CODEWITHSUNDEEP
pythonnumpy
kurro
kurro

Reputation: 85

Python one-liner for a confusion/contingency matrix needed

I want to write a one-liner to calculate a confusion/contingency matrix M (square matrix with either dimension equal to the number of classes ) that counts the cases presented in two vectors of lenght n: Ytrue and Ypredicted. Obiously the following does not work using python and numpy:

error = N.array([error[x,y]+1 for x, y in zip(Ytrue,Ypredicted)]).reshape((n,n))

Any hint to create a one-liner matrix confusion calculator?

Upvotes: 0

Views: 2802

Answers (2)

tkamishima
tkamishima

Reputation: 56

If NumPy is newer or equal than 1.6 and Ytrue and Ypred are NumPy arrays, this code works

np.bincount(n * (Ytrue - 1) + (Ypred -1), minlength=n*n).reshape(n, n)

Upvotes: 2

jedwards
jedwards

Reputation: 30240

error = N.array([zip(Ytrue,Ypred).count(x) for x in itertools.product(classes,repeat=2)]).reshape(n,n)

or

error = N.array([z.count(x) for z in [zip(Ytrue,Ypred)] for x in itertools.product(classes,repeat=2)]).reshape(n,n)

The latter being more efficient but possibly more confusing.

import numpy as N
import itertools

Ytrue = [1,1,1,1,1,1,1,1,
         2,2,2,2,2,2,2,2,
         3,3,3,3,3,3,3,3]
Ypred = [1,1,2,1,2,1,3,1,
         2,2,2,2,2,2,2,2,
         3,3,2,2,2,1,1,1]

classes = list(set(Ytrue))
n = len(classes)

error = N.array([zip(Ytrue,Ypred).count(x) for x in itertools.product(classes,repeat=2)]).reshape(n,n)
print error

error = N.array([z.count(x) for z in [zip(Ytrue,Ypred)] for x in itertools.product(classes,repeat=2)]).reshape(n,n)
print error

Which produces

[[5 2 1]
 [0 8 0]
 [3 3 2]]

[[5 2 1]
 [0 8 0]
 [3 3 2]]

Upvotes: 4

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