CODEWITHSUNDEEP
pythonalgorithmsetsimilarity
kmace
kmace

Reputation: 2044

Compute the similarity between two lists

I'd like to compute the similarity between two lists of various lengths.

eg:

listA = ['apple', 'orange', 'apple', 'apple', 'banana', 'orange'] # (length = 6)
listB = ['apple', 'orange', 'grapefruit', 'apple'] # (length = 4)

as you can see, a single item can appear multiple times in a list, and the lengths are of different sizes.

I've already thought of comparing the frequencies of each item, but that does not encompass the size of each list (a list that is simply twice another list should be similar, but not perfectly similar)

eg2:

listA = ['apple', 'apple', 'orange', 'orange']
listB = ['apple', 'orange']
similarity(listA, listB) # should NOT equal 1

So I basically want to encompass the size of the lists, and the distribution of items in the list.

Any ideas?

Upvotes: 14

Views: 37690

Answers (3)

Martijn Pieters
Martijn Pieters

Reputation: 1121416

Use collections.Counter() perhaps; those are multi-sets, or bags, in datatype parlance:

from collections import Counter

counterA = Counter(listA)
counterB = Counter(listB)

Now you can compare these by entries or frequencies:

>>> counterA
Counter({'apple': 3, 'orange': 2, 'banana': 1})
>>> counterB
Counter({'apple': 2, 'orange': 1, 'grapefruit': 1})
>>> counterA - counterB
Counter({'orange': 1, 'apple': 1, 'banana': 1})
>>> counterB - counterA
Counter({'grapefruit': 1})

You can calculate their cosine similarity using:

import math

def counter_cosine_similarity(c1, c2):
    terms = set(c1).union(c2)
    dotprod = sum(c1.get(k, 0) * c2.get(k, 0) for k in terms)
    magA = math.sqrt(sum(c1.get(k, 0)**2 for k in terms))
    magB = math.sqrt(sum(c2.get(k, 0)**2 for k in terms))
    return dotprod / (magA * magB)

Which gives:

>>> counter_cosine_similarity(counterA, counterB)
0.8728715609439696

The closer to 1 that value, the more similar the two lists are.

The cosine similarity is one score you can calculate. If you care about the length of the list, you can calculate another; if you keep that score between 0.0 and 1.0 as well you can multiply the two values for a final score between -1.0 and 1.0.

For example, to take relative lengths into account you could use:

def length_similarity(c1, c2):
    lenc1 = sum(c1.itervalues())
    lenc2 = sum(c2.itervalues())
    return min(lenc1, lenc2) / float(max(lenc1, lenc2))

and then combine into a function that takes the lists as inputs:

def similarity_score(l1, l2):
    c1, c2 = Counter(l1), Counter(l2)
    return length_similarity(c1, c2) * counter_cosine_similarity(c1, c2)

For your two example lists, that results in:

>>> similarity_score(['apple', 'orange', 'apple', 'apple', 'banana', 'orange'], ['apple', 'orange', 'grapefruit', 'apple'])
0.5819143739626463
>>> similarity_score(['apple', 'apple', 'orange', 'orange'], ['apple', 'orange'])
0.4999999999999999

You can mix in other metrics as needed.

Upvotes: 32

Computernerd
Computernerd

Reputation: 7766

I believe what you are looking for is to counting the number of inversions in an array The question has your answer: Counting inversions in an array

Upvotes: 1

Vigneshwaren
Vigneshwaren

Reputation: 1393

From a theoretical point of view : I recommend you look up cosine similarity http://en.wikipedia.org/wiki/Cosine_similarity

You may have to modify to fit your scheme, but the idea of cosine similarity is great.

Upvotes: 1

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