CODEWITHSUNDEEP
pythonpandasnumpy
Pluss
Pluss

Reputation: 33

comparing a matrix and a vector

Is there an easier vectorized way to achieve the same result as this?

import numpy as np
tempY = np.zeros((10,10))
y = np.array([0,0,2,3,4,5,5,8,8,2]).reshape(10,1)
for index in range(y.size):
    tempY[y[index].squeeze(), index] = 1

print(tempY)

output:

[[1. 1. 0. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 1. 0. 0. 0. 0. 0. 0. 1.]
 [0. 0. 0. 1. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 1. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 1. 1. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 1. 1. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]]

Upvotes: 2

Views: 51

Answers (1)

sammywemmy
sammywemmy

Reputation: 28644

You could pass it as rows, columns to index and replace the values :

tempY[y.ravel(), np.arange(y.size)] = 1

tempY

array([[1., 1., 0., 0., 0., 0., 0., 0., 0., 0.],
       [0., 0., 0., 0., 0., 0., 0., 0., 0., 0.],
       [0., 0., 1., 0., 0., 0., 0., 0., 0., 1.],
       [0., 0., 0., 1., 0., 0., 0., 0., 0., 0.],
       [0., 0., 0., 0., 1., 0., 0., 0., 0., 0.],
       [0., 0., 0., 0., 0., 1., 1., 0., 0., 0.],
       [0., 0., 0., 0., 0., 0., 0., 0., 0., 0.],
       [0., 0., 0., 0., 0., 0., 0., 0., 0., 0.],
       [0., 0., 0., 0., 0., 0., 0., 1., 1., 0.],
       [0., 0., 0., 0., 0., 0., 0., 0., 0., 0.]])

Upvotes: 1

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