Dictionaries: keys, retrieval and word-frequency tracing

Board scope and conventions

AQA: direct key/value operations and a complete original Python word-frequency example. The activity's ASCII tokenisation is a stated exercise convention, not a universal language-processing rule.

Shared foundations and optional extensions are labelled separately; this page does not claim complete coverage of the qualification.

Use keys deliberately

In Python, counts['kite'] retrieves the value associated with the string key 'kite'. It does not request the second or third stored entry. A list uses positional indices; a dictionary's chosen keys carry meaning.

Use counts['kite'] = 2 to insert a new key or replace its existing value. Use 'kite' in counts to test membership by key and del counts['kite'] to delete it after checking that it exists. Direct retrieval or deletion of an absent key raises KeyError.

Choose a missing-key rule

counts.get(word, 0) returns a stored value or default zero without creating an entry. The later assignment counts[word] = previous + 1 performs the update. A default of zero is appropriate for counting, but not automatically appropriate for a missing account balance or identifier.

Keys are unique within a dictionary. Python keys must be hashable; strings and integers are suitable examples. Dictionary behaviour does not require you to assume every operation scans a sorted list, and average constant-time lookup is not an unconditional worst-case guarantee.

Declare case and punctuation handling

This exercise converts ASCII text to lowercase and treats consecutive a–z letters as words. 'Red', 'red,' and 'RED' therefore contribute to the same key. The original text remains unchanged.

This simple rule splits apostrophes and does not correctly model all languages or names with accents. A different task might preserve punctuation, distinguish case or apply Unicode-aware tokenisation. Choose rules from requirements, not from convenience.

Implement the update in Python

After tokenisation, the same short loop handles both a new and repeated key. Sorting entries for display is a separate presentation step; it does not change the counts.

import re

def frequencies(text):
    counts = {}
    for word in re.findall(r'[a-z]+', text.lower()):
        counts[word] = counts.get(word, 0) + 1
    return counts

print(sorted(frequencies('Red kite, blue kite; red cloud.').items()))
# [('blue', 1), ('cloud', 1), ('kite', 2), ('red', 2)]