Python text search question [closed]

Question

I wonder, if you open a text file in Python. And then you'd like to search of words containing a number of letters.

Say you type in 6 different letters (a,b,c,d,e,f) you want to search. You'd like to find words matching at least 3 letters. Each letter can only appear once in a word. And the letter 'a' always has to be containing.

How should the code look like for this specific kind of search?

Answer 1

Let's see...

return [x for x in document.split()
        if 'a' in x and sum((1 if y in 'abcdef' else 0 for y in x)) >= 3]

split with no parameters acts as a "words" function, splitting on any whitespace and removing words that contain no characters. Then you check if the letter 'a' is in the word. If 'a' is in the word, you use a generator expression that goes over every letter in the word. If the letter is inside of the string of available letters, then it returns a 1 which contributes to the sum. Otherwise, it returns 0. Then if the sum is 3 or greater, it keeps it. A generator is used instead of a list comprehension because sum will accept anything iterable and it stops a temporary list from having to be created (less memory overhead).

It doesn't have the best access times because of the use of in (which on a string should have an O(n) time), but that generally isn't a very big problem unless the data sets are huge. You can optimize that a bit to pack the string into a set and the constant 'abcdef' can easily be a set. I just didn't want to ruin the nice one liner.

EDIT: Oh, and to improve time on the if portion (which is where the inefficiencies are), you could separate it out into a function that iterates over the string once and returns True if the conditions are met. I would have done this, but it ruined my one liner.

EDIT 2: I didn't see the "must have 3 different characters" part. You can't do that in a one liner. You can just take the if portion out into a function.

def is_valid(word, chars):
    count = 0
    for x in word:
        if x in chars:
            count += 1
            chars.remove(x)
    return count >= 3 and 'a' not in chars

def parse_document(document):
    return [x for x in document.split() if is_valid(x, set('abcdef'))]

This one shouldn't have any performance problems on real world data sets.