Python Project: Expense Tracker

What you are building

You record expenses as pairs: a category and an amount. The program prints a table of every row, a total for each category, the grand total, and the single largest spend.

Work in the Python editor at /try. Leave HTML, C, and C++ for their own tracks. The numbers are listed in the script so Run prints the report at once.

Skills used

• A list of (category, amount) tuples
• A loop that prints a table of rows
• A dict that sums amounts per category
• sum for the grand total
• max with a key to find the largest row

Store rows and print a table

A tuple is enough for one spend: the category does not change, the amount does not change. Keep all rows in one list. Print a header, then each pair, then a count of rows.

expenses = [
    ("Food", 12.50),
    ("Transit", 3.20),
    ("Food", 8.75),
    ("Books", 24.00),
]

print(f"{'category':<12} {'amount':>8}")
for category, amount in expenses:
    print(f"{category:<12} {amount:>8.2f}")
print("Rows:", len(expenses))

Two Food rows stay two rows. Do not merge them yet. The table is the ledger. Totals come next, from the same list.

Totals by category

Walk the list once. For each category, add the amount to a dict. Missing keys start at 0. The grand total is the sum of the amounts, not a second pass over the dict, unless you prefer that.

expenses = [
    ("Food", 12.50),
    ("Transit", 3.20),
    ("Food", 8.75),
    ("Books", 24.00),
]

by_cat = {}
for category, amount in expenses:
    by_cat[category] = by_cat.get(category, 0) + amount

print(by_cat)
print("Grand total:", sum(amount for _, amount in expenses))

Complete program

The full report prints the ledger, the category totals, the largest single row, and the grand total. Run it at /try and check the Books line: 24.00 is the largest spend.

expenses = [
    ("Food", 12.50),
    ("Transit", 3.20),
    ("Food", 8.75),
    ("Books", 24.00),
    ("Transit", 3.20),
    ("Food", 15.10),
]

def print_table(rows):
    print(f"{'category':<12} {'amount':>8}")
    print("-" * 21)
    for category, amount in rows:
        print(f"{category:<12} {amount:>8.2f}")

by_cat = {}
for category, amount in expenses:
    by_cat[category] = by_cat.get(category, 0) + amount

largest = max(expenses, key=lambda row: row[1])
grand = sum(amount for _, amount in expenses)

print("Expense tracker")
print()
print_table(expenses)
print()
print("Totals by category")
print(f"{'category':<12} {'total':>8}")
print("-" * 21)
for category in sorted(by_cat):
    print(f"{category:<12} {by_cat[category]:>8.2f}")
print()
print("Largest spend:", largest[0], f"{largest[1]:.2f}")
print("Grand total:  ", f"{grand:.2f}")
print("Row count:    ", len(expenses))

Food is 12.50 + 8.75 + 15.10. Transit is 3.20 twice. Sorted category names keep the second table stable from run to run.

Common mistakes

• Using max(expenses) without a key. Tuples compare the category string first, so Books would not always win.
• Rounding each row before you add. Round when you print. Sum the original amounts.
• Updating by_cat with = instead of +. The second Food row would replace the first instead of adding to it.
• Printing money with print(amount) only. 8.75 is fine; 3.2 looks better as 3.20 with .2f.
• Mixing currencies in one list with no tag. This project assumes one unit, such as dollars.

How to extend / Practice tasks

Keep the six demo rows until the 66.75 total matches.

1. Add ("Food", 4.15) and confirm the Food total becomes 40.50 and the grand total becomes 70.90.
2. Print the category with the highest total, not only the highest single row. Usemax(by_cat, key=by_cat.get).
3. Filter the ledger to one category before printing: a helperonly(rows, name) that returns matching tuples, then print Food only.