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Stochastic Gradient Decent vs. Gradient Decent for x**2 function

I would like to understand the difference between SGD and GD on the easiest example of function: y=x**2
The function of GD is here:
def gradient_descent(
gradient, start, learn_rate, n_iter=50, tolerance=1e-06
):
vector = start
for _ in range(n_iter):
diff = -learn_rate * gradient(vector)
if np.all(np.abs(diff) <= tolerance):
break
vector += diff
return vector
And in order to find the minimum of x**2 function we should do next (the answer is almost 0, which is correct):
gradient_descent(gradient=lambda v: 2 * x, start=10.0, learn_rate=0.2)
How i understood, in the classical GD the gradient is calculated precisely from all the data points. What is "all data points" in the implementation that i showed above?
And further, how we should modernize this function in order to call it SGD (SGD uses one single data point for calculating gradient. Where is "one single point" in the gradient_descent function?)

The function minimized in your example does not depend on any data, so it is not helpful to illustrate the difference between GD and SGD.
Consider this example:
import numpy as np
rng = np.random.default_rng(7263)
y = rng.normal(loc=10, scale=4, size=100)
def loss(y, mean):
return 0.5 * ((y-mean)**2).sum()
def gradient(y, mean):
return (mean - y).sum()
def mean_gd(y, learning_rate=0.005, n_iter=15, start=0):
"""Estimate the mean of y using gradient descent"""
mean = start
for i in range(n_iter):
mean -= learning_rate * gradient(y, mean)
print(f'Iter {i} mean {mean:0.2f} loss {loss(y, mean):0.2f}')
return mean
def mean_sgd(y, learning_rate=0.005, n_iter=15, start=0):
"""Estimate the mean of y using stochastic gradient descent"""
mean = start
for i in range(n_iter):
rng.shuffle(y)
for single_point in y:
mean -= learning_rate * gradient(single_point, mean)
print(f'Iter {i} mean {mean:0.2f} loss {loss(y, mean):0.2f}')
return mean
mean_gd(y)
mean_sgd(y)
y.mean()
Two (very simplistic) versions of GD and SGD are used to estimate the mean of a random sample y. Estimating the mean is achieved by minimizing the squared loss.
As you understood correctly, in GD each update uses the gradient computed on the whole dataset and in SGD we look at a single random point at a time.

Numeric Gradient Descent in python

I wrote this code to get the gradient descent of a vector function .
Where: f is the function, X0 is the starting point and eta is the step size.
It is essentially made up of two parts, first that obtains the gradient of the function and the second, which iterates over x, subtracting the gradient.
the problem is that you usually have trouble converging on some functions, for example:
if we take , the gradient descent does not converge to [20,25]
Something I need to change or add?
def descenso_grad(f,X0,eta):
def grad(f,X):
import numpy as np
def partial(g,k,X):
h=1e-9
Y=np.copy(X)
X[k-1]=X[k-1]+h
dp=(g(X)-g(Y))/h
return dp
grd=[]
for i in np.arange(0,len(X)):
ai=partial(f,i+1,X)
grd.append(ai)
return grd
#iterations
i=0
while True:
i=i+1
X0=X0-eta*np.array(grad(f,X0))
if np.linalg.norm(grad(f,X0))<10e-8 or i>400: break
return X0

Your gradient descent implementation is a good basic implementation, but your gradient sometimes oscillate and exploses. First we should precise that your gradient descent does not always diverge. For some combinations of eta and X0, it actually converges.
But first let me suggest a few edits to the code:
The import numpy as np statement should be at the top of your file, not within a function. In general, any import statement should be at the beginning of the code so that they are executed only once
It is better not to write nested functions but to separate them: you can write the partial function outside of the gradfunction, and the gradfunction outside of the descendo_grad function. it is better for debugging.
I strongly recommend to pass parameters such as the learning rate (eta), the number of steps (steps) and the tolerance (set to 10e-8 in your code, or 1e-7) as parameters to the descendo_grad function. This way you will be able to compare their influence on the result.
Anyway, here is the implementation of your code I will use in this answer:
import numpy as np
def partial(g, k, X):
h = 1e-9
Y = np.copy(X)
X[k - 1] = X[k - 1] + h
dp = (g(X) - g(Y)) / h
return dp
def grad(f, X):
grd = []
for i in np.arange(0, len(X)):
ai = partial(f, i + 1, X)
grd.append(ai)
return grd
def descenso_grad(f,X0,eta, steps, tolerance=1e-7):
#iterations
i=0
while True:
i=i+1
X0=X0-eta*np.array(grad(f,X0))
if np.linalg.norm(grad(f,X0))<tolerance or i>steps: break
return X0
def f(X):
return (X[0]-20)**4 + (X[1]-25)**4
Now, about the convergence. I said that your implementation didn't always diverge. Indeed:
X0 = [2, 30]
eta = 0.001
steps = 400
xmin = descenso_grad(f, X0, eta, steps)
print(xmin)
Will print [20.55359068 25.55258024]
But:
X0 = [2, 0]
eta = 0.001
steps = 400
xmin = descenso_grad(f, X0, eta, steps)
print(xmin)
Will actually diverge to [ 2.42462695e+01 -3.54879793e+10]
1) What happened
Your gradient is actually oscillating aroung the y axis. Let's compute the gradient of f at X0 = [2, 0]:
print(grad(f, X0))
We get grad(f, X0) = [-23328.00067961216, -62500.01024454831], which is quite high but in the right direction.
Now let's compute the next step of gradient descent:
eta = 0.001
X1=X0-eta*np.array(grad(f,X0))
print(X1)
We get X1 = [25.32800068 62.50001025]. We can see that on the x axis, we actually get closer to the minimal, but on the y axis, the gradient descent jumped to the other side of the minimal and went even further from it. Indeed, X0[1] was at a disctance of 25 from the minimal (X0[1] - Xmin[1] = 25) at its left while X0[1] is now at a distance of 65-25 = 40 but on its right*. Since the curve drawn by f has a simple U shape around the y axis, the value taken by f in X1 will be higher than before (to simplify, we ignore the influence of the x coordinate).
If we look at the next steps, we can clearly see the exploding oscillations around the minimal:
X0 = [2, 0]
eta = 0.001
steps = 10
#record values taken by X[1] during gradient descent
curve_y = [X0[1]]
i = 0
while True:
i = i + 1
X0 = X0 - eta * np.array(grad(f, X0))
curve_y.append(X0[1])
if np.linalg.norm(grad(f, X0)) < 10e-8 or i > steps: break
print(curve_y)
We get [0, 62.50001024554831, -148.43710232226067, 20719.6258707022, -35487979280.37413]. We can see that X1 gets further and further from the minimal while oscillating around it.
In order to illustrate this, let's assume that the value along the x axis is fixed, and look only at what happens on the y axis. The picture shows in black the oscillations of the function's values taken at each step of the gradient descent (synthetic data for the purpose of illustrating only). The gradient descent takes us further from the minimal at each step because the update value is too large:
Note that the gradient descent we gave as an example makes only 5 steps while we programmed 10 steps. This is because when the values taken by the function are too high, python does not succeed to make the difference between f(X[1]) and f(X[1]+h), so it computes a gradient equal to zero:
x = 24 # for the example
y = -35487979280.37413
z = f([x, y+h]) - f([x, y])
print(z)
We get 0.0. This issue is about the computer's computation precision, but we will get back to this later.
So, these oscillations are due to the combination of:
the very high value of the partial gradient with regards to the y axis
a too big value of eta that does not compensate the exploding gradient in the update.
If this is true, we might converge if we use a smaller learning rate. Let's check:
X0 = [2, 0]
# divide eta by 100
eta = 0.0001
steps = 400
xmin = descenso_grad(f, X0, eta, steps)
print(xmin)
We will get [18.25061287 23.24796497]. We might need more steps but we are converging this time!!
2) How to avoid that?
A) In your specific case
Since the function shape is simple and it has no local minimas or no saddle points, we can avoid this issue by simply clipping the gradient value. This means that we define a maximum value for the norm of the gradients:
def grad_clipped(f, X, clip):
grd = []
for i in np.arange(0, len(X)):
ai = partial(f, i + 1, X)
if ai<0:
ai = max(ai, -1*clip)
else:
ai = min(ai, clip)
grd.append(ai)
return grd
def descenso_grad_clipped(f,X0,eta, steps, clip=100, tolerance=10e-8):
#iterations
i=0
while True:
i=i+1
X0=X0-eta*np.array(grad_clipped(f,X0, clip))
if np.linalg.norm(grad_clipped(f,X0, clip))<tolerance or i>steps: break
return X0
Let's test it using the diverging example:
X0 = [2, 0]
eta = 0.001
steps = 400
clip=100
xmin = descenso_grad_clipped(f, X0, eta, clip, steps)
print(xmin)
This time we are converging: [19.31583901 24.20307188]. Note that this can slow the process since the gradient descend will take smaller steps. Here we can get closer to the real minimum by increasing the number of steps.
Note that this technique also solves the numerical calculous issue we faced when the function's value was too high.
B) In general
In general, there are a lot of caveats the gradient descent allgorithms try to avoid (exploding or vanishing gradients, saddle points, local minimas...). Backpropagation algorithms like Adam, RMSprop, Adagrad, etc try to avoid these caveats.
I am not going to dwelve into the details because this would deserve a whole article, however here are two resources you can use (I suggest to read them in the order given) to deepen your understanding of the topic:
A good article on towardsdatascience.com explaining the basics of gradients descents and its most common flaws
An overview of gradient descent algorithms

Linear regression using gradient descent algorithm, getting unexpected results

I'm trying to create a function which returns the value of θ0 & θ1 of hypothesis function of linear regression. But I'm getting different results for different initial (random) values of θ0 & θ1.
What's wrong in the code?
training_data_set = [[1, 1], [2, 3], [4, 3], [3, 2], [5, 5]]
initial_theta = [1, 0]
def gradient_descent(data, theta0, theta1):
def h(x, theta0, theta1):
return theta0 + theta1 * x
m = len(data)
alpha = 0.01
for n in range(m):
cost = 0
for i in range(m):
cost += (h(data[i][0], theta0, theta1) - data[i][1])**2
cost = cost/(2*m)
error = 0
for i in range(m):
error += h(data[i][0], theta0, theta1) - data[i][1]
theta0 -= alpha*error/m
theta1 -= alpha*error*data[n][0]/m
return theta0, theta1
for i in range(5):
initial_theta = gradient_descent(training_data_set, initial_theta[0], initial_theta[1])
final_theta0 = initial_theta[0]
final_theta1 = initial_theta[1]
print(f'theta0 = {final_theta0}\ntheta1 = {final_theta1}')
Output:
When initial_theta = [0, 0]
theta0 = 0.27311526522692103
theta1 = 0.7771301328221445
When initial_theta = [1, 1]
theta0 = 0.8829506006170339
theta1 = 0.6669442287905096

Convergence
You've run five iterations of gradient descent over just 5 training samples with a (probably reasonable) learning rate of 0.01. That is not expected to bring you to a "final" answer of your problem - you'd need to do many iterations of gradient descent just like you implemented, repeating the process until your thetas converge to a stable value. Then it'd make sense to compare the resulting values.
Replace the 5 in for i in range(5) with 5000 and then look at what happens. It might be illustrative to plot the decrease of the error rate / cost function to see how fast the process converges to a solution.

This is not a problem rather a very usual thing. For that you need to understand how gradient decent works.
Every time you randomly initialise your parameters the hypothesis starts it's journey from a random place. With every iteration it updates the parameters so that the cost function converges. In your case as you have ran your gradient decent just for 5 iteration, for different initialisation it ends up with too much different results. Try higher iterations you will see significant similarity even with different initialisation. If i could use visualisation that would be helpful for you.

Here is how I see gradient descent: imagine that you are high up on a rocky mountainside in the fog. Because of the fog, you cannot see the fastest path down the mountain. So, you look around your feet and go down based on what you see nearby. After taking a step, you look around your feet again, and take another step. Sometimes this will trap you in a small low spot where you cannot see any way down (a local minimum) and sometimes this will get you safely to the bottom of the mountain (global minimum). Starting from different random locations on the foggy mountainside might trap you in different local minima, though you might find your way down safely if the random starting location is good.

Maximum likelihood linear regression tensorflow

After I implemented a LS estimation with gradient descent for a simple linear regression problem, I'm now trying to do the same with Maximum Likelihood.
I used this equation from wikipedia. The maximum has to be found.
train_X = np.random.rand(100, 1) # all values [0-1)
train_Y = train_X
X = tf.placeholder("float", None)
Y = tf.placeholder("float", None)
theta_0 = tf.Variable(np.random.randn())
theta_1 = tf.Variable(np.random.randn())
var = tf.Variable(0.5)
hypothesis = tf.add(theta_0, tf.mul(X, theta_1))
lhf = 1 * (50 * np.log(2*np.pi) + 50 * tf.log(var) + (1/(2*var)) * tf.reduce_sum(tf.pow(hypothesis - Y, 2)))
op = tf.train.GradientDescentOptimizer(0.01).minimize(lhf)
This code works, but I still have some questions about it:
If I change the lhf function from 1 * to -1 * and minimize -lhf (according to the equation), it does not work. But why?
The value for lhf goes up and down during optimization. Shouldn't it only change in one direction?
The value for lhf sometimes is a NaN during optimization. How can I avoid that?
In the equation, σ² is the variance of the error (right?). My values are perfectly on a line. Why do I get a value of var above 100?

The symptoms in your question indicate a common problem: the learning rate or step size might be too high for the problem.
The zig-zag behaviour, where the function to be maximized goes up and down, is usual when the learning rate is too high. Specially when you get NaNs.
The simplest solution is to lower the learning rate, by dividing your current learning rate by 10 until the learning curve is smooth and there are no NaNs or up-down behavior.
As you are using TensorFlow you can also try AdamOptimizer as this adjust the learning rate dynamically as you train.

What determines whether my Python gradient descent algorithm converges?

I've implemented a single-variable linear regression model in Python that uses gradient descent to find the intercept and slope of the best-fit line (I'm using gradient descent rather than computing the optimal values for intercept and slope directly because I'd eventually like to generalize to multiple regression).
The data I am using are below. sales is the dependent variable (in dollars) and temp is the independent variable (degrees celsius) (think ice cream sales vs temperature, or something similar).
sales temp
215 14.20
325 16.40
185 11.90
332 15.20
406 18.50
522 22.10
412 19.40
614 25.10
544 23.40
421 18.10
445 22.60
408 17.20
And this is my data after it has been normalized:
sales temp
0.06993007 0.174242424
0.326340326 0.340909091
0 0
0.342657343 0.25
0.515151515 0.5
0.785547786 0.772727273
0.529137529 0.568181818
1 1
0.836829837 0.871212121
0.55011655 0.46969697
0.606060606 0.810606061
0.51981352 0.401515152
My code for the algorithm:
import numpy as np
import pandas as pd
from scipy import stats
class SLRegression(object):
def __init__(self, learnrate = .01, tolerance = .000000001, max_iter = 10000):
# Initialize learnrate, tolerance, and max_iter.
self.learnrate = learnrate
self.tolerance = tolerance
self.max_iter = max_iter
# Define the gradient descent algorithm.
def fit(self, data):
# data : array-like, shape = [m_observations, 2_columns]
# Initialize local variables.
converged = False
m = data.shape[0]
# Track number of iterations.
self.iter_ = 0
# Initialize theta0 and theta1.
self.theta0_ = 0
self.theta1_ = 0
# Compute the cost function.
J = (1.0/(2.0*m)) * sum([(self.theta0_ + self.theta1_*data[i][1] - data[i][0])**2 for i in range(m)])
print('J is: ', J)
# Iterate over each point in data and update theta0 and theta1 on each pass.
while not converged:
diftemp0 = (1.0/m) * sum([(self.theta0_ + self.theta1_*data[i][1] - data[i][0]) for i in range(m)])
diftemp1 = (1.0/m) * sum([(self.theta0_ + self.theta1_*data[i][1] - data[i][0]) * data[i][1] for i in range(m)])
# Subtract the learnrate * partial derivative from theta0 and theta1.
temp0 = self.theta0_ - (self.learnrate * diftemp0)
temp1 = self.theta1_ - (self.learnrate * diftemp1)
# Update theta0 and theta1.
self.theta0_ = temp0
self.theta1_ = temp1
# Compute the updated cost function, given new theta0 and theta1.
new_J = (1.0/(2.0*m)) * sum([(self.theta0_ + self.theta1_*data[i][1] - data[i][0])**2 for i in range(m)])
print('New J is: %s') % (new_J)
# Test for convergence.
if abs(J - new_J) <= self.tolerance:
converged = True
print('Model converged after %s iterations!') % (self.iter_)
# Set old cost equal to new cost and update iter.
J = new_J
self.iter_ += 1
# Test whether we have hit max_iter.
if self.iter_ == self.max_iter:
converged = True
print('Maximum iterations have been reached!')
return self
def point_forecast(self, x):
# Given feature value x, returns the regression's predicted value for y.
return self.theta0_ + self.theta1_ * x
# Run the algorithm on a data set.
if __name__ == '__main__':
# Load in the .csv file.
data = np.squeeze(np.array(pd.read_csv('sales_normalized.csv')))
# Create a regression model with the default learning rate, tolerance, and maximum number of iterations.
slregression = SLRegression()
# Call the fit function and pass in the data.
slregression.fit(data)
# Print out the results.
print('After %s iterations, the model converged on Theta0 = %s and Theta1 = %s.') % (slregression.iter_, slregression.theta0_, slregression.theta1_)
# Compare our model to scipy linregress model.
slope, intercept, r_value, p_value, slope_std_error = stats.linregress(data[:,1], data[:,0])
print('Scipy linear regression gives intercept: %s and slope = %s.') % (intercept, slope)
# Test the model with a point forecast.
print('As an example, our algorithm gives y = %s given x = .87.') % (slregression.point_forecast(.87)) # Should be about .83.
print('The true y-value for x = .87 is about .8368.')
I'm having trouble understanding exactly what allows the algorithm to converge versus return values that are completely wrong. Given learnrate = .01, tolerance = .0000000001, and max_iter = 10000, in combination with normalized data, I can get the gradient descent algorithm to converge. However, when I use the un-normalized data, the smallest I can make the learning rate without the algorithm returning NaN is .005. This brings changes in the cost function from iteration to iteration down to around 614, but I can't get it to go any lower.
Is it definitely a requirement of this type of algorithm to have normalized data, and if so, why? Also, what would be the best way to take a novel x-value in non-normalized form and plug it into the point forecast, given that the algorithm needs normalized values? For instance, if I were going to deliver this algorithm to a client so they could make predictions of their own (I'm not, but for the sake of argument..), wouldn't I want them to simply be able to plug in the un-normalized x-value?
All and all, playing around with the tolerance, max_iter, and learnrate gives me non-convergent results the majority of the time. Is this normal, or are there flaws in my algorithm that are contributing to this issue?

Given learnrate = .01, tolerance = .0000000001, and max_iter = 10000, in combination with normalized data, I can get the gradient descent algorithm to converge. However, when I use the un-normalized data, the smallest I can make the learning rate without the algorithm returning NaN is .005
That's kind of to be expected the way you have your algorithm set up.
The normalization of the data makes it so the y-intercept of the best fit is around 0.0. Otherwise, you could have a y-intercept thousands of units off of the starting guess, and you'd have to trek there before you ever really started the optimization part.
Is it definitely a requirement of this type of algorithm to have normalized data, and if so, why?
No, absolutely not, but if you don't normalize, you should pick a starting point more intelligently (you're starting at (m,b) = (0,0)). Your learnrate may also be too small if you don't normalize your data, and same with your tolerance.
Also, what would be the best way to take a novel x-value in non-normalized form and plug it into the point forecast, given that the algorithm needs normalized values?
Apply whatever transformation that you applied to the original data to get the normalized data to your new x-value. (The code for normalization is outside of what you have shown). If this test point fell within the (minx,maxx) range of your original data, once transformed, it should fall within 0 <= x <= 1. Once you have this normalized test point, plug it into your theta equation of a line (remember, your thetas are m,b of the y-intercept form of the equation of a line).
All and all, playing around with the tolerance, max_iter, and learnrate gives me non-convergent results the majority of the time.
For a well-formed problem, if you're in fact diverging it often means your step size is too large. Try lowering it.
If it's simply not converging before it hits the max iterations, that could be a few issues:
Your step size is too small,
Your tolerance is too small,
Your max iterations is too small,
Your starting point is poorly chosen
In your case, using the non normalized data results in your starting point of (0,0) being very far off (the (m,b) of the non-normalized data is around (-159, 30) while the (m,b) of your normalized data is (0.10,0.79)), so most if not all of your iterations are being used just getting to the area of interest.
The problem with this is that by increasing the step size to get to the area of interest faster also makes it less-likely to find convergence once it gets there.
To account for this, some gradient descent algorithms have dynamic step size (or learnrate) such that large steps are taken at the beginning, and smaller ones as it nears convergence.
It may also be helpful for you to keep a history of of the theta pairs throughout the algorithm, then plot them. You'll be able to see the difference immediately between using normalized and non-normalized input data.