在python中实现梯度下降

Question

I am trying to implement gradient descent in python. Though my code is returning result by I think results I am getting are completely wrong.

Here is the code I have written:

import numpy as np
import pandas

dataset = pandas.read_csv('D:\ML Data\house-prices-advanced-regression-techniques\\train.csv')

X = np.empty((0, 1),int)
Y = np.empty((0, 1), int)

for i in range(dataset.shape[0]):
  X = np.append(X, dataset.at[i, 'LotArea'])
  Y = np.append(Y, dataset.at[i, 'SalePrice'])

X = np.c_[np.ones(len(X)), X]
Y = Y.reshape(len(Y), 1)

def gradient_descent(X, Y, theta, iterations=100, learningRate=0.000001):
  m = len(X)
  for i in range(iterations):
    prediction = np.dot(X, theta)
    theta = theta - (1/m) * learningRate * (X.T.dot(prediction - Y))

  return theta

  theta = np.random.randn(2,1)
  theta = gradient_descent(X, Y, theta)
  print('theta',theta)

The result I get after running this program is:

theta [[-5.23237458e+228] [-1.04560188e+233]]

Which are very high values. Can someone point out the mistake I have made in implementation.

Also, 2nd problem is I have to set value of learning rate very low (in this case i have set to 0.000001) to work other wise program throws an error.

Please help me in diagnosis the problem.

Answer 1

try to reduce the learning rate with iteration otherwise it wont be able to reach the optimal lowest.try this

import numpy as np
import pandas

dataset = pandas.read_csv('start.csv')

X = np.empty((0, 1),int)
Y = np.empty((0, 1), int)

for i in range(dataset.shape[0]):
  X = np.append(X, dataset.at[i, 'R&D Spend'])
  Y = np.append(Y, dataset.at[i, 'Profit'])

X = np.c_[np.ones(len(X)), X]
Y = Y.reshape(len(Y), 1)

def gradient_descent(X, Y, theta, iterations=50, learningRate=0.01):
  m = len(X)
  for i in range(iterations):
    prediction = np.dot(X, theta)
    theta = theta - (1/m) * learningRate * (X.T.dot(prediction - Y))
    learningRate/=10;

  return theta

theta = np.random.randn(2,1)
theta = gradient_descent(X, Y, theta)
print('theta',theta)