop() method with no arguments removes and returns the last element of a list. The loop iterates over each of the four sublists (i = 0, 1, 2, 3), popping the final element of each: 4, 8, 12, 16. With end='' suppressing newlines, these print…

Question

What is the expected output of the following code?
data = [
[1, 2, 3, 4],
[5, 6, 7, 8],
[9, 10, 11, 12],
[13, 14, 15, 16]
]
for i in range(0, 4):
print(data[i].pop(), end='')

Options

A15913
B1234
C13 14 15 16
D481216

How the community answered

(43 responses)

A
2% (1)
B
7% (3)
C
16% (7)
D
74% (32)

Explanation

D is correct because Python's .pop() method with no arguments removes and returns the last element of a list. The loop iterates over each of the four sublists (i = 0, 1, 2, 3), popping the final element of each: 4, 8, 12, 16. With end='' suppressing newlines, these print concatenated as 481216.

Why the distractors fail:

A (15913) is what you'd get from .pop(0), which removes the first element (1, 5, 9, 13) - confusing pop() with pop(0).
B (1234) would result from accessing only the first sublist data[0] repeatedly, ignoring the loop variable i.
C (13 14 15 16) would require printing the entire last row, not one element per sublist - and the spaces don't match end=''.

Memory tip: Think of .pop() like a stack - it always removes from the end (LIFO order). If you want the front, you must explicitly pass index 0 with .pop(0).

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