Dictionaries

Python 3.12 · ✓ verified by execution on 2026-07-23

Unlike sequences, which are indexed by a range of numbers, dictionaries are indexed by keys, which can be any immutable type

tel = {'jack': 4098, 'sape': 4139}
tel['guido'] = 4127
print(tel['jack'])

Output

Your output

Dictionary Constructors

The dict() constructor builds dictionaries directly from sequences of key-value pairs:

print(dict([('sape', 4139), ('guido', 4127)]))

Output

{'sape': 4139, 'guido': 4127}

Your output

Deleting Items

It is also possible to delete a key:value pair with del.

tel = {'jack': 4098, 'sape': 4139}
del tel['sape']
print(list(tel))

Output

['jack']

Your output

Missing Keys

It is an error to extract a value using a non-existent key.

tel = {'jack': 4098}
try:
    print(tel['guido'])
except KeyError as e:
    print(repr(e))

Output

KeyError('guido')

Your output

To prevent this error, you can use the .get() method.

tel = {'jack': 4098}
print(tel.get('guido', 'Not Found'))

Output

Not Found

Your output

Dictionary Iteration

When looping through dictionaries, the key and corresponding value can be retrieved at the same time using the items() method.

knights = {'gallahad': 'the pure', 'robin': 'the brave'}
for k, v in knights.items():
    print(k, v)

Check yourself

Lists can be used as keys in a dictionary.

Reveal answer

Lists are mutable and therefore cannot be used as dictionary keys. Attempting to do so raises a TypeError. — Lists are mutable and therefore cannot be used as dictionary keys. Attempting to do so raises a TypeError.

Iterating over a dictionary directly with 'for item in my_dict' returns the values.

Reveal answer

Iterating over a dictionary directly yields its keys, not its values. — Iterating over a dictionary directly yields its keys, not its values.

Dictionaries in Python 3.12 are unordered.

Reveal answer

Since Python 3.7, dictionaries formally preserve the insertion order of their keys. — Since Python 3.7, dictionaries formally preserve the insertion order of their keys.

Challenges

Challenge 1 +15 XP

Write a function count_words(words) that takes a list of strings and returns a dictionary where keys are words and values are the number of times they appear.

Test 1 — expects "{'apple': 2, 'banana': 1} \n"

Need a hint? (−25% XP)

Use .get(word, 0) to fetch the current count or 0 if it doesn't exist, then add 1.

Show solution (0 XP)

def count_words(words):
    counts = {}
    for word in words:
        counts[word] = counts.get(word, 0) + 1
    return counts
print(count_words(['apple', 'banana', 'apple']))

Challenge 2 +15 XP

Write a function merge_profiles(p1, p2) that takes two dictionaries and returns a new dictionary combining both. If keys overlap, p2 should win.

Test 1 — expects "{'name': 'jack', 'age': 40} \n"

Need a hint? (−25% XP)

Copy the first dictionary, then .update() it with the second.

Show solution (0 XP)

def merge_profiles(p1, p2):
    merged = p1.copy()
    merged.update(p2)
    return merged
print(merge_profiles({'name': 'jack'}, {'age': 40}))