Grouping Similar Strings - python

I'm trying to analyze a bunch of search terms, so many that individually they don't tell much. That said, I'd like to group the terms because I think similar terms should have similar effectiveness. For example,
Term Group
NBA Basketball 1
Basketball NBA 1
Basketball 1
Baseball 2
It's a contrived example, but hopefully it explains what I'm trying to do. So then, what is the best way to do what I've described? I thought the nltk may have something along those lines, but I'm only barely familiar with it.
Thanks

You'll want to cluster these terms, and for the similarity metric I recommend Dice's Coefficient at the character-gram level. For example, partition the strings into two-letter sequences to compare (term1="NB", "BA", "A ", " B", "Ba"...).
nltk appears to provide dice as nltk.metrics.association.BigramAssocMeasures.dice(), but it's simple enough to implement in a way that'll allow tuning. Here's how to compare these strings at the character rather than word level.
import sys, operator
def tokenize(s, glen):
g2 = set()
for i in xrange(len(s)-(glen-1)):
g2.add(s[i:i+glen])
return g2
def dice_grams(g1, g2): return (2.0*len(g1 & g2)) / (len(g1)+len(g2))
def dice(n, s1, s2): return dice_grams(tokenize(s1, n), tokenize(s2, n))
def main():
GRAM_LEN = 4
scores = {}
for i in xrange(1,len(sys.argv)):
for j in xrange(i+1, len(sys.argv)):
s1 = sys.argv[i]
s2 = sys.argv[j]
score = dice(GRAM_LEN, s1, s2)
scores[s1+":"+s2] = score
for item in sorted(scores.iteritems(), key=operator.itemgetter(1)):
print item
When this program is run with your strings, the following similarity scores are produced:
./dice.py "NBA Basketball" "Basketball NBA" "Basketball" "Baseball"
('NBA Basketball:Baseball', 0.125)
('Basketball NBA:Baseball', 0.125)
('Basketball:Baseball', 0.16666666666666666)
('NBA Basketball:Basketball NBA', 0.63636363636363635)
('NBA Basketball:Basketball', 0.77777777777777779)
('Basketball NBA:Basketball', 0.77777777777777779)
At least for this example, the margin between the basketball and baseball terms should be sufficient for clustering them into separate groups. Alternatively you may be able to use the similarity scores more directly in your code with a threshold.

Related

how can i refactor my python code to decrease the time complexity

this code takes 9 sec which is very long time, i guess the problem in 2 loop in my code
for symptom in symptoms:
# check if the symptom is mentioned in the user text
norm_symptom = symptom.replace("_"," ")
for combin in list_of_combinations:
print(getSimilarity([combin, norm_symptom]))
if getSimilarity([combin,norm_symptom])>0.25:
if symptom not in extracted_symptoms:
extracted_symptoms.append(symptom)
i tried to use zip like this:
for symptom, combin in zip(symptoms,list_of_combinations):
norm_symptom = symptom.replace("_"," ")
if (getSimilarity([combin, norm_symptom]) > 0.25 and symptom not in extracted_symptoms):
extracted_symptoms.append(symptom)
Indeed, you're algorithm is slow because of the 2 nested loops.
It performs with big O N*M (see more here https://www.freecodecamp.org/news/big-o-notation-why-it-matters-and-why-it-doesnt-1674cfa8a23c/)
N being the lenght of symptoms
and M being the list_of_combinations
What can takes time also is the computation getSimilarity, what is this operation ?
Use a dict to store the results of getSimilarity for each combination and symptom. This way, you can avoid calling getSimilarity multiple times for the same combination and symptom. This way it will be more efficient, thus faster.
import collections
similarity_results = collections.defaultdict(dict)
for symptom in symptoms:
norm_symptom = symptom.replace("_"," ")
for combin in list_of_combinations:
# Check if the similarity has already been computed
if combin in similarity_results[symptom]:
similarity = similarity_results[symptom][combin]
else:
similarity = getSimilarity([combin, norm_symptom])
similarity_results[symptom][combin] = similarity
if similarity > 0.25:
if symptom not in extracted_symptoms:
extracted_symptoms.append(symptom)
Update:
Alternatively You could use an algorithm based on the Levenshtein distance, which is a measure of the minimum number of single-character edits (insertions, deletions, or substitutions) required to transform one string into another. Python-Levenshtein library does that.
import Levenshtein
def getSimilarity(s1, s2):
distance = Levenshtein.distance(s1, s2)
return 1.0 - (distance / max(len(s1), len(s2)))
extracted_symptoms = []
for symptom, combin in zip(symptoms, list_of_combinations):
norm_symptom = symptom.replace("_", " ")
if (getSimilarity(combin, norm_symptom) > 0.25) and (symptom not in extracted_symptoms):
extracted_symptoms.append(symptom)

How to count the top 150 words and remove common words from 2 lists?

This code below is to find out the top 150 words which appeared the most in 2 strings.
pwords = re.findall(r'\w+',p)
ptop150words=Counter(pwords).most_common(150)
sorted(ptop150words)
nwords = re.findall(r'\w+',n)
ntop150words=Counter(nwords).most_common(150)
sorted(ntop150words)
This code below is to remove the common words which appeared in the 2 strings.
def new(ntopwords,ptopwords):
for i in ntopwords[:]:
if i in potopwords:
ntopwords.remove(i)
ptopwords.remove(i)
print(i)
However, there is no output for print(i). what is wrong?
Most likely your indentation.
new(negativetop150words,positivetop150words):
for i in negativetop150words[:]:
if i in positivetop150words:
negativetop150words.remove(i)
positivetop150words.remove(i)
print(i)
You could rely on set methods. Once you have both lists, you convert them to sets. The common subset is the intersection of the 2 sets, and you can simply take the difference from both original sets:
positiveset = set(positivewords)
negativeset = set(negativewords)
commons = positiveset & negativeset
positivewords = sorted(positiveset - commons)
negativewords = sorted(negativeset - commons)
commonwords = sorted(commons)
The code you posted does not call the function new(negativetop150words, positivetop150words) Also per Jesse's comment, the print(i) command is outside the function. Here's the code that worked for me:
import re
from collections import Counter
def new(negativetop150words, positivetop150words):
for i in negativetop150words[:]:
if i in positivetop150words:
negativetop150words.remove(i)
positivetop150words.remove(i)
print(i)
return negativetop150words, positivetop150words
positive = 'The FDA is already fairly gung-ho about providing this. It receives about 1,000 applications a year and approves all but 1%. The agency makes sure there is sound science behind the request, and no obvious indication that the medicine would harm the patient.'
negative = 'Thankfully these irritating bits of bureaucracy have been duly dispatched. This victory comes courtesy of campaigning work by a libertarian think-tank, the Goldwater Institute, based in Arizona. It has been pushing right-to-try legislation for around four years, and it can now be found in 40 states. Speaking about the impact of these laws on patients, Arthur Caplan, a professor of bioethics at NYU School of Medicine in New York, says he can think of one person who may have been helped.'
positivewords = re.findall(r'\w+', positive)
positivetop150words = Counter(positivewords).most_common(150)
sorted(positivetop150words)
negativewords = re.findall(r'\w+', negative)
negativetop150words = Counter(negativewords).most_common(150)
words = new(negativewords, positivewords)
This prints:
a
the
It
and
about
the

Why does my association model find subgroups in a dataset when there shouldn't any?

I give a lot of information on the methods that I used to write my code. If you just want to read my question, skip to the quotes at the end.
I'm working on a project that has a goal of detecting sub populations in a group of patients. I thought this sounded like the perfect opportunity to use association rule mining as I'm currently taking a class on the subject.
I there are 42 variables in total. Of those, 20 are continuous and had to be discretized. For each variable, I used the Freedman-Diaconis rule to determine how many categories to divide a group into.
def Freedman_Diaconis(column_values):
#sort the list first
column_values[1].sort()
first_quartile = int(len(column_values[1]) * .25)
third_quartile = int(len(column_values[1]) * .75)
fq_value = column_values[1][first_quartile]
tq_value = column_values[1][third_quartile]
iqr = tq_value - fq_value
n_to_pow = len(column_values[1])**(-1/3)
h = 2 * iqr * n_to_pow
retval = (column_values[1][-1] - column_values[1][1])/h
test = int(retval+1)
return test
From there I used min-max normalization
def min_max_transform(column_of_data, num_bins):
min_max_normalizer = preprocessing.MinMaxScaler(feature_range=(1, num_bins))
data_min_max = min_max_normalizer.fit_transform(column_of_data[1])
data_min_max_ints = take_int(data_min_max)
return data_min_max_ints
to transform my data and then I simply took the interger portion to get the final categorization.
def take_int(list_of_float):
ints = []
for flt in list_of_float:
asint = int(flt)
ints.append(asint)
return ints
I then also wrote a function that I used to combine this value with the variable name.
def string_transform(prefix, column, index):
transformed_list = []
transformed = ""
if index < 4:
for entry in column[1]:
transformed = prefix+str(entry)
transformed_list.append(transformed)
else:
prefix_num = prefix.split('x')
for entry in column[1]:
transformed = str(prefix_num[1])+'x'+str(entry)
transformed_list.append(transformed)
return transformed_list
This was done to differentiate variables that have the same value, but appear in different columns. For example, having a value of 1 for variable x14 means something different from getting a value of 1 in variable x20. The string transform function would create 14x1 and 20x1 for the previously mentioned examples.
After this, I wrote everything to a file in basket format
def create_basket(list_of_lists, headers):
#for filename in os.listdir("."):
# if filename.e
if not os.path.exists('baskets'):
os.makedirs('baskets')
down_length = len(list_of_lists[0])
with open('baskets/dataset.basket', 'w') as basketfile:
basket_writer = csv.DictWriter(basketfile, fieldnames=headers)
for i in range(0, down_length):
basket_writer.writerow({"trt": list_of_lists[0][i], "y": list_of_lists[1][i], "x1": list_of_lists[2][i],
"x2": list_of_lists[3][i], "x3": list_of_lists[4][i], "x4": list_of_lists[5][i],
"x5": list_of_lists[6][i], "x6": list_of_lists[7][i], "x7": list_of_lists[8][i],
"x8": list_of_lists[9][i], "x9": list_of_lists[10][i], "x10": list_of_lists[11][i],
"x11": list_of_lists[12][i], "x12":list_of_lists[13][i], "x13": list_of_lists[14][i],
"x14": list_of_lists[15][i], "x15": list_of_lists[16][i], "x16": list_of_lists[17][i],
"x17": list_of_lists[18][i], "x18": list_of_lists[19][i], "x19": list_of_lists[20][i],
"x20": list_of_lists[21][i], "x21": list_of_lists[22][i], "x22": list_of_lists[23][i],
"x23": list_of_lists[24][i], "x24": list_of_lists[25][i], "x25": list_of_lists[26][i],
"x26": list_of_lists[27][i], "x27": list_of_lists[28][i], "x28": list_of_lists[29][i],
"x29": list_of_lists[30][i], "x30": list_of_lists[31][i], "x31": list_of_lists[32][i],
"x32": list_of_lists[33][i], "x33": list_of_lists[34][i], "x34": list_of_lists[35][i],
"x35": list_of_lists[36][i], "x36": list_of_lists[37][i], "x37": list_of_lists[38][i],
"x38": list_of_lists[39][i], "x39": list_of_lists[40][i], "x40": list_of_lists[41][i]})
and I used the apriori package in Orange to see if there were any association rules.
rules = Orange.associate.AssociationRulesSparseInducer(patient_basket, support=0.3, confidence=0.3)
print "%4s %4s %s" % ("Supp", "Conf", "Rule")
for r in rules:
my_rule = str(r)
split_rule = my_rule.split("->")
if 'trt' in split_rule[1]:
print 'treatment rule'
print "%4.1f %4.1f %s" % (r.support, r.confidence, r)
Using this, technique I found quite a few association rules with my testing data.
THIS IS WHERE I HAVE A PROBLEM
When I read the notes for the training data, there is this note
...That is, the only
reason for the differences among observed responses to the same treatment across patients is
random noise. Hence, there is NO meaningful subgroup for this dataset...
My question is,
why do I get multiple association rules that would imply that there are subgroups, when according to the notes I shouldn't see anything?
I'm getting lift numbers that are above 2 as opposed to the 1 that you should expect if everything was random like the notes state.
Supp Conf Rule
0.3 0.7 6x0 -> trt1
Even though my code runs, I'm not getting results anywhere close to what should be expected. This leads me to believe that I messed something up, but I'm not sure what it is.
After some research, I realized that my sample size is too small for the number of variables that I have. I would need a way larger sample size in order to really use the method that I was using. In fact, the method that I tried to use was developed with the assumption that it would be run on databases with hundreds of thousands or millions of rows.

How to compare clusters?

Hopefully this can be done with python! I used two clustering programs on the same data and now have a cluster file from both. I reformatted the files so that they look like this:
Cluster 0:
Brucellaceae(10)
Brucella(10)
abortus(1)
canis(1)
ceti(1)
inopinata(1)
melitensis(1)
microti(1)
neotomae(1)
ovis(1)
pinnipedialis(1)
suis(1)
Cluster 1:
Streptomycetaceae(28)
Streptomyces(28)
achromogenes(1)
albaduncus(1)
anthocyanicus(1)
etc.
These files contain bacterial species info. So I have the cluster number (Cluster 0), then right below it 'family' (Brucellaceae) and the number of bacteria in that family (10). Under that is the genera found in that family (name followed by number, Brucella(10)) and finally the species in each genera (abortus(1), etc.).
My question: I have 2 files formatted in this way and want to write a program that will look for differences between the two. The only problem is that the two programs cluster in different ways, so two cluster may be the same, even if the actual "Cluster Number" is different (so the contents of Cluster 1 in one file might match Cluster 43 in the other file, the only different being the actual cluster number). So I need something to ignore the cluster number and focus on the cluster contents.
Is there any way I could compare these 2 files to examine the differences? Is it even possible? Any ideas would be greatly appreciated!
Given:
file1 = '''Cluster 0:
giant(2)
red(2)
brick(1)
apple(1)
Cluster 1:
tiny(3)
green(1)
dot(1)
blue(2)
flower(1)
candy(1)'''.split('\n')
file2 = '''Cluster 18:
giant(2)
red(2)
brick(1)
tomato(1)
Cluster 19:
tiny(2)
blue(2)
flower(1)
candy(1)'''.split('\n')
Is this what you need?
def parse_file(open_file):
result = []
for line in open_file:
indent_level = len(line) - len(line.lstrip())
if indent_level == 0:
levels = ['','','']
item = line.lstrip().split('(', 1)[0]
levels[indent_level - 1] = item
if indent_level == 3:
result.append('.'.join(levels))
return result
data1 = set(parse_file(file1))
data2 = set(parse_file(file2))
differences = [
('common elements', data1 & data2),
('missing from file2', data1 - data2),
('missing from file1', data2 - data1) ]
To see the differences:
for desc, items in differences:
print desc
print
for item in items:
print '\t' + item
print
prints
common elements
giant.red.brick
tiny.blue.candy
tiny.blue.flower
missing from file2
tiny.green.dot
giant.red.apple
missing from file1
giant.red.tomato
So just for help, as I see lots of different answers in the comment, I'll give you a very, very simple implementation of a script that you can start from.
Note that this does not answer your full question but points you in one of the directions in the comments.
Normally if you have no experience I'd argue to go a head and read up on Python (which i'll do anyways, and i'll throw in a few links in the bottom of the answer)
On to the fun stuffs! :)
class Cluster(object):
'''
This is a class that will contain your information about the Clusters.
'''
def __init__(self, number):
'''
This is what some languages call a constructor, but it's not.
This method initializes the properties with values from the method call.
'''
self.cluster_number = number
self.family_name = None
self.bacteria_name = None
self.bacteria = []
#This part below isn't a part of the class, this is the actual script.
with open('bacteria.txt', 'r') as file:
cluster = None
clusters = []
for index, line in enumerate(file):
if line.startswith('Cluster'):
cluster = Cluster(index)
clusters.append(cluster)
else:
if not cluster.family_name:
cluster.family_name = line
elif not cluster.bacteria_name:
cluster.bacteria_name = line
else:
cluster.bacteria.append(line)
I wrote this as dumb and overly simple as I could without any fancy stuff and for Python 2.7.2
You could copy this file into a .py file and run it directly from command line python bacteria.py for example.
Hope this helps a bit and don't hesitate to come by our Python chat room if you have any questions! :)
http://learnpythonthehardway.org/
http://www.diveintopython.net/
http://docs.python.org/2/tutorial/inputoutput.html
check if all elements in a list are identical
Retaining order while using Python's set difference
You have to write some code to parse the file. If you ignore the cluster, you should be able to distinguish between family, genera and species based on indentation.
The easiest way it to define a named tuple:
import collections
Bacterium = collections.namedtuple('Bacterium', ['family', 'genera', 'species'])
You can make in instance of this object like this:
b = Bacterium('Brucellaceae', 'Brucella', 'canis')
Your parser should read a file line by line, and set the family and genera. If it then finds a species, it should add a Bacterium to a list;
with open('cluster0.txt', 'r') as infile:
lines = infile.readlines()
family = None
genera = None
bacteria = []
for line in lines:
# set family and genera.
# if you detect a bacterium:
bacteria.append(Bacterium(family, genera, species))
Once you have a list of all bacteria in each file or cluster, you can select from all the bacteria like this:
s = [b for b in bacteria if b.genera == 'Streptomycetaceae']
Comparing two clusterings is not trivial task and reinventing the wheel is unlikely to be successful. Check out this package which has lots of different cluster similarity metrics and can compare dendrograms (the data structure you have).
The library is called CluSim and can be found here:
https://github.com/Hoosier-Clusters/clusim/
After learning so much from Stackoverflow, finally I have an opportunity to give back! A different approach from those offered so far is to relabel clusters to maximize alignment, and then comparison becomes easy. For example, if one algorithm assigns labels to a set of six items as L1=[0,0,1,1,2,2] and another assigns L2=[2,2,0,0,1,1], you want these two labelings to be equivalent since L1 and L2 are essentially segmenting items into clusters identically. This approach relabels L2 to maximize alignment, and in the example above, will result in L2==L1.
I found a soution to this problem in "Menéndez, Héctor D. A genetic approach to the graph and spectral clustering problem. MS thesis. 2012." and below is an implementation in Python using numpy. I'm relatively new to Python, so there may be better implementations, but I think this gets the job done:
def alignClusters(clstr1,clstr2):
"""Given 2 cluster assignments, this funciton will rename the second to
maximize alignment of elements within each cluster. This method is
described in in Menéndez, Héctor D. A genetic approach to the graph and
spectral clustering problem. MS thesis. 2012. (Assumes cluster labels
are consecutive integers starting with zero)
INPUTS:
clstr1 - The first clustering assignment
clstr2 - The second clustering assignment
OUTPUTS:
clstr2_temp - The second clustering assignment with clusters renumbered to
maximize alignment with the first clustering assignment """
K = np.max(clstr1)+1
simdist = np.zeros((K,K))
for i in range(K):
for j in range(K):
dcix = clstr1==i
dcjx = clstr2==j
dd = np.dot(dcix.astype(int),dcjx.astype(int))
simdist[i,j] = (dd/np.sum(dcix!=0) + dd/np.sum(dcjx!=0))/2
mask = np.zeros((K,K))
for i in range(K):
simdist_vec = np.reshape(simdist.T,(K**2,1))
I = np.argmax(simdist_vec)
xy = np.unravel_index(I,simdist.shape,order='F')
x = xy[0]
y = xy[1]
mask[x,y] = 1
simdist[x,:] = 0
simdist[:,y] = 0
swapIJ = np.unravel_index(np.where(mask.T),simdist.shape,order='F')
swapI = swapIJ[0][1,:]
swapJ = swapIJ[0][0,:]
clstr2_temp = np.copy(clstr2)
for k in range(swapI.shape[0]):
swapj = [swapJ[k]==i for i in clstr2]
clstr2_temp[swapj] = swapI[k]
return clstr2_temp

Find the similarity metric between two strings

How do I get the probability of a string being similar to another string in Python?
I want to get a decimal value like 0.9 (meaning 90%) etc. Preferably with standard Python and library.
e.g.
similar("Apple","Appel") #would have a high prob.
similar("Apple","Mango") #would have a lower prob.
There is a built in.
from difflib import SequenceMatcher
def similar(a, b):
return SequenceMatcher(None, a, b).ratio()
Using it:
>>> similar("Apple","Appel")
0.8
>>> similar("Apple","Mango")
0.0
Solution #1: Python builtin
use SequenceMatcher from difflib
pros:
native python library, no need extra package.
cons: too limited, there are so many other good algorithms for string similarity out there.
example :
>>> from difflib import SequenceMatcher
>>> s = SequenceMatcher(None, "abcd", "bcde")
>>> s.ratio()
0.75
Solution #2: jellyfish library
its a very good library with good coverage and few issues.
it supports:
- Levenshtein Distance
- Damerau-Levenshtein Distance
- Jaro Distance
- Jaro-Winkler Distance
- Match Rating Approach Comparison
- Hamming Distance
pros:
easy to use, gamut of supported algorithms, tested.
cons: not native library.
example:
>>> import jellyfish
>>> jellyfish.levenshtein_distance(u'jellyfish', u'smellyfish')
2
>>> jellyfish.jaro_distance(u'jellyfish', u'smellyfish')
0.89629629629629637
>>> jellyfish.damerau_levenshtein_distance(u'jellyfish', u'jellyfihs')
1
I think maybe you are looking for an algorithm describing the distance between strings. Here are some you may refer to:
Hamming distance
Levenshtein distance
Damerau–Levenshtein distance
Jaro–Winkler distance
TheFuzz is a package that implements Levenshtein distance in python, with some helper functions to help in certain situations where you may want two distinct strings to be considered identical. For example:
>>> fuzz.ratio("fuzzy wuzzy was a bear", "wuzzy fuzzy was a bear")
91
>>> fuzz.token_sort_ratio("fuzzy wuzzy was a bear", "wuzzy fuzzy was a bear")
100
You can create a function like:
def similar(w1, w2):
w1 = w1 + ' ' * (len(w2) - len(w1))
w2 = w2 + ' ' * (len(w1) - len(w2))
return sum(1 if i == j else 0 for i, j in zip(w1, w2)) / float(len(w1))
Note, difflib.SequenceMatcher only finds the longest contiguous matching subsequence, this is often not what is desired, for example:
>>> a1 = "Apple"
>>> a2 = "Appel"
>>> a1 *= 50
>>> a2 *= 50
>>> SequenceMatcher(None, a1, a2).ratio()
0.012 # very low
>>> SequenceMatcher(None, a1, a2).get_matching_blocks()
[Match(a=0, b=0, size=3), Match(a=250, b=250, size=0)] # only the first block is recorded
Finding the similarity between two strings is closely related to the concept of pairwise sequence alignment in bioinformatics. There are many dedicated libraries for this including biopython. This example implements the Needleman Wunsch algorithm:
>>> from Bio.Align import PairwiseAligner
>>> aligner = PairwiseAligner()
>>> aligner.score(a1, a2)
200.0
>>> aligner.algorithm
'Needleman-Wunsch'
Using biopython or another bioinformatics package is more flexible than any part of the python standard library since many different scoring schemes and algorithms are available. Also, you can actually get the matching sequences to visualise what is happening:
>>> alignment = next(aligner.align(a1, a2))
>>> alignment.score
200.0
>>> print(alignment)
Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-
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App-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-el
Package distance includes Levenshtein distance:
import distance
distance.levenshtein("lenvestein", "levenshtein")
# 3
You can find most of the text similarity methods and how they are calculated under this link: https://github.com/luozhouyang/python-string-similarity#python-string-similarity
Here some examples;
Normalized, metric, similarity and distance
(Normalized) similarity and distance
Metric distances
Shingles (n-gram) based similarity and distance
Levenshtein
Normalized Levenshtein
Weighted Levenshtein
Damerau-Levenshtein
Optimal String Alignment
Jaro-Winkler
Longest Common Subsequence
Metric Longest Common Subsequence
N-Gram
Shingle(n-gram) based algorithms
Q-Gram
Cosine similarity
Jaccard index
Sorensen-Dice coefficient
Overlap coefficient (i.e.,Szymkiewicz-Simpson)
The builtin SequenceMatcher is very slow on large input, here's how it can be done with diff-match-patch:
from diff_match_patch import diff_match_patch
def compute_similarity_and_diff(text1, text2):
dmp = diff_match_patch()
dmp.Diff_Timeout = 0.0
diff = dmp.diff_main(text1, text2, False)
# similarity
common_text = sum([len(txt) for op, txt in diff if op == 0])
text_length = max(len(text1), len(text2))
sim = common_text / text_length
return sim, diff
BLEUscore
BLEU, or the Bilingual Evaluation Understudy, is a score for comparing
a candidate translation of text to one or more reference translations.
A perfect match results in a score of 1.0, whereas a perfect mismatch
results in a score of 0.0.
Although developed for translation, it can be used to evaluate text
generated for a suite of natural language processing tasks.
Code:
import nltk
from nltk.translate import bleu
from nltk.translate.bleu_score import SmoothingFunction
smoothie = SmoothingFunction().method4
C1='Text'
C2='Best'
print('BLEUscore:',bleu([C1], C2, smoothing_function=smoothie))
Examples: By updating C1 and C2.
C1='Test' C2='Test'
BLEUscore: 1.0
C1='Test' C2='Best'
BLEUscore: 0.2326589746035907
C1='Test' C2='Text'
BLEUscore: 0.2866227639866161
You can also compare sentence similarity:
C1='It is tough.' C2='It is rough.'
BLEUscore: 0.7348889200874658
C1='It is tough.' C2='It is tough.'
BLEUscore: 1.0
Textdistance:
TextDistance – python library for comparing distance between two or more sequences by many algorithms. It has Textdistance
30+ algorithms
Pure python implementation
Simple usage
More than two sequences comparing
Some algorithms have more than one implementation in one class.
Optional numpy usage for maximum speed.
Example1:
import textdistance
textdistance.hamming('test', 'text')
Output:
1
Example2:
import textdistance
textdistance.hamming.normalized_similarity('test', 'text')
Output:
0.75
Thanks and Cheers!!!
There are many metrics to define similarity and distance between strings as mentioned above. I will give my 5 cents by showing an example of Jaccard similarity with Q-Grams and an example with edit distance.
The libraries
from nltk.metrics.distance import jaccard_distance
from nltk.util import ngrams
from nltk.metrics.distance import edit_distance
Jaccard Similarity
1-jaccard_distance(set(ngrams('Apple', 2)), set(ngrams('Appel', 2)))
and we get:
0.33333333333333337
And for the Apple and Mango
1-jaccard_distance(set(ngrams('Apple', 2)), set(ngrams('Mango', 2)))
and we get:
0.0
Edit Distance
edit_distance('Apple', 'Appel')
and we get:
2
And finally,
edit_distance('Apple', 'Mango')
and we get:
5
Cosine Similarity on Q-Grams (q=2)
Another solution is to work with the textdistance library. I will provide an example of Cosine Similarity
import textdistance
1-textdistance.Cosine(qval=2).distance('Apple', 'Appel')
and we get:
0.5
Adding the Spacy NLP library also to the mix;
#profile
def main():
str1= "Mar 31 09:08:41 The world is beautiful"
str2= "Mar 31 19:08:42 Beautiful is the world"
print("NLP Similarity=",nlp(str1).similarity(nlp(str2)))
print("Diff lib similarity",SequenceMatcher(None, str1, str2).ratio())
print("Jellyfish lib similarity",jellyfish.jaro_distance(str1, str2))
if __name__ == '__main__':
#python3 -m spacy download en_core_web_sm
#nlp = spacy.load("en_core_web_sm")
nlp = spacy.load("en_core_web_md")
main()
Run with Robert Kern's line_profiler
kernprof -l -v ./python/loganalysis/testspacy.py
NLP Similarity= 0.9999999821467294
Diff lib similarity 0.5897435897435898
Jellyfish lib similarity 0.8561253561253562
However the time's are revealing
Function: main at line 32
Line # Hits Time Per Hit % Time Line Contents
==============================================================
32 #profile
33 def main():
34 1 1.0 1.0 0.0 str1= "Mar 31 09:08:41 The world is beautiful"
35 1 0.0 0.0 0.0 str2= "Mar 31 19:08:42 Beautiful is the world"
36 1 43248.0 43248.0 99.1 print("NLP Similarity=",nlp(str1).similarity(nlp(str2)))
37 1 375.0 375.0 0.9 print("Diff lib similarity",SequenceMatcher(None, str1, str2).ratio())
38 1 30.0 30.0 0.1 print("Jellyfish lib similarity",jellyfish.jaro_distance(str1, str2))
Here's what i thought of:
import string
def match(a,b):
a,b = a.lower(), b.lower()
error = 0
for i in string.ascii_lowercase:
error += abs(a.count(i) - b.count(i))
total = len(a) + len(b)
return (total-error)/total
if __name__ == "__main__":
print(match("pple inc", "Apple Inc."))
Python3.6+=
No Libuary Imported
Works Well in most scenarios
In stack overflow, when you tries to add a tag or post a question, it bring up all relevant stuff. This is so convenient and is exactly the algorithm that I am looking for. Therefore, I coded a query set similarity filter.
def compare(qs, ip):
al = 2
v = 0
for ii, letter in enumerate(ip):
if letter == qs[ii]:
v += al
else:
ac = 0
for jj in range(al):
if ii - jj < 0 or ii + jj > len(qs) - 1:
break
elif letter == qs[ii - jj] or letter == qs[ii + jj]:
ac += jj
break
v += ac
return v
def getSimilarQuerySet(queryset, inp, length):
return [k for tt, (k, v) in enumerate(reversed(sorted({it: compare(it, inp) for it in queryset}.items(), key=lambda item: item[1])))][:length]
if __name__ == "__main__":
print(compare('apple', 'mongo'))
# 0
print(compare('apple', 'apple'))
# 10
print(compare('apple', 'appel'))
# 7
print(compare('dude', 'ud'))
# 1
print(compare('dude', 'du'))
# 4
print(compare('dude', 'dud'))
# 6
print(compare('apple', 'mongo'))
# 2
print(compare('apple', 'appel'))
# 8
print(getSimilarQuerySet(
[
"java",
"jquery",
"javascript",
"jude",
"aja",
],
"ja",
2,
))
# ['javascript', 'java']
Explanation
compare takes two string and returns a positive integer.
you can edit the al allowed variable in compare, it indicates how large the range we need to search through. It works like this: two strings are iterated, if same character is find at same index, then accumulator will be added to a largest value. Then, we search in the index range of allowed, if matched, add to the accumulator based on how far the letter is. (the further, the smaller)
length indicate how many items you want as result, that is most similar to input string.
I have my own for my purposes, which is 2x faster than difflib SequenceMatcher's quick_ratio(), while providing similar results. a and b are strings:
score = 0
for letters in enumerate(a):
score = score + b.count(letters[1])

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