A Row Is Just a Dictionary

A record is one thing you know several facts about: a student, an order, a sensor reading. In Python it is almost always a dictionary, because a dictionary lets you name the facts instead of remembering their positions.

student = {"name": "Amara", "track": "python", "chapters": 18} # The same thing as a tuple. Technically fine, practically hostile:student = ("Amara", "python", 18)

Both hold three values. Only one of them still makes sense in six months, when student[2] could plausibly be chapters, minutes, or the year they joined. The dictionary version costs a few more characters and buys you the ability to read your own code.

Put those records in a list and you have a table: the list gives you rows in order, each dictionary gives you named columns. There is no table type involved, and none is needed.

students = [    {"name": "Amara", "track": "python", "chapters": 18, "minutes": 164},    {"name": "Ben", "track": "ml", "chapters": 6, "minutes": 71},    {"name": "Chidi", "track": "python", "chapters": 24, "minutes": 231},]

This shape is worth recognising on sight, because it is what almost every API hands back and what almost every CSV reader produces. Learn to work with it directly and a large amount of real data work stops needing a library at all.

The instinct from other languages is to walk the list by index and collect what matches. Python asks the question of each row instead, and the comprehension you already know is the whole mechanism.

students = [    {"name": "Amara", "track": "python", "chapters": 18, "minutes": 164},    {"name": "Ben", "track": "ml", "chapters": 6, "minutes": 71},    {"name": "Chidi", "track": "python", "chapters": 24, "minutes": 231},] finished = [s for s in students if s["chapters"] >= 18] # Two conditions read exactly as they sound:python_finishers = [    s for s in students    if s["track"] == "python" and s["chapters"] >= 18] print([s["name"] for s in finished])print([s["name"] for s in python_finishers])

Try itLower both thresholds from 18 to 5 and run it again.

No index, no counter, and no chance of stopping one row early. What comes back is a new list holding the same dictionaries — not copies of them. Change a field on a filtered row and the original table sees it, which is usually what you want and occasionally a surprise.

A key function receives one whole row and returns the single value to order by. The rows are never taken apart, which is why the same one-liner works no matter how many columns a record has.

students = [    {"name": "Amara", "track": "python", "chapters": 18, "minutes": 164},    {"name": "Ben", "track": "ml", "chapters": 6, "minutes": 71},    {"name": "Chidi", "track": "python", "chapters": 24, "minutes": 231},] by_minutes = sorted(students, key=lambda s: s["minutes"], reverse=True) # Two columns at once: track first, then most chapters within each track.ranked = sorted(students, key=lambda s: (s["track"], -s["chapters"])) print([s["name"] for s in by_minutes])print([s["name"] for s in ranked])

Try itRemove the minus sign in front of s["chapters"] and run it again.

Returning a tuple from the key sorts by the first element, then breaks ties with the second. The minus sign flips just that one field, which is the trick that lets you sort one column ascending and another descending in a single pass.

Pulling one column out and reducing it is two steps written as one line, and the generator expression means the intermediate list is never built.

total = sum(s["minutes"] for s in students)average = total / len(students)longest = max(students, key=lambda s: s["minutes"]) print(longest["name"])# Chidi

This shape works until one of three conditions applies. Know the boundary before you hit it at speed.

The middle case shows the limit: if you keep scanning the whole list to find one student by name, you do not need a bigger library — you need a dictionary keyed by name instead, which the last chapter of this part is about.