Python function returning a 4x4 matrix instead of a floating number like in an equivalent Octave function in a Linear Regression problem

1

I am trying to translate code from Octave to Python, and I am stuck. I am aware they are libraries out there such as scikit-learn etc., but for my own learnin,g I would like to be able to implement the cost function from scratch. Furthermore, I managed to solve some issues by looking into numpy documentation, but I am still stuck with one last one. The function doesn't return the expected result. I am providing everything you need to reproduce below. Nothing more is needed. The python function returns a 4x4 matrices instead of a float like it does in Octave and I get this:

Actual results: an numpy.ndarray

Expected results: a floating number

  • The function in computeCost(X, y, theta) in Octave:
function J = computeCost(X, y, theta)
    m = length(y);
    J = (((X*theta)-y)'*((X*theta)-y))/(2*m);
end
  • The function compute_cost(X, y, theta) in Python:
from numpy import dot


def compute_cost(X, y, theta):
    m = len(y)
    ans = (X * theta) - y
    product = dot((1/(2*m)), dot(ans, ans.T))
    return product
  • The tests:
import unittest
import numpy as np
from ml.ml_utils import compute_cost


class TestComputeCost(unittest.TestCase):
    data = np.loadtxt('../data/data1.txt', delimiter=',')
    x = data[:, 0]
    y = data[:, 1]
    m = len(y)
    X = np.vstack((np.ones(m), x))

    def test_compute_cost_with_theta_zeros(self):
        theta = np.zeros((2, 1), dtype=int)
        result = compute_cost(self.X, self.y, theta)
        self.assertEqual(round(result, 2), 32.07,
                         'Result is wrong!')

    def test_compute_cost_with_theta_values(self):
        theta = np.array([[-1], [2]])
        result = compute_cost(self.X, self.y, theta)
        self.assertEqual(round(result, 2), 54.24,
                         'Result is wrong!')


if __name__ == '__main__':
    unittest.main()

  • data1.txt
6.1101,17.592
5.5277,9.1302
8.5186,13.662
7.0032,11.854
5.8598,6.8233
8.3829,11.886
7.4764,4.3483
8.5781,12
6.4862,6.5987
5.0546,3.8166
5.7107,3.2522
14.164,15.505
5.734,3.1551
8.4084,7.2258
5.6407,0.71618
5.3794,3.5129
6.3654,5.3048
5.1301,0.56077
6.4296,3.6518
7.0708,5.3893
6.1891,3.1386
20.27,21.767
5.4901,4.263
6.3261,5.1875
5.5649,3.0825
18.945,22.638
12.828,13.501
10.957,7.0467
13.176,14.692
22.203,24.147
5.2524,-1.22
6.5894,5.9966
9.2482,12.134
5.8918,1.8495
8.2111,6.5426
7.9334,4.5623
8.0959,4.1164
5.6063,3.3928
12.836,10.117
6.3534,5.4974
5.4069,0.55657
6.8825,3.9115
11.708,5.3854
5.7737,2.4406
7.8247,6.7318
7.0931,1.0463
5.0702,5.1337
5.8014,1.844
11.7,8.0043
5.5416,1.0179
7.5402,6.7504
5.3077,1.8396
7.4239,4.2885
7.6031,4.9981
6.3328,1.4233
6.3589,-1.4211
6.2742,2.4756
5.6397,4.6042
9.3102,3.9624
9.4536,5.4141
8.8254,5.1694
5.1793,-0.74279
21.279,17.929
14.908,12.054
18.959,17.054
7.2182,4.8852
8.2951,5.7442
10.236,7.7754
5.4994,1.0173
20.341,20.992
10.136,6.6799
7.3345,4.0259
6.0062,1.2784
7.2259,3.3411
5.0269,-2.6807
6.5479,0.29678
7.5386,3.8845
5.0365,5.7014
10.274,6.7526
5.1077,2.0576
5.7292,0.47953
5.1884,0.20421
6.3557,0.67861
9.7687,7.5435
6.5159,5.3436
8.5172,4.2415
9.1802,6.7981
6.002,0.92695
5.5204,0.152
5.0594,2.8214
5.7077,1.8451
7.6366,4.2959
5.8707,7.2029
5.3054,1.9869
8.2934,0.14454
13.394,9.0551
5.4369,0.61705

Kader DJEHAF

Posted 2020-12-26T15:31:52.457

Reputation: 11

Answers

1

Change the compute_cost function to the following to match dimensions for matrix multiplications:

def compute_cost(X, y, theta):
    m = len(y)
    ans = np.dot(X.T, theta).T - y
    product = np.dot((1/(2*m)), np.dot(ans, ans.T))
    return product[0, 0]

Result for the data given:

32.0727338775
54.242455082

PS: one can simplify the transpose operations further..

Nikos M.

Posted 2020-12-26T15:31:52.457

Reputation: 819

0

In Octave, matrix multiplication is done with just *, but in Python you need @ (* will give coordinate-wise product, broadcasting shapes if needed). So it seems your ans definition is incorrect. (I haven't checked for further issues...)

Ben Reiniger

Posted 2020-12-26T15:31:52.457

Reputation: 7 097