Count the right thing

Task 1 · L6 p. 1. Setup used by every later task:

import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
from statistics import mean, median

1. Registrations versus students

students = pd.DataFrame({"id": ["A", "B"],
                         "mode": ["Full time", "Part time"]})
enrolments = pd.DataFrame([
    ["A", "C1", "T1", 60], ["A", "C2", "T1", 70],
    ["A", "C3", "T1", 80], ["B", "C1", "T1", 40]
], columns=["id", "course", "term", "grade"])
enrolments["passed"] = enrolments["grade"] >= 50

passes = enrolments["passed"].sum()
n_reg = len(enrolments)
print("Registration pass fraction:", passes, "/", n_reg, "=", passes / n_reg)
print("Registration pass percentage:", 100 * passes / n_reg)

students_passed = enrolments.groupby("id")["passed"].any().sum()
n_students = enrolments["id"].nunique()
print("Students passing at least one course:", students_passed, "/", n_students,
      "=", 100 * students_passed / n_students, "%")

Output: 3 / 4 = 0.75, 75.0, and 1 / 2 = 50.0 %.

Why both are correct. They answer different questions with different units. 75% is the share of registrations that passed (denominator: 4 course registrations). 50% is the share of students who passed at least one course (denominator: 2 students). Student A contributes three passing registrations but only one passing student.

2. Attach student information without multiplying rows

joined = enrolments.merge(students, on="id", how="left",
                          validate="many_to_one", indicator=True)
print("Rows before and after:", len(enrolments), len(joined))
print("Unmatched:", joined["_merge"].ne("both").sum())
print("students id unique:", students["id"].is_unique)
print("Duplicate registrations:",
      enrolments.duplicated(["id", "course", "term"]).sum())

Output: 4 4, Unmatched: 0, True, Duplicate registrations: 0.

Explanation. In enrolments one row is one course registration in one term, so A legitimately appears three times; the natural key is (id, course, term), and it has no duplicates. In students one row is one student, so id must be unique. If A appeared twice in students, every A registration would match two student rows and the join would grow from 4 to 7 rows, silently inflating counts. validate="many_to_one" would raise an error instead.

A complete question for the TA: "Among the two synthetic students registered in term T1, what share of course registrations had a final grade of at least 50, and what share of students passed at least one course?" It names the population (these students), period (T1) and unit (registration versus student). Student IDs are labels; their digits are not quantities to average.