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How to label multi-word entities?

I'm quite new to data analysis (and Python in general), and I'm currently a bit stuck in my project.
For my NLP-task I need to create training data, i.e. find specific entities in sentences and label them. I have multiple csv files containing the entities I am trying to find, many of them consisting of multiple words. I have tokenized and lemmatized the unlabeled sentences with spaCy and loaded them into a pandas.DataFrame.
My main problem is: how do I now compare the tokenized sentences with the entity-lists and label the (often multi-word) entities? Having around 0.5 GB of sentences, I don't think it is feasible to just for-loop every sentence and then for-loop every entity in every class-list and do a simple substring-search. Is there any smart way to use pandas.Series or DataFrame to do this labeling?
As mentioned, I don't really have any experience regarding pandas/numpy etc. and after a lot of web searching I still haven't seemed to find the answer to my problem
Say that this is a sample of finance.csv, one of my entity lists:
"Frontwave Credit Union",
"St. Mary's Bank",
"Center for Financial Services Innovation",
...
And that this is a sample of sport.csv, another one of my entity lists:
"Christiano Ronaldo",
"Lewis Hamilton",
...
And an example (dumb) sentence:
"Dear members of Frontwave Credit Union, any credit demanded by Lewis Hamilton is invalid, said Ronaldo"
The result I'd like would be something like a table of tokens with the matching entity labels (with IOB labeling):
"Dear "- O
"members" - O
"of" - O
"Frontwave" - B-FINANCE
"Credit" - I-FINANCE
"Union" - I-FINANCE
"," - O
"any" - O
...
"Lewis" - B-SPORT
"Hamilton" - I-SPORT
...
"said" - O
"Ronaldo" - O

Use:
FINANCE = ["Frontwave Credit Union",
"St. Mary's Bank",
"Center for Financial Services Innovation"]
SPORT = [
"Christiano Ronaldo",
"Lewis Hamilton",
]
FINANCE = '|'.join(FINANCE)
sent = pd.DataFrame({'sent': ["Dear members of Frontwave Credit Union, any credit demanded by Lewis Hamilton is invalid, said Ronaldo"]})
home = sent['sent'].str.extractall(f'({FINANCE})')
def labeler(row, group):
l = len(row.split())
return [f'I-{group}' if i !=0 else f'B-{group}' for i in range(l)]
home[0].apply(labeler, group='FINANCE').explode()

Python - Search large list of keywords in a large unstructured data

I have large bag of keywords that I want to search in large unstructured data and auto-tag this content.
First few steps that I took was to eliminate to pre-process the data like: removing stop words, punctuation, case-sentivity, check high-frequency words and delete them if I don't need. I adapted few things from this article: https://medium.com/analytics-vidhya/automated-keyword-extraction-from-articles-using-nlp-bfd864f41b34.
After the pre-processing, I am un-sure how I should be searching the keywords and how should I be auto-tagging them in a efficient way. Are there any good machine learning tutorials, algorithms, and/or Natural Language Processing that can do such task much better than what I am doing now?
Here is an example:
Bag of keywords:
keywords_to_search = ['Harry Potter','LOTR','Lord of the Rings','Secret Garden','Pinocchio'] # Some example, I have over 100K list of keywords to search
l1 = ['abc','def','ghi','jkl']
l2 = ['Pinocchio and Harry Potter is a famous children\'s book..','LOTR was written by J. R. R. Tolkien','Fordo Baggins is a character in Lord of the Rings Book Series.','blank']
df = pd.DataFrame({'some_col':l1,'text':l2})
Try:
df[df.text.str.contains('|'.join(keywords_to_search))] # for searching ... very slow for large unstructured texts
Looking for this:
some_col text tags
abc Pinocchio and Harry Potter is a famous children's book. Pinocchio,Harry Potter
def LOTR was written by J. R. R. Tolkien LOTR
ghi Fordo Baggins is a character in Lord of the Ri... Lord of the Rings

How to Find Company Names in Text Using Python

I have a list of properly-formatted company names, and I am trying to find when those companies appear in a document. The problem is that they are unlikely to appear in the document exactly as they do in the list. For example, Visa Inc may appear as Visa or American Airlines Group Inc may appear as American Airlines.
How would I go about iterating over the entire contents of the document and then return the properly formatted company name when a close match is found?
I have tried both fuzzywuzzy and difflib.get_close_matches, but the problem is it looks at each individual word rather than clusters of words:
from fuzzywuzzy import process
from difflib import get_close_matches
company_name = ['American Tower Inc', 'American Airlines Group Inc', 'Atlantic American Corp', 'American International Group']
text = 'American Tower is one company. American Airlines is another while there is also Atlantic American Corp but we cannot forget about American International Group Inc.'
#using fuzzywuzzy
for word in text.split():
print('- ' + word+', ', ', '.join(map(str,process.extractOne(word, company_name))))
#using get_close_matches
for word in text.split():
match = get_close_matches(word, company_name, n=1, cutoff=.4)
print(match)

I was working on a similar problem. Fuzzywuzzy internally uses difflib and both of them perform slowly on large datasets.
Chris van den Berg's pipeline converts company names into vectors of 3-grams using a TF-IDF matrix and then compares the vectors using cosine similarity.
The pipeline is quick and gives accurate results for partially matched strings too.

For that type of task I use a record linkage algorithm, it will find those clusters for you with the help of ML. You will have to provide some actual examples so the algorithm can learn to label the rest of your dataset properly.
Here is some info:
https://pypi.org/project/pandas-dedupe/
Cheers,

Object Standarization Using NLTK

I'm new to NLP and to Python.
I'm trying to use object standardization to replace abbreviations with their full meaning. I found code online and altered it to test it out on a wikipedia exert. but all the code does is print out the original text. Can any one help out a newbie in need?
heres the code:
import nltk
lookup_dict = {'EC': 'European Commission', 'EU': 'European Union', "ECSC": "European Coal and Steel Commuinty",
"EEC": "European Economic Community"}
def _lookup_words(input_text):
words = input_text.split()
new_words = []
for word in words:
if word.lower() in lookup_dict:
word = lookup_dict[word.lower()]
new_words.append(word)
new_text = " ".join(new_words)
print(new_text)
return new_text
_lookup_words(
"The High Authority was the supranational administrative executive of the new European Coal and Steel Community ECSC. It took office first on 10 August 1952 in Luxembourg. In 1958, the Treaties of Rome had established two new communities alongside the ECSC: the eec and the European Atomic Energy Community (Euratom). However their executives were called Commissions rather than High Authorities")
Thanks in advance, any help is appreciated!

In your case, the lookup dict has the abbreviations for EC and ECSC amongs the words found in your input sentence. Calling split splits the input based on whitespace. But your sentence has the words ECSC. and ECSC: ,ie these are the tokens obtained post splitting as opposed to ECSC thus you are not able to map the input. I would suggest to do some depunctuation and run it again.

Figure out if a business name is very similar to another one - Python

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")