After this lesson, you will be able to:

• Choose lists, tuples, sets, and dictionaries for different jobs in one program.
• Remove duplicates while preserving their first-seen order.
• Combine records with dictionary lookups and handle missing information safely.
• Build a multi-step collection report using normalization, filtering, grouping, and a final decision.

Choosing a data structure is a design decision. Use a list when order and repeated values matter; a tuple for a fixed record; a set for uniqueness and membership; and a dictionary for key-based lookup.

A set alone is not an ordered result. When you want the first occurrence of each item in its original order, use a set to remember what you have seen and a list to collect the results. This combines the strengths of both structures.

You can also connect collections: loop through a list of records, use a dictionary to look up details by key, and add one transformed result to a new list. For a multi-step report, separate the work into normalization, validation, counting, and presentation. Return the finished data when writing a reusable function.

No new advanced Python syntax is required here. The challenge is choosing the right representation and combining techniques from this unit and Python Basics for Programmers.

1. Choose collections by purpose

steps = ["build", "test", "build"]
release = ("v3", "eu")
tags = {"api", "worker", "api"}
ports = {"api": 8080, "worker": 9000}
print(len(steps))
print(release[0])
print(len(tags))
print(ports["worker"])

Output:

3
v3
2
9000

Each collection expresses a different need: sequence, fixed record, uniqueness, or lookup.

2. Preserve first-seen order when deduplicating

raw = ["api", "worker", "api", "cache"]
seen = set()
unique = []
for name in raw:
    if name not in seen:
        seen.add(name)
        unique.append(name)
print(unique)

Output:

['api', 'worker', 'cache']

The list retains order; the set remembers which names were already added.

3. Join records with dictionary lookups

jobs = [{"name": "search"}, {"name": "backup"}]
owners = {"search": "Maya"}
labels = []
for job in jobs:
    labels.append(f"{job['name']}: {owners.get(job['name'], 'unassigned')}")
print(labels)

Output:

['search: Maya', 'backup: unassigned']

The dictionary provides an optional lookup for each record.

4. Derive a report from records

results = [("api", "passed"), ("worker", "failed"), ("cache", "passed")]
counts = {"passed": 0, "failed": 0}
for service, state in results:
    if state in counts:
        counts[state] += 1
print(f"Passed: {counts['passed']}")
print(f"Failed: {counts['failed']}")

Output:

Passed: 2
Failed: 1

The fixed count dictionary keeps reporting predictable even when a category has zero events.

1. A set cannot promise the order you want to show in a report.
2. Counting events and counting unique services are different questions.
3. Using if mapping.get(key) to check existence drops valid falsy values.
4. A function should return its final result instead of printing when callers need to reuse it.