I am doing a school project with logistic regression for binary classification and I am getting pretty bad results (accuracy around 50%). I had to do this classification from scratch so I am worried there is a problem with my implementation.
I fit a model with a single predictor and plotted the resulting logistic curve and noticed that it was centered around 0, outside the range of my data (the predictor ranges from 50 - 90) . Because of this, only the asymptote of the logistic curve is within the range of my data and therefore is classifying all examples into the same class.
I thought that using an intercept weight would address this problem but in my case it has not. I also thought about normalizing my data and centering it around 0 but I was hoping there would be a more direct approach.
Any suggestions?
EDIT - Here is my implementation
class LogisticRegression:
def __init__(self, alpha, iters, intercept=True):
self.alpha = alpha
self.iters = iters
self.weights = None
self.intercept = intercept
def sigmoid(self, z):
return 1.0 / (1 + np.exp(-z))
def add_intercept(self, X):
intercept = np.ones((X.shape[0], 1))
return np.concatenate((intercept, X), axis=1)
def cost(self, h, y):
return (-y * np.log(h) - (1 - y) * np.log(1 - h)).mean()
def fit(self, X, y):
if self.intercept:
X = self.add_intercept(X)
self.weights = np.zeros(X.shape[1])
for i in range(self.iters):
z = np.dot(X, self.weights)
h = self.sigmoid(z)
gradient = np.dot(X.T, (h - y)) / len(y)
self.weights -= self.alpha * gradient
EDIT - link to data: https://raw.githubusercontent.com/efosler/cse5522data/master/height_vs_weight.csv
The target is whether a person plays basketball or not, and the feature I used for the single regression is height.
I'm thinking now that I do have to normalize my data anyway because I am getting some overflow with np.exp() in the multiple regression cases.
When I make a scatterplot of Height versus Basketball, the data does not appear to lie on any type of sigmoidal or logistic curve. Based on this data set, my thought is that Height alone is insufficient to make a prediction on Basketball.
Related
I would like to generate a polynomial 'fit' to the cluster of colored pixels in the image here
(The point being that I would like to measure how much that cluster approximates an horizontal line).
I thought of using grabit or something similar and then treating this as a cloud of points in a graph. But is there a quicker function to do so directly on the image file?
thanks!
Here is a Python implementation. Basically we find all (xi, yi) coordinates of the colored regions, then set up a regularized least squares system where the we want to find the vector of weights, (w0, ..., wd) such that yi = w0 + w1 xi + w2 xi^2 + ... + wd xi^d "as close as possible" in the least squares sense.
import numpy as np
import matplotlib.pyplot as plt
def rgb2gray(rgb):
return np.dot(rgb[...,:3], [0.299, 0.587, 0.114])
def feature(x, order=3):
"""Generate polynomial feature of the form
[1, x, x^2, ..., x^order] where x is the column of x-coordinates
and 1 is the column of ones for the intercept.
"""
x = x.reshape(-1, 1)
return np.power(x, np.arange(order+1).reshape(1, -1))
I_orig = plt.imread("2Md7v.jpg")
# Convert to grayscale
I = rgb2gray(I_orig)
# Mask out region
mask = I > 20
# Get coordinates of pixels corresponding to marked region
X = np.argwhere(mask)
# Use the value as weights later
weights = I[mask] / float(I.max())
# Convert to diagonal matrix
W = np.diag(weights)
# Column indices
x = X[:, 1].reshape(-1, 1)
# Row indices to predict. Note origin is at top left corner
y = X[:, 0]
We want to find vector w that minimizes || Aw - y ||^2
so that we can use it to predict y = w . x
Here are 2 versions. One is a vanilla least squares with l2 regularization and the other is weighted least squares with l2 regularization.
# Ridge regression, i.e., least squares with l2 regularization.
# Should probably use a more numerically stable implementation,
# e.g., that in Scikit-Learn
# alpha is regularization parameter. Larger alpha => less flexible curve
alpha = 0.01
# Construct data matrix, A
order = 3
A = feature(x, order)
# w = inv (A^T A + alpha * I) A^T y
w_unweighted = np.linalg.pinv( A.T.dot(A) + alpha * np.eye(A.shape[1])).dot(A.T).dot(y)
# w = inv (A^T W A + alpha * I) A^T W y
w_weighted = np.linalg.pinv( A.T.dot(W).dot(A) + alpha * \
np.eye(A.shape[1])).dot(A.T).dot(W).dot(y)
The result
# Generate test points
n_samples = 50
x_test = np.linspace(0, I_orig.shape[1], n_samples)
X_test = feature(x_test, order)
# Predict y coordinates at test points
y_test_unweighted = X_test.dot(w_unweighted)
y_test_weighted = X_test.dot(w_weighted)
# Display
fig, ax = plt.subplots(1, 1, figsize=(10, 5))
ax.imshow(I_orig)
ax.plot(x_test, y_test_unweighted, color="green", marker='o', label="Unweighted")
ax.plot(x_test, y_test_weighted, color="blue", marker='x', label="Weighted")
fig.legend()
fig.savefig("curve.png")
For simple straight line fit, set the argument order of feature to 1. You can then use the gradient of the line to get a sense of how close it is to a horizontal line (e.g., by checking the angle of its slope).
It is also possible to set this to any degree of polynomial you want. I find that degree 3 looks pretty good. In this case, the 6 times the absolute value of the coefficient corresponding to x^3 (w_unweighted[3] or w_weighted[3]) is one measure of the curvature of the line.
See A measure for the curvature of a quadratic polynomial in Matlab for additional details.
I am trying to implement a convolutional neural network from scratch and I am not able to figure out how to perform (vectorized)operations on multi-channel images like rgb, which have 3 dimensions. On following the articles and tutorials such as this CS231n tutorial , it's pretty clear to implement a network for a single input as the input layer will be a 3d matrix but there are always multiple data points in a dataset. so, I cannot figure out how to implement these networks for vectorized operation on entire datsets.
I have implemented a network which takes a 3d matrix as input but now I have realized that It will not work on entire dataset but I will have to propagate one input at a time.I don't really know whether conv nets are vectorized over entire dataset or not .But if they are, how can I vectorize my convolutional network for multi-channel images ?
If I got your question right, you're basically asking how to do convolutional layer for a mini-batch, which will be a 4-D tensor.
To put it simply, you want to treat each input in a batch independently and apply convolution to each one. It's fairly straightforward to code without vectorization using a loop.
A vectorization implementation is often based on im2col technique, which basically transforms the 4-D input tensor into a giant matrix and performs a matrix multiplication. Here's an implementation of a forward pass using numpy.lib.stride_tricks in python:
import numpy as np
def conv_forward(x, w, b, stride, pad):
N, C, H, W = x.shape
F, _, HH, WW = w.shape
# Check dimensions
assert (W + 2 * pad - WW) % stride == 0, 'width does not work'
assert (H + 2 * pad - HH) % stride == 0, 'height does not work'
# Pad the input
p = pad
x_padded = np.pad(x, ((0, 0), (0, 0), (p, p), (p, p)), mode='constant')
# Figure out output dimensions
H += 2 * pad
W += 2 * pad
out_h = (H - HH) / stride + 1
out_w = (W - WW) / stride + 1
# Perform an im2col operation by picking clever strides
shape = (C, HH, WW, N, out_h, out_w)
strides = (H * W, W, 1, C * H * W, stride * W, stride)
strides = x.itemsize * np.array(strides)
x_stride = np.lib.stride_tricks.as_strided(x_padded,
shape=shape, strides=strides)
x_cols = np.ascontiguousarray(x_stride)
x_cols.shape = (C * HH * WW, N * out_h * out_w)
# Now all our convolutions are a big matrix multiply
res = w.reshape(F, -1).dot(x_cols) + b.reshape(-1, 1)
# Reshape the output
res.shape = (F, N, out_h, out_w)
out = res.transpose(1, 0, 2, 3)
out = np.ascontiguousarray(out)
return out
Note that it uses some non-trivial features of linear algebra library, which are implemented in numpy, but may be not in your library.
BTW, you generally don't want to push the entire data set as one batch - split it into several batches.
Vectorized implementation of gradient descent
for iter = 1:num_iters
theta = theta - (alpha / m) * X' * (X * theta - y);
J_history(iter) = computeCostMulti(X, y, theta);
end
Implementation of computeCostMulti()
function J = computeCostMulti(X, y, theta)
m = length(y);
J = 0;
J = 1 / (2 * m) * (X * theta - y)' * (X * theta - y);
Normal equation implementation
theta = pinv(X' * X) * X' * y;
These two implementations converge to different values of theta for the same values of X and y. The Normal Equation gives the right answer but Gradient descent gives a wrong answer.
Is there anything wrong with the implementation of Gradient Descent?
I suppose that when you use gradient descent, you first process your input using feature scaling. That is not done with the normal equation method (as feature scaling is not required), and that should result in a different theta. If you use your models to make predictions they should come up with the same result.
It doesn't matter. As you're not making feature scaling to use the normal equation, you'll discover that the prediction is the same
Nobody promised you that gradient descent with fixed step size will converge by num_iters iterations even to a local optimum. You need to iterate until some well defined convergency criteria are met (e.g. gradient is close to zero).
If you have normalized the training data before gradient descent, you should also do it with your input data for the prediction. Concretely, your new input data should be like:
[1, (x-mu)/sigma]
where:
- 1 is the bias term
- mu is the mean of the training data
- sigma is the standard deviation of the training data
I am trying to move on from simple linear single-variable gradient descent into something more advanced: best polynomial fit for a set of points. I created a simple octave test script which allows me to visually set the points in a 2D space, then start the gradient dsecent algorithm and see how it gradually approaches the best fit.
Unfortunately, it doesn't work as good as it did with the simple single-variable linear regression: the results I get ( when I get them ) are inconsistent with the polynome I expect!
Here is the code:
dim=5;
h = figure();
axis([-dim dim -dim dim]);
hold on
index = 1;
data = zeros(1,2);
while(1)
[x,y,b] = ginput(1);
if( length(b) == 0 )
break;
endif
plot(x, y, "b+");
data(index, :) = [x y];
index++;
endwhile
y = data(:, 2);
m = length(y);
X = data(:, 1);
X = [ones(m, 1), data(:,1), data(:,1).^2, data(:,1).^3 ];
theta = zeros(4, 1);
iterations = 100;
alpha = 0.001;
J = zeros(1,iterations);
for iter = 1:iterations
theta -= ( (1/m) * ((X * theta) - y)' * X)' * alpha;
plot(-dim:0.01:dim, theta(1) + (-dim:0.01:dim).*theta(2) + (-dim:0.01:dim).^2.*theta(3) + (-dim:0.01:dim).^3.*theta(4), "g-");
J(iter) = sum( (1/m) * ((X * theta) - y)' * X);
end
plot(-dim:0.01:dim, theta(1) + (-dim:0.01:dim).*theta(2) + (-dim:0.01:dim).^2.*theta(3) + (-dim:0.01:dim).^3.*theta(4), "r-");
figure()
plot(1:iter, J);
I continuously get wrong results, even though it would seem that J is minimized correctly. I checked the plotting function with the normal equation ( which works correctly of course, and although I believe the error lies somewhere in the theta equation, I cannot figure out what it.
i implemented your code and it seems to be just fine, the reason that you do not have the results that you want is that Linear regression or polynomial regression in your case suffers from local minimum when you try to minimize the objective function. The algorithm traps in local minimum during execution. i implement your code changing the step (alpha) and i saw that with smaller step it fits the data better but still you are trapping in local minimum.
Choosing random initialization point of thetas every time i am trapping in a different local minimum. If you are lucky you will find a better initial points for theta and fit the data better. I think that there are some algorithms that finds the best initial points.
Below i attach the results for random initial points and the results with Matlab's polyfit.
In the above plot replace "Linear Regression with Polynomial Regression", type error.
If you observe better the plot, you will see that by chance (using rand() ) i chose some initial points that leaded me to the best data fitting comparing the other initial points.... i am showing that with a pointer.
I implemented a gradient descent algorithm to minimize a cost function in order to gain a hypothesis for determining whether an image has a good quality. I did that in Octave. The idea is somehow based on the algorithm from the machine learning class by Andrew Ng
Therefore I have 880 values "y" that contains values from 0.5 to ~12. And I have 880 values from 50 to 300 in "X" that should predict the image's quality.
Sadly the algorithm seems to fail, after some iterations the value for theta is so small, that theta0 and theta1 become "NaN". And my linear regression curve has strange values...
here is the code for the gradient descent algorithm:
(theta = zeros(2, 1);, alpha= 0.01, iterations=1500)
function [theta, J_history] = gradientDescent(X, y, theta, alpha, num_iters)
m = length(y); % number of training examples
J_history = zeros(num_iters, 1);
for iter = 1:num_iters
tmp_j1=0;
for i=1:m,
tmp_j1 = tmp_j1+ ((theta (1,1) + theta (2,1)*X(i,2)) - y(i));
end
tmp_j2=0;
for i=1:m,
tmp_j2 = tmp_j2+ (((theta (1,1) + theta (2,1)*X(i,2)) - y(i)) *X(i,2));
end
tmp1= theta(1,1) - (alpha * ((1/m) * tmp_j1))
tmp2= theta(2,1) - (alpha * ((1/m) * tmp_j2))
theta(1,1)=tmp1
theta(2,1)=tmp2
% ============================================================
% Save the cost J in every iteration
J_history(iter) = computeCost(X, y, theta);
end
end
And here is the computation for the costfunction:
function J = computeCost(X, y, theta) %
m = length(y); % number of training examples
J = 0;
tmp=0;
for i=1:m,
tmp = tmp+ (theta (1,1) + theta (2,1)*X(i,2) - y(i))^2; %differenzberechnung
end
J= (1/(2*m)) * tmp
end
If you are wondering how the seemingly complex looking for loop can be vectorized and cramped into a single one line expression, then please read on. The vectorized form is:
theta = theta - (alpha/m) * (X' * (X * theta - y))
Given below is a detailed explanation for how we arrive at this vectorized expression using gradient descent algorithm:
This is the gradient descent algorithm to fine tune the value of θ:
Assume that the following values of X, y and θ are given:
m = number of training examples
n = number of features + 1
Here
m = 5 (training examples)
n = 4 (features+1)
X = m x n matrix
y = m x 1 vector matrix
θ = n x 1 vector matrix
xi is the ith training example
xj is the jth feature in a given training example
Further,
h(x) = ([X] * [θ]) (m x 1 matrix of predicted values for our training set)
h(x)-y = ([X] * [θ] - [y]) (m x 1 matrix of Errors in our predictions)
whole objective of machine learning is to minimize Errors in predictions. Based on the above corollary, our Errors matrix is m x 1 vector matrix as follows:
To calculate new value of θj, we have to get a summation of all errors (m rows) multiplied by jth feature value of the training set X. That is, take all the values in E, individually multiply them with jth feature of the corresponding training example, and add them all together. This will help us in getting the new (and hopefully better) value of θj. Repeat this process for all j or the number of features. In matrix form, this can be written as:
This can be simplified as:
[E]' x [X] will give us a row vector matrix, since E' is 1 x m matrix and X is m x n matrix. But we are interested in getting a column matrix, hence we transpose the resultant matrix.
More succinctly, it can be written as:
Since (A * B)' = (B' * A'), and A'' = A, we can also write the above as
This is the original expression we started out with:
theta = theta - (alpha/m) * (X' * (X * theta - y))
i vectorized the theta thing...
may could help somebody
theta = theta - (alpha/m * (X * theta-y)' * X)';
I think that your computeCost function is wrong.
I attended NG's class last year and I have the following implementation (vectorized):
m = length(y);
J = 0;
predictions = X * theta;
sqrErrors = (predictions-y).^2;
J = 1/(2*m) * sum(sqrErrors);
The rest of the implementation seems fine to me, although you could also vectorize them.
theta_1 = theta(1) - alpha * (1/m) * sum((X*theta-y).*X(:,1));
theta_2 = theta(2) - alpha * (1/m) * sum((X*theta-y).*X(:,2));
Afterwards you are setting the temporary thetas (here called theta_1 and theta_2) correctly back to the "real" theta.
Generally it is more useful to vectorize instead of loops, it is less annoying to read and to debug.
If you are OK with using a least-squares cost function, then you could try using the normal equation instead of gradient descent. It's much simpler -- only one line -- and computationally faster.
Here is the normal equation:
http://mathworld.wolfram.com/NormalEquation.html
And in octave form:
theta = (pinv(X' * X )) * X' * y
Here is a tutorial that explains how to use the normal equation: http://www.lauradhamilton.com/tutorial-linear-regression-with-octave
While not scalable like a vectorized version, a loop-based computation of a gradient descent should generate the same results. In the example above, the most probably case of the gradient descent failing to compute the correct theta is the value of alpha.
With a verified set of cost and gradient descent functions and a set of data similar with the one described in the question, theta ends up with NaN values just after a few iterations if alpha = 0.01. However, when set as alpha = 0.000001, the gradient descent works as expected, even after 100 iterations.
Using only vectors here is the compact implementation of LR with Gradient Descent in Mathematica:
Theta = {0, 0}
alpha = 0.0001;
iteration = 1500;
Jhist = Table[0, {i, iteration}];
Table[
Theta = Theta -
alpha * Dot[Transpose[X], (Dot[X, Theta] - Y)]/m;
Jhist[[k]] =
Total[ (Dot[X, Theta] - Y[[All]])^2]/(2*m); Theta, {k, iteration}]
Note: Of course one assumes that X is a n * 2 matrix, with X[[,1]] containing only 1s'
This should work:-
theta(1,1) = theta(1,1) - (alpha*(1/m))*((X*theta - y)'* X(:,1) );
theta(2,1) = theta(2,1) - (alpha*(1/m))*((X*theta - y)'* X(:,2) );
its cleaner this way, and vectorized also
predictions = X * theta;
errorsVector = predictions - y;
theta = theta - (alpha/m) * (X' * errorsVector);
If you remember the first Pdf file for Gradient Descent form machine Learning course, you would take care of learning rate. Here is the note from the mentioned pdf.
Implementation Note: If your learning rate is too large, J(theta) can di-
verge and blow up', resulting in values which are too large for computer
calculations. In these situations, Octave/MATLAB will tend to return
NaNs. NaN stands fornot a number' and is often caused by unde ned
operations that involve - infinity and +infinity.