trjhtr

For coding practice / interview exercises, I'd like to know if there's an optimizaton I can make to the following, where I "clean" a given word to remove punctuation or other characters that are not within "a" to "z".

There are some great answers here to remove punctuation from a string, so my question today is not the best way how to do this, but instead whether there is an optimization I can make to my 3 lines of code below in the word_count_engine function? Can I do this in 1 or 2 lines or make the code more efficient so it doesn't loop over the list twice (i.e. with 2 list comprehensions)?

def clean(word):
 returnword = ""
 for letter in word.lower():
 if letter >= 'a' and letter <='z':
 # not out of bounds
 returnword += letter
 return returnword


def word_count_engine(document):

 words = document.split() # if there are extra spaces, split() still filters empty words out FYI
 words = [clean(word) for word in words] # a word like "$33!" will result in an empty string though
 words = [word for word in words if word] # so filter out empty strings and get the final list of clean words

document = "Practice makes perfect. you'll only get Perfect by practice. just practice! $544 test"

Since Python strings are immutable, appending one character at a time using += is inefficient. You end up allocating a new string, copying all of the old string, then writing one character.

Instead, clean() should be written like this:

def clean(word):
 return ''.join(letter for letter in word.lower() if 'a' <= letter <= 'z')

Note that Python supports double-ended inequalities.

The name of your word_count_engine function poorly describes what it does. In fact, the function doesn't print or return anything, so it's all dead code. If I had to rewrite it, though, I'd say:

words = [word for word in map(clean, document.split()) if word]

Also consider replacing all of this code with a simple regular expression substitution.