So far we have tried to classify flowers from the Iris dataset. Classification tasks assume that we have a set of features and a set of class labels that we would like our models to learn to predict.
Regression tasks don't assume categorical labels. In regression tasks we again have some set of features but the value we want to predict isn't a label anymore but some value from a continuous domain. Therefore the the output domain of the model is continuous.
You can for example imagine the quadratic equation
An example of a regression task might be to predict what
In this assignment we will be using a altered version of the Iris dataset were we have ditched the class label targets and replaced them with the last feature column (representing the pedal length of the flower). You can load this data with tools.load_regression_iris.
We are going to use the Gaussian basis functions to predict real valued target variables of our data. $$ \phi_k(x) = \frac{1}{(2\pi)^{D/2}} \frac{1}{|\Sigma_k|^{1/2}} e^{-\frac{1}{2} (x-\mu_k)^T \Sigma_k^{-1} (x-\mu_k)} $$
We control the shift of the basis function using the mean vector
Create a function mvn_basis(features, mu, sigma) that applies the multivariate normal basis function on the set of features. You should use scipy.stats.multivariate_normal to create your multivariate gaussians.
Example inputs and outputs:
Load the data
X, t = load_regression_iris()
N, D = X.shape
We need to define how many basis functions we are going to use and determine the mean vectors that we are using. We also have to define the variance, sigma. We do this arbitrarily.
M, sigma = 10, 10
mu = np.zeros((M, D))
for i in range(D):
mmin = np.min(X[i, :])
mmax = np.max(X[i, :])
mu[:, i] = np.linspace(mmin, mmax, M)
fi = mvn_basis(X, mu, sigma)
Output
fi: [[0.00081185 0.00100599 0.00119015 ... 0.00137739 0.00123433 0.00105608]
[0.00095701 0.0011535 0.00132743 ... 0.0013378 0.00116613 0.00097051]
[0.00099061 0.00118896 0.00136247 ... 0.00134443 0.00116697 0.00096711]
...
[0.00022494 0.00033425 0.00047422 ... 0.00136119 0.00146281 0.0015009 ]
[0.00022415 0.00033669 0.00048286 ... 0.00146273 0.00158896 0.001648 ]
[0.00031179 0.00045031 0.00062097 ... 0.00154612 0.00161495 0.00161053]]
Plot the output of each basis function, using the same parameters as above, as a function of the features. You should plot all the outputs onto the same plot. Include your plot as plot_1_2 in your PDF document.
A single basis function output is shown below
Create a function max_likelihood_linreg(fi, targets, lam) that estimates the maximum likelihood values for the linear regression model.
Example inputs and outputs:
fi = mvn_basis(X, mu, sigma) # same as before
lamda = 0.001
wml = max_likelihood_linreg(fi, t, lamda)
Output:
wml: [ 3.04661879 8.98364452 17.87352659 29.86144389 44.59162487
61.12887842 78.01047234 93.44289728 105.61212037 113.03406275]
Finally create a function linear_model(features, mu, sigma, w) that predicts targets given the weights from your max_likelihood_linreg, the basis functions, defined by mu and sigma, and the features.
Example inputs and outputs:
wml = max_likelihood_linreg(fi, t, lamda) # as before
prediction = linear_model(X, mu, sigma, wml)
Output:
prediction = [0.72174704 0.71462572 0.72048698 0.75382486 0.73161981 0.7494642
0.7502493 0.74274217 0.73390818 0.73639504 0.71611075 0.77254612
...
0.77988157 0.60555882 0.61733184 0.62902646 0.75304099 0.57770296
0.61914859 0.64629346 0.67533913 0.67677248 0.72763469 0.76905279]
This question should be answered in your PDF document
- How good are these predictions?
- Use plots to show the prediction accuracy, either by plotting the actual values vs predicted values or the mean-square-error.
This is a completely open ended independent section. You are free to test out the methods in the assignment on different datasets, change the models, change parameters, make visualizations, analyze the models, or anything you desire.
Upload a PDF document with your experiments and results. Include the code you wrote specifically for this independent section as well.