2.3 Dictionaries

Big ideas

• A dictionary maps keys to values: ages = {"Ana": 19, "Ben": 20}. Look up with ages["Ana"], add or change with ages["Cy"] = 18, remove with del ages["Ben"] or pop.
• A missing key raises KeyError; ages.get("Zed", 0) returns a default instead. "Ana" in ages tests keys, not values.
• Loop over ages (keys), ages.values() or ages.items() for key–value pairs. sorted(ages) gives the keys in order.
• The counting pattern: counts[word] = counts.get(word, 0) + 1. Grouping: a dictionary whose values are lists.
• Keys must be immutable (strings, numbers, tuples). Values can be anything, including lists and other dictionaries.

Code you should know

ages = {"Ana": 19, "Ben": 20}
ages["Cy"] = 18
ages["Ana"] += 1
print(ages["Ana"], len(ages), "Ben" in ages, 20 in ages)
print(ages.get("Zed", "unknown"))
for name in sorted(ages):
    print(name, ages[name])
for name, age in ages.items():
    if age >= 20:
        print(name, "is 20+")

20 3 True False
unknown
Ana 20
Ben 20
Cy 18
Ana is 20+
Ben is 20+

text = "the cat and the hat and the bat"
counts = {}
for word in text.split():
    counts[word] = counts.get(word, 0) + 1
print(counts)
best = max(counts, key=counts.get)
print(best, counts[best])

{'the': 3, 'cat': 1, 'and': 2, 'hat': 1, 'bat': 1}
the 3

pairs = [("math", "Ana"), ("cs", "Ben"), ("math", "Cy")]
by_class = {}
for subject, name in pairs:
    if subject not in by_class:
        by_class[subject] = []
    by_class[subject].append(name)
print(by_class)
print(by_class["math"][1])

{'math': ['Ana', 'Cy'], 'cs': ['Ben']}
Cy

Worked example

d = {"a": 1, "b": 2}
d["c"] = d["a"] + d["b"]
d["a"] = 10
print(len(d), d["c"], sorted(d.values()))

Common mistakes

• d[key] on a missing key crashes; use get or check with in first.
• Assuming in looks at values.
• Trying to use a list as a key (TypeError: unhashable).
• Looping with for k, v in d: instead of d.items().