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This quiz was administered in-person. It was closed-book and
closed-note; students were not allowed to use the DSC
10 Reference Sheet. Students had 20 minutes to work on
the quiz.
This quiz covered Lectures 8-11 of the Winter 2024 offering
of DSC 10.
⚠️ PDF version available here .
The DataFrame flights contains information about recent
flights, with each row representing a specific flight and the following
columns:
"flight_num" (str): The unique code of the flight,
consisting of a two-character airline designator followed by 1 to 4
digits (e.g., "UA1989")."airline" (str): The airline name (e.g.,
"United")."departure" (str): The code for the airport from which
the flight departs (e.g., "SAN")."arrival" (str): The code for the airport at which the
flight arrives (e.g., "LAX").Suppose the function first_two takes as input a string
and returns a string with the first two characters only. For example
first_two("panda") is "pa". Using this
function, write an expression that evaluates to a Series containing the
two-character airline designator for each flight in
flights.
Answer:
flights.get("flight_num").apply(first_two)
flights.get("flight_num") selects the
"flight_num" column as a Series, with values like
"UA1989". Calling .apply(first_two) runs the
first_two function on every value in that Series, replacing
each flight number with just its first two characters — the airline
designator.
The average score on this problem was 78%.
Fill in the blanks below so that grouped is a DataFrame
showing how many flights on each airline departed from each airport.
grouped = flights.groupby(___(x)___).___(y)___.get(["flight_num"])Answer (x): ["departure", "airline"] or
["airline", "departure"]
“How many flights on each airline departed from each airport” asks
for a count for every combination of departure airport
and airline. Grouping by a list of both columns creates one group — one
row in the output — for each distinct (departure, airline) pair that
appears in flights. The order of the two columns in the
list doesn’t change which pairs exist, only how the resulting index is
arranged.
The average score on this problem was 82%.
Answer (y): count()
.count() counts the number of flights that fall into
each group. .get(["flight_num"]) then keeps just the
"flight_num" column (wrapped in a list so the result stays
a DataFrame), which holds that count for every (departure, airline)
pair.
The average score on this problem was 89%.
Suppose the expression
grouped.shape[0] == flights.shape[0] evaluates to
True. Select all true statements below.
No two flights in flights were on the same airline.
No two flights in flights had the same departure and
arrival airports.
Among all flights in flights that were on the same
airline, no two had the same departure airport.
Among all flights in flights from the same departure
airport, no two were on the same airline.
Answer: Option 3 and Option 4
grouped has one row per distinct (departure, airline)
pair, and grouped.shape[0] == flights.shape[0] means there
are exactly as many such pairs as there are flights. Since no row can be
grouped twice, this means every flight has a unique (departure,
airline) pair — no two flights share both the same departure
airport and the same airline.
Option 1: False. This would mean every flight is on a different airline, which is much stronger than what we’re told. Two flights can share an airline, as long as they don’t also share a departure airport.
Option 2: False. Arrival airport isn’t one of the
columns we grouped by, so this condition doesn’t follow from
grouped.shape[0] == flights.shape[0] at all.
Option 3: True. If two flights were on the same airline and had the same departure airport, they’d share the same (departure, airline) pair, which we just said can’t happen. So within a single airline, every flight departs from a different airport.
Option 4: True. This is the same fact, viewed the other way: if two flights departed from the same airport and were on the same airline, they’d again share a pair. So among flights from the same airport, every flight is on a different airline.
The average score on this problem was 80%.
Suppose we have another DataFrame more_flights which
contains the same columns as flights, but different rows.
Define merged as follows.
merged = flights.merge(more_flights, on = "airline")Suppose that in merged, there are 108 flights where the
airline is "United", and in more_flights,
there are 12 flights where the airline is "United". If
flights has 15 rows in total, how many of these rows are
not for "United" flights? Give your answer
as an integer.
Answer: 6
When we merge on "airline", every "United"
row in flights is paired with every "United"
row in more_flights. So the number of "United"
rows in merged is
(\text{\# United rows in } \texttt{flights}) \times (\text{\# United rows in } \texttt{more\_flights}).
We’re told this product is 108, and
that more_flights has 12
"United" rows, so
(\text{\# United rows in } \texttt{flights}) \times 12 = 108 \implies \text{\# United rows in } \texttt{flights} = 9.
Since flights has 15
rows in total, the number that are not
"United" flights is
15 - 9 = \boxed{6}.
The average score on this problem was 64%.
Fill in the blanks below so that prob evaluates to the
probability that a randomly selected flight from the
flights DataFrame arrives at "SAN" given that
it departs from "LAX".
san = flights.get("arrival") == "SAN"
lax = flights.get("departure") == "LAX"
prob = flights[san___(x)___lax].shape[0] / ___(y)___Answer (x): &
The probability of arriving at "SAN"
given departure from "LAX" is the fraction
of LAX departures that go to SAN — that is, flights where
both san and lax are
True. Since san and lax are
Boolean Series, we combine them element-wise with & to
find flights satisfying both conditions at once.
The average score on this problem was 66%.
Answer (y): flights[lax].shape[0]
A conditional probability is (# satisfying both conditions) / (#
satisfying the condition we’re given). The numerator,
flights[san & lax].shape[0], counts flights that both
depart from LAX and arrive at SAN. The denominator should be the number
of flights departing from LAX — the event we’re conditioning on — which
is flights[lax].shape[0].
The average score on this problem was 62%.
Write an expression that evaluates to the probability that a randomly
selected flight from the flights DataFrame is
not on the airline "Delta".
Answer:
1 - flights[flights.get("airline") == "Delta"].shape[0] / flights.shape[0]
or
flights[flights.get("airline") != "Delta"].shape[0] / flights.shape[0]
or (flights.get("airline") != "Delta").mean()
flights[flights.get("airline") == "Delta"].shape[0] / flights.shape[0]
is the probability that a random flight is on Delta. By
the complement rule, the probability it’s not on Delta
is 1 minus that.
Alternatively, we can count the “not Delta” flights directly:
flights.get("airline") != "Delta" is a Boolean Series
that’s True for every non-Delta flight, so
flights[flights.get("airline") != "Delta"].shape[0] / flights.shape[0]
divides that count by the total number of flights.
The last expression uses a shortcut: taking .mean() of a
Boolean Series gives the proportion of True values (since
True and False act as 1 and 0),
which here is exactly the proportion of flights that are not on
Delta.
The average score on this problem was 76%.
Use the function defined below to answer the questions on the right.
def discount(duration, price):
if duration > 6:
price = price * 0.9
if duration > 4:
price = price * 0.95
elif duration > 3:
price = price * 0.98
else:
price = price * 0.99
price = price - 5
return priceWhat is the output of discount(7, 100)?
100 * 0.9 * 0.95 * 0.98 - 5
100 * 0.9 * 0.95 - 5
100 * 0.9 - 5
100 * 0.9
Answer: Option 2
discount has two separate conditional
blocks, not one long chain — the second
if/elif/else is independent of
the first if. Calling discount(7, 100) runs
both blocks:
duration > 6 is 7 > 6,
which is True, so price = 100 * 0.9.duration > 4 is 7 > 4,
which is True, so this branch runs (and the
elif/else are skipped):
price = (100 * 0.9) * 0.95.price = price - 5.So the output is 100 * 0.9 * 0.95 - 5, which is Option
2.
The average score on this problem was 72%.
Find x and y such that
discount(x, y) evaluates to 97 * 0.98 - 5.
Answer (x): anything greater than 3 and
less than or equal to 4
The target expression 97 * 0.98 - 5 only has
one multiplier, 0.98, before the
- 5. That 0.98 only comes from the
elif duration > 3 branch of the second block, so the
first if duration > 6 block must not
have multiplied price by anything — meaning it didn’t run,
so duration must be 6 or less. The
elif duration > 3 branch running (instead of the
if duration > 4 branch before it, or the
else) means duration is greater than
3 but not greater than 4. Together, that’s
3 < duration <= 4.
The average score on this problem was 95%.
Answer (y): 97
Since the first block never runs, price enters the
second block unchanged, so the 0.98 multiplies the original
input price directly: price * 0.98 - 5 must
equal 97 * 0.98 - 5, so price = 97.
The average score on this problem was 95%.