cs2370 Notes: 22 Dictionaries

New data type: Dictionaries #

legs = {
    'dog': 4,
    'chicken': 2,
    'cat': 4,
    'spider': 8,
}

print(legs['dog'])

legs.keys())
legs.values()
'dog' in legs
'giraffe' in legs
legs.get('giraffe', 4)
legs.get('lobster', 4)

Note: Dictionary keys may print in insertion order, but generally it’s best to think of dictionaries as being unordered.

Not a new capability:

legs_list = [
  ('dog', 4),
  ('chicken', 2),
  ('cat', 4),
  ('spider', 8),
]

def get(xs, key):
    for (kk, vv) in xs:
        if kk == key:
            return vv
    raise Exception("key not found")

print(get(legs_list, 'dog'))

How many steps does it take to look up a key in the list?

Dictionaries give us two benefits over association lists:

• Nicer syntax
• Faster (can lookup in 1 step rather than having to loop)

Standard Patterns #

def dict1(dd: dict[int, int], ...) -> None:
   dd[key]
   ...

def process(noises: dict[str, str]):
    for key in noises.keys():
        ... noises[key] ...

def produce() -> dict[str, str]:
    info = {}
    info['color'] = 'blue'
    return info

Repeats #

Design a method that takes a list of numbers and returns a list of how many times the number in that position appeared in the lsit.

Example: [7, 7, 1, 7, 2, 2, 7, 1, 3] -> [4, 4, 2, 4, 2, 2, 4, 2, 1]

• version 1: nested loop
• version 2: two loops and a dictionary

How many times through a loop in each case?

Next example:

• Calculate a recipt total for a list of (item, count, price each) tuples.