Python program to print even length words in a string

In Python, you can find and print all words in a string that have even length (divisible by 2). This is useful for text processing tasks where you need to filter words based on their character count.

For example, in the string "A big cube was found inside the box", the words "cube" (length 4) and "inside" (length 6) have even lengths.

Basic Approach

The simplest method involves splitting the string into words and checking each word's length ?

def print_even_length_words(text):
    # Split the string into individual words
    words = text.split()
    
    for word in words:
        # Check if word length is even
        if len(word) % 2 == 0:
            print(word)

# Example usage
sample = "this is a test string"
print_even_length_words(sample)

this
is
test
string

Using filter() with Lambda

Python's filter() function provides a more functional approach ?

def print_even_length_words(text):
    words = text.split()
    even_words = filter(lambda x: len(x) % 2 == 0, words)
    
    for word in even_words:
        print(word)

# Example usage
print_even_length_words("this is a test string")

this
is
test
string

Using List Comprehension

List comprehension offers a concise Python-style solution ?

def print_even_length_words(text):
    even_words = [word for word in text.split() if len(word) % 2 == 0]
    
    for word in even_words:
        print(word)

# Example usage
print_even_length_words("this is a test string")

this
is
test
string

One-liner Approach

You can combine unpacking with print() for a single line solution ?

def print_even_length_words(text):
    print(*[word for word in text.split() if len(word) % 2 == 0], sep='\n')

# Example usage
print_even_length_words("this is a test string")

this
is
test
string

Handling Punctuation

When dealing with sentences containing punctuation, you may need to clean the text first ?

def print_even_length_words(text):
    # Remove common punctuation
    cleaned_text = text.replace('.', '').replace(',', '').replace('!', '').replace('?', '')
    words = cleaned_text.split()
    
    for word in words:
        if len(word) % 2 == 0:
            print(word)

# Example with punctuation
sample = "Hello world! This is a test sentence."
print_even_length_words(sample)

world
This
is
test

Returning Results as List

Sometimes you might want to return the results instead of printing them ?

def get_even_length_words(text):
    return [word for word in text.split() if len(word) % 2 == 0]

# Example usage
text = "Python programming is fun and educational"
even_words = get_even_length_words(text)
print("Even length words:", even_words)
print("Count:", len(even_words))

Even length words: ['programming', 'is', 'fun', 'and']
Count: 4

Comparison of Methods

Conclusion

Use list comprehension for the most Pythonic approach to filter even-length words. For simple cases, a basic loop is perfectly readable and efficient. Remember to handle punctuation when processing real-world text data.