Python Project: To-Do List

What you are building

You keep a list of task titles as strings. New work is appended. Finishing a task moves it to a second list with pop. Deleting a task also uses pop, but the title is not kept. At the end you print what is still open.

Paste the examples into /try and run them. That editor is Python. It will not render HTML and it will not compile C or C++. The demo never calls input(), so the full script prints in one click.

Skills used

• A list of strings for open tasks
• append to add at the end
• pop(index) to remove by position
• enumerate so printed lines show a number the user can quote
• A second list for completed titles

Add and show

Start empty. Append three titles. Print each with its index. Indexes start at 0, the same numberpop will use later.

todo = []
todo.append("Write the report")
todo.append("Call the lab")
todo.append("Buy milk")

for i, title in enumerate(todo):
    print(i, title)

print("Count:", len(todo))

append never overwrites an earlier item. That is the point of a list: the old titles stay, the new one lands at the end.

Complete and delete by index

Completing is a move: take the title out of todo and put it on done. Deleting is a drop: take it out and ignore it. Both use pop. After a pop, later indexes shift down by one.

todo = ["Write the report", "Call the lab", "Buy milk"]
done = []

title = todo.pop(1)
done.append(title)
print("Completed:", title)
print("Open:", todo)
print("Done:", done)

removed = todo.pop(1)
print("Deleted:", removed)
print("Open now:", todo)

Complete program

Helpers wrap add, show, complete, and delete. A short script of calls stands in for a menu. Run it at/try and read the remaining list at the bottom.

def show(label, items):
    print(label)
    if not items:
        print("  (none)")
        return
    for i, title in enumerate(items):
        print(f"  {i}: {title}")

def add(todo, title):
    todo.append(title)
    print("Added:", title)

def complete(todo, done, index):
    if index < 0 or index >= len(todo):
        print("No task at index", index)
        return
    title = todo.pop(index)
    done.append(title)
    print("Completed:", title)

def delete(todo, index):
    if index < 0 or index >= len(todo):
        print("No task at index", index)
        return
    title = todo.pop(index)
    print("Deleted:", title)

todo = []
done = []

add(todo, "Write the report")
add(todo, "Call the lab")
add(todo, "Buy milk")
print()
show("Open tasks", todo)

print()
complete(todo, done, 1)
print()
show("Open tasks", todo)

print()
delete(todo, 1)
print()
show("Open tasks", todo)
show("Completed", done)

print()
print("Remaining:")
for title in todo:
    print("-", title)

Walk the log: three adds, complete index 1 (Call the lab), then delete the new index 1 (Buy milk). Remaining is Write the report. Completed holds Call the lab.

Common mistakes

• Using remove(title) when two tasks share a name. remove drops the first match only. Indexes are exact.
• Calling pop() with no argument. That removes the last item, not the one you numbered.
• Printing 1-based numbers and then popping that same number. If you show 1, 2, 3, you mustpop(n - 1).
• Completing with todo[index] = "done". The title is then lost and you cannot list finished work.
• Forgetting the bounds check. pop(5) on a short list raises IndexError and the rest of the script never runs.

How to extend / Practice tasks

Three changes you can make in the complete program.

1. Add a fourth task "Email the draft" and complete index 0 first. Confirm the remaining order is Call the lab, Buy milk, Email the draft.
2. Print a 1-based menu (1:, 2:) but keep pop zero-based. Convert withindex = choice - 1 in complete and delete.
3. Refuse empty titles: if title.strip() is empty, print a message and skipappend. Test it by calling add(todo, " ").