I have a set of text files with blurbs of text and I need to search these for a particular keyword such that a set of words before and/or after the keyword (i.e. phrases) are returned along with a count of the phrases across the files. For example, contents of a few of files are:
File 1: This is a great day. I wish I could go to a beautiful green park today but unfortunately, we are in a lockdown!
File 2: Having a beautiful green park close to your house is great.
File 3: I visited a green park today. My friend also visited a green park today.
So if I search for the keyword park, I'm looking for the output to be a set of phrases (let's say one word before & after park), ranked based on how many times the phrase occurs across files. So in this example, the output should be:
green park today: 2
green park close: 1
Is there a way I can achieve this in Python, maybe using some NLP libraries or even without them. I have some code in my post here but that doesn't solve the purpose (I'll perhaps delete that post once I get a response to this one).
Thank you
Based on your expected output above, it looks like you only want to add one to the count for a single phrase per file (even if it appears several times in the same file). Below is an example of how you can do this without any special NLP libraries, just defining "words" as chains of non-space characters delimited by spaces (I'm assuming you know how to read text from a file so leaving that part out).
from collections import Counter
str1 = "This is a great day. I wish I could go to a beautiful green park today but unfortunately, we are in a lockdown!"
str2 = "Having a beautiful green park close to your house is great."
str3 = "I visited a green park today. My friend also visited a green park today."
str1_words = ["START"] + str1.split(" ") + ["END"]
str2_words = ["START"] + str2.split(" ") + ["END"]
str3_words = ["START"] + str3.split(" ") + ["END"]
print(str1_words)
all_phrases = []
SEARCH_WORD = "park"
for words in [str1_words, str2_words, str3_words]:
phrases = []
for i in range(1, len(words) - 1):
if words[i] == SEARCH_WORD:
phrases.append(" ".join(words[i-1:i+2]))
# Only count each phrase once for this text
phrases = set(phrases)
all_phrases.extend(phrases)
phrase_count = Counter(all_phrases)
print(phrase_count.most_common())
The output is:
[('green park today', 1), ('green park close', 1), ('green park today.', 1)]
This perfectly demonstrates the problem with the definition of a "word" above - punctuation is treated as part of the word. For a better way to do it, look into the NLTK library, specifically methods for "word tokenization".
Hopefully the above gives you an idea of how to get started on this.
I have the following sentences:
sent_1 = 'The cat caught the mouse.'
sent_2 = 'The cat caught and killed the mouse.'
Now I want to know who did what to whoom. Spacy's noun_chunks work perfectly in the first case, indicating "The cat" as the "nsubj" with the chunk.root.head.text being "caught". Likewise, "the mouse" is correctly classified as being the "dobj" with again "caught" as chunk.root.head.text. So it is easy to match these two.
However, in the second case, the nsubj gets "caught" as its chunk.root.head.text while the dobj gets "killed", whereas they actually would belong together. Is there a way to account for this kind of cases?
In the second case 'killed' is the head of the 'the mouse' as it is the text connecting the noun chunk to the rest of the phrase. From the spacy documentation:
Root text: The original text of the word connecting the noun chunk to the rest of the parse.
https://spacy.io/usage/linguistic-features#noun-chunks
N.b. that link has a very similar example to yours - a sentence with multiple noun chunks with different roots. ('Autonomous cars shift insurance liability toward manufacturers')
To answer your question, if you want 'caught' to be found as the head in both instances, then really what you're asking for is to recursively check the head of the tree for each noun_chunk... something like this:
nlp = spacy.load('en_core_web_sm')
doc = nlp('The cat caught and killed the mouse.')
[x.root.head.head for x in doc.noun_chunks]
which avails:
[caught, caught]
N.b, this works for your example but if you needed to handle arbitrary sentences then you'd need to do something a bit more sophisticated, i.e. actually recursing the tree. e.g.
def get_head(x):
return x.head if x.head.head == x.head else get_head(x.head)
resulting in:
doc2 = nlp("Autonomous cars shift insurance liability toward manufacturers away from everyday users") # adapted from the spacy example with an additional NC 'everyday users' added
In [17]: [get_head(x.root.head) for x in doc.noun_chunks]
In [187]: [caught, caught]
In [18]: [get_head(x.root.head) for x in doc2.noun_chunks]
Out[18]: [shift, shift, shift, shift]
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
I'm trying to extract ONLY one string that contains $ character. The input based on output that I extracted using BeautifulSoup.
Code
price = [m.split() for m in re.findall(r"\w+/$(?:\s+\w+/$)*", soup_content.find('blockquote', { "class": "postcontent restore" }).text)]
Input
For Sale is my Tag Heuer Carrera Calibre 6 with box and papers and extras.
39mm
47 ish lug to lug
19mm in between lugs
Pretty thin but not sure exact height. Likely around 12mm (maybe less)
I've owned it for about 2 years. I absolutely love the case on this watch. It fits my wrist and sits better than any other watch I've ever owned. I'm selling because I need cash and other pieces have more sentimental value
I am the second owner, but the first barely wore it.
It comes with barely worn blue leather strap, extra suede strap that matches just about perfectly and I'll include a blue Barton Band Elite Silicone.
I also purchased an OEM bracelet that I personally think takes the watch to a new level. This model never came with a bracelet and it was several hundred $ to purchase after the fact.
The watch was worn in rotation and never dropped or knocked around.
The watch does have hairlines, but they nearly all superficial. A bit of time with a cape cod cloth would take care of a lot it them. The pics show the imperfections in at "worst" possible angle to show the nature of scratches.
The bracelet has a few desk diving marks, but all in all, the watch and bracelet are in very good shape.
Asking $2000 obo. PayPal shipped. CONUS.
It's a big hard to compare with others for sale as this one includes the bracelet.
The output should be like this.
2000
You don't need a regex. Instead you can iterate over lines and over each word to check for starting with '$' and extract the word:
[word[1:] for line in s.split('\n') for word in line.split() if word.startswith('$') and len(word) > 1]
where s is your paragraph.
which outputs:
['2000']
Since this is very simple you don't need a regex solution, this should sufice:
words = text.split()
words_with_dollar = [word for word in words if '$' in word]
print(words_with_dollar)
>>> ['$', '$2000']
If you don't want the dollar sign alone, you can add a filter like this:
words_with_dollar = [word for word in words if '$' in word and '$' != word]
print(words_with_dollar)
>>> ['$2000']
I would do something like that (provided input is the string you wrote above)-
price_start = input.find('$')
price = input[price_start:].split(' ')[0]
IF there is only 1 occurrence like you said.
Alternative- you could use regex like that-
price = re.findall('\S*\$\S*\d', input)[0]
price = price.replace('$', '')
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")