Winter 2026 Quiz 1

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This quiz was administered in-person. Students were allowed a double-sided sheet of handwritten notes. Students had 30 minutes to work on the quiz.

This quiz covered Lectures 1-6 of the Winter 2026 offering of DSC 10.

Note (groupby / pandas 2.0): Pandas 2.0+ no longer silently drops columns that can’t be aggregated after a groupby, so code written for older pandas may behave differently or raise errors. In these practice materials we use .get() to select the column(s) we want after .groupby(...).mean() (or other aggregations) so that our solutions run on current pandas. On real exams you will not be penalized for omitting .get() when the old behavior would have produced the same answer.

You are working as a student assistant in the UCSD Registrar’s Office, and your job is to analyze enrollment data to understand student preferences and improve course scheduling.

The courses DataFrame (first few rows are shown below) contains information about courses offered this quarter. Each row represents a unique course offering. For each course, you have the "Course" code (str), "Department" (str), "Professor" name (str), "Enrollment" (int), "Capacity" (int), "Units" (int), and whether the course is "Online" (bool).

Assume that we have already run import babypandas as bpd and import numpy as np.

Problem 1

Your supervisor asks you to create a welcome message for new students. The code below generates this message and stores it in the variable message. In the box, write in the value of the message variable after the code executes. Write clearly!

school = "ucsd"
year = "2026"
message = "hello"
message = school.upper() + " " + year
message = message.replace("SD", " San Diego")

Answer: "UC San Diego 2026"

Trace through each line:

  • Lines 1–3: school = "ucsd", year = "2026", message = "hello".
  • Line 4: message = school.upper() + " " + year. school.upper() is "UCSD". Concatenating gives "UCSD" + " " + "2026" → "UCSD 2026".
  • Line 5: message.replace("SD", " San Diego") replaces the substring "SD" inside "UCSD 2026" with " San Diego" (note the leading space in the replacement string), yielding "UC San Diego 2026".

Problem 2

For each expression below, write the output that Python produces, or write error if the expression produces an error.


Problem 2.1

str(courses.take(np.arange(4)).shape[0]) * 2

Answer: "44"

courses.take(np.arange(4)) keeps the first four rows, so .shape[0] is 4. Then str(4) is "4", and "4" * 2 repeats the string twice → "44".


Problem 2.2

courses.get("Enrollment").iloc[0] != courses.get("Enrollment").loc[0]

Answer: False

.iloc[0] and .loc[0] both refer to the first row (label 0), so they read the same enrollment value. Comparing a value to itself with != is always False (e.g. 120 != 120).


Problem 2.3

courses.shape[1] / 7

Answer: 1.0

courses.shape[1] is the number of columns, which is 7 for this DataFrame. So 7 / 7 is 1.0 (division in Python 3 returns a float; 1. is also acceptable).


Problem 3

The Registrar wants to calculate what percentage of each course’s capacity is currently filled. Fill in the blanks to add a new column called "Percent_Full" to the courses DataFrame that contains this percentage (as a number between 0 and 100).

courses = courses.assign(Percent_Full = ___(a)___ / ___(b)___ * 100)


Problem 3.1

(a):

Answer: courses.get("Enrollment") — current enrollment in each row (numerator of the fraction).


Problem 3.2

(b):

Answer: courses.get("Capacity")

Percent full is (enrollment ÷ capacity) × 100. Blank (a) is current enrollment per row; blank (b) is max capacity per row. Division is element-wise (row by row), then multiply by 100. For example, 120 enrolled and 150 capacity gives (120/150) × 100 = 80.


Problem 4

With more students requesting online courses, the Registrar wants to know how many online courses are offered by the CSE department. Select all expressions that correctly evaluate to an int that is the total number of online CSE courses in courses.

Answer: Select courses.groupby("Department").sum().get("Online").loc["CSE"] and ((courses.get("Department") == "CSE") & (courses.get("Online"))).sum().

We need an int equal to the number of courses that are both CSE and online.

  • First option: filters to those rows but returns a DataFrame, not an int.
  • Second option: groupby("Department").sum() aggregates; for boolean columns, .sum() counts True values. .get("Online").loc["CSE"] is the count of online courses in CSE — an integer. Correct.
  • Third option: ((...) & (...)) is a boolean Series with one entry per row; .shape[0] is the length of that Series (total number of courses), not how many are True. Incorrect.
  • Fourth option: summing the boolean Series counts True values (online CSE rows). Correct.
  • “None of the above” is wrong because two options work.

Problem 5

For a report to the Academic Senate, you need to analyze course enrollment patterns. Write or finish one line of Python code that produces each quantity described.


Problem 5.1

(a) A DataFrame consisting of "Department" as the index, and a single column containing the mean "Enrollment" of all courses within each "Department".

courses.get(["Department", "Enrollment"]).___________________________

Answer: groupby("Department").mean() — full one-liner:

courses.get(["Department", "Enrollment"]).groupby("Department").mean()

Explanation: .get(["Department", "Enrollment"]) keeps only those columns. .groupby("Department") groups rows by department. .mean() computes the mean of numeric columns per group — here mean "Enrollment" — producing a DataFrame with "Department" as the index and one column of means.


Problem 5.2

(b) A scatter plot of the data in the courses DataFrame with "Capacity" on the x-axis and "Enrollment" on the y-axis.

Answer: courses.plot(kind="scatter", x="Capacity", y="Enrollment")

kind="scatter" draws a scatter plot; x and y name the columns for the horizontal and vertical axes.


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