Lists: Keeping Many Values Together

Why this matters

So far each variable has held one value. Real programs handle many: every name in a class, every price in a basket, every message in an inbox. A list holds many values in order, in one variable.

The idea

colors = ["red", "green", "blue"]
print(colors[0])        # red   (counting starts at 0)
print(colors[-1])       # blue  (-1 is the last item)
print(len(colors))      # 3

colors.append("pink")   # add to the end
colors.remove("green")  # remove the first "green"
print("red" in colors)  # True
print(colors)           # ['red', 'blue', 'pink']

Square brackets make a list, and [0], [1], … pick items by index (position). .append() and .remove() change the list itself.

One more tool: "milk eggs bread".split() turns text into a list of words: ['milk', 'eggs', 'bread'].

Try it

Make a list of three of your favourite foods. Print the middle one using its index. Then try print(colors[5]) on the list above. You will meet IndexError: list index out of range, which just means “there is no item at that position”.

Your turn

The starter reads a shopping list typed as words separated by spaces and turns it into a list called items. Leave those two lines alone. Then:

1. Add "apples" to the end of the list.
2. If "eggs" is in the list, remove it.
3. Print the first item, the last item and how many items are left, exactly like this:

First: milk
Last: apples
Items: 3

Recap

• A list is an ordered collection written with square brackets; items are found by index, starting at 0.
• .append() adds, .remove() deletes, len() counts, and in checks.
• text.split() makes a list of words from a string.

Common mistakes

• Counting from 1. The first item is [0], not [1]. Going too far gives IndexError.
• Removing something that is not there raises ValueError. Check with in first.
• Expecting append to return the list. It changes the list and returns nothing, so items = items.append("x") makes items equal to None.