Dictionaries

Look it up by name

A dictionarydictionaryA collection that finds each value by its key, written {key: value, ...}. Each key appears only once. (or dict) pairs each key with a value. You write it in curly brackets, with a colon between each key and its value:

contacts = {
    "Abel": "0900 000 001",
    "Hana": "0900 000 002",
    "Selam": "0900 000 003",
}
print(contacts["Hana"])
print(len(contacts))

contacts["Hana"] looks like indexing a list, but the thing in the brackets is a key, not a position. Change it to look up Selam.

The same square brackets add a new pair, or change the value of a key that's already there. del removes a pair:

contacts = {"Abel": "0900 000 001", "Hana": "0900 000 002", "Selam": "0900 000 003"}
contacts["Dawit"] = "0900 000 004"
contacts["Hana"] = "0900 000 222"
del contacts["Selam"]
print(contacts)

A key can only appear once, so giving "Hana" a new number replaced the old one. Notice the order: a dict keeps its pairs in the order they were added, and Dawit, added last, comes last.

When the key isn't there

Looking up a key that isn't in the dict stops the program:

in asks about keys. Predict this:

cups = {"abol": 1, "tona": 2, "bereka": 3}
print("tona" in cups)
print(2 in cups)

True
False

The other way is get(). It returns the value if the key is there, and None if it isn't. Give it a second value and it returns that instead of None:

contacts = {"Abel": "0900 000 001", "Hana": "0900 000 002"}
print(contacts.get("Abel"))
print(contacts.get("Meron"))
print(contacts.get("Meron", "not saved"))

Going through a dict

A for loop over a dict goes through its keys. .values() gives the values, and .items() gives each pair as a tuple, ready to unpack (lesson 4.3):

contacts = {"Abel": "0900 000 001", "Hana": "0900 000 002"}
for name in contacts:
    print(name)
for number in contacts.values():
    print(number)
for name, number in contacts.items():
    print(f"{name}: {number}")

.items() is the one you'll use most. Add a contact and run it again.

Counting with a dict: the tally plan

How many times does each word appear in a list? Keep a dict of counts: the word is the key, its count is the value. For each word, get the count so far (0 if the word is new) and add 1. This is the tally plan:

{'buna': 2, 'shai': 1}

counts is a whole group of counters, one per word. At step 4, "buna" isn't a key yet, so get returns the default 0 and the count becomes 1. At step 8 it's already 1, so it becomes 2. Without the default, counts["buna"] + 1 would be a KeyError the first time. Add "buna" to the list once more and predict the new count before you run it.

words = ["buna", "shai", "buna"]
counts = {}
for word in words:
    counts[word] = counts.get(word, 0) + 1
print(counts)