I'm trying to loop a pattern of numbers using a For loop in matlab / octave
The pattern I'm looking for is
40,80,160,320,280,200 and then 1 is added to each one so the pattern would look like this:
40,80,160,320,280,200,41,81,161,321,281,201,42,82,162,322,282,202
I tried using a for loop below
clear all
numL_tmp=[40;80;160;320;280;200]
numL=[numL_tmp];
for ii=1:length(numL_tmp)
for jj=1:4
numL=[numL;numL_tmp(ii,1)+jj]
end
end
But I get
40,80,160,320,280,200,41,42,81,82,161,162,321,322,281,282,201,202
How can I fix this?
For the problem stated, nested loops are unnecessary. You could simply do the following:
clear all;
numL_tmp=[40;80;160;320;280;200];
numL = numL_tmp;
for ii=1:2
numL = [numL;numL_tmp+ii];
end
numL
This would yield:
numL =
40
80
160
320
280
200
41
81
161
321
281
201
42
82
162
322
282
202
This works because MATLAB recognizes the piece of code numL_tmp+ii as something equivalent to numL_tmp + ii*ones(size(numL_tmp)).
You can avoid loops completely:
N = 3;
numL = kron(ones(N,1),numL_tmp) + kron((0:N-1)',ones(numel(numL_tmp),1));
There are easier ways to do it, but the fundamental problem with your code is that the inner and outer loops in the wrong order. See what happens if you leave your code as is, but simply interchange the order of the two loops:
...
numL=[numL_tmp];
for jj=1:4
for ii=1:length(numL_tmp)
numL=[numL;numL_tmp(ii,1)+jj]
end
end
Related
I drew lines at 100,100,20 intervals as shown in the picture.
How do I create code using (for) or (while)? Please let me know.
xline(100,'b-');
xline(200,'b-');
xline(220,'b-');
xline(320,'b-');
xline(420,'b-');
xline(440,'b-');
xline(540,'b-');
It it's not a constant spacing between x coordinates of your line you'll have to define a matrix then call for it's element :
x_matrix = [100 200 220 320 420 440 540];
for i = 1:length(x_matrix)
xline(x_matrix(i),'b-');
end
PS: I wouldn't recommend to use plot function in a while loop. If the while loop is executed too much times or even if matlab is stuck in it, your computer won't like...
I'm having surprisingly difficult time to figure out something which appears so simple. I have two known coordinates on a graph, (X1,Y1) and (X2,Y2). What I'm trying to identify are the coordinates for (X3,Y3).
I thought of using sin and cos but once I get here my brain stops working. I know that
sin O = y/R
cos O = x/R
so I thought of simply importing in the length of the line (in this case it was 2) and use the angles which are known. Seems very simple but for the life of me, my brain won't wrap around this.
The reason I need this is because I'm trying to print a line onto an image using poly2mask in matlab. The code has to work in the 2D space as I will be building movies using the line.
X1 = [134 134 135 147 153 153 167]
Y1 = [183 180 178 173 164 152 143]
X2 = [133 133 133 135 138 143 147]
Y2 = [203 200 197 189 185 173 163]
YZdist = 2;
for aa = 1:length(X2)
XYdis(aa) = sqrt((x2(aa)-x1(aa))^2 + (Y2(aa)-Y1(aa))^2);
X3(aa) = X1(aa) * tan(XYdis/YZdis);
Y3(aa) = Y1(aa) * tan(XYdis/YZdis);
end
polmask = poly2mask([Xdata X3],[Ydata Y3],50,50);
one approach would be to first construct a vector l connection points (x1,y1) and (x2,y2), rotate this vector 90 degrees clockwise and add it to the point (x2,y2).
Thus l=(x2-x1, y2-y1), its rotated version is l'=(y2-y1,x1-x2) and therefore the point of interest P=(x2, y2) + f*(y2-y1,x1-x2), where f is the desired scaling factor. If the lengths are supposed to be the same, then f=1 and thus P=(x2 + y2-y1, y2 + x1-x2).
I'm in the process of coding what I'm learning about Linear Regression from the coursera Machine Learning course (MATLAB). There was a similar post that I found here, but I don't seem to be able to understand everything. Perhaps because my fundamentals in Machine Learning are a bit weak.
The problem I'm facing is that, for some data... both gradient descent (GD) and the Closed Form Solution (CFS) give the same hypothesis line. However, on one particular dataset, the results are different. I've read something about, that, if the data is singular, then results should be the same. However, I have no idea how to check whether or not my data is singular.
I will try to illustrate the best I can:
1) Firstly, here is the MATLAB code adapted from here. For the given dataset, everything turned out good where both GD and the CFS gave similar results.
The Dataset
X Y
2.06587460000000 0.779189260000000
2.36840870000000 0.915967570000000
2.53999290000000 0.905383540000000
2.54208040000000 0.905661380000000
2.54907900000000 0.938988900000000
2.78668820000000 0.966847400000000
2.91168250000000 0.964368240000000
3.03562700000000 0.914459390000000
3.11466960000000 0.939339440000000
3.15823890000000 0.960749710000000
3.32759440000000 0.898370940000000
3.37931650000000 0.912097390000000
3.41220060000000 0.942384990000000
3.42158230000000 0.966245780000000
3.53157320000000 1.05265000000000
3.63930020000000 1.01437910000000
3.67325370000000 0.959694260000000
3.92564620000000 0.968537160000000
4.04986460000000 1.07660650000000
4.24833480000000 1.14549780000000
4.34400520000000 1.03406250000000
4.38265310000000 1.00700090000000
4.42306020000000 0.966836480000000
4.61024430000000 1.08959190000000
4.68811830000000 1.06344620000000
4.97773330000000 1.12372390000000
5.03599670000000 1.03233740000000
5.06845360000000 1.08744520000000
5.41614910000000 1.07029880000000
5.43956230000000 1.16064930000000
5.45632070000000 1.07780370000000
5.56984580000000 1.10697580000000
5.60157290000000 1.09718750000000
5.68776170000000 1.16486030000000
5.72156020000000 1.14117960000000
5.85389140000000 1.08441560000000
6.19780260000000 1.12524930000000
6.35109410000000 1.11683410000000
6.47970330000000 1.19707890000000
6.73837910000000 1.20694620000000
6.86376860000000 1.12510460000000
7.02233870000000 1.12356720000000
7.07823730000000 1.21328290000000
7.15142320000000 1.25226520000000
7.46640230000000 1.24970650000000
7.59738740000000 1.17997060000000
7.74407170000000 1.18972990000000
7.77296620000000 1.30299340000000
7.82645140000000 1.26011340000000
7.93063560000000 1.25622670000000
My MATLAB code:
clear all; close all; clc;
x = load('ex2x.dat');
y = load('ex2y.dat');
m = length(y); % number of training examples
% Plot the training data
figure; % open a new figure window
plot(x, y, '*r');
ylabel('Height in meters')
xlabel('Age in years')
% Gradient descent
x = [ones(m, 1) x]; % Add a column of ones to x
theta = zeros(size(x(1,:)))'; % initialize fitting parameters
MAX_ITR = 1500;
alpha = 0.07;
for num_iterations = 1:MAX_ITR
thetax = x * theta;
% for theta_0 and x_0
grad0 = (1/m) .* sum( x(:,1)' * (thetax - y));
% for theta_0 and x_0
grad1 = (1/m) .* sum( x(:,2)' * (thetax - y));
% Here is the actual update
theta(1) = theta(1) - alpha .* grad0;
theta(2) = theta(2) - alpha .* grad1;
end
% print theta to screen
theta
% Plot the hypothesis (a.k.a. linear fit)
hold on
plot(x(:,2), x*theta, 'ob')
% Plot using the Closed Form Solution
plot(x(:,2), x*((x' * x)\x' * y), '--r')
legend('Training data', 'Linear regression', 'Closed Form')
hold off % don't overlay any more plots on this figure''
[EDIT: Sorry for the wrong labeling... It's not Normal Equation, but Closed Form Solution. My mistake]
The results for this code is as shown below (Which is peachy :D Same results for both GD and CFS) -
Now, I am testing my code with another dataset. The URL for the dataset is here - GRAY KANGAROOS. I converted it to CSV and read it into MATLAB. Note that I did scaling (divided by the maximum, since if I didn't do that, no hypothesis line appears at all and the thetas come out as Not A Number (NaN) in MATLAB).
The Gray Kangaroo Dataset:
X Y
609 241
629 222
620 233
564 207
645 247
493 189
606 226
660 240
630 215
672 231
778 263
616 220
727 271
810 284
778 279
823 272
755 268
710 278
701 238
803 255
855 308
838 281
830 288
864 306
635 236
565 204
562 216
580 225
596 220
597 219
636 201
559 213
615 228
740 234
677 237
675 217
629 211
692 238
710 221
730 281
763 292
686 251
717 231
737 275
816 275
The changes I made to the code to read in this dataset
dataset = load('kangaroo.csv');
% scale?
x = dataset(:,1)/max(dataset(:,1));
y = dataset(:,2)/max(dataset(:,2));
The results that came out was like this: [EDIT: Sorry for the wrong labeling... It's not Normal Equation, but Closed Form Solution. My mistake]
I was wondering if there is any explanation for this discrepancy? Any help would be much appreciate. Thank you in advance!
I haven't run your code, but let me trow you some theory:
If your code is right (it looks like it): Increase MAX_ITER and it will look better.
Gradient descend is not ensured to converge at MAX_ITER, and actually gradient descend is a quite slow method (convergence-wise).
The convergence of Gradient descend for a "standard" convex function (like the one you try to solve) looks like this (from the Internets):
Forget, about iteration number, as it depedns in the problem, and focus in the shape. What may be happening is that your maxiter falls somewhere like "20" in this image. Thus your result is good, but not the best!
However, solving the normal equations directly will give you the minimums square error solution. (I assume normal equation you mean x=(A'*A)^(-1)*A'*b). The problem is that there are loads of cases where you can not store A in memory, or in an ill-posed problem, the normal equation will lead to ill-conditioned matrices that will be numerically unstable, thus gradient descend is used.
more info
I think I figured it out.
I immaturely thought that a maximum iteration of 1500 was enough. I tried with a higher value (i.e. 5k and 10k), and both algorithms started to give the similar solution. So my main issue was the number of iterations. It needed more iteration to properly converge for that dataset :D
Does someone know how to make a graph similar to this one with matlab?
To me it seems like a continuous stacked bar plot.
I did not manage to download the same data so I used other ones.
I tried the following code:
clear all
filename = 'C:\Users\andre\Desktop\GDPpercapitaconstant2000US.xlsx';
sheet = 'Data';
xlRange = 'AP5:AP259'; %for example
A = xlsread(filename,sheet,xlRange);
A(isnan(A))=[]; %remove NaNs
%create four subsets
A1=A(1:70);
A2=A(71:150);
A3=A(151:180);
A4=A(181:end);
edges=80:200:8000; %bins of the distributions
[n_A1,xout_A1] = histc(A1,edges); %distributions of the first subset
[n_A2,xout_A2] = histc(A2,edges);
[n_A3,xout_A3] = histc(A3,edges);
[n_A4,xout_A4] = histc(A4,edges);
%make stacked bar plot
for ii=1:numel(edges)
y(ii,:) = [n_A1(ii) n_A2(ii) n_A3(ii) n_A4(ii)];
end
bar(y,'stacked', 'BarWidth', 1)
and obtained this:
It is not so bad.. Maybe with other data it would look nicer... but I was wondering if someone has better ideas. Maybe it is possible to adapt fitdist in a similar way?
First, define the x axis. If you want it to follow the rules of bar, then use:
x = 0.5:numel(edges)-0.5;
Then use area(x,y), which produces a filled/stacked area plot:
area(x,y)
And if you want the same colors as the example you posted at the top, define the colormap and call colormap as:
map = [
218 96 96
248 219 138
253 249 199
139 217 140
195 139 217
246 221 245
139 153 221]/255;
colormap(map)
(It may not be exactly as the one you posted, but I got it quite close I think. Also, not all colors are shown in the result below as there are only 4 parameters, but all colors are defined)
Result:
I am smoothing an image and some forums gave me this.
fstr = #(a) median(a(:));
smooth_img = nlfilter(A,[50 50],fstr);
Is it going to find the median of 50x50 block and move to next 50x 50 block?
I mean the block is from pixel 1 to 50, in the next iteration it goes to 51 to 100 or 1 to 50 then to 2 to 51 and so on?
Thank you.
nlfilter() is a sliding filter, so the latter is correct, i.e. 1:50, 2:51, 3:52, etc..
The function blockproc() works in a blockwise manner, i.e. 1:50, 51:100, etc.. if that is what you need