kitchenprinzessin kitchenprinzessin - 6 months ago 89
Python Question

How to compute jaccard similarity from a pandas dataframe

I have a dataframe as follows: the shape of the frame is (1510, 1399). The columns represents products, the rows represents the values (0 or 1) assigned by an user for a given product. How can I can compute a jaccard_similarity_score?

enter image description here

I created a placeholder dataframe listing product vs. product

data_ibs = pd.DataFrame(index=data_g.columns,columns=data_g.columns)


I am not sure how to iterate though data_ibs to compute similarities.

for i in range(0,len(data_ibs.columns)) :
# Loop through the columns for each column
for j in range(0,len(data_ibs.columns)) :
.........

Answer Source

Short and vectorized (fast) answer:

Use 'hamming' from the pairwise distances of scikit learn:

from sklearn.metrics.pairwise import pairwise_distances
jac_sim = 1 - pairwise_distances(df.T, metric = "hamming")
# optionally convert it to a DataFrame
jac_sim = pd.DataFrame(jac_sim, index=df.columns, columns=df.columns)

Explanation:

Assume this is your dataset:

import pandas as pd
import numpy as np
np.random.seed(0)
df = pd.DataFrame(np.random.binomial(1, 0.5, size=(100, 5)), columns=list('ABCDE'))
print(df.head())

   A  B  C  D  E
0  1  1  1  1  0
1  1  0  1  1  0
2  1  1  1  1  0
3  0  0  1  1  1
4  1  1  0  1  0

Using sklearn's jaccard_similarity_score, similarity between column A and B is:

from sklearn.metrics import jaccard_similarity_score
print(jaccard_similarity_score(df['A'], df['B']))
0.43

This is the number of rows that have the same value over total number of rows, 100.

As far as I know, there is no pairwise version of the jaccard_similarity_score but there are pairwise versions of distances.

However, SciPy defines Jaccard distance as follows:

Given two vectors, u and v, the Jaccard distance is the proportion of those elements u[i] and v[i] that disagree where at least one of them is non-zero.

So it excludes the rows where both columns have 0 values. jaccard_similarity_score doesn't. Hamming distance, on the other hand, is inline with the similarity definition:

The proportion of those vector elements between two n-vectors u and v which disagree.

So if you want to calculate jaccard_similarity_score, you can use 1 - hamming:

from sklearn.metrics.pairwise import pairwise_distances
print(1 - pairwise_distances(df.T, metric = "hamming"))

array([[ 1.  ,  0.43,  0.61,  0.55,  0.46],
       [ 0.43,  1.  ,  0.52,  0.56,  0.49],
       [ 0.61,  0.52,  1.  ,  0.48,  0.53],
       [ 0.55,  0.56,  0.48,  1.  ,  0.49],
       [ 0.46,  0.49,  0.53,  0.49,  1.  ]])

In a DataFrame format:

jac_sim = 1 - pairwise_distances(df.T, metric = "hamming")
jac_sim = pd.DataFrame(jac_sim, index=df.columns, columns=df.columns)

      A     B     C     D     E
A  1.00  0.43  0.61  0.55  0.46
B  0.43  1.00  0.52  0.56  0.49
C  0.61  0.52  1.00  0.48  0.53
D  0.55  0.56  0.48  1.00  0.49
E  0.46  0.49  0.53  0.49  1.00

You can do the same by iterating over combinations of columns but it will be much slower.

import itertools
sim_df = pd.DataFrame(np.ones((5, 5)), index=df.columns, columns=df.columns)
for col_pair in itertools.combinations(df.columns, 2):
    sim_df.loc[col_pair] = sim_df.loc[tuple(reversed(col_pair))] = jaccard_similarity_score(df[col_pair[0]], df[col_pair[1]])
print(sim_df)
      A     B     C     D     E
A  1.00  0.43  0.61  0.55  0.46
B  0.43  1.00  0.52  0.56  0.49
C  0.61  0.52  1.00  0.48  0.53
D  0.55  0.56  0.48  1.00  0.49
E  0.46  0.49  0.53  0.49  1.00
Recommended from our users: Dynamic Network Monitoring from WhatsUp Gold from IPSwitch. Free Download