Dictionary Examples

Note

Source: Adapted from the C# edition (dictionaries/dictionaryexamples.rst). The word-count example is a direct Python translation of the C# count_words program. collections.Counter is a Python-specific addition with no C# equivalent; it is introduced as a modern convenience.

Word Count

Counting how often each word appears in a text is a classic dictionary application. We process a string word by word, updating a count for each word seen:

def word_count(text):
    counts = {}
    for word in text.lower().split():
        word = word.strip(".,!?;:\"'")   # remove punctuation
        if word:
            counts[word] = counts.get(word, 0) + 1
    return counts

text = "four score and seven years ago our fathers four score"
counts = word_count(text)
for word in sorted(counts):
    print(f"{word:10s}  {counts[word]}")

Output:

ago            1
and            1
fathers        1
four           2
our            1
score          2
seven          1
years          1

counts.get(word, 0) returns the current count or 0 if the word has not been seen yet — a cleaner alternative to checking word in counts first.

Inverting a Dictionary

To build a reverse lookup (value → key) from a one-to-one dictionary:

e2sp = {"one": "uno", "two": "dos", "three": "tres"}
sp2e = {v: k for k, v in e2sp.items()}
print(sp2e)

Output:

{'uno': 'one', 'dos': 'two', 'tres': 'three'}

This dict comprehension is the dictionary analogue of a list comprehension.

Using collections.Counter

Python’s collections.Counter is a specialised dict subclass that counts hashable objects automatically. It is cleaner than writing the accumulation loop by hand:

from collections import Counter

text = "four score and seven years ago our fathers four score"
counts = Counter(text.lower().split())
print(counts.most_common(3))

Output:

[('four', 2), ('score', 2), ('ago', 1)]

most_common(n) returns the n most frequent elements as a list of (element, count) tuples.

Grouping Data

A dictionary whose values are lists is a natural way to group items:

words = ["apple", "ant", "bear", "bee", "cat"]
by_letter = {}
for word in words:
    letter = word[0]
    if letter not in by_letter:
        by_letter[letter] = []
    by_letter[letter].append(word)

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

Output:

a: ['apple', 'ant']
b: ['bear', 'bee']
c: ['cat']

dict.setdefault(key, default) can simplify the inner if:

by_letter.setdefault(letter, []).append(word)