Remove repeating characters from words
I was wondering what is the best way to convert something like "haaaaapppppyyy" to "haappyy".
Basically, when parsing slang, people sometimes repeat characters for added emphasis.
I was wondering what the best way to do this is? Using set() doesn't work because the order of the letters is obviously important.
Any ideas? I'm using Python + nltk.
It can be done using regular expressions:
>>> import re >>> re.sub(r'(.)\1+', r'\1\1', "haaaaapppppyyy") 'haappyy'
(.)\1+ repleaces any character (.) followed by one or more of the same character (because of the backref \1 it must be the same) by twice the character.
You can squash multiple occurrences of letters with itertools.groupby:
>>> ''.join(c for c, _ in groupby("haaaaapppppyyy")) 'hapy'
Similarly, you can get haappyy from groupby with
>>> ''.join(''.join(s)[:2] for _, s in groupby("haaaaapppppyyy")) 'haappyy'
You should do it without reduce or regexps:
>>> s = 'hhaaaaapppppyyy' >>> ''.join(['' if i>1 and e==s[i-2] else e for i,e in enumerate(s)]) 'haappyy'
The number of repetitions are hardcoded to >1 and -2 above. The general case:
>>> reps = 1 >>> ''.join(['' if i>reps-1 and e==s[i-reps] else e for i,e in enumerate(s)]) 'hapy'
This is one way of doing it (limited to the obvious constraint that python doesn't speak english).
>>> s="haaaappppyy" >>> reduce(lambda x,y: x+y if x[-2:]!=y*2 else x, s, "") 'haappyy'
For the thing you mentioned about the set(), you can use collections.OrderedDict to maintain the order of the letters. So use:
text = "happy" print(list(OrderedDict.fromkeys(text)))
which will give you:
['h', 'a', 'p', 'y']