Given a list of *arbitrary* words, how to remove all words with double letters?

Question

Similar to this post, but I want to perform this task on an arbitrary list or words supplied by an external source, which may change according to user input. For example, I may have:

input = ['annotate','color','october','cellular','wingding','appeasement','sorta']

and the output ought to be

output = ['color','october','wingding','sorta']

Any help would be appreciated!

Answer 1

You can use a regular expression as simple as this. This answer extends the post you mentioned to fulfill your requirements.

import re

arr = ['annotate','color','october','cellular','wingding','appeasement','sorta']
result = [w for w in arr if not re.search(r'(\w)\1', w)]

print(result)

It works for word characters, which includes: a-z, A-Z, 0-9, _.

Answer 2

You can use regular expression to to this.

list(filter(lambda x: not  re.search(r'(\w)\1', x), input))

Answer 3

Use nested loops to check each word for double letters:

This solution may be easier to understand than a RegEx based solution.

words = ['annotate','color','october','cellular','wingding','appeasement','sorta']

for index, word in enumerate(words):
    last = None
    for letter in word:
        if letter == last:
            del words[index]  # Double letters, so delete from the list.
        last = letter
print(words)

Output:

['color', 'october', 'wingding', 'sorta']

This code simply checks each word for double letters, and removes it from the words list if it does have double letters. Note that issues will arise if you name a variable input.

Answer 4

You can compare adjacent characters by zipping a word with itself. This makes for a fairly succinct list comprehension:

words = ['annotate','color','october','cellular','wingding','appeasement','sorta']

[w for w in words if all(a != b for a, b in zip(w, w[1:]))]    
# ['color', 'october', 'wingding', 'sorta']

zip(w, w[1:]) will make tuples of letters like [('a', 'n'),('n', 'n'), ('n', 'o')...] which you can then compare, any that are equal indicate the same letter in a row.

Answer 5

import re

def filter_words(input_list):
    output_list = []
    pattern = r'(.)\1'  # Regex pattern to match consecutive identical characters
    
    for word in input_list:
        if not re.search(pattern, word):
            output_list.append(word)
    
    return output_list

# Example usage:
input = ['annotate', 'color', 'october', 'cellular', 'wingding', 'appeasement', 'sorta']
output = filter_words(input)
print(output)

Output:

['color', 'october', 'wingding', 'sorta']

Answer 6

Still without using Regex, a variation of the answer using nested loops can be constructed using a Python language feature, the use of the else clause in loops.

In Python, loop statements can have an else clause that is executed when the loop finishes by exhausting the iterable but is not executed when the loop is terminated by a break statement. This allow the following algorithm:

input = ['annotate','color','october','cellular','wingding','appeasement','sorta', 'watt']
output = []

for word in input:                            
    for idx, char in enumerate(word[1:]):     # Iterate over the enumeration from the second to the last character of the word
        if word[idx] == char:                 # Check if the char is equal to the previous character in the word 
            break                             # It interrupts the loop and the `else` clause is not executed
    else:
        output.append(word)
      
print(output)
#['color', 'october', 'wingding', 'sorta']

^{Try it online}

• See also Python built-in function enumeration()