Dictionaries and records: give your data fields names

Learning lesson · Reviewed October 2, 2026. Allow additional time for the practice. Prerequisites: lists, conditions, and functions. Python 3; no extra packages.

Position-based data is easy to misread. In ["A01", "Python", 80], you need to remember what each position means. A record such as {"learner": "A01", "course": "Python", "score": 80} makes the fields visible. A list can hold many such records. All examples in this lesson are fictional; identifiers refer to invented learners.

Keys describe fields; values hold observations

record = {"learner": "A01", "course": "Python", "score": 80}
print(record["course"])  # Python
record["score"] = 82
print(record["score"])  # 82
print(record.get("comment"))  # None

Brackets access a required key and raise KeyError if it is absent. get() can return a default for a missing key. That default is a decision, not proof that the missing value was actually observed. For required fields, letting a missing-key error reveal a broken record may be more useful than quietly inventing a replacement.

Define the unit of observation

Here one record is one submitted assessment. A learner may submit more than one assessment. Counting records therefore counts submissions, not necessarily unique people. This distinction becomes important when someone labels a grouped total “learners.” Decide what one row represents before interpreting any summary.

records = [
    {"learner": "A01", "course": "Python", "score": 80},
    {"learner": "A02", "course": "Python", "score": 0},
    {"learner": "A03", "course": "Data", "score": 70},
    {"learner": "A04", "course": "Data", "score": None},
]
groups = {}
missing = 0
for record in records:
    score = record["score"]
    if score is None:
        missing += 1
        continue
    course = record["course"]
    if course not in groups:
        groups[course] = {"total": 0, "count": 0}
    groups[course]["total"] += score
    groups[course]["count"] += 1

for course, stats in groups.items():
    print(course, stats["count"], stats["total"] / stats["count"])
print("Missing:", missing)
# Python 2 40.0
# Data 1 70.0
# Missing: 1

The continue moves to the next record after recording a missing score. Zero remains in the Python group, so its mean is (80 + 0) / 2 = 40. Replacing score is None with if not score would incorrectly skip zero as well. A short condition can hide a large analytical choice.

Inspect what the summary leaves out

The Data group has one valid score and one missing score. Its mean of 70 says nothing about the missing submission. The Python group has two valid scores with a wide difference. Neither number proves that one course is better: these invented samples are tiny, and the assessments may not be comparable. A useful report includes counts and limitations alongside averages.

Practice: count submissions by course

Build a dictionary of submission counts using all four records, including the missing score. Expected result: Python has two submissions and Data has two. Explain why this count differs from the number of valid scores used in each mean.

counts = {}
for record in records:
    course = record["course"]
    counts[course] = counts.get(course, 0) + 1
print(counts)  # {'Python': 2, 'Data': 2}

Before moving on

Be able to say what one record represents, which keys are required, and how zero differs from missing. Next, you will validate text before converting it so that malformed data does not silently enter this grouped calculation.