Python essentials for data: records, lists and dictionaries

A quick recap

You already know variables, if, for loops and functions from the Python course. Two built-in types matter most for data. A list holds items in order, written in square brackets: [250, 900, 120]. A dictionary (dict) holds pairs of key and value, written in curly brackets: {"city": "Pune", "amount": 250}. You read a value by its key: row["city"] gives "Pune".

A table as a list of dictionaries

Why combine them? Because a table is a list of rows, and each row is a set of named values. So a list of dicts is a table: each dict is a row (also called a record), and the keys are the column names. This is the same shape you get from a CSV file or a JSON API, which you will meet in the next topic.

orders = [
    {"id": 1, "city": "Pune", "amount": 250},
    {"id": 2, "city": "Delhi", "amount": 900},
    {"id": 3, "city": "Pune", "amount": 120},
    {"id": 4, "city": "Chennai", "amount": 640},
]
print(orders[1]["city"])   # Delhi
print(len(orders))         # 4

List comprehensions: filter and transform

A list comprehension builds a new list from an old one in a single line. The pattern is [what_to_keep for item in items if condition]. The if part is optional. It is shorter than a for loop with append, and it does not change the original list.

big = [o for o in orders if o["amount"] >= 500]
print([o["id"] for o in big])          # [2, 4]

amounts = [o["amount"] for o in orders]
print(amounts)                         # [250, 900, 120, 640]

with_gst = [round(a * 1.18, 2) for a in amounts]
print(with_gst)                        # [295.0, 1062.0, 141.6, 755.2]

A dict comprehension works the same way with curly brackets: {o["id"]: o["amount"] for o in orders} gives {1: 250, 2: 900, 3: 120, 4: 640}, a quick lookup from id to amount.

Sorting by a column

sorted() returns a new sorted list. Python cannot guess how to compare two dicts, so you pass key: a small function that takes one row and returns the value to sort by. lambda makes such a one-line function without a name: lambda o: o["amount"] means "given o, return its amount". reverse=True sorts from largest to smallest.

by_amount = sorted(orders, key=lambda o: o["amount"], reverse=True)
print([o["id"] for o in by_amount])   # [2, 4, 1, 3]

print(max(orders, key=lambda o: o["amount"])["city"])   # Delhi
print(min(orders, key=lambda o: o["amount"])["id"])     # 3

Python's sort is stable: rows with equal keys stay in their original order. That lets you sort by two columns at once with a tuple key, such as key=lambda o: (o["city"], -o["amount"]) for city A to Z and, inside each city, the biggest amount first.

zip and enumerate

zip walks over two or more lists side by side, pairing the first items, then the second items, and so on. enumerate gives you a counter next to each item, which is handy for ranks and row numbers.

names = ["Asha", "Ravi", "Meena"]
marks = [78, 91, 65]
print(dict(zip(names, marks)))   # {'Asha': 78, 'Ravi': 91, 'Meena': 65}

for rank, o in enumerate(by_amount, start=1):
    print(rank, o["city"], o["amount"])
# 1 Delhi 900
# 2 Chennai 640
# 3 Pune 250
# 4 Pune 120

Common mistakes

• Writing rows = rows.sort(...). list.sort() sorts in place and returns None, so rows becomes None. Fix: use rows.sort(...) on its own line, or new_rows = sorted(rows, ...).
• A KeyError when a row is missing a column. Fix: use row.get("city"), which gives None instead of crashing, or row.get("city", "Unknown") for a default.
• Sorting numbers stored as text: sorted(["100", "20", "3"]) gives ['100', '20', '3'] because text is compared letter by letter. Fix: convert with int() or float() first, or use key=int.
• Changing a list while looping over it (removing rows inside the loop). Fix: build a new list with a comprehension instead.