string包含细长的单词

Question

My string is: "sooo dear how areeeee youuuuuu"

I want to check if the words in the string are elongated or not.

Elongated means: if the number of characters in the word is repeated more than twice so for example, too is not elongated but tooo is elongated.

>>> import itertools
>>> my_str = 'soooo hiiiii whyyyy done'
>>> print [[g[0], sum(1 for _ in g[1])] for g in itertools.groupby(my_str)]
[['s', 1], ['o', 4], [' ', 1], ['h', 1], ['i', 5], [' ', 1], ['w', 1], ['h', 1], 
['y', 4], [' ', 1], ['d', 1], ['o', 1], ['n', 1], ['e', 1]]

I want to display that sooo, areeeee and youuuuuu are elongated. I did individual character count but I want to check for every word to see if its elongated or not.

Answer 1

A regex comes to mind:

>>> my_str = 'soooo hiiiii whyyyy done'
>>> import re
>>> regex = re.compile(r"(.)\1{2}")
>>> [word for word in my_str.split() if regex.search(word)]
['soooo', 'hiiiii', 'whyyyy']

Explanation:

(.)    # Match any character, capture it in group number 1
\1{2}  # Try to match group number 1 here, twice.

Note that this algorithm will also find some unelongated words like countessship or laparohysterosalpingooophorectomy , but I guess those false positives are rare :)

Answer 2

You can use:

def get_groups(word):
    return [list(g) for k, g in itertools.groupby(word)]

print [word for word in my_str.split(' ') if any(len(x) > 2 for x in get_groups(word))]

Here's how it works: get_groups turns a word into groups. So 'sooo' becomes [['s'], ['o', 'o', 'o']] .

We then filter all words from the given string if the length of any of the groups is more than two. This means you'll end up with all words that have three or more consecutive characters.

Answer 3

you have to check by the sequence and compare length, without importing anything :

>>> filter(lambda word: len([letter for index,letter in enumerate(word) if index ==0 or word[index-1] != letter ]) == len( word), my_str.split(" "))
['done']

>>> filter(lambda word: len([letter for index,letter in enumerate(word) if index ==0 or word[index-1] != letter ]) != len( word), my_str.split(" "))
['soooo', 'hiiiii', 'whyyyy']

or import itertools and doing it with groupby :

>>> filter(lambda word: len([letter for letter,gp in itertools.groupby(word) ]) == len( word), my_str.split(" "))
['done']

>>> filter(lambda word: len([letter for letter,gp in itertools.groupby(word) ]) != len( word), my_str.split(" "))
['soooo', 'hiiiii', 'whyyyy']

this last solution permit tu use also ifilter instead of filter and iter on every good or bad words . useful for stream or very big string