We Keep Coding

Check how many words from a given list occur in list of text/strings

I have a list of text data which contains reviews, something likes this:
1. 'I have bought several of the Vitality canned dog food products and have found them all to be of good quality. The product looks more like a stew than a processed meat and it smells better. My Labrador is finicky and she appreciates this product better than most.'
2. 'Product arrived labeled as Jumbo Salted Peanuts...the peanuts were actually small sized unsalted. Not sure if this was an error or if the vendor intended to represent the product as "Jumbo".',
3. 'This is a confection that has been around a few centuries. It is a light, pillowy citrus gelatin with nuts - in this case Filberts. And it is cut into tiny squares and then liberally coated with powdered sugar. And it is a tiny mouthful of heaven. Not too chewy, and very flavorful. I highly recommend this yummy treat. If you are familiar with the story of C.S. Lewis\' "The Lion, The Witch, and The Wardrobe" - this is the treat that seduces Edmund into selling out his Brother and Sisters to the Witch.
I have a seperate list of words which I want to know exists in the these reviews:
['food','science','good','buy','feedback'....]
I want to know which of these words are present in the review and select reviews which contains certain number of these words. For example, lets say only select reviews which contains atleast 3 of the words from this list, so it displays all those reviews, but also show which of those were encountered in the review while selecting it.
I have the code for selecting reviews containing at least 3 of the words, but how do I get the second part which tells me which words exactly were encountered. Here is my initial code:
keywords = list(words)
text = list(df.summary.values)
sentences=[]
for element in text:
if len(set(keywords)&set(element.split(' '))) >=3:
sentences.append(element)

To answer the second part, allow me to revisit how to approach the first part. A handy approach here is to cast your review strings into sets of word strings.
Like this:
review_1 = "I have bought several of the Vitality canned dog food products and"
review_1 = set(review_1.split(" "))
Now the review_1 set contains one of every word. Then take your list of words, convert it to a set, and do an intersection.
words = ['food','science','good','buy','feedback'....]
words = set(['food','science','good','buy','feedback'....])
matches = review_1.intersection(words)
The resulting set, matches, contains all the words that are common. The length of this is the number of matches.
Now, this does not work if you cared about how many of each word matches. For example, if the word "food" is found twice in the review and "science" is found once, does that count as matching three words?
If so, let me know via comment and I can write some code to update the answer to include that scenario.
EDIT: Updating to include comment question
If you want to keep a count of how many times each word repeats, then hang onto the review list. Only cast it to set when performing the intersection. Then, use the 'count' list method to count the number of times each match appears in the review. In the example below, I use a dictionary to store the results.
review_1 = "I have bought several of the Vitality canned dog food products and"
words = ['food','science','good','buy','feedback'....]
words = set(['food','science','good','buy','feedback'....])
matches = set(review_1).intersection(words)
match_counts = dict()
for match in matches:
match_counts[match] = words.count(match)

You can use set intersection for finding the common words:
def filter_reviews(data, *, trigger_words = frozenset({'food', 'science', 'good', 'buy', 'feedback'})):
for review in data:
words = review.split() # use whatever method is appropriate to get the words
common = trigger_words.intersection(words)
if len(common) >= 3:
yield review, common

How to find and rank all prefixes in a list of strings?

I have a list of strings and I want to find popular prefixes. The prefixes are special in that they occur as strings in the input list.
I found a similar question here but the answers are geared to find the one most common prefix:
Find *most* common prefix of strings - a better way?
While my problem is similar, it differs in that I need to find all popular prefixes. Or to maybe state it a little simplistically, rank prefixes from most common to least.
As an example, consider the following list of strings:
in, india, indian, indian flag, bull, bully, bullshit
Prefixes rank:
in - 4 times
india - 3 times
bull - 3 times
...and so on. Please note - in, bull, india are all present in the input list.
The following are not valid prefixes:
ind
bu
bul
...since they do not occur in the input list.
What data structure should I be looking at to model my solution? I'm inclined to use a "trie" with a counter on each node that tracks how many times has that node been touched during the creation of the trie.
All suggestions are welcome.
Thanks.
p.s. - I love python and would love if someone could post a quick snippet that could get me started.

words = [ "in", "india", "indian", "indian", "flag", "bull", "bully", "bullshit"]
Result = sorted([ (sum([ w.startswith(prefix) for w in words ]) , prefix ) for prefix in words])[::-1]
it goes through every word as a prefix and checks how many of the other words start with it and then sorts the result. the[::-1] simply reverses that order

If we know the length of the prefix (say 3)
from nltk import FreqDist
suffixDist=FreqDist()
for word in vocabulary:
suffixDist[word[-3:]] +=1
commonSuffix=[suffix for (suffix,count) in suffixDist.most_common(150) ]
print(commonSuffix)

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

I have extracted the list of sentences from a document. I am pre-processing this list of sentences to make it more sensible. I am faced with the following problem
I have sentences such as "more recen t ly the develop ment, wh ich is a po ten t "
I would like to correct such sentences using a look up dictionary? to remove the unwanted spaces.
The final output should be "more recently the development, which is a potent "
I would assume that this is a straight forward task in preprocessing text? I need help with some pointers to look for such approaches. Thanks.

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.

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.

--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 :
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

I would recommend stripping away the spaces and looking for dictionary words to break it down into. There are a few things you can do to make it more accurate. To make it get the first word in text with no spaces, try taking the entire string, and going through dictionary words from a file (you can download several such files from http://wordlist.sourceforge.net/), the longest ones first, than taking off letters from the end of the string you want to segment. If you want it to work on a big string, you can make it automatically take off letters from the back so that the string you are looking for the first word in is only as long as the longest dictionary word. This should result in you finding the longest words, and making it less likely to do something like classify "asynchronous" as "a synchronous". Here is an example that uses raw input to take in the text to correct and a dictionary file called dictionary.txt:
dict = open("dictionary.txt",'r') #loads a file with a list of words to break string up into
words = raw_input("enter text to correct spaces on: ")
words = words.strip() #strips away spaces
spaced = [] #this is the list of newly broken up words
parsing = True #this represents when the while loop can end
while parsing:
if len(words) == 0: #checks if all of the text has been broken into words, if it has been it will end the while loop
parsing = False
iterating = True
for iteration in range(45): #goes through each of the possible word lengths, starting from the biggest
if iterating == False:
break
word = words[:45-iteration] #each iteration, the word has one letter removed from the back, starting with the longest possible number of letters, 45
for line in dict:
line = line[:-1] #this deletes the last character of the dictionary word, which will be a newline. delete this line of code if it is not a newline, or change it to [1:] if the newline character is at the beginning
if line == word: #this finds if this is the word we are looking for
spaced.append(word)
words = words[-(len(word)):] #takes away the word from the text list
iterating = False
break
print ' '.join(spaced) #prints the output
If you want it to be even more accurate, you could try using a natural language parsing program, there are several available for python free online.

Here's something really basic:
chunks = []
for chunk in my_str.split():
chunks.append(chunk)
joined = ''.join(chunks)
if is_word(joined):
print joined,
del chunks[:]
# deal with left overs
if chunks:
print ''.join(chunks)
I assume you have a set of valid words somewhere that can be used to implement is_word. You also have to make sure it deals with punctuation. Here's one way to do that:
def is_word(wd):
if not wd:
return False
# Strip of trailing punctuation. There might be stuff in front
# that you want to strip too, such as open parentheses; this is
# just to give the idea, not a complete solution.
if wd[-1] in ',.!?;:':
wd = wd[:-1]
return wd in valid_words

You can iterate through a dictionary of words to find the best fit. Adding the words together when a match is not found.
def iterate(word,dictionary):
for word in dictionary:
if words in possibleWord:
finished_sentence.append(words)
added = True
else:
added = False
return [added,finished_sentence]
sentence = "more recen t ly the develop ment, wh ich is a po ten t "
finished_sentence = ""
sentence = sentence.split()
for word in sentence:
added,new_word = interate(word,dictionary)
while True:
if added == False:
word += possible[sentence.find(possibleWord)]
iterate(word,dictionary)
else:
break
finished_sentence.append(word)
This should work. For the variable dictionary, download a txt file of every single english word, then open it in your program.

my index.py file be like
from wordsegment import load, segment
load()
print(segment('morerecentlythedevelopmentwhichisapotent'))
my index.php file be like
<html>
<head>
<title>py script</title>
</head>
<body>
<h1>Hey There!Python Working Successfully In A PHP Page.</h1>
<?php
$python = `python index.py`;
echo $python;
?>
</body>
</html>
Hope this will work

Discovering Poetic Form with NLTK and CMU Dict

Edit: This code has been worked on and released as a basic module: https://github.com/hyperreality/Poetry-Tools
I'm a linguist who has recently picked up python and I'm working on a project which hopes to automatically analyze poems, including detecting the form of the poem. I.e. if it found a 10 syllable line with 0101010101 stress pattern, it would declare that it's iambic pentameter. A poem with 5-7-5 syllable pattern would be a haiku.
I'm using the following code, part of a larger script, but I have a number of problems which are listed below the program:
corpus in the script is simply the raw text input of the poem.
import sys, getopt, nltk, re, string
from nltk.tokenize import RegexpTokenizer
from nltk.util import bigrams, trigrams
from nltk.corpus import cmudict
from curses.ascii import isdigit
...
def cmuform():
tokens = [word for sent in nltk.sent_tokenize(corpus) for word in nltk.word_tokenize(sent)]
d = cmudict.dict()
text = nltk.Text(tokens)
words = [w.lower() for w in text]
regexp = "[A-Za-z]+"
exp = re.compile(regexp)
def nsyl(word):
lowercase = word.lower()
if lowercase not in d:
return 0
else:
first = [' '.join([str(c) for c in lst]) for lst in max(d[lowercase])]
second = ''.join(first)
third = ''.join([i for i in second if i.isdigit()]).replace('2', '1')
return third
#return max([len([y for y in x if isdigit(y[-1])]) for x in d[lowercase]])
sum1 = 0
for a in words:
if exp.match(a):
print a,nsyl(a),
sum1 = sum1 + len(str(nsyl(a)))
print "\nTotal syllables:",sum1
I guess that the output that I want would be like this:
1101111101
0101111001
1101010111
The first problem is that I lost the line breaks during the tokenization, and I really need the line breaks to be able to identify form. This should not be too hard to deal with though. The bigger problems are that:
I can't deal with non-dictionary words. At the moment I return 0 for them, but this will confound any attempt to identify the poem, as the syllabic count of the line will probably decrease.
In addition, the CMU dictionary often says that there is stress on a word - '1' - when there is not - '0 - . Which is why the output looks like this: 1101111101, when it should be the stress of iambic pentameter: 0101010101
So how would I add some fudging factor so the poem still gets identified as iambic pentameter when it only approximates the pattern? It's no good to code a function that identifies lines of 01's when the CMU dictionary is not going to output such a clean result. I suppose I'm asking how to code a 'partial match' algorithm.

Welcome to stack overflow. I'm not that familiar with Python, but I see you have not received many answers yet so I'll try to help you with your queries.
First some advice: You'll find that if you focus your questions your chances of getting answers are greatly improved. Your post is too long and contains several different questions, so it is beyond the "attention span" of most people answering questions here.
Back on topic:
Before you revised your question you asked how to make it less messy. That's a big question, but you might want to use the top-down procedural approach and break your code into functional units:
split corpus into lines
For each line: find the syllable length and stress pattern.
Classify stress patterns.
You'll find that the first step is a single function call in python:
corpus.split("\n");
and can remain in the main function but the second step would be better placed in its own function and the third step would require to be split up itself, and would probably be better tackled with an object oriented approach. If you're in academy you might be able to convince the CS faculty to lend you a post-grad for a couple of months and help you instead of some workshop requirement.
Now to your other questions:
Not loosing line breaks: as #ykaganovich mentioned, you probably want to split the corpus into lines and feed those to the tokenizer.
Words not in dictionary/errors: The CMU dictionary home page says:
Find an error? Please contact the developers. We will look at the problem and improve the dictionary. (See at bottom for contact information.)
There is probably a way to add custom words to the dictionary / change existing ones, look in their site, or contact the dictionary maintainers directly.
You can also ask here in a separate question if you can't figure it out. There's bound to be someone in stackoverflow that knows the answer or can point you to the correct resource.
Whatever you decide, you'll want to contact the maintainers and offer them any extra words and corrections anyway to improve the dictionary.
Classifying input corpus when it doesn't exactly match the pattern: You might want to look at the link ykaganovich provided for fuzzy string comparisons. Some algorithms to look for:
Levenshtein distance: gives you a measure of how different two strings are as the number of changes needed to turn one string into another. Pros: easy to implement, Cons: not normalized, a score of 2 means a good match for a pattern of length 20 but a bad match for a pattern of length 3.
Jaro-Winkler string similarity measure: similar to Levenshtein, but based on how many character sequences appear in the same order in both strings. It is a bit harder to implement but gives you normalized values (0.0 - completely different, 1.0 - the same) and is suitable for classifying the stress patterns. A CS postgrad or last year undergrad should not have too much trouble with it ( hint hint ).
I think those were all your questions. Hope this helps a bit.

To preserve newlines, parse line by line before sending each line to the cmu parser.
For dealing with single-syllable words, you probably want to try both 0 and 1 for it when nltk returns 1 (looks like nltk already returns 0 for some words that would never get stressed, like "the"). So, you'll end up with multiple permutations:
1101111101
0101010101
1101010101
and so forth. Then you have to pick ones that look like a known forms.
For non-dictionary words, I'd also fudge it the same way: figure out the number of syllables (the dumbest way would be by counting the vowels), and permutate all possible stresses. Maybe add some more rules like "ea is a single syllable, trailing e is silent"...
I've never worked with other kinds of fuzzying, but you can check https://stackoverflow.com/questions/682367/good-python-modules-for-fuzzy-string-comparison for some ideas.

This is my first post on stackoverflow.
And I'm a python newbie, so please excuse any deficits in code style.
But I too am attempting to extract accurate metre from poems.
And the code included in this question helped me, so I post what I came up with that builds on that foundation. It is one way to extract the stress as a single string, correct with a 'fudging factor' for the cmudict bias, and not lose words that are not in the cmudict.
import nltk
from nltk.corpus import cmudict
prondict = cmudict.dict()
#
# parseStressOfLine(line)
# function that takes a line
# parses it for stress
# corrects the cmudict bias toward 1
# and returns two strings
#
# 'stress' in form '0101*,*110110'
# -- 'stress' also returns words not in cmudict '0101*,*1*zeon*10110'
# 'stress_no_punct' in form '0101110110'
def parseStressOfLine(line):
stress=""
stress_no_punct=""
print line
tokens = [words.lower() for words in nltk.word_tokenize(line)]
for word in tokens:
word_punct = strip_punctuation_stressed(word.lower())
word = word_punct['word']
punct = word_punct['punct']
#print word
if word not in prondict:
# if word is not in dictionary
# add it to the string that includes punctuation
stress= stress+"*"+word+"*"
else:
zero_bool=True
for s in prondict[word]:
# oppose the cmudict bias toward 1
# search for a zero in array returned from prondict
# if it exists use it
# print strip_letters(s),word
if strip_letters(s)=="0":
stress = stress + "0"
stress_no_punct = stress_no_punct + "0"
zero_bool=False
break
if zero_bool:
stress = stress + strip_letters(prondict[word][0])
stress_no_punct=stress_no_punct + strip_letters(prondict[word][0])
if len(punct)>0:
stress= stress+"*"+punct+"*"
return {'stress':stress,'stress_no_punct':stress_no_punct}
# STRIP PUNCTUATION but keep it
def strip_punctuation_stressed(word):
# define punctuations
punctuations = '!()-[]{};:"\,<>./?##$%^&*_~'
my_str = word
# remove punctuations from the string
no_punct = ""
punct=""
for char in my_str:
if char not in punctuations:
no_punct = no_punct + char
else:
punct = punct+char
return {'word':no_punct,'punct':punct}
# CONVERT the cmudict prondict into just numbers
def strip_letters(ls):
#print "strip_letters"
nm = ''
for ws in ls:
#print "ws",ws
for ch in list(ws):
#print "ch",ch
if ch.isdigit():
nm=nm+ch
#print "ad to nm",nm, type(nm)
return nm
# TESTING results
# i do not correct for the '2'
line = "This day (the year I dare not tell)"
print parseStressOfLine(line)
line = "Apollo play'd the midwife's part;"
print parseStressOfLine(line)
line = "Into the world Corinna fell,"
print parseStressOfLine(line)
"""
OUTPUT
This day (the year I dare not tell)
{'stress': '01***(*011111***)*', 'stress_no_punct': '01011111'}
Apollo play'd the midwife's part;
{'stress': "0101*'d*01211***;*", 'stress_no_punct': '010101211'}
Into the world Corinna fell,
{'stress': '01012101*,*', 'stress_no_punct': '01012101'}

efficient algorithm for finding co occurrence matrix of phrases

I have a list L of around 40,000 phrases and a document of around 10 million words. what I want to check is which pair of these phrases co occur within a window of 4 words. For example, consider L=["brown fox","lazy dog"]. The document contains the words "a quick brown fox jumps over the lazy dog". I want to see, how many times brown fox and lazy dog appears within an window of four words and store that in a file. I have following code for doing this:
content=open("d.txt","r").read().replace("\n"," ");
for i in range(len(L)):
for j in range(i+1,len(L)):
wr=L[i]+"\W+(?:\w+\W+){1,4}"+L[j]
wrev=L[j]+"\W+(?:\w+\W+){1,4}"+L[i]
phrasecoccur=len(re.findall(wr, content))+len(re.findall(wrev,content))
if (phrasecoccur>0):
f.write(L[i]+", "+L[j]+", "+str(phrasecoccur)+"\n")
Essentially, for each pair of phrases in the list L, I am checking in the document content that how many times these phrases appear within an window of 4 words. However, this method is computationally inefficient when the list L is pretty large, like 40K elements. Is there a better way of doing this?

You could use something similar to the Aho-Corasick string matching algorithm. Build the state machine from your list of phrases. Then start feeding words into the state machine. Whenever a match occurs, the state machine will tell you which phrase matched and at what word number. So your output would be something like:
"brown fox", 3
"lazy dog", 8
etc.
You can either capture all of the output and post-process it, or you can process the matches as they're found.
It takes a little time to build the state machine (a few seconds for 40,000 phrases), but after that it's linear in the number of input tokens, number of phrases, and number of matches.
I used something similar to match 50 million YouTube video titles against the several million song titles and artist names in the MusicBrainz database. Worked great. And very fast.

It should be possible to assemble your 40000 phrases into a big regular expression pattern, and use that to match against your document. It might not be as fast as something more job-specific, but it does work. Here's how I'd do it:
import re
class Matcher(object):
def __init__(self, phrases):
phrase_pattern = "|".join("(?:{})".format(phrase) for phrase in phrases)
gap_pattern = r"\W+(?:\w+\W+){0,4}?"
full_pattern = "({0}){1}({0})".format(phrase_pattern, gap_pattern)
self.regex = re.compile(full_pattern)
def match(self, doc):
return self.regex.findall(doc) # or use finditer to generate match objs
Here's how you can use it:
>>> L = ["brown fox", "lazy dog"]
>>> matcher = Matcher(L)
>>> doc = "The quick brown fox jumps over the lazy dog."
>>> matcher.match(doc)
[('brown fox', 'lazy dog')]
This solution does have a few limitations. One is that it won't detect overlapping pairs of phrases. So in the example, if you added the phrase "jumps over" to the phrase list, you would still only get one matched pair, ("brown fox", "jumps over"). It would miss both ("brown fox", "lazy dog") and ("jumps over", "lazy dog"), since they include some of the same words.

Expanding on Joel's answer, your iterator could be something like this:
def doc_iter(doc):
words=doc[0:4]
yield words
for i in range(3,len(doc)):
words=words[1:]
words.append(doc[i])
yield words
put your phrases in a dict and use the iterator over the doc, checking the phrases at each iteration. This should give you performance between O(n) and O(n*log(n)).