Dictionaries

A list finds things by position: "give me item number 2". A dictionary finds things by name: "give me the age". It stores key-value pairs, so each piece of data has a label.

Creating dictionaries

Use curly braces, with key: value pairs separated by commas:

person = {
    "name": "Ada",
    "age": 36,
    "language": "Python",
}
 
empty = {}
print(len(person))   # 3

Keys are usually strings or numbers, and each key can appear only once. Values can be anything, including lists and other dictionaries.

Accessing and modifying values

Look up a value by putting its key in square brackets:

print(person["name"])   # Ada

print(person.get("email"))             # None
print(person.get("email", "unknown"))  # unknown

.get(key, default) returns the default instead of crashing.

Assigning to a key either updates an existing entry or adds a new one:

person["age"] = 37                # update
person["email"] = "ada@mail.com"  # add
del person["language"]            # delete
print(person)
# {'name': 'Ada', 'age': 37, 'email': 'ada@mail.com'}

Check whether a key exists with in:

print("name" in person)    # True
print("phone" in person)   # False

Dictionary methods

stock = {"apple": 5, "pear": 2}
print(list(stock.keys()))     # ['apple', 'pear']
print(list(stock.values()))   # [5, 2]

Iterating over dictionaries

A plain for loop goes through the keys. Use .items() to get the key and value together, unpacked like a tuple:

stock = {"apple": 5, "pear": 2, "fig": 12}
 
for item in stock:
    print(item)
 
for item, qty in stock.items():
    print(f"{item}: {qty}")

Output of the second loop:

apple: 5
pear: 2
fig: 12

Dictionaries remember the order in which keys were added. If you want them in a different order, loop over sorted(stock) to go through the keys alphabetically or numerically.

A very common pattern: counting

counts = {}
for letter in "banana":
    counts[letter] = counts.get(letter, 0) + 1
 
print(counts)   # {'b': 1, 'a': 3, 'n': 2}

counts.get(letter, 0) returns 0 the first time a letter shows up, so you never need to check whether the key exists first.

Nested dictionaries

Values can be dictionaries (or lists) too, which lets you model richer data:

students = {
    "ada": {"age": 36, "grades": [90, 85]},
    "alan": {"age": 41, "grades": [70, 95, 88]},
}
 
print(students["ada"]["age"])         # 36
print(students["alan"]["grades"][1])  # 95
students["ada"]["grades"].append(100)

Read it from left to right: first get the "ada" dictionary, then get the "age" inside it.

Lists as values are great for grouping. Here we collect words under the first letter of each word:

groups = {}
for word in ["apple", "avocado", "banana"]:
    letter = word[0]
    if letter not in groups:
        groups[letter] = []
    groups[letter].append(word)
 
print(groups)   # {'a': ['apple', 'avocado'], 'b': ['banana']}

Now try it yourself. The next step is three puzzles using dictionaries to count, look up, and group data.