Insert the correct snippet to convert the t tuple to a dictionary named d. Expected output: {'A': 1, 'B': 2, 'C': 3} Code: t = (('A', 1), ('B', 2), ('C', 3)) insert code here print(d)

The correct answer is B. d = dict(t). Option B is correct because Python's built-in dict() function accepts an iterable of key-value pairs - exactly what t is: a tuple of 2-element tuples. Each inner tuple ('A', 1) becomes a key-value pair in the resulting dictionary. Why the distractors fail: A (t >> d.dict) - >>…

Question

Insert the correct snippet to convert the t tuple to a dictionary named d. Expected output:
{'A': 1, 'B': 2, 'C': 3}
Code:
t = (('A', 1), ('B', 2), ('C', 3))
insert code here
print(d)

Options

At >> d.dict
Bd = dict(t)
Cd.dict(t)
Dd = t(dict)

How the community answered

(20 responses)

A
5% (1)
B
80% (16)
C
10% (2)
D
5% (1)

Explanation

Option B is correct because Python's built-in dict() function accepts an iterable of key-value pairs - exactly what t is: a tuple of 2-element tuples. Each inner tuple ('A', 1) becomes a key-value pair in the resulting dictionary.

Why the distractors fail:

A (t >> d.dict) - >> is a bitwise right-shift operator, not a conversion method, and d doesn't exist yet so d.dict would raise a NameError.
C (d.dict(t)) - d hasn't been defined, and even if it had, dictionaries have no .dict() method; this would raise AttributeError.
D (d = t(dict)) - this tries to call the tuple t like a function with dict as an argument, which raises TypeError since tuples aren't callable.

Memory tip: Think of dict() as a factory that "zips up" pairs into a dictionary - if your data is already shaped as (key, value) pairs, dict() handles the conversion directly, just like list() or tuple() convert other iterables.

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