Dictionaries inside dictionaries, and grouping

What you will be able to do

After this lesson you can

• store a record of several named parts in a dictionary,
• put a dictionary or a list inside a dictionary,
• look up a value that is deep inside, and
• group items by a key.

A dictionary as a record

A student has a name, an age and some marks. With a list, you would have to remember that position 0 is the name, position 1 the age, and so on. With a dictionary, each part has a name: {"age": 12, "marks": [72, 85, 64]}. This is a record. The code that reads it is easy to read too: student["age"].

A dictionary inside a dictionary

The value of a pair can be another dictionary, or a list. So you can keep one record for each student, with the student's name as the key.

A hint for long expressions: read the name of each key aloud, one by one. "The students, Ravi, the marks, the third mark." If you lose your way, print the middle steps: print(students["Ravi"]).

Grouping: a list as the value

A second very useful pattern: one key, many items. For example, a list of words, grouped by their first letter.

At the end, groups is {"a": ["apple", "avocado", "apricot"], "b": ["banana", "blueberry"], "c": ["cherry"]}. The last loop shows the groups in the order of the letters, with sorted.

Without step 2, the first word of each letter would give a KeyError at the append, because the key and its list do not exist yet.

Each bar is the length of one list. The dictionary itself has one key for each letter seen so far.

Which structure?

When data has several levels, draw it first. A school has classes. A class has students. A student has marks. The structure can be a dictionary of dictionaries of lists. But a structure that is too deep is hard to read. If you find yourself writing four levels of brackets, think about whether a simpler form, or a function that hides the lookup, would be clearer.

Common mistakes

• A KeyError in the middle of a long lookup. Check each level: if name in students:.
• Forgetting to make the empty list the first time. Then append fails.
• Putting the same list object in two places. Remember the lesson on copies: [] for each new key, made fresh.
• Mixing up the levels: students["age"] instead of students["Asha"]["age"].
• Keys with different spelling: "asha" and "Asha" are two records.

Practice

Start with two quick questions. Each New numbers gives you a fresh one, so you can practise as long as you like.

Then the labs and the problems:

1. Warm-up. Open the lab Group by length. You group words by how long they are, and the checker confirms the groups.
2. Build. Open the problem Look inside a record and the problem Group the words.
3. Stretch. Open the problem The highest mark and who has it again, for the nested records of this lesson, where each student has a list of marks.

Remember these

• A value can be a list or a dictionary. A record keeps named parts together.
• Read a deep lookup from left to right, one key at a time.
• Grouping: if the key is new, make an empty list. Then append.
• Counting: the value is a number. Grouping: the value is a list.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. What is {"a": {"b": 5}}["a"]["b"]?

2. Why must we make the empty list the first time we group?

3. How do you add a new student record to students?