Sunday, December 16, 2012

Using matplotlib to plot the answer to ps1

Now that you know how to solve the first problem in Problem set 1, you have to graph the answer. I used matplotlib. The website for this plotting module can be found here.

Here's a copy of code to plot the data for ps 1:

#Plots x1 vs x2 for Problem 1 of the Stanford machine learning class

from pylab import *
import numpy as np
import numpy.linalg
from math import *


# we need to show when h=.5 as a line to separate the data. h=.5 when
# theta transpose x =0. so 0=theta0 +theta1*x1+theta2*x2
# solving for x2 give -theta0/theta2-theta1/theta2*x1

# Use different colors and markers to plot x1 vs x2 if y=0 or y=1

for i in range(0,98):
    if y[i]==0:
    if y[i]==1:



I used a for x1 and b as the calculated value for x2. Plotting a vs b gives that beautiful straight line across the plot.

One of my previous posts talked about expecting the unexpected in numpy. It happens again in this little short program. This line:
was not what I though it was. I thought this gave me a list of values between the minimum of x1 and the maximum of x1. I was wrong. Here is what this actually gives:
  print a
[ 0.57079941  7.7054006   1.          0.01      ]

If you want a list of values, you need to use this code:


This gives 25 values between the minimum and maximum of your data set.

The rest of the code is very straight forward. Just set up a loop. If y=1, the data plots as a red circle. If y=0, the data plots as a blue x.

I know I have posted this graph before but it is so pretty, I'm going to post it again:

Wednesday, December 5, 2012

Code for the Newton's method problem

As promised, here is my code for Newton's method. Note that I am using Numpy but not pandas. A friend has lent me this book so I am currently working on learning what pandas can do for me.

The first thing you have to do is import numpy. I import as np so all of my numpy commands begin with np. For some reason, the linear algebra module does not load unless I specifically call it with an import command. I need this to invert the Hessian matrix.
import numpy as np
import numpy.linalg
from math import *

Here I read in the data. The size of x along axis=0 (rows) gives the number of training examples. In order to get the intercept term, I have to put a column of ones in front of the x data. In order to get the column of ones where I wanted it, I had to use the insert command. Then I checked the number of columns.
m=np.size(x,0) #number of training examples
n=np.size(x,1) #number of columns

This piece of code sets up a vector of zeros for theta to start the iterations. I do the same thing in the code below for the Hessian matrix and gradient vector. Recall that everything in Python starts at zero so I have an n x 1 vector for the gradient instead of n+1 x 1 like in Matlab.


 Note the reshape command for theta when I am calculating h. For some reason, I couldn't get theta to transpose.
# set up for 10 iterations
for i in range(0,10):
    print "Theta just before h is ", theta

This part of the code is pretty self explanatory and looks very much like the matlab code from the answer sheet. However, note the outer command in the calculation for H. Let's examine this a little more closely.

Here's an example. In this example, a is an example of the lprime vector with one variable (ignore the intercept) and 4 training examples. b is the same vector. Notice they are both 4x1, so you cannot use matrix multiplication.
>>> print a
>>> b=np.array([[1],[2],[3],[4]])
>>> print b

Traceback (most recent call last):
  File "<pyshell#23>", line 1, in <module>,b)
ValueError: objects are not aligned

However, look at what happens when you use the outer command:
>>> d=np.outer(a,b)
>>> print d
[[ 1  2  3  4]
 [ 2  4  6  8]
 [ 3  6  9 12]
 [ 4  8 12 16]]

You get a 4x4 matrix. Where did this come from? The outer product gives you
out[i, j] = a[i] * b[j]
This is what you need for Hessian matrix. It took me a really, really long time to find this. You're welcome.
So here is the rest of the code:
# This calculates the values for lprime vector which will be summed below
    for j in range(0,m):


The next post will show the code for the plot.