Python Program to Count the Frequency of Words in a String

Counts how many times each word appears in a sentence, with a dictionary that has the words as keys and their counts as values.

Word frequency is how many times each word appears in a piece of text. In the tongue twister "red lorry yellow lorry red lorry yellow lorry", lorry appears 4 times, and red and yellow twice each. A dictionary suits the job: each word is a key, and its value is how many times it has been seen so far.

Program

def word_frequency(text):
    """How many times each word appears in text, as a dict."""
    counts = {}
    for word in text.lower().split():
        counts[word] = counts.get(word, 0) + 1
    return counts


sentence = "Red lorry yellow lorry red lorry yellow lorry"
print("Word counts:", word_frequency(sentence))

Open in the editor Download word_frequency.py

Output

Word counts: {'red': 2, 'lorry': 4, 'yellow': 2}

How it works

Visualize Step through the program line by line: each step below shows up as its line runs.

1. counts starts as an empty dictionary.
2. lower() makes Red and red the same word, and split() breaks the text into a list of words at the spaces. The loop takes the words one by one.
3. counts.get(word, 0) is the word's count so far, or 0 when the word hasn't been seen yet, where counts[word] would stop with a KeyError (dictionaries). Adding 1 and storing it back counts this one.
4. The function returns the finished dictionary.
5. The program counts the words of the tongue twister and prints the dictionary.

The first five words:

The last three words take lorry to 4 and yellow to 2. A dictionary keeps its keys in the order they were added, so the words print in the order they first appear.

Other ways to write it

With collections.Counter

Counter from the collections module counts a list in one step. It's a dictionary with extras, such as most_common(), which lists the words from the most frequent down.

from collections import Counter

def word_frequency(text):
    return Counter(text.lower().split())

counts = word_frequency("the cat and the hat and the bat")
print(counts.most_common(2))

With an if statement

Without get(), the loop checks whether the word is already a key: if it is, its count goes up by 1; if not, it starts at 1.

def word_frequency(text):
    counts = {}
    for word in text.lower().split():
        if word in counts:
            counts[word] += 1
        else:
            counts[word] = 1
    return counts

print(word_frequency("To be or not to be"))

Try it yourself

• Add punctuation, as in "Red lorry, yellow lorry.": lorry, and lorry. count as different words. Remove it with word.strip(".,!?") before counting.
• Print each word and its count on a line of its own, with a for loop over counts.items().
• Find the most frequent word with max(counts, key=counts.get).
• Count letters instead of words, by looping over the text itself instead of text.split().

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