fixing words with spaces using a dictionary look up in python?

Your goal is to improve text, not necessarily to make it perfect; so the approach you outline makes sense in my opinion. I would keep it simple and use a "greedy" approach: Start with the first fragment and stick pieces to it as long as the result is in the dictionary; if the result is not, spit out what you have so far and start over with the next fragment. Yes, occasionally you'll make a mistake with cases like the me thod, so if you'll be using this a lot, you could look for something more sophisticated. However, it's probably good enough.

Mainly what you require is a large dictionary. If you'll be using it a lot, I would encode it as a "prefix tree" (a.k.a. trie), so that you can quickly find out if a fragment is the start of a real word. The nltk provides a Trie implementation.

Since this kind of spurious word breaks are inconsistent, I would also extend my dictionary with words already processed in the current document; you may have seen the complete word earlier, but now it's broken up.

Take a look at word or text segmentation. The problem is to find the most probable split of a string into a group of words. Example:

 thequickbrownfoxjumpsoverthelazydog

The most probable segmentation should be of course:

 the quick brown fox jumps over the lazy dog

Here's an article including prototypical source code for the problem using Google Ngram corpus:

• http://jeremykun.com/2012/01/15/word-segmentation/

The key for this algorithm to work is access to knowledge about the world, in this case word frequencies in some language. I implemented a version of the algorithm described in the article here:

• https://gist.github.com/miku/7279824

Example usage:

$ python segmentation.py t hequi ckbrownfoxjum ped
thequickbrownfoxjumped
['the', 'quick', 'brown', 'fox', 'jumped']

Using data, even this can be reordered:

$ python segmentation.py lmaoro fll olwt f pwned
lmaorofllolwtfpwned
['lmao', 'rofl', 'lol', 'wtf', 'pwned']

Note that the algorithm is quite slow - it's prototypical.

Another approach using NLTK:

• http://web.archive.org/web/20160123234612/http://www.winwaed.com:80/blog/2012/03/13/segmenting-words-and-sentences/

As for your problem, you could just concatenate all string parts you have to get a single string and the run a segmentation algorithm on it.

--Solution 1:

Lets think of these chunks in your sentence as beads on an abacus, with each bead consisting of a partial string, the beads can be moved left or right to generate the permutations. The position of each fragment is fixed between two adjacent fragments. In current case, the beads would be :

(more)(recen)(t)(ly)(the)(develop)(ment,)(wh)(ich)(is)(a)(po)(ten)(t)

This solves 2 subproblems:

a) Bead is a single unit,so We do not care about permutations within the bead i.e. permutations of "more" are not possible.

b) The order of the beads is constant, only the spacing between them changes. i.e. "more" will always be before "recen" and so on.

Now, generate all the permutations of these beads , which will give output like :

morerecentlythedevelopment,which is a potent
morerecentlythedevelopment,which is a poten t
morerecentlythedevelop ment, wh ich is a po tent
morerecentlythedevelop ment, wh ich is a po ten t
morerecentlythe development,whichisapotent

Then score these permutations based on how many words from your relevant dictionary they contain, most correct results can be easily filtered out. more recently the development, which is a potent will score higher than morerecentlythedevelop ment, wh ich is a po ten t

Code which does the permutation part of the beads:

import re

def gen_abacus_perms(frags):
    if len(frags) == 0:
        return []
    if len(frags) == 1:
        return [frags[0]]

    prefix_1 = "{0}{1}".format(frags[0],frags[1])
    prefix_2 = "{0} {1}".format(frags[0],frags[1])
    if len(frags) == 2:
        nres = [prefix_1,prefix_2]
        return nres

    rem_perms = gen_abacus_perms(frags[2:])
    res = ["{0}{1}".format(prefix_1, x ) for x in rem_perms] + ["{0} {1}".format(prefix_1, x ) for x in rem_perms] +  \
["{0}{1}".format(prefix_2, x ) for x in rem_perms] + ["{0} {1}".format(prefix_2 , x ) for x in rem_perms]
    return res



broken = "more recen t ly the develop ment, wh ich is a po ten t"
frags = re.split("\s+",broken)
perms = gen_abacus_perms(frags)
print("\n".join(perms))

demo:http://ideone.com/pt4PSt

---

--Solution#2:

I would suggest an alternate approach which makes use of text analysis intelligence already developed by folks working on similar problems and having worked on big corpus of data which depends on dictionary and grammar .e.g. search engines.

I am not well aware of such public/paid apis, so my example is based on google results.

Lets try to use google :

1. You can keep putting your invalid terms to Google, for multiple passes, and keep evaluating the results for some score based on your lookup dictionary. here are two relevant outputs by using 2 passes of your text :

This outout is used for a second pass :

Which gives you the conversion as ""more recently the development, which is a potent".

To verify the conversion, you will have to use some similarity algorithm and scoring to filter out invalid / not so good results.

One raw technique could be using a comparison of normalized strings using difflib.

>>> import difflib
>>> import re
>>> input = "more recen t ly the develop ment, wh ich is a po ten t "
>>> output = "more recently the development, which is a potent "
>>> input_norm = re.sub(r'\W+', '', input).lower()
>>> output_norm = re.sub(r'\W+', '', output).lower()
>>> input_norm
'morerecentlythedevelopmentwhichisapotent'
>>> output_norm
'morerecentlythedevelopmentwhichisapotent'
>>> difflib.SequenceMatcher(None,input_norm,output_norm).ratio()
1.0