Example: Word Frequencies

Now that we have seen dictionary syntax let's put it to work and solve the problem of counting the frequencies of the words in a string. The input will be a string, and the desired output is a display of each word and the number of times it occurred in the string. We'll use a dictionary for our counters where each key-value pair in the dictionary will consist of a word and the number of times it occurred.

Pseudocode

The pseudocode description might be,

get the string
break the string into a list of words
initialize a dictionary of counters
loop through the list of words
    if the word is in the dictionary
        increment its counter by 1
    otherwise
        set its counter to 1
loop through the entries in the dictionary
    for each entry display the key (the word) and the value (the count)

Python

A straight translation into Python gives,

# WordFrequencies.py
# Get the string:
test_string = '''The programmer, who needs clarity, who must talk all day to
    a machine that demands declarations, hunkers down into a low-grade annoyance.
    It is here that the stereotype of the programmer,
    sitting in a dim room, growling from behind Coke cans, has its origins.
    The disorder of the desk, the floor; the yellow Post-It notes everywhere;
    the whiteboards covered with scrawl:
    all this is the outward manifestation of the messiness of human thought.
    The messiness cannot go into the program;
    it piles up around the programmer.
    ~ Ellen Ullman
'''

# Break the string into a list of words:
words = test_string.split()
# Initialize a dictionary of counters:
word_counts = {}

# Loop through the list of words:
for word in words:
    # If the word is in the dictionary:
    if word in word_counts:
        # Increment its counter by 1:
        word_counts[word] = word_counts[word] + 1
    # Otherwise:
    else:
        # Set its counter to 1:
        word_counts[word] = 1

# Loop through the entries in the dictionary:
for word in word_counts:
    # Display the key value pair:
    print(word, ':', word_counts[word])

Refinements

When you run it and examine the output you will see that while this is a useful core there are some erroneous entries in the output. Among them:

The : 3
floor; : 1
annoyance. : 1
the : 10
~ : 1

The problems here are that,

• 'The' and 'the' are the same word and so their counts should be combined.
• The trailing punctuation on 'floor;' and 'annoyance.' should be removed since it is not part of the word, and could mislead counts since 'floor;' and 'floor,' would be separate entries.
• '\~' is not a word at all and so should not be counted.

Application of some string methods can fix some of these problems,

• We can use the lower method to convert strings to all lower case.
• We can use strip to remove trailing (and leading) punctuation characters.
• We can use isalpha to test that the string is all letters (and thus eliminate '\~').

The body of the loop then becomes,

for word in words:
    w = word.lower()
    w = w.strip('.,;:\'"?!()') # Notice the \ to escape the ' inside the string
    if w.isalpha():
        if w in word_counts:
            word_counts[w] = word_counts[w] + 1
        else:
            word_counts[w] = 1

Of course isalpha is too blunt an instrument since we now lose the hyphenated words, but we could deal with that too (perhaps by writing our own little function to check that a string contains only letters and hyphens!), though it is a matter of details we will not pursue.