Perceptron training in Matlab - matlab

I am trying to create a simple perceptron training function in MATLAB. I want to return the weights when no errors are found.
Here is the data I want to classify.
d = rand(10,2);
figure
labels = ones(10,1);
diff = d(:,1) + d(:,2) - 1;
labels( diff + 1 >= 1) = 2;
pathWidth = 0.05;
labels( abs(diff) < pathWidth) = [];
d(abs(diff) < pathWidth,:) = [];
plot(d(labels == 1,1),d(labels == 1,2),'k.','MarkerSize',10)
plot(d(labels == 2,1),d(labels == 2,2),'r.','MarkerSize',10)
It produces a labelled data set, where the division between the two classes (red, black) are more visible if you increase the number of points for d.
For my perceptron function I pass the data (d) and labels. I have 3 inputs, the x values, y values and bias which is one. Each input has a random weight between 0 and 1 assigned. Note that the dataset d I have named Z in the perceptron function.
I did use a sigmoid activation function but that would run once through the while loop and always return true after that, the sigmoid function also gave me values of either inf or 1. Below I am using just a threshold activation but it seems to continually loop and not returning my weights. I think the problem may lie in the if-statement below
if(v >= 0 && labels(i) == 1 || v < 0 && labels(i) == 2)
Perceptron function:
function perceptron(data,labels)
sizea = numel(data(:,1));
weights = rand(sizea,3);
Z = data(:,:)
eta = 0.5;
errors = 1;
count = 0;
while errors > 0
errors = 0;
v = sum((1*weights(1,1)) + (Z(:,1)*weights(1,2)) + (Z(:,2)*weights(1,3)));
if v >= 1
v = 1;
else
v = 0;
end
count = count + 1
for i = 1:sizea % for each object in dataset
if(v == 1 && labels(i) == 1 || v == 0 && labels(i) == 2)
errors = 1;
weights(1,1) = weights(1,1) - (2*v-1) * eta * 1;
weights(1,2) = weights(1,2) - (2*v-1) * eta * Z(i,1);
weights(1,3) = weights(1,3) - (2*v-1) * eta * Z(i,2);
v = sum((1*weights(1,1)) + (Z(:,1)*weights(1,2)) + (Z(:,2)*weights(1,3)));
if v >= 1
v = 1;
else
v = 0;
end
end
end
end

There are two major problems in your code:
You need to update v in your loop every time your weights vectors are updated.
It seems like you have 10 training set. So you have to update v sequentially in the loop instead of simultaneously. Keep iterating for every training set, update weights, then use the new weights to calculate the v for the next training set, so on and so forth, until there is no error (errors = 0 in your case).
Minor issue:
if(v >= 0 && labels(i) == 1 || v < 0 && labels(i) == 2)
should be
if(v == 1 && labels(i) == 1 || v == 0 && labels(i) == 2)
You may refer to this example to get more details of the algorithm.

Related

Nested for loop error or indexing error in MATLAB

I have created this code from scratch. I want to make a plot and/or histogram of my "Observed" and "State" (these are 2 matrices). Some problem occurs at the 200th iteration, where my State matrix just becomes all 0's, there is no data being input into the State matrix. Can anyone troubleshoot the code? My possible states are {1,2,3}.
UPDATE:
When I adjust my n value, it adjusts how much of length T it will fill. So, n=5, only runs for 1/5 of T and n=1, run for entire length of T. I need an nxT matrix at the end (5X1000). The problem lies in the way I setup my for loops.
I still cannot solve the error though.
%Initialize A,pi,T
N = 3; # of states
%A is transition prob matrix
A = [.99,.005,.005;.005,.990,.005;.005,.005,.990];
%pi is initial state vector
pi = [1/3,1/3,1/3];
%T is # of observations per simulation
T = 1000;
%n is # of simulations
n = 5;
%Allocate space for the state matrix
State = zeros(n,T);
Observe = zeros(n,T);
%Create dummy emission matrix, must be row stochastic
B = ones(n,T)./T;
%loop over # of simulations
for i=1:1:n
x = rand(1);
if x <= (1/3)
State(i,1) = 1;
elseif x > (1/3) && x <= (2/3)
State(i,1) = 2;
else
State(i,1) = 3;
end
if State(i,1) == 1
b = -1;
elseif State(i,1) == 2
b = 0;
else
b = 1;
end
Observe(i,1)= normrnd(b,1);
for k=2:1:T
%Possible state 1,2,3
State(k) = randsample(N, 1, true, A(State(k-1),:));
if State == 1
c = -1;
elseif State == 2
c = 0;
else
c = 1;
end
Observe(i,k)= normrnd(c,1);
end
end
State(k) = randsample(N, 1, true, A(State(k-1),:));
This line is missing index (i) in position 1 inside State(k-1). It should be:
State(i,k) = randsample(N, 1, true, A(State(i,k-1),:));

Output 1, 0.5, or 0 depending if a matrix elements are prime, 1, or neither

I am sending a matrix to my function modifikuj, where I want to replace the elements of the matrix with:
1 if element is a prime number
0 if element is a composite number
0.5 if element is 1
I dont understand why it is not working. I just started with MATLAB, and I created this function:
function B = modifikuj(A)
[n,m] = size(A);
for i = 1:n
for j = 1:m
prost=1;
if (A(i,j) == 1)
A(i,j) = 0.5;
else
for k = 2:(A(i,j))
if(mod(A(i,j),k) == 0)
prost=0;
end
end
if(prost==1)
A(i,j)=1;
else
A(i,j)=0;
end
end
end
end
With
A = [1,2;3,4];
D = modifikuj(A);
D should be:
D=[0.5, 1; 1 0];
In MATLAB you'll find you can often avoid loops, and there's plenty of built in functions to ease your path. Unless this is a coding exercise where you have to use a prescribed method, I'd do the following one-liner to get your desired result:
D = isprime( A ) + 0.5*( A == 1 );
This relies on two simple tests:
isprime( A ) % 1 if prime, 0 if not prime
A == 1 % 1 if == 1, 0 otherwise
Multiplying the 2nd test by 0.5 gives your desired condition for when the value is 1, since it will also return 0 for the isprime test.
You are not returning anything from the function. The return value is supposed to be 'B' according to your code but this is not set. Change it to A.
You are looping k until A(i,j) which is always divisible by itself, loop to A(i,j)-1
With the code below I get [0.5,1;1,0].
function A = modifikuj(A)
[n,m] = size(A);
for i = 1:n
for j = 1:m
prost=1;
if (A(i,j) == 1)
A(i,j) = 0.5;
else
for k = 2:(A(i,j)-1)
if(mod(A(i,j),k) == 0)
prost=0;
end
end
if(prost==1)
A(i,j)=1;
else
A(i,j)=0;
end
end
end
end
In addition to #EuanSmith's answer. You can also use the in built matlab function in order to determine if a number is prime or not.
The following code will give you the desired output:
A = [1,2;3,4];
A(A==1) = 0.5; %replace 1 number with 0.5
A(isprime(A)) = 1; %replace prime number with 1
A(~ismember(A,[0.5,1])) = 0; %replace composite number with 0
I've made the assumption that the matrice contains only integer.
If you only want to learn, you can also preserve the for loop with some improvement since the function mod can take more than 1 divisor as input:
function A = modifikuj(A)
[n,m] = size(A);
for i = 1:n
for j = 1:m
k = A(i,j);
if (k == 1)
A(i,j) = 0.5;
else
if all(mod(k,2:k-1)) %check each modulo at the same time.
A(i,j)=1;
else
A(i,j)=0;
end
end
end
end
And you can still improve the prime detection:
2 is the only even number to test.
number bigger than A(i,j)/2 are useless
so instead of all(mod(k,2:k-1)) you can use all(mod(k,[2,3:2:k/2]))
Note also that the function isprime is a way more efficient primality test since it use the probabilistic Miller-Rabin algorithme.

Multi valued function graph in matlab

This is my first script in matlab. ( I cannot use functions)
Let's say I have a vector of time instants
t = [0:999]*1e-5; %vector of time instants
And my function is m
This is the part where it is implemented:
if (t >= 0)
if t <= to/3
m = 1;
elseif (t <= 2*to/3)
m = -2;
end
else
m = 0;
end
As I realised, m has only one value equal to 1.
How is this possible to have a 1x1000 value here? Where for values of t from 0 to to/3 -> m = 1, to/3 to 2*to/3 -> m = -2and else m=0
You can apply a function on each element of an array using arrayfun like the following:
arrayfun(#(x) m(x), t)
%or
arrayfun(#m, t)
You can find the details here. Also, you can implement your function like the following:
function result = m(t)
result = t;
result(t >= 0 && t <= to/3) = 1;
result(t > to/3 && t <= 2*to/3) = -2;
result(t < 0) = 0;
then call the function m on t such as m(t).

Iterate over a vector in matlab

So I'm working on a program. I need to iterate over a vector similar to this sample = [0; 0; 0; 1; 1; 1 ; 0]. I was thinking about using loops with accumulators to build a new 2d array where column 1 is how many 0s or 1s appear and column 2 is for which token it is. But I'm new to the syntax of matlab and checking the docs i mostly see slicing. Any ideas about building the new matrix are welcome.
sample vector and output below
arr = [0; 0; 0; 1; 1; 1; 0];
tokenizeSignal(arr)
ans =
3 0
3 1
1 0
Proposed strategy (array contains only 1 and 0) :
Initialize 2 counters
count_0 = 0;
count_1 = 0;
Iterate over the whole array arr
arr = [0; 0; 0; 1; 1; 1; 0];
[n,m] = size(arr); %m is espected equal to 1
y = arr[1,1]; %first element of the array, we need a reference
Start the loop and read the current element
for i=1:n
x = arr[i,1];
start to count how many zeros or how many ones from the last group if the last element was zero or one
if (x == 0 && y == 0)
count_0 = count_0 + 1;
count_1 = 0;
else if (x == 1 && y == 1)
count_1 = count_1 + 1;
count_0 = 0;
end
print every time the value will change (last term of a "sequence")
if (x != arr[1,i+1] && count_1 > 0 && i<n)
print(count_1, '1');
else if (x != arr[1,i+1] && count_0 > 0 && i<n)
print(count_0, '0');
end
update the values and close the FOR loop
y = x;
end
out of the cycle, print the last time
if (count_1 > 0)
print(count_1, '1');
else if (count_0 >)
print(count_0, '0');
end
Of course you can change the print with store the values in a proper array.

Only allow fixed number of random points in each tile

Following up on a previous question, I have code that I think should limit the number of randomly generated points in each quadrant of the total tile; however, it is not working.
n = 4;
used = [];
k = 0;
a1_count = 0;
a2_count = 0;
a3_count = 0;
a4_count = 0;
min = 1;
max = 1;
while k<n
x = rand*2;
y = rand*2;
notvalid = 0;
if (0 <= x) && (x <= 1) && (0 <= y) && (y <= 1)
a1_count = a1_count + 1;
end
if (1 < x) && (x <= 2) && (0 <= y) && (y <= 1)
a2_count = a2_count + 1;
end
if (0 <= x) && (x <= 1) && (1 < y) && (y <= 2)
a3_count = a3_count + 1;
end
if (1 < x) && (x <= 2) && (1 < y) && (y <= 2)
a4_count = a4_count + 1;
end
%%%
if (min <= a1_count) && (a1_count <= max) && (min <= a2_count) && (a2_count <= max)...
&& (min <= a3_count) && (a3_count <= max) && (min <= a4_count) && (a4_count <= max)
notvalid=1;
end
if notvalid
continue
end
used(end+1,:) = [x;y];
k = k+1;
end
I wish to generate 4 random points, and have one in each quadrant of the total area. To do this, I have a maximum and minimum number of points in each quadrant (in this case 1), and an if statement to check that the count for each tile falls within the min and max. If it doesn't, then notvalid = 0 and the loop should begin again. This function doesn't seem to work however, as the loop finishes with 4 points total and is completely random (all the counts should = 1).
Can anyone spot where I'm going wrong?
I may be missing something, but the easiest approach would probably be something like
Select N random numbers within the x/y range of the first grid cell
Repeat for all grid cells
Here is some basic code that should create N random x/y points per grid cell
% Define the grid (for demonstration purposes)
dx = 1; dy = 1;
xrange = 0:dx:2;
yrange = 0:dy:2;
% Number of points per cell
N = 1;
[lowerx, lowery] = meshgrid(xrange(1:end-1), yrange(1:end-1));
% Store all those random numbers in a cell
data = cell(size(lowerx));
for k = 1:numel(lowerx);
% Generate 4 random points within the x/y range
xcoord = (rand(N, 1) * dx) + lowerx(k);
ycoord = (rand(N, 1) * dy) + lowery(k);
data{k} = [xcoord, ycoord];
end
disp(data)
data =
[1x2 double] [1x2 double]
[1x2 double] [1x2 double]
EDIT
To address your question directly using the code that you have provided, the logic in the code in your question is a little wonky. I have rewritten your while loop to be a little clearer so we can talk through it.
while k < n
x = rand * 2;
y = rand * 2;
if y >= 0 && y < 1
if x >= 0 && x < 1
a1_count = a1_count + 1;
else
a2_count = a2_count + 1;
end
else
if x >= 0 && x < 1
a3_count = a3_count + 1;
else
a4_count = a4_count + 1;
end
end
counts = [a1_count, a2_count, a3_count, a4_count];
notValid = all(counts >= minimum) && all(counts <= maximum);
if notValid
continue;
end
used(end+1,:) = [x;y];
k = k+1;
end
So the biggest thing is your notValid check. If you actually look at what you're checking (that all your *_count variables are within the pre-specified limits), I believe that if all of those conditions are true, then the current point is valid; however you state just the opposite.
Then you basically say, that if the current x y is valid, then add it to the used list. Well this logic is fine except that you define validity backwards as I stated before.
Ok so that aside, let's look at when you think that a point is not valid. Well, then you (correctly) go to the next iteration, but you never decrement the *_count variable. So say you had 1 point in quadrant 1 already and the second iteration through the loop it's in quadrant 1 again. Well you'd add 1 to a1_count and then see that it isn't valid (a1_count exceeds max) and go to the next loop, but a1_count stays at 2 despite really only having 1 because you just rejected it.
Now, all of that aside. Let's consider the first time through your loop and look at your validity check. Since you only add one point. Your validity check can never pass (if implemented correctly) because all *_count variables except the one that was just incremented will be less than the min.
So I think what happened is you probably did the validity check correctly at first, ended up with an infite while loop, and then negated that check, didn't get an infinite loop, but as a result got an incorrect solution.
The solution that you are getting currently, is literally the first 4 times through the while loop due to the incorrect logic.
If you really like your current approach, we can clean up the code to be correct.
n = 4;
used = zeros(0,2);
minimum = 1;
maximum = 1;
counts = [0 0 0 0];
while true
x = rand * 2;
y = rand * 2;
if y >= 0 && y < 1
if x >= 0 && x < 1
quadrant = 1;
else
quadrant = 2;
end
else
if x >= 0 && x < 1
quadrant = 3;
else
quadrant = 4;
end
end
% Check to see if we can even add this point
if counts(quadrant) + 1 > maximum
continue;
end
counts(quadrant) = counts(quadrant) + 1;
used = cat(1, used, [x, y]);
isComplete = all(counts >= minimum & counts <= maximum) && ...
size(used, 1) == n;
if isComplete
break
end
end