Unit 1 • Lesson 6

Counting and Grouping

Overview

A dictionary is how you count things, and how you split a table into groups.

Beginner 20–25 min

← Previous Unit 1 test

What You Will Learn in This Lesson

By the end of this lesson, you will be able to:

Count how often each value appears with a dictionary and .get.
Use collections.Counter for the same job, and read its most common values.
Group rows into lists by a key with setdefault.
Summarise each group — the split-apply-combine pattern behind every group-by.

Why This Matters

“How many in each category?” and “what is the average per group?” are the two questions asked of nearly every table. The dictionary pattern in this lesson is exactly what pandas’ groupby does in unit 7: split the rows by a key, apply a summary to each part, combine the answers. Writing it once by hand is what makes group-by stop looking like magic.

Step 1: Counting with a dictionary

The key is the thing being counted; the value is how many times you have seen it.

The first sighting problem

counts[s] += 1 raises KeyError the first time s appears, because there is nothing to add to yet. counts.get(s, 0) + 1 treats an unseen key as zero.

subjects = ["maths", "computing", "maths", "art", "maths"]

counts = {}
for s in subjects:
    counts[s] = counts.get(s, 0) + 1
print(counts)

Counter does it for you

collections.Counter is a dictionary built for counting. most_common(n) answers “which values are most frequent?” directly.

from collections import Counter

subjects = ["maths", "computing", "maths", "art", "maths"]
counts = Counter(subjects)
print(counts)
print(counts.most_common(1))

Mini Practice #1: Most common answer

Try It Yourself

Count the survey answers with Counter and print the two most common.

from collections import Counter

answers = ["yes", "no", "yes", "maybe", "yes", "no"]

# The two most common answers:

Press Run to see output

Step 2: Grouping is counting with lists

Same shape, except the value is a list you append to instead of a number you add to.

Split the rows by a key

groups.setdefault(key, []) returns the list for that key, creating an empty one on the first sighting. Appending to what it returns fills the group.

rows = [("maths", 92), ("computing", 88), ("maths", 79)]

groups = {}
for subject, score in rows:
    groups.setdefault(subject, []).append(score)
print(groups)

Apply a summary to each group

Once the scores are split, each group is an ordinary list and every summary from the last lesson works on it.

groups = {"maths": [92, 79], "computing": [88]}

for subject, scores in groups.items():
    print(subject, len(scores), sum(scores) / len(scores))

Combine into one answer

A dictionary comprehension collects the per-group results into a single table: one key per group, one summary per key.

rows = [("maths", 92), ("computing", 88), ("maths", 79)]

groups = {}
for subject, score in rows:
    groups.setdefault(subject, []).append(score)

means = {s: sum(v) / len(v) for s, v in groups.items()}
print(means)

Key idea: Split, apply, combine. In unit 7 this whole block becomes df.groupby("subject")["score"].mean(), and the three words are how that line is explained.

Mini Practice #2: Highest per group

Try It Yourself

Group the rows by subject, then print the highest score in each subject.

rows = [("maths", 92), ("computing", 88), ("maths", 79), ("computing", 95)]

groups = {}
# split the scores by subject

# print each subject with its highest score

Press Run to see output

Step 3: Reading the result honestly

A per-group summary invites comparison, and comparisons need the group sizes next to them.

Report the count with the mean

A group of one student with 95 “beats” a group of forty with 90. Printing the count beside each mean shows which comparison is worth making.

groups = {"maths": [90, 91, 89, 90], "art": [95]}

for subject, scores in sorted(groups.items()):
    print(f"{subject}: mean {sum(scores) / len(scores):.1f} from {len(scores)} scores")

Predict, then check: Keep each group’s denominator

Art has two results and maths has one. Predict both averages without combining the groups.

rows = [("art", 0), ("maths", 90), ("art", 60)]
groups = {}
for subject, score in rows:
    groups.setdefault(subject, []).append(score)
for subject, scores in groups.items():
    print(subject, len(scores), round(sum(scores) / len(scores), 1))

Show the output and explanation

art 2 30.0
maths 1 90.0

Each group has its own total and count. Dividing art’s total by all three rows would give 20.0, which is not its mean. Keeping the count beside the average also shows that maths has only one observation.

Try one change: Add ("maths", 30). Predict the new count and mean for maths; art should stay unchanged.

Go deeper: Keep each group’s state separate

Counting tracks one number per key. Grouping stores the values belonging to each key so you can compute summaries within each group. A per-group mean needs that group’s total and count, not the total number of rows in the dataset.

Trace the example

Predict the output before opening the explanation.

rows = [("A", 2), ("B", 9), ("A", 6)]
groups = {}
for name, score in rows:
    groups.setdefault(name, []).append(score)
print(groups["A"])
print(sum(groups["A"]) / len(groups["A"]))

Show the output and trace

[2, 6]
4.0

The first A creates its list with 2. B gets a different list containing 9. The final A appends 6 to A’s existing list. A’s mean divides its sum of 8 by its own count of 2, producing 4.

A mistake to watch for

Dividing A’s total by len(rows) would mix a group total with a whole-dataset count. Replacing groups[name] with [score] on every iteration would lose earlier values instead of accumulating them.

Transfer challenge

Add ("B", 3). Print each group and its mean. Check that changing B’s observations does not change A’s mean.

Change the example below to solve the challenge. Predict the result, run your code, and compare. This practice is ungraded.

rows = [("A", 2), ("B", 9), ("A", 6)]
groups = {}
for name, score in rows:
    groups.setdefault(name, []).append(score)
print(groups["A"])
print(sum(groups["A"]) / len(groups["A"]))

Press Run to see output

Compare with a worked solution

rows = [("A", 2), ("B", 9), ("A", 6), ("B", 3)]
groups = {}
for name, score in rows:
    groups.setdefault(name, []).append(score)
for name, scores in groups.items():
    print(name, sum(scores) / len(scores))

Run both versions. Explain any difference in their output, then change one input and predict the new result.

Step 4: When to Use It

How many of each

Categories, answers, subjects. Counter or a dictionary with .get gives the frequency table.

Counter(subjects).most_common(3)

A summary per group

When the question says “per” or “by”: split into lists with setdefault, then summarise each list.

When Not to Use It

Do not compare group averages without their sizes. A group of one can top the table by accident — report the count next to every mean.

End-of-Lesson Exercises

Exercise 1: Rows per subject

Write count_subjects(rows) that takes a list of (subject, score) pairs and returns a dictionary of how many rows each subject has. Count every row, including a zero score, and return an empty dictionary for no rows.

Start with an empty dictionary and walk the rows. counts.get(subject, 0) + 1 handles both the first sighting and every one after it.

def count_subjects(rows):
    # Return {subject: how many rows}
    return {}

print(count_subjects([("maths", 92), ("computing", 88), ("maths", 79)]))

Press Run to try it, then Check my work to mark it.

Exercise 2: Mean per subject

Write mean_by_subject(rows) that takes (subject, score) pairs and returns a dictionary of each subject’s mean score, rounded to one decimal place.

Split first: groups.setdefault(subject, []).append(score). Then combine: {s: round(sum(v) / len(v), 1) for s, v in groups.items()}.

def mean_by_subject(rows):
    # {subject: mean score, rounded to 1 place}
    return {}

print(mean_by_subject([("maths", 92), ("computing", 88), ("maths", 79)]))

Press Run to try it, then Check my work to mark it.

← Prev Unit 1 test