calculate distance matrix with mixed categorical and numerics

问题

I have a data frame with a mixture of numeric (15 fields) and categorical (5 fields) data.

I can create a complete distance matrix of the numeric fields following create distance matrix using own calculation pandas

I want to include the categorical fields as well.

Using as template:

import scipy
from scipy.spatial import distance_matrix
from scipy.spatial.distance import squareform
from scipy.spatial.distance import pdist
df2=pd.DataFrame({'col1':[1,2,3,4],'col2':[5,6,7,8],'col3':['cat','cat','dog','bird']})
df2
pd.DataFrame(squareform(pdist(df2.values, lambda u, v: np.sqrt((w*(u-v)**2).sum()))), index=df2.index, columns=df2.index)

in the squareform calculation, I would like to include the test np.where(u[2]==v[2], 0, 10) (as well as with the other categorical columns)

Hpw do I modify the lambda function to carry out this test as well

Here, the distance between [0,1]

= sqrt((2-1)^2 + (6-5)^2 + (cat - cat)^2)
= sqrt(1 + 1 + 0)

and the distance between [0,2]

= sqrt((3-1)^2 + (7-5)^2 + (dog - cat)^2)
= sqrt(4 + 4 + 100)

etc.

Can anyone suggest how I can implement this algorithm?

回答1:

import pandas as pd
import numpy as np
from scipy.spatial.distance import pdist, squareform

df2 = pd.DataFrame({'col1':[1,2,3,4],'col2':[5,6,7,8],'col3':['cat','cat','dog','bird']})

def fun(u,v):
    const = 0 if u[2] == v[2] else 10
    return np.sqrt((u[0]-v[0])**2 + (u[1]-v[1])**2 + const**2)

pd.DataFrame(squareform(pdist(df2.values, fun)), index=df2.index, columns=df2.index)

Result:

           0          1          2          3
0   0.000000   1.414214  10.392305  10.862780
1   1.414214   0.000000  10.099505  10.392305
2  10.392305  10.099505   0.000000  10.099505
3  10.862780  10.392305  10.099505   0.000000

来源：https://stackoverflow.com/questions/57868339/calculate-distance-matrix-with-mixed-categorical-and-numerics

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distance