I have a use case where I want to match one list of words with a list of sentences and bring the most relevant sentences
I am working in python. What I have already tried is using KMeans where we cluster our set of documents into the clusters and then predict the sentence that in which structure it resides. But in my case I have already available list of words available.
def getMostRelevantSentences():
Sentences = ["This is the most beautiful place in the world.",
"This man has more skills to show in cricket than any other game.",
"Hi there! how was your ladakh trip last month?",
"Isn’t cricket supposed to be a team sport? I feel people should decide first whether cricket is a team game or an individual sport."]
words = ["cricket","sports","team","play","match"]
#TODO: now this should return me the 2nd and last item from the Sentences list as the words list mostly matches with them
So from the above code I want to return the sentences which are closely matching with the words provided. I don't want to use the supervised machine learning here. Any help will be appreciated.
So finally I have used this super library called gensim to generate the similarity.
import gensim
from nltk.tokenize import word_tokenize
def getSimilarityScore(raw_documents, words):
gen_docs = [[w.lower() for w in word_tokenize(text)]
for text in raw_documents]
dictionary = gensim.corpora.Dictionary(gen_docs)
corpus = [dictionary.doc2bow(gen_doc) for gen_doc in gen_docs]
tf_idf = gensim.models.TfidfModel(corpus)
sims = gensim.similarities.Similarity('/usr/workdir',tf_idf[corpus],
num_features=len(dictionary))
query_doc_bow = dictionary.doc2bow(words)
query_doc_tf_idf = tf_idf[query_doc_bow]
return sims[query_doc_tf_idf]
You can use this method as:
Sentences = ["This is the most beautiful place in the world.",
"This man has more skills to show in cricket than any other game.",
"Hi there! how was your ladakh trip last month?",
"Isn’t cricket supposed to be a team sport? I feel people should decide first whether cricket is a team game or an individual sport."]
words = ["cricket","sports","team","play","match"]
words_lower = [w.lower() for w in words]
getSimilarityScore(Sentences,words_lower)
I wrote some basic program and I want extract lemma for Polish words what is very important since Polish use inflection.
I was create simple program to load data and convert word into lemma but not know how to do it:
from nltk import corpus
pl = corpus.pl196x
print dir(pl)
print iter(pl.tagged_words()).next()
For example I want do such lemmatization (ignore that some lemmatization can be ambiguous - it is normal in Polish):
kot, kota, kota, kotu, kotem, kocie, kocie == kot (singular male)
kotka, kotki, kotkę, kotce, kotką, kotce, kotka == kot (singular female!)
kociątko, kociątka, kociątko, kociątku, kociątkiem, kociątku, kociątko == kot (singular neutral)
etc. (plural males, plural females, plural neutrals)
How to use pl196x to do such job in or nltk.
Source data for corpus contains such inflections and lemmas - so it is possible but how to access it:
<w id="pu147125" lemma="kot" ana="SSNA---------P">kot</w>
<w id="pr021633" lemma="kot" ana="SSAA---------P">kota</w>
etc.
I don't know if it is possible in NLTK itself. When I had similar problem I had used Morfeusz for that (http://sgjp.pl/morfeusz/). Although I have used only C version of the library, there is a Python binding for it (http://jwilk.net/software/python-morfeusz).
This question already has answers here:
How to get synonyms from nltk WordNet Python
(8 answers)
Closed 7 years ago.
The code to get the synonyms of a word in python is say:
from nltk.corpus import wordnet
dog = wordnet.synset('dog.n.01')
print dog.lemma_names
>>['dog', 'domestic_dog', 'Canis_familiaris']
However dog.n.02 gives different words. For any words i can't know how many words there may be. How can i return all of the synonyms for a word?
Using wn.synset('dog.n.1').lemma_names is the correct way to access the synonyms of a sense. It's because a word has many senses and it's more appropriate to list synonyms of a particular meaning/sense. To enumerate words with similar meanings, possibly you can also look at the hyponyms.
Sadly, the size of Wordnet is very limited so there are very few lemma_names available for each senses.
Using Wordnet as a dictionary/thesarus is not very apt per se, because it was developed as an inventory of sense/meaning rather than a inventory of words. However you can use access the a particular sense and several (not a lot) related words to the sense. One can use Wordnet as a:
Dictionary: given a word, what are the different meaning of the word
for i,j in enumerate(wn.synsets('dog')):
print "Meaning",i, "NLTK ID:", j.name
print "Definition:",j.definition
Thesarus: given a word, what are the different words for each meaning of the word
for i,j in enumerate(wn.synsets('dog')):
print "Meaning",i, "NLTK ID:", j.name
print "Definition:",j.definition
print "Synonyms:", ", ".join(j.lemma_names)
print
Ontology: given a word, what are the hyponyms (i.e. sub-types) and hypernyms (i.e. super-types).
for i,j in enumerate(wn.synsets('dog')):
print "Meaning",i, "NLTK ID:", j.name
print "Hypernyms:", ", ".join(list(chain(*[l.lemma_names for l in j.hypernyms()])))
print "Hyponyms:", ", ".join(list(chain(*[l.lemma_names for l in j.hyponyms()])))
print
[Ontology Output]
Meaning 0 NLTK ID: dog.n.01
Hypernyms words domestic_animal, domesticated_animal, canine, canid
Hyponyms puppy, Great_Pyrenees, basenji, Newfoundland, Newfoundland_dog, lapdog, poodle, poodle_dog, Leonberg, toy_dog, toy, spitz, pooch, doggie, doggy, barker, bow-wow, cur, mongrel, mutt, Mexican_hairless, hunting_dog, working_dog, dalmatian, coach_dog, carriage_dog, pug, pug-dog, corgi, Welsh_corgi, griffon, Brussels_griffon, Belgian_griffon
Meaning 1 NLTK ID: frump.n.01
Hypernyms: unpleasant_woman, disagreeable_woman
Hyponyms:
Meaning 2 NLTK ID: dog.n.03
Hypernyms: chap, fellow, feller, fella, lad, gent, blighter, cuss, bloke
Hyponyms:
Meaning 3 NLTK ID: cad.n.01
Hypernyms: villain, scoundrel
Hyponyms: perisher
Note this other answer:
>>> wn.synsets('small')
[Synset('small.n.01'),
Synset('small.n.02'),
Synset('small.a.01'),
Synset('minor.s.10'),
Synset('little.s.03'),
Synset('small.s.04'),
Synset('humble.s.01'),
Synset('little.s.07'),
Synset('little.s.05'),
Synset('small.s.08'),
Synset('modest.s.02'),
Synset('belittled.s.01'),
Synset('small.r.01')]
Keep in mind that in your code you were trying to get the lemmas, but that's one level too deep for what you want. The synset level is about meaning, while the lemma level gives you words. In other words:
In WordNet (and I’m speaking of English WordNet here, though I think
the ones in other langauges are similarly organized) a lemma has
senses. Specifically, a lemma (that is, a base word form that is
indexed in WordNet) has exactly as many senses as the number of
synsets that it participates in. Conversely, and as you say, synsets
contain one more more lemmas, which means that multiple lemmas (words)
can represent the same sense, or meaning.
Also have a look at the NLTK's WordNet how to for a few more ways of exploring around a meaning or a word.
The documentation suggests
wordnet.synsets('dog')
to get all synsets for dog.
I'm working with a large database of businesses.
I'd like to be able to compare two business names for similarity to see if they possibly might be duplicates.
Below is a list of business names that should test as having a high probability of being duplicates, what is a good way to go about this?
George Washington Middle Schl
George Washington School
Santa Fe East Inc
Santa Fe East
Chop't Creative Salad Co
Chop't Creative Salad Company
Manny and Olga's Pizza
Manny's & Olga's Pizza
Ray's Hell Burger Too
Ray's Hell Burgers
El Sol
El Sol de America
Olney Theatre Center for the Arts
Olney Theatre
21 M Lounge
21M Lounge
Holiday Inn Hotel Washington
Holiday Inn Washington-Georgetown
Residence Inn Washington,DC/Dupont Circle
Residence Inn Marriott Dupont Circle
Jimmy John's Gourmet Sandwiches
Jimmy John's
Omni Shoreham Hotel at Washington D.C.
Omni Shoreham Hotel
I've recently done a similar task, although I was matching new data to existing names in a database, rather than looking for duplicates within one set. Name matching is actually a well-studied task, with a number of factors beyond what you'd consider for matching generic strings.
First, I'd recommend taking a look at a paper, How to play the “Names Game”: Patent retrieval comparing different heuristics by Raffo and Lhuillery. The published version is here, and a PDF is freely available here. The authors provide a nice summary, comparing a number of different matching strategies. They consider three stages, which they call parsing, matching, and filtering.
Parsing consists of applying various cleaning techniques. Some examples:
Standardizing lettercase (e.g., all lowercase)
Standardizing punctuation (e.g., commas must be followed by spaces)
Standardizing whitespace (e.g., converting all runs of whitespace to single spaces)
Standardizing accented and special characters (e.g., converting accented letters to ASCII equivalents)
Standardizing legal control terms (e.g., converting "Co." to "Company")
In my case, I folded all letters to lowercase, replaced all punctuation with whitespace, replaced accented characters by unaccented counterparts, removed all other special characters, and removed legal control terms from the beginning and ends of the names following a list.
Matching is the comparison of the parsed names. This could be simple string matching, edit distance, Soundex or Metaphone, comparison of the sets of words making up the names, or comparison of sets of letters or n-grams (letter sequences of length n). The n-gram approach is actually quite nice for names, as it ignores word order, helping a lot with things like "department of examples" vs. "examples department". In fact, comparing bigrams (2-grams, character pairs) using something simple like the Jaccard index is very effective. In contrast to several other suggestions, Levenshtein distance is one of the poorer approaches when it comes to name matching.
In my case, I did the matching in two steps, first with comparing the parsed names for equality and then using the Jaccard index for the sets of bigrams on the remaining. Rather than actually calculating all the Jaccard index values for all pairs of names, I first put a bound on the maximum possible value for the Jaccard index for two sets of given size, and only computed the Jaccard index if that upper bound was high enough to potentially be useful. Most of the name pairs were still dissimilar enough that they weren't matches, but it dramatically reduced the number of comparisons made.
Filtering is the use of auxiliary data to reject false positives from the parsing and matching stages. A simple version would be to see if matching names correspond to businesses in different cities, and thus different businesses. That example could be applied before matching, as a kind of pre-filtering. More complicated or time-consuming checks might be applied afterwards.
I didn't do much filtering. I checked the countries for the firms to see if they were the same, and that was it. There weren't really that many possibilities in the data, some time constraints ruled out any extensive search for additional data to augment the filtering, and there was a manual checking planned, anyway.
I'd like to add some examples to the excellent accepted answer. Tested in Python 2.7.
Parsing
Let's use this odd name as an example.
name = "THE | big,- Pharma: LLC" # example of a company name
We can start with removing legal control terms (here LLC). To do that, there is an awesome cleanco Python library, which does exactly that:
from cleanco import cleanco
name = cleanco(name).clean_name() # 'THE | big,- Pharma'
Remove all punctuation:
name = name.translate(None, string.punctuation) # 'THE big Pharma'
(for unicode strings, the following code works instead (source, regex):
import regex
name = regex.sub(ur"[[:punct:]]+", "", name) # u'THE big Pharma'
Split the name into tokens using NLTK:
import nltk
tokens = nltk.word_tokenize(name) # ['THE', 'big', 'Pharma']
Lowercase all tokens:
tokens = [t.lower() for t in tokens] # ['the', 'big', 'pharma']
Remove stop words. Note that it might cause problems with companies like On Mars will be incorrectly matched to Mars, because On is a stopword.
from nltk.corpus import stopwords
tokens = [t for t in tokens if t not in stopwords.words('english')] # ['big', 'pharma']
I don't cover accented and special characters here (improvements welcome).
Matching
Now, when we have mapped all company names to tokens, we want to find the matching pairs. Arguably, Jaccard (or Jaro-Winkler) similarity is better than Levenstein for this task, but is still not good enough. The reason is that it does not take into account the importance of words in the name (like TF-IDF does). So common words like "Company" influence the score just as much as words that might uniquely identify company name.
To improve on that, you can use a name similarity trick suggested in this awesome series of posts (not mine). Here is a code example from it:
# token2frequency is just a word counter of all words in all names
# in the dataset
def sequence_uniqueness(seq, token2frequency):
return sum(1/token2frequency(t)**0.5 for t in seq)
def name_similarity(a, b, token2frequency):
a_tokens = set(a.split())
b_tokens = set(b.split())
a_uniq = sequence_uniqueness(a_tokens)
b_uniq = sequence_uniqueness(b_tokens)
return sequence_uniqueness(a.intersection(b))/(a_uniq * b_uniq) ** 0.5
Using that, you can match names with similarity exceeding certain threshold. As a more complex approach, you can also take several scores (say, this uniqueness score, Jaccard and Jaro-Winkler) and train a binary classification model using some labeled data, which will, given a number of scores, output if the candidate pair is a match or not. More on this can be found in the same blog post.
You could use the Levenshtein distance, which could be used to measure the difference between two sequences (basically an edit distance).
Levenshtein Distance in Python
def levenshtein_distance(a,b):
n, m = len(a), len(b)
if n > m:
# Make sure n <= m, to use O(min(n,m)) space
a,b = b,a
n,m = m,n
current = range(n+1)
for i in range(1,m+1):
previous, current = current, [i]+[0]*n
for j in range(1,n+1):
add, delete = previous[j]+1, current[j-1]+1
change = previous[j-1]
if a[j-1] != b[i-1]:
change = change + 1
current[j] = min(add, delete, change)
return current[n]
if __name__=="__main__":
from sys import argv
print levenshtein_distance(argv[1],argv[2])
There is great library for searching for similar/fuzzy strings for python: fuzzywuzzy. It's a nice wrapper library upon mentioned Levenshtein distance measuring.
Here how your names could be analysed:
#!/usr/bin/env python
from fuzzywuzzy import fuzz
names = [
("George Washington Middle Schl",
"George Washington School"),
("Santa Fe East Inc",
"Santa Fe East"),
("Chop't Creative Salad Co",
"Chop't Creative Salad Company"),
("Manny and Olga's Pizza",
"Manny's & Olga's Pizza"),
("Ray's Hell Burger Too",
"Ray's Hell Burgers"),
("El Sol",
"El Sol de America"),
("Olney Theatre Center for the Arts",
"Olney Theatre"),
("21 M Lounge",
"21M Lounge"),
("Holiday Inn Hotel Washington",
"Holiday Inn Washington-Georgetown"),
("Residence Inn Washington,DC/Dupont Circle",
"Residence Inn Marriott Dupont Circle"),
("Jimmy John's Gourmet Sandwiches",
"Jimmy John's"),
("Omni Shoreham Hotel at Washington D.C.",
"Omni Shoreham Hotel"),
]
if __name__ == '__main__':
for pair in names:
print "{:>3} :: {}".format(fuzz.partial_ratio(*pair), pair)
>>> 79 :: ('George Washington Middle Schl', 'George Washington School')
>>> 100 :: ('Santa Fe East Inc', 'Santa Fe East')
>>> 100 :: ("Chop't Creative Salad Co", "Chop't Creative Salad Company")
>>> 86 :: ("Manny and Olga's Pizza", "Manny's & Olga's Pizza")
>>> 94 :: ("Ray's Hell Burger Too", "Ray's Hell Burgers")
>>> 100 :: ('El Sol', 'El Sol de America')
>>> 100 :: ('Olney Theatre Center for the Arts', 'Olney Theatre')
>>> 90 :: ('21 M Lounge', '21M Lounge')
>>> 79 :: ('Holiday Inn Hotel Washington', 'Holiday Inn Washington-Georgetown')
>>> 69 :: ('Residence Inn Washington,DC/Dupont Circle', 'Residence Inn Marriott Dupont Circle')
>>> 100 :: ("Jimmy John's Gourmet Sandwiches", "Jimmy John's")
>>> 100 :: ('Omni Shoreham Hotel at Washington D.C.', 'Omni Shoreham Hotel')
Another way of solving such kind of problems could be Elasticsearch, which also supports fuzzy searches.
I searched for "python edit distance" and this library came as the first result: http://www.mindrot.org/projects/py-editdist/
Another Python library that does the same job is here: http://pypi.python.org/pypi/python-Levenshtein/
An edit distance represents the amount of work you need to carry out to convert one string to another by following only simple -- usually, character-based -- edit operations. Every operation (substition, deletion, insertion; sometimes transpose) has an associated cost and the minimum edit distance between two strings is a measure of how dissimilar the two are.
In your particular case you may want to order the strings so that you find the distance to go from the longer to the shorter and penalize character deletions less (because I see that in many cases one of the strings is almost a substring of the other). So deletion shouldn't be penalized a lot.
You could also make use of this sample code: http://norvig.com/spell-correct.html
This a bit of an update to Dennis comment. That answer was really helpful as was the links he posted but I couldn't get them to work right off. After trying the Fuzzy Wuzzy search I found this gave me a bunch better set of answers. I have a large list of merchants and I just want to group them together. Eventually I'll have a table I can use to try some machine learning to play around with but for now this takes a lot of the effort out of it.
I only had to update his code a little bit and add a function to create the tokens2frequency dictionary. The original article didn't have that either and then the functions didn't reference it correctly.
import pandas as pd
from collections import Counter
from cleanco import cleanco
import regex
import nltk
from nltk.corpus import stopwords
nltk.download('stopwords')
# token2frequency is just a Counter of all words in all names
# in the dataset
def sequence_uniqueness(seq, token2frequency):
return sum(1/token2frequency[t]**0.5 for t in seq)
def name_similarity(a, b, token2frequency):
a_tokens = set(a)
b_tokens = set(b)
a_uniq = sequence_uniqueness(a, token2frequency)
b_uniq = sequence_uniqueness(b, token2frequency)
if a_uniq==0 or b_uniq == 0:
return 0
else:
return sequence_uniqueness(a_tokens.intersection(b_tokens), token2frequency)/(a_uniq * b_uniq) ** 0.5
def parse_name(name):
name = cleanco(name).clean_name()
#name = name.translate(None, string.punctuation)
name = regex.sub(r"[[:punct:]]+", "", name)
tokens = nltk.word_tokenize(name)
tokens = [t.lower() for t in tokens]
tokens = [t for t in tokens if t not in stopwords.words('english')]
return tokens
def build_token2frequency(names):
alltokens = []
for tokens in names.values():
alltokens += tokens
return Counter(alltokens)
with open('marchants.json') as merchantfile:
merchants = pd.read_json(merchantfile)
merchants = merchants.unique()
parsed_names = {merchant:parse_name(merchant) for merchant in merchants}
token2frequency = build_token2frequency(parsed_names)
grouping = {}
for merchant, tokens in parsed_names.items():
grouping[merchant] = {merchant2: name_similarity(tokens, tokens2, token2frequency) for merchant2, tokens2 in parsed_names.items()}
filtered_matches = {}
for merchant in pcard_merchants:
filtered_matches[merchant] = {merchant1: ratio for merchant1, ratio in grouping[merchant].items() if ratio >0.3 }
This will give you a final filtered list of names and the other names they match up to. It's the same basic code as the other post just with a couple of missing pieces filled in. This also is run in Python 3.8
Consider using the Diff-Match-Patch library. You'd be interested in the Diff process - applying a diff on your text can give you a good idea of the differences, along with a programmatic representation of them.
What you can do is separate the words by whitespaces, commas, etc. and then you you count the number of words it have in common with another name and you add a number of words thresold before it is considered "similar".
The other way is to do the same thing, but take the words and splice them for each caracters. Then for each words you need to compare if letters are found in the same order (from both sides) for an x amount of caracters (or percentage) then you can say that the word is similar too.
Ex: You have sqre and square
Then you check by caracters and find that sqre are all in square and in the same order, then it's a similar word.
The algorithms that are based on the Levenshtein distance are good (not perfect) but their main disadvantage is that they are very slow for each comparison and concerning the fact that you would have to compare every possible combination.
Another way of working out the problem would be, to use embedding or bag of words to transform each company name (after some cleaning and prepossessing ) into a vector of numbers. And after that you apply an unsupervised or supervised ML method depending on what is available.
I created matchkraft (https://github.com/MatchKraft/matchkraft-python). It works on top of fuzzy-wuzzy and you can fuzzy match company names in one list.
It is very easy to use. Here is an example in python:
from matchkraft import MatchKraft
mk = MatchKraft('<YOUR API TOKEN HERE>')
job_id = mk.highlight_duplicates(name='Stackoverflow Job',
primary_list=[
'George Washington Middle Schl',
'George Washington School',
'Santa Fe East Inc',
'Santa Fe East',
'Rays Hell Burger Too',
'El Sol de America',
'microsoft',
'Olney Theatre',
'El Sol'
]
)
print (job_id)
mk.execute_job(job_id=job_id)
job = mk.get_job_information(job_id=job_id)
print (job.status)
while (job.status!='Completed'):
print (job.status)
time.sleep(10)
job = mk.get_job_information(job_id=job_id)
results = mk.get_results_information(job_id=job_id)
if isinstance(results, list):
for r in results:
print(r.master_record + ' --> ' + r.match_record)
else:
print("No Results Found")