Compare groups fairly · Data Playground

The question

Compare study groups using attendees with known scores only.

df Group comparison

Your inputs

`df`

One row is one learner’s submission. Attendance and a known score are required; study hours may be unknown.

Field meanings and units

submission_id: Attendance-submission identifier.
student: Learner label.
group: Comparison group.
hours: Study duration in hours.
score: Assessment score in points.
attended: Whether the learner attended.
date: Calendar date supplied as YYYY-MM-DD text unless the task states it is invalid.

`df` · first 5 of 18 rows
`submission_id`	`student`	`group`	`hours`	`score`	`attended`	`date`
S001	Learner 1	A	2	55	false	2026-04-01
S002	Learner 2	B	5	78	true	2026-04-01
S003	Learner 3	A	3	68	true	2026-04-02
S004	Learner 4	B	7	None	true	2026-04-02
S005	Learner 5	A	4	72	false	2026-04-03

View all 18 rows

`df` · complete input
`submission_id`	`student`	`group`	`hours`	`score`	`attended`	`date`
S001	Learner 1	A	2	55	false	2026-04-01
S002	Learner 2	B	5	78	true	2026-04-01
S003	Learner 3	A	3	68	true	2026-04-02
S004	Learner 4	B	7	None	true	2026-04-02
S005	Learner 5	A	4	72	false	2026-04-03
S006	Learner 6	B	6	85	true	2026-04-03
S007	Learner 7	A	8	None	true	2026-04-04
S008	Learner 8	B	1	49	true	2026-04-04
S009	Learner 9	A	2	56	false	2026-04-05
S010	Learner 10	B	5	79	true	2026-04-05
S011	Learner 11	A	3	69	true	2026-04-06
S012	Learner 12	B	7	91	true	2026-04-06
S013	Learner 13	A	4	73	false	2026-04-07
S014	Learner 14	B	6	86	true	2026-04-07
S015	Learner 15	A	8	95	true	2026-04-08
S016	Learner 16	B	1	50	true	2026-04-08
S017	Learner 17	A	2	57	false	2026-04-09
S018	Learner 18	B	5	80	true	2026-04-09

The editable setup in Your Python creates these inputs when you run it.

Your task · Compare groups fairly

Compare study groups using attendees with known scores only.

eligibleDataFramecountsSeriesmeansSeries

Deliverables

Make the outputs listed below. Use each name exactly as shown. Then choose Check answer.

Python name key

Teal outlines name table columns. Violet left rules name Python objects. Dotted rules mark syntax, functions and parameters. Use each name exactly as shown.

Words in this task

eligible: rows that meet this task’s stated conditions

Attendees with scoresOutputeligibleDataFrame
1. Filter df to rows where attended is True and score is present.
2. Store them in eligible, with every original column and row order preserved.
Group countsOutputcountsSeries
1. Store the number of rows in eligible for each group in counts.
Mean scoresOutputmeansSeries
1. Store the mean score of rows in eligible for each group in means.

Data policy Keep all supplied source tables unchanged. Work on separate outputs.

Help Hints & explained solution

Choose the amount of help you need. Opening one hint does not reveal the others.

Hint 1 — Think

Are the counts and averages describing the same learners?

Hint 2 — Tools

notna(), groupby(), count() and mean()

Revisit: Combine conditions · Summarise groups · Answer one numerical question

Hint 3 — Approach

Use one eligible population for both group sizes and mean scores.

Explained solution

Add these lines below # Your work in the editor. The supplied input setup is already there.

eligible = df[df.attended & df.score.notna()]
counts = eligible.groupby('group')['score'].count()
means = eligible.groupby('group')['score'].mean()
means

View supplied setup

This code is already at the top of the editor.

import pandas as pd

# Supplied input: df (18 rows)
df = pd.DataFrame(
    [
        ('S001', 'Learner 1', 'A', 2, 55, False, '2026-04-01'),
        ('S002', 'Learner 2', 'B', 5, 78, True, '2026-04-01'),
        ('S003', 'Learner 3', 'A', 3, 68, True, '2026-04-02'),
        ('S004', 'Learner 4', 'B', 7, None, True, '2026-04-02'),
        ('S005', 'Learner 5', 'A', 4, 72, False, '2026-04-03'),
        ('S006', 'Learner 6', 'B', 6, 85, True, '2026-04-03'),
        ('S007', 'Learner 7', 'A', 8, None, True, '2026-04-04'),
        ('S008', 'Learner 8', 'B', 1, 49, True, '2026-04-04'),
        ('S009', 'Learner 9', 'A', 2, 56, False, '2026-04-05'),
        ('S010', 'Learner 10', 'B', 5, 79, True, '2026-04-05'),
        ('S011', 'Learner 11', 'A', 3, 69, True, '2026-04-06'),
        ('S012', 'Learner 12', 'B', 7, 91, True, '2026-04-06'),
        ('S013', 'Learner 13', 'A', 4, 73, False, '2026-04-07'),
        ('S014', 'Learner 14', 'B', 6, 86, True, '2026-04-07'),
        ('S015', 'Learner 15', 'A', 8, 95, True, '2026-04-08'),
        ('S016', 'Learner 16', 'B', 1, 50, True, '2026-04-08'),
        ('S017', 'Learner 17', 'A', 2, 57, False, '2026-04-09'),
        ('S018', 'Learner 18', 'B', 5, 80, True, '2026-04-09'),
    ],
    columns=['submission_id', 'student', 'group', 'hours', 'score', 'attended', 'date'],
)

Why these steps exist

One eligibility mask gives both summaries the same population. Otherwise a larger count could include unscored learners.
The mean excludes missing scores; a zero-filled score would change the question being answered.

Another valid approach

Named group aggregations can calculate both outputs together; expose the requested separate named results afterwards.