I tried delta compression in MATLAB. I tried to use array instead of for loop but encountered a problem during the decompression. No syntax errors but not able to get back the original stream. Please help me. Here is my code:
clear all;
close all;
m = [20,3,55,11,222,555,6,98,0,46];
subplot(3,1,1);plot(m);title('Raw Data');
delta(1) = m(1);
i = [1:(length(m)-1)];
delta(i+1) = m(i+1)-m(i);
subplot(3,1,2);plot(delta);title('Delta Encoding')
j =[1:(length(delta)-1)];
delta_decode(1) = delta(1);
delta_decode(j+1)=delta(j+1)+delta(j);
subplot(3,1,3);plot(delta_decode);title('Delta Decoding')
So why is your decoding not working, lets have a look on the math
Lets assume we have a sequence of N numbers X1,X2,...,XN
The variable delta holds the following information
delta= delta(1), delta(2),delta(3), ...,delta(N)
delta= X1 , X2-X1 , X3-X2 , ..., XN - X(N-1)
So what you are doing right now is to add always two entries which will result in the following:
delta(1)+delta(2), delta(2)+delta(3), delta(3)+delta(4)
X1+X2-X1, X2-X1+X3-X2, X3-X2+X4-X3,...
Summarizing it:
X2, X3-X1, X4-X2
So you see this not what you want, to restore / decode the real values you need the accumulation of all previous information. Thats why you have to add everything
This means:
X2= X1 + delta(2)=X1+X2-X1=X2
X3= X1 + delta(2)+delta(3)=X2+delta(3)=X2+X3-X2=X3
X4= X1 + delta(2)+delta(3)+delta(4)=X2+delta(3)+delta(4)=X3 +delta(4)=X3+X4-X3=X4
and so on ...
As mentioned in the comments, you can achieve that without a for-loop by using cumsum(delta)
Related
I am trying to code something in Matlab and it involves a lot of accessing elements in vectors. Below is a snippet of code that I am working on:
x(1)=1;
for i=2:18
x(i)=0;
end
for i=1:18
y(i)=1;
end
for i = 0:262124
x(i+18+1) = x(i+7+1) + mod(x(i+1),2);
y(i+18+1) = y(i+10+1) + y(i+7+1) + y(i+5+1) + mod(y(i+1), 2);
end
% n can be = 0, 1, 2,..., 262142
n = 2;
for i = 0: 262142
z(i+1) = x(mod(i+n+1, 262143)); %error: Subscript indices must either be real positive integers or logicals.
end
In the last "for" loop where I am initialising vector z(), I get an error saying: "Subscript indices must either be real positive integers or logicals." However, when I do not suppres z(i+1) by ommiting the semi colon, the program is able to run, and I can see the values of z in the workspace. Why is this?
The code I am writing in Matlab is based upon the series of instructions shown in the image below. However, I can't seem to track down my error which leads to me not being able to access the elements of x() (without not suppressing the output of z()).
I appreciate any ideas :-) Thank you!
The code breaks at that loop last iteration because , for i=262140 you get
(mod(i+n+1, 262143)) = 0
so you cant access x(0) in matlab. the first elements of any variable is x(1).
In addition, and not related to your question, this code doesn't use the advantages matlab has, instead of
for i=2:18
x(i)=0;
end
you can just write:
x(2:18)=0;
etc
So after beating my head against the wall for a few hours, I looked online for a solution to my problem, and it worked great. I just want to know what caused the issue with the way I was originally going about it.
here are some more details. The input is a 20x20px image from the MNIST datset, and there are 5000 samples, so X, or A1 is 5000x400. There are 25 nodes in the single hidden layer. The output is a one hot vector of 0-9 digits. y (not Y, which is the one hot encoding of y) is a 5000x1 vector with the value of 1-10.
Here was my original code for the cost function:
Y = zeros(m, num_labels);
for i = 1:m
Y(i, y(i)) = 1;
endfor
H = sigmoid(Theta2*[ones(1,m);sigmoid(Theta1*[ones(m, 1) X]'))
J = (1/m) * sum(sum((-Y*log(H]))' - (1-Y)*log(1-H]))')))
But then I found this:
A1 = [ones(m, 1) X];
Z2 = A1 * Theta1';
A2 = [ones(size(Z2, 1), 1) sigmoid(Z2)];
Z3 = A2*Theta2';
H = A3 = sigmoid(Z3);
J = (1/m)*sum(sum((-Y).*log(H) - (1-Y).*log(1-H), 2));
I see that this may be slightly cleaner, but what functionally causes my original code to get 304.88 and the other to get ~ 0.25? Is it the element wise multiplication?
FYI, this is the same problem as this question if you need the formal equation written out.
Thanks for any help I can get! I really want to understand where I'm going wrong
Transfer from the comments:
With a quick look, in J = (1/m) * sum(sum((-Y*log(H]))' - (1-Y)*log(1-H]))'))) there is definetely something going on with the parenthesis, but probably on how you pasted it here, not with the original code as this would throw an error when you run it. If I understand correctly and Y, H are matrices, then in your 1st version Y*log(H) is matrix multiplication while in the 2nd version Y.*log(H) is an entrywise multiplication (not matrix-multiplication, just c(i,j)=a(i,j)*b(i,j) ).
Update 1:
In regards to your question in the comment.
From the first screenshot, you represent each value yk(i) in the entry Y(i,k) of the Y matrix and each value h(x^(i))k as H(i,k). So basically, for each i,k you want to compute Y(i,k) log(H(i,k)) + (1-Y(i,k)) log(1-H(i,k)). You can do it for all the values together and store the result in matrix C. Then C = Y.*log(H) + (1-Y).*log(1-H) and each C(i,k) has the above mentioned value. This is an operation .* because you want to do the operation for each element (i,k) of each matrix (in contrast to multiplying the matrices which is totally different). Afterwards, to get the sum of all the values inside the 2D dimensional matrix C, you use the octave function sum twice: sum(sum(C)) to sum both columnwise and row-wise (or as # Irreducible suggested, just sum(C(:))).
Note there may be other errors as well.
I have this task to create a script that acts similarly to normcdf on matlab.
x=linspace(-5,5,1000); %values for x
p= 1/sqrt(2*pi) * exp((-x.^2)/2); % THE PDF for the standard normal
t=cumtrapz(x,p); % the CDF for the standard normal distribution
plot(x,t); %shows the graph of the CDF
The problem is when the t values are assigned to 1:1000 instead of -5:5 in increments. I want to know how to assign the correct x values, that is -5:5,1000 to the t values output? such as when I do t(n) I get the same result as normcdf(n).
Just to clarify: the problem is I cannot simply say t(-5) and get result =1 as I would in normcdf(1) because the cumtrapz calculated values are assigned to x=1:1000 instead of -5 to 5.
Updated answer
Ok, having read your comment; here is how to do what you want:
x = linspace(-5,5,1000);
p = 1/sqrt(2*pi) * exp((-x.^2)/2);
cdf = cumtrapz(x,p);
q = 3; % Query point
disp(normcdf(q)) % For reference
[~,I] = min(abs(x-q)); % Find closest index
disp(cdf(I)) % Show the value
Sadly, there is no matlab syntax which will do this nicely in one line, but if you abstract finding the closest index into a different function, you can do this:
cdf(findClosest(x,q))
function I = findClosest(x,q)
if q>max(x) || q<min(x)
warning('q outside the range of x');
end
[~,I] = min(abs(x-q));
end
Also; if you are certain that the exact value of the query point q exists in x, you can just do
cdf(x==q);
But beware of floating point errors though. You may think that a certain range outght to contain a certain value, but little did you know it was different by a tiny roundoff erorr. You can see that in action for example here:
x1 = linspace(0,1,1000); % Range
x2 = asin(sin(x1)); % Ought to be the same thing
plot((x1-x2)/eps); grid on; % But they differ by rougly 1 unit of machine precision
Old answer
As far as I can tell, running your code does reproduce the result of normcdf(x) well... If you want to do exactly what normcdf does them use erfc.
close all; clear; clc;
x = linspace(-5,5,1000);
cdf = normcdf(x); % Result of normcdf for comparison
%% 1 Trapezoidal integration of normal pd
p = 1/sqrt(2*pi) * exp((-x.^2)/2);
cdf1 = cumtrapz(x,p);
%% 2 But error function IS the integral of the normal pd
cdf2 = (1+erf(x/sqrt(2)))/2;
%% 3 Or, even better, use the error function complement (works better for large negative x)
cdf3 = erfc(-x/sqrt(2))/2;
fprintf('1: Mean error = %.2d\n',mean(abs(cdf1-cdf)));
fprintf('2: Mean error = %.2d\n',mean(abs(cdf2-cdf)));
fprintf('3: Mean error = %.2d\n',mean(abs(cdf3-cdf)));
plot(x,cdf1,x,cdf2,x,cdf3,x,cdf,'k--');
This gives me
1: Mean error = 7.83e-07
2: Mean error = 1.41e-17
3: Mean error = 00 <- Because that is literally what normcdf is doing
If your goal is not not to use predefined matlab funcitons, but instead to calculate the result numerically (i.e. calculate the error function) then it's an interesting challange which you can read about for example here or in this stats stackexchange post. Just as an example, the following piece of code calculates the error function by implementing eq. 2 form the first link:
nerf = #(x,n) (-1)^n*2/sqrt(pi)*x.^(2*n+1)./factorial(n)/(2*n+1);
figure(1); hold on;
temp = zeros(size(x)); p =[];
for n = 0:20
temp = temp + nerf(x/sqrt(2),n);
if~mod(n,3)
p(end+1) = plot(x,(1+temp)/2);
end
end
ylim([-1,2]);
title('\Sigma_{n=0}^{inf} ( 2/sqrt(pi) ) \times ( (-1)^n x^{2*n+1} ) \div ( n! (2*n+1) )');
p(end+1) = plot(x,cdf,'k--');
legend(p,'n = 0','\Sigma_{n} 0->3','\Sigma_{n} 0->6','\Sigma_{n} 0->9',...
'\Sigma_{n} 0->12','\Sigma_{n} 0->15','\Sigma_{n} 0->18','normcdf(x)',...
'location','southeast');
grid on; box on;
xlabel('x'); ylabel('norm. cdf approximations');
Marcin's answer suggests a way to find the nearest sample point. It is easier, IMO, to interpolate. Given x and t as defined in the question,
interp1(x,t,n)
returns the estimated value of the CDF at x==n, for whatever value of n. But note that, for values outside the computed range, it will extrapolate and produce unreliable values.
You can define an anonymous function that works like normcdf:
my_normcdf = #(n)interp1(x,t,n);
my_normcdf(-5)
Try replacing x with 0.01 when you call cumtrapz. You can either use a vector or a scalar spacing for cumtrapz (https://www.mathworks.com/help/matlab/ref/cumtrapz.html), and this might solve your problem. Also, have you checked the original x-values? Is the problem with linspace (i.e. you are not getting the correct x vector), or with cumtrapz?
I am trying to optimize my code and am not sure how and if I would be able to vectorize this particular section??
for base_num = 1:base_length
for sub_num = 1:base_length
dist{base_num}(sub_num) = sqrt((x(base_num) - x(sub_num))^2 + (y(base_num) - y(sub_num))^2);
end
end
The following example provides one method of vectorization:
%# Set example parameters
N = 10;
X = randn(N, 1);
Y = randn(N, 1);
%# Your loop based solution
Dist1 = cell(N, 1);
for n = 1:N
for m = 1:N
Dist1{n}(m) = sqrt((X(n) - X(m))^2 + (Y(n) - Y(m))^2);
end
end
%# My vectorized solution
Dist2 = sqrt(bsxfun(#minus, X, X').^2 + bsxfun(#minus, Y, Y').^2);
Dist2Cell = num2cell(Dist2, 2);
A quick speed test at N = 1000 has the vectorized solution running two orders of magnitude faster than the loop solution.
Note: I've used a second line in my vectorized solution to mimic your cell array output structure. Up to you whether you want to include it or two combine it into one line etc.
By the way, +1 for posting code in the question. However, two small suggestions for the future: 1) When posting to SO, use simple variable names - especially for loop subscripts - such as I have in my answer. 2) It is nice when we can copy and paste example code straight into a script and run it without having to do any changes or additions (again such as in my answer). This allows us to converge on a solution more rapidly.
I’m currently a Physics student and for several weeks have been compiling data related to ‘Quantum Entanglement’. I’ve now got to a point where I have to plot my data (which should resemble a cos² graph - and does) to a sort of “best fit” cos² graph. The lab script says the following:
A more precise determination of the visibility V (this is basically how 'clean' the data is) follows from the best fit to the measured data using the function:
f(b) = A/2[1-Vsin(b-b(center)/P)]
Granted this probably doesn’t mean much out of context, but essentially A is the amplitude, b is an angle and P is the periodicity. Hence this is also a “wave” like the experimental data I have found.
From this I understand, as previously mentioned, I am making a “best fit” curve. However, I have been told that this isn’t possible with Excel and that the best approach is Matlab.
I know intermediate JavaScript but do not know Matlab and was hoping for some direction.
Is there a tutorial I can read for this? Is it possible for someone to go through it with me? I really have no idea what it entails, so any feed back would be greatly appreciated.
Thanks a lot!
Initial steps
I guess we should begin by getting a representation in Matlab of the function that you're trying to model. A direct translation of your formula looks like this:
function y = targetfunction(A,V,P,bc,b)
y = (A/2) * (1 - V * sin((b-bc) / P));
end
Getting hold of the data
My next step is going to be to generate some data to work with (you'll use your own data, naturally). So here's a function that generates some noisy data. Notice that I've supplied some values for the parameters.
function [y b] = generateData(npoints,noise)
A = 2;
V = 1;
P = 0.7;
bc = 0;
b = 2 * pi * rand(npoints,1);
y = targetfunction(A,V,P,bc,b) + noise * randn(npoints,1);
end
The function rand generates random points on the interval [0,1], and I multiplied those by 2*pi to get points randomly on the interval [0, 2*pi]. I then applied the target function at those points, and added a bit of noise (the function randn generates normally distributed random variables).
Fitting parameters
The most complicated function is the one that fits a model to your data. For this I use the function fminunc, which does unconstrained minimization. The routine looks like this:
function [A V P bc] = bestfit(y,b)
x0(1) = 1; %# A
x0(2) = 1; %# V
x0(3) = 0.5; %# P
x0(4) = 0; %# bc
f = #(x) norm(y - targetfunction(x(1),x(2),x(3),x(4),b));
x = fminunc(f,x0);
A = x(1);
V = x(2);
P = x(3);
bc = x(4);
end
Let's go through line by line. First, I define the function f that I want to minimize. This isn't too hard. To minimize a function in Matlab, it needs to take a single vector as a parameter. Therefore we have to pack our four parameters into a vector, which I do in the first four lines. I used values that are close, but not the same, as the ones that I used to generate the data.
Then I define the function I want to minimize. It takes a single argument x, which it unpacks and feeds to the targetfunction, along with the points b in our dataset. Hopefully these are close to y. We measure how far they are from y by subtracting from y and applying the function norm, which squares every component, adds them up and takes the square root (i.e. it computes the root mean square error).
Then I call fminunc with our function to be minimized, and the initial guess for the parameters. This uses an internal routine to find the closest match for each of the parameters, and returns them in the vector x.
Finally, I unpack the parameters from the vector x.
Putting it all together
We now have all the components we need, so we just want one final function to tie them together. Here it is:
function master
%# Generate some data (you should read in your own data here)
[f b] = generateData(1000,1);
%# Find the best fitting parameters
[A V P bc] = bestfit(f,b);
%# Print them to the screen
fprintf('A = %f\n',A)
fprintf('V = %f\n',V)
fprintf('P = %f\n',P)
fprintf('bc = %f\n',bc)
%# Make plots of the data and the function we have fitted
plot(b,f,'.');
hold on
plot(sort(b),targetfunction(A,V,P,bc,sort(b)),'r','LineWidth',2)
end
If I run this function, I see this being printed to the screen:
>> master
Local minimum found.
Optimization completed because the size of the gradient is less than
the default value of the function tolerance.
A = 1.991727
V = 0.979819
P = 0.695265
bc = 0.067431
And the following plot appears:
That fit looks good enough to me. Let me know if you have any questions about anything I've done here.
I am a bit surprised as you mention f(a) and your function does not contain an a, but in general, suppose you want to plot f(x) = cos(x)^2
First determine for which values of x you want to make a plot, for example
xmin = 0;
stepsize = 1/100;
xmax = 6.5;
x = xmin:stepsize:xmax;
y = cos(x).^2;
plot(x,y)
However, note that this approach works just as well in excel, you just have to do some work to get your x values and function in the right cells.