Library Hard

What does this code print?

import json
print(json.dumps({1: "a", 2: "b"}))

  • {"1": "a", "2": "b"}

  • {1: "a", 2: "b"}

  • raises TypeError

  • [[1, "a"], [2, "b"]]

Answer

JSON object keys must be strings, so json.dumps coerces integer keys to their string form. A later json.loads will not turn them back into ints, so the round trip changes the key type.

What does this code print?

import copy
inner = [1, 2]
a = [inner, inner]
b = copy.deepcopy(a)
b[0].append(3)
print(b)

  • [[1, 2, 3], [1, 2, 3]]

  • [[1, 2, 3], [1, 2]]

  • [[1, 2], [1, 2]]

  • [[1, 2, 3, 3], [1, 2]]

Answer

Both elements of a are the same list object. deepcopy uses a memo so that shared identity is preserved in the copy: b[0] and b[1] are still the same object, so appending to one shows up in both.

What does this code print?

from itertools import groupby
data = [1, 1, 2, 1, 1]
print([k for k, g in groupby(data)])

  • [1, 2, 1]

  • [1, 2]

  • [1, 1, 2, 1, 1]

  • [1, 2, 1, 1]

Answer

groupby only collapses consecutive equal items; it does not sort first. The two runs of 1 are separated by a 2, so they form distinct groups and the keys are [1, 2, 1].

What does this code print?

from collections import Counter
c = Counter(a=3, b=1) - Counter(a=1, b=2)
print(dict(c))

  • {'a': 2}

  • {'a': 2, 'b': -1}

  • {'a': 2, 'b': 0}

  • {'a': 4, 'b': 3}

Answer

The - operator on Counter keeps only results with a positive count. a becomes 2 and stays; b would be -1 and is dropped entirely. Use .subtract if you need to keep zero/negative counts.

What does this code print?

from datetime import datetime, timedelta
print(datetime(2020, 3, 8, 12) + timedelta(days=1))

  • 2020-03-09 12:00:00

  • 2020-03-09 13:00:00

  • 2020-03-09 11:00:00

  • 2020-03-08 12:00:00

Answer

A naive datetime has no time zone, so timedelta arithmetic is pure wall-clock math with no DST adjustment. Adding exactly one day keeps the time at 12:00:00 even though March 8, 2020 was a US DST transition.

What happens when this runs?

from functools import lru_cache
@lru_cache
def f(x):
    return len(x)
f([1, 2, 3])

  • raises TypeError

  • prints 3

  • raises KeyError

  • returns 3 with no error

Answer

lru_cache stores arguments as dictionary keys, so every argument must be hashable. A list is unhashable, so the wrapper raises TypeError: unhashable type: 'list' before the body runs.

What does this code print?

import re
print(re.findall(r"a.*?b", "axbxxab"))

  • ['axb', 'ab']

  • ['axbxxab']

  • ['axbxxab', 'ab']

  • ['ab']

Answer

.*? is non-greedy, matching as few characters as possible before the next b. The first match stops at the first b (axb); matching resumes after it and finds ab. A greedy .* would swallow the whole string.

What does this code print?

from functools import cmp_to_key
def cmp(a, b):
    return b - a
print(sorted([3, 1, 2], key=cmp_to_key(cmp)))

  • [3, 2, 1]

  • [1, 2, 3]

  • [2, 1, 3]

  • raises TypeError

Answer

cmp_to_key adapts an old-style comparison function to a key. Returning b - a is positive when a < b, which orders larger values first, producing a descending sort [3, 2, 1].

What does this code print?

import math
print(math.nan == math.nan)

  • False

  • True

  • raises ValueError

  • nan

Answer

By the IEEE-754 rule, NaN is not equal to anything, including itself, so math.nan == math.nan is False. Use math.isnan to test for NaN.

What does this code print?

from decimal import Decimal, ROUND_HALF_UP
print(Decimal("2.675").quantize(Decimal("0.01"), rounding=ROUND_HALF_UP))

  • 2.68

  • 2.67

  • 2.675

  • 2.7

Answer

Decimal("2.675") is exactly 2.675, so rounding half up to two places gives 2.68. The float round(2.675, 2) returns 2.67 because 2.675 cannot be represented exactly in binary.

What does this code print?

from collections import namedtuple
P = namedtuple("P", "x y")
p = P(1, 2)
q = p._replace(y=99)
print(p, q)

  • P(x=1, y=2) P(x=1, y=99)

  • P(x=1, y=99) P(x=1, y=99)

  • P(x=1, y=2) P(x=99, y=2)

  • raises AttributeError

Answer

Named tuples are immutable, so _replace returns a new instance with the given fields changed and leaves the original untouched. p still has y=2 while q has y=99.

What does this code print?

from itertools import cycle, islice
print(list(islice(cycle("AB"), 5)))

  • ['A', 'B', 'A', 'B', 'A']

  • ['A', 'B', 'A', 'B', 'A', 'B']

  • ['A', 'B']

  • ['A', 'B', 'A', 'B']

Answer

cycle("AB") repeats A, B endlessly; islice(..., 5) takes the first five items, giving ['A', 'B', 'A', 'B', 'A'].

What does this code print?

from collections import defaultdict
d = defaultdict(int)
_ = d["missing"]
print(dict(d))

  • {'missing': 0}

  • {}

  • raises KeyError

  • {'missing': None}

Answer

Merely reading a missing key on a defaultdict triggers the factory and inserts it. So d["missing"] creates the key with value 0 as a side effect, and the dict is no longer empty.

What does this code print?

import json
print(json.dumps(float("nan")))

  • NaN

  • raises ValueError

  • null

  • nan

Answer

By default Python's json emits the non-standard token NaN (also Infinity/-Infinity), which strict JSON parsers reject. Passing allow_nan=False would instead raise ValueError.

What does this code print?

import re
print(re.split(r"(\d)", "a1b2"))

  • ['a', '1', 'b', '2', '']

  • ['a', 'b', '']

  • ['a', 'b']

  • ['a', '1', 'b', '2']

Answer

When the split pattern contains a capturing group, the captured delimiters are kept in the result. The trailing empty string comes from the split right after the final 2.

What does this code print?

from dataclasses import dataclass, field
@dataclass
class C:
    items: list = field(default_factory=list)
a = C()
a.items.append(1)
print(C().items)

  • []

  • [1]

  • [1, 1]

  • raises ValueError

Answer

default_factory=list calls list() fresh for each new instance, so a and the new C() do not share a list. Mutating a.items leaves the new instance's items empty. A bare mutable default would be rejected at class creation.

What does this code print?

from fractions import Fraction
print(Fraction(0.1).limit_denominator(10))

  • 1/10

  • 1/8

  • 3602879701896397/36028797018963968

  • 0.1

Answer

Fraction(0.1) captures the exact binary value of the float, a huge ratio. limit_denominator(10) finds the closest fraction with denominator at most 10, which is 1/10.

What does this code print?

from itertools import groupby
words = ["apple", "avocado", "banana", "cherry", "citrus"]
print({k: len(list(g)) for k, g in groupby(words, key=lambda w: w[0])})

  • {'a': 2, 'b': 1, 'c': 2}

  • {'a': 2, 'b': 1, 'c': 1}

  • {'a': 1, 'b': 1, 'c': 1}

  • {'apple': 1, 'banana': 1}

Answer

The list is already ordered by first letter, so groupby cleanly groups the two a words, one b word, and two c words. The dict comprehension counts each group, giving {'a': 2, 'b': 1, 'c': 2}.

What does this code print?

from datetime import datetime
d = datetime.strptime("2020-12-25", "%Y-%m-%d")
print(d.strftime("%A"))

  • Friday

  • Saturday

  • Thursday

  • December

Answer

strptime parses the string into a datetime, and %A formats the full weekday name (locale default English). December 25, 2020 fell on a Friday.

What does this code print?

from collections import OrderedDict
a = OrderedDict([("x", 1), ("y", 2)])
b = OrderedDict([("y", 2), ("x", 1)])
print(a == b)

  • False

  • True

  • raises TypeError

  • None

Answer

Comparing two OrderedDict objects is order-sensitive, so a and b differ despite having the same items. (Comparing an OrderedDict to a plain dict would ignore order and be True.)