Dictionaries: Look It Up 🗂️

Keys and values

pirate = {
    "name": "Guybrush",
    "role": "pirate",
    "insults": 7,
    "has_ship": False,
}

print(pirate["name"])
print(pirate["insults"])
print(len(pirate))

Guybrush
7
4

Curly braces, key: value pairs, commas between. You look things up by key, not by position, which is why a dictionary stays readable when a tuple of eight fields does not.

Adding, changing, removing

pirate = {"name": "Guybrush", "insults": 7}

pirate["ship"] = "Sea Monkey"      # add
pirate["insults"] += 1             # change
print(pirate)

del pirate["ship"]                 # remove
print(pirate)

removed = pirate.pop("insults")    # remove and give back
print(removed, pirate)

{'name': 'Guybrush', 'insults': 8, 'ship': 'Sea Monkey'}
{'name': 'Guybrush', 'insults': 8}
8 {'name': 'Guybrush'}

Assigning to a key that does not exist creates it. Assigning to one that does replaces it. There is no separate "add" and "update", which is one less thing to remember.

The missing key problem

pirate = {"name": "Guybrush"}
print(pirate["ship"])

KeyError: 'ship'

Three ways to handle it, in order of how often you want them:

pirate = {"name": "Guybrush"}

print(pirate.get("ship"))                     # None instead of an explosion
print(pirate.get("ship", "no ship yet"))      # your own default
print("ship" in pirate)                       # just ask

if "name" in pirate:
    print(f"Captain {pirate['name']}")

None
no ship yet
False
Captain Guybrush

Looping

scores = {"Guybrush": 95, "Elaine": 88, "Otis": 72}

for name in scores:
    print(name)

print("---")

for name, score in scores.items():
    print(f"{name:10} {score}")

print("---")

print(list(scores.keys()))
print(list(scores.values()))
print(sum(scores.values()))

Guybrush
Elaine
Otis
---
Guybrush   95
Elaine     88
Otis       72
---
['Guybrush', 'Elaine', 'Otis']
[95, 88, 72]
255

Looping over a dictionary gives you the keys. Almost always you want .items(), which gives you both as a tuple that the for line unpacks for you.

Sorting a dictionary

scores = {"Guybrush": 95, "Elaine": 88, "Otis": 72, "Meathook": 91}

for name, score in sorted(scores.items(), key=lambda pair: pair[1], reverse=True):
    print(f"{score:3}  {name}")

 95  Guybrush
 91  Meathook
 88  Elaine
 72  Otis

lambda pair: pair[1] is a tiny throwaway function meaning "the second part of each pair", so we sort by score rather than by name. Lambdas get a proper treatment in Lesson 37; for now, read it as "sort by this bit".

Counting things: the classic use

text = "the rubber chicken with a pulley in the middle"

counts = {}
for word in text.split():
    counts[word] = counts.get(word, 0) + 1

for word, n in counts.items():
    if n > 1:
        print(f"{word}: {n}")

the: 2

counts.get(word, 0) + 1 is the counting idiom: "whatever it was, or zero if new, plus one". Memorise it. The standard library also has a purpose-built tool:

from collections import Counter

text = "the rubber chicken with a pulley in the middle"
counts = Counter(text.split())

print(counts.most_common(3))
print(counts["the"])
print(counts["banana"])      # missing keys are 0, not an error

[('the', 2), ('rubber', 1), ('chicken', 1)]
2
0

What can be a key?

valid = {
    "text": 1,
    42: 2,
    3.5: 3,
    True: 4,
    ("x", "y"): 5,       # tuples are fine: they cannot change
}
print(valid[("x", "y")])

# {["x", "y"]: 5}       # TypeError: unhashable type: 'list' 

5

Keys must be immutable. The reason is mechanical: a dictionary finds things instantly by computing a number from the key (a hash) and using it as an address. If the key could change afterwards, the address would be wrong and the value would be lost. Lists can change, so lists cannot be keys. Tuples cannot, so they can.

Nested dictionaries: the shape of real data

game = {
    "title": "The Secret of Monkey Island",
    "year": 1990,
    "characters": {
        "Guybrush": {"role": "hero", "insults": 8},
        "LeChuck": {"role": "villain", "insults": 3},
    },
    "islands": ["Melee", "Monkey"],
}

print(game["title"])
print(game["characters"]["Guybrush"]["insults"])
print(game["islands"][0])

for name, info in game["characters"].items():
    print(f"{name:10} {info['role']:8} {info['insults']} insults")

The Secret of Monkey Island
8
Melee
Guybrush   hero     8 insults
LeChuck    villain  3 insults

Dictionaries containing dictionaries containing lists is exactly the shape of JSON, which is how essentially every web API on earth sends data. When you call an AI model in Level 6, this is what comes back. Get comfortable here and Level 5 becomes easy.

book = {
    "Elaine": "555-0100",
    "Guybrush": "555-0199",
    "Otis": "555-0110",
}

print(book.get("Elaine"))
print(book.get("LeChuck", "not in the book"))

book["Meathook"] = "555-0123"

for name in sorted(book):
    print(f"{name:10} {book[name]}")

555-0100
not in the book
Elaine     555-0100
Guybrush   555-0199
Meathook   555-0123
Otis       555-0110

word = "Mississippi"

counts = {}
for letter in word.lower():
    counts[letter] = counts.get(letter, 0) + 1

for letter in sorted(counts):
    print(f"{letter}: {counts[letter]}")

i: 4
m: 1
p: 2
s: 4

original = {"a": 1, "b": 2, "c": 3}

flipped = {}
for key, value in original.items():
    flipped[value] = key

print(flipped)

# and the catch
clash = {"a": 1, "b": 1}
flipped_clash = {}
for key, value in clash.items():
    flipped_clash[value] = key
print(flipped_clash)

{1: 'a', 2: 'b', 3: 'c'}
{1: 'b'}