I've got several distribution objects in Matlab like this one:
% gravity [m/s2]
Uncer.Param.gravity.LB = 9.801;
Uncer.Param.gravity.UB = 9.867;
Uncer.Param.gravity.pd = makedist('Uniform','Lower',Uncer.Param.gravity.LB, 'Upper', Uncer.Param.gravity.UB);
Uncer.Param.gravity.value=0;
I know I can generate a random sample with the random-function, but I want to generate a sample made of quasi-random-numbers (Sobol).
I get a matrix filled with these quasi-random-numbers like this:
set = net(sobolset(countParameter*2, 'Skip',1), countSimulation);
And I know furthermore, that I can interpolate the distribution values with the function interp1 and the corresponding CDF.
The problem is that my matrix-dimensions are round about 1000x20 and the interpolation would cost a ton ammount of time.
Is there a faster way to do this?
The search has come to an end and was way to simple: icdf
Related
I 've found the best fitting of a variable distribution (D(:,2)) using the function "allfitdist". Now i want to save this result in a structure and then i want to randomly sample 10000 times from this result. I'm using this code:
[Ddg2 PDdg2] = allfitdist(D(:,2),'cdf')
My(2).result = PDdg2{1,1} %generalized pareto
output = random(My(2).result,10000)
Something is weard because in the output i get a really big matrix. Maybe i'm wrong in the third raw of the code, when i randomly sample from this distribution.
Someone can help me?
The documentation of random says:
R = random(___,sz1,...,szN) or R = random(___,[sz1,...,szN]) generates a sz1-by-⋯-by-szN array of random numbers from the specified probability distribution using input arguments...
...
If you specify a single value sz1, then R is a square matrix of size sz1.
You have specified sz1 as 10000 which is a single value and hence your output matrix is 10000×10000.
So the solution is:
output = random(pd,1,10000);
I find examples the best way to demonstrate my question. I generate some data, add some random noise, and fit it to get back my chosen "generator" value...
x = linspace(0.01,1,50);
value = 3.82;
y = exp(-value.*x);
y = awgn(y,30);
options = optimset('MaxFunEvals',1000,'MaxIter',1000,'TolFun',1e-10,'Display','off');
model = #(p,x) exp(-p(1).*x);
startingVals = [5];
lb = [1];
ub = [10];
[fittedValue] = lsqcurvefit(model,startingVals,x,y,lb,ub,options)
fittedGraph = exp(-fittedValue.*x);
plot(x,y,'o');
hold on
plot(x,fittedGraph,'r-');
In this new example, I have generated the same data but this time added much more noise to the first 15 points. Because it is random sometimes it works out okay, but after a few runs I get a good example that illustrates my problem. Same code, except for these lines added under value = 3.82
y = exp(-value.*x);
y(1:15) = awgn(y(1:15),5);
y(15:end) = awgn(y(15:end),30);
As you can see, it has clearly not given a good fit to where the data seems reliable, because it is fitting from points 1-50. What I want to do is say, okay MATLAB, I can see we have some noisy data but it seems decent over a range, only fit your exponential from points 15 to the end. I could go back to my code and update it to do this, but I will be batch fitting graphs like this where each one will have different ranges of 'good' data.
So what I am after is a GUI callback mechanisms that allows me to click on two circles from the data and have them change color or something, which indicates the lsqcurvefit will only fit over that range. Internally all it has to change is inside the lsqcurvefit call e.g.
x(16:end),y(16:end)
But the range should update depending on the starting and ending circles I have clicked.
I hope my question is clear. Thanks.
You could use ginput to select the two points for your min and max in the plot.
[x,y]=ginput(2);
%this returns you the x and y coordinates of two points clicked after each other
%the min point is assumed to be clicked first
min=[x(1) y(1)];
max=[x(2) y(2)];
then you could fit your curve with the coordinates for min and max I guess.
You could also switch between a rightclick for the min and a leftclick for the max etc.
Hope this helps you.
I'm interested in understanding the variety of zeroes that a given function produces with the ultimate goal of identifying the what frequencies are passed in high/low pass filters. My idea is that finding the lowest value zero of a filter will identify the passband for a LPF specifically. I'm attempting to use the [hz,hp,ht] = zplane(z,p) function to do so.
The description for that function reads "returns vectors of handles to the zero lines, hz". Could someone help me with what a vector of a handle is and what I do with one to be able to find the various zeros?
For example, a simple 5-point running average filter:
runavh = (1/5) * ones(1,5);
using zplane(runavh) gives an acceptable pole/zero plot, but running the [hz,hp,ht] = zplane(z,p) function results in hz=175.1075. I don't know what this number represents and how to use it.
Many thanks.
Using the get command, you can find out things about the data.
For example, type G=get(hz) to get a list of properties of the zero lines. Then the XData is given by G.XData, i.e. X=G.XData.
Alternatively, you can only pull out the data you want
X=get(hz,'XData')
Hope that helps.
As seen in the code below, I am currently generating random numbers from a Normal Distribution and am selecting the ones within the -3*sigma and 3*sigma interval. However, I now want to generate numbers such that there is a higher probability that I select numbers from outside the -3*sigma and 3*sigma interval. For eg. A number from [-4*sigma -3*sigma) should have 35% probability of being chosen and same for [3*sigma 4*sigma). Basically, I'll be calling this function several times and am wondering if there is a way for me to select a higher proportion of random numbers from the "tails" of the normal distribution, without actually altering the shape of the normal distribution. I'm struggling to do this.
function [new_E11, new_E22] = elasticmodulusrng()
new_E11 = normrnd(136e9,9.067e9,[1 1]);
new_E22 = normrnd(8.9e9,2.373e9,[1 1]);
while new_E11<=-3*9.067e9 && new_E11>=3*9.067e9
new_E11 = normrnd(136e9,9.067e9,[1 1]);
end
while new_E11<=-3*2.373e9 && new_E11>=3*2.373e9
new_E22 = normrnd(8.9e9,2.373e9,[1 1]);
end
Thanks
The question does not make much sense, as pointed out by Jojo: this is not a normal distribution anymore.
What you could do is to create your own Probability density functions pdf and draw from it.
For instance,
N = pdf('Normal',-5:0.2:5,0,1);
gives you the normal PDF with a good resolution.
You could alter it, say
Z = N;
Z(5:15)=3*Z(5:15);
Z(35:45)=3*Z(35:45);
and use Direct Methods, Inversion Methods, or Acceptance-Rejection Methods as explained here
There is an implementation in the FileExchange:
http://www.mathworks.com/matlabcentral/fileexchange/27590-simple-rejection-sampling
What is the difference between hist and imhist functions in Matlab? I have a matrix of color levels values loaded from image with imread and need to count entropy value of the image using histogram.
When using imhist the resulting matrix contains zeros in all places except the last one (lower-right) which contains some high value number (few thousands or so).
Because that output seems to be wrong, I have tried to use hist instead of imhist and the resulting values are much better, the matrix is fulfilled with correct-looking values instead of zeros.
However, according to the docs, imhist should be better in this case and hist should give weird results..
Unfortunately I am not good at Matlab, so I can not provide you with better problem description. I can add some other information in the future, though.
So I will try to better explain my problem..I have an image, for which I should count entropy and few other values (how much bytes it will take to save that image,..). I wrote this function and it works pretty well
function [entropy, bytes_image, bytes_coding] = entropy_single_pixels(im)
im = double(im);
histg = hist(im);
histg(histg==0) = [];
nzhist = histg ./ numel(im);
entropy = -sum(nzhist.*log2(nzhist));
bytes_image = (entropy*(numel(im))/8);
bytes_coding = 2*numel(unique(im));
fprintf('ENTROPY_VALUE:%s\n',num2str(entropy));
fprintf('BYTES_IMAGE:%s\n',num2str(bytes_image));
fprintf('BYTES_CODING:%s\n',num2str(bytes_coding));
end
Then I have to count the same, but I have to make "pairs" from pixels which are below each other. So I have only half the rows and the same count of columns. I need to express every unique pixel pair as a different number, so I multiplied the first one by 1000 and added the second one to it... Subsequently I need to actually apply the same function as in the first example, but that is the time, when I am getting weird numbers from the imhist function. When using hist, it seems to be OK, but I really don't think that behavior is correct, so that must be my error somewhere. I actually understand pretty good, to what I want to do, or at least I hope so, but unfortunately Matlab makes all that kind of hard for me :)
hist- compute histogram(count number of occurance of each pixel) in color image.........
imhist- compute histogram in two dimensional image.
Use im2double instead of double if you want to use imhist. The imhist function expects double or single-precision data to be in the [0,1] data range, which is why you see everything in the last bin of the histogram.