after using gnuplot for years and experiencing many user-related issues, I thought I'd finally know how to fit a function to a dataset.
Today I tried to fit a simple
y = m * x² + b
function. However, gnuplot did not change my 'm' value. It does change 'b' to the correct value however.
I have my dataset uploaded and here is my gnuplot script with which I'm trying to fit, maybe someone can reproduce this on his machine and confirm, that it is not a fault of my computer but some kind of faulty code in the script, or it may even be a bug (I highly doubt that).
set xtics 0.000001
set format x '%10.1E'
set xrange [0:2E-07]
#fit
f(x)=a*(x**2)+b
a=380812
b=1
fit [0:2E-07] f(x) 'GDAMitte1.txt' using ($1+7.6E-06):2 via a,b
plot 'GDAMitte1.txt' using ($1+7.6E-06):2, f(x)
I've pasted the dataset here: http://www.heypasteit.com/clip/29LU
I'd be very thankful for an answer to that, even if it's just a confirmation, that it doesn't fit on your machine as well. Thank you.
Btw: The initial value I've set is pretty much the one it has to be after the fit, but it's not as exact of course. Should be good enough though for gnuplot to get where to go to.
This is because two parameters are of greatly different magnitude, check help fit tips.
You should replace the function with one that has a prefactor built in:
f(x) = a *1e5 * (x**2)+b
a=3.8 # instead of 380000
b=1
fit ....
From gnuplot version 5.0 on, gnuplot by default internally prescales all parameters, so this problem with calculating the residuals should no longer occur for any function, provided your initial values are not off too much.
Related
I'm newer to Mathematica language, and I'm having a big issue in graphing a set of points. It goes as follows:
f[w_] = expr.1
calculations = Table[expr.1, {w,0,numtimes}]
omegas = Table[i,{i,0,numtimes}]
orderedpairs = Transpose[{calculations,omegas}]
ListPlot[orderedpairs]
This returns a graph with just one point rather than numtimes amount of points, and it doesn't match the first point in the dataset. I've tried a listplot command for the two lists seperately, like
Listplot[{orderedpairs[[i]],omegas[[i]]},{i,0,numtimes}]
but i get an error that says "the expression i cannot be used as a part specification."
The data set is in the form x+iy, where x and y are real numbers. If I could get some help, I would appreciate it greatly.
Plot the real by itself
ListPlot[Re[orderedpairs]]
Hello all :) I'm pretty new to Optimization and barely understand it (was about ready to slit my wrist after figuring out how to write Objective Functions without any formal learning on the matter), and need a little help on a work project.
How would I go about setting a logical constraint when using the Optimization Toolbox, fmincon specifically (using Trust Region Reflective algorithm)?
I am optimizing 5 values (lets call it matrix OptMat), and I want to optimize with the constraint such that
max(OptMat)/min(OptMat) > 10
I assume this will optimize the 5 values of OptMat as low as possible, while keeping the above constraint in mind so that if a set of values for OptMat is found with a lower OF in which it breaks the constraint it will NOT report those values and instead report the next lowest OF where OptMat values meet the above constraint?
For the record, my lower bounds are [0,0,0,0,0]. I'm not sure how to enter it into upper bounds as it only accepts doubles and that would be logical. I tried the Active Set Algorithm and that enabled the Nonlinear Constraint Function box and I think I'm on the right track with that. If so, I'm not sure what the syntax for entering my desired constraint. Another method^that ^may ^or ^may ^not ^work I could think of is using this as an Upper Boundary.
[min(OptMat)*10, min(OptMat)*10, min(OptMat)*10, min(OptMat)*10, min(OptMat)*10]
Again, I'm using the GUI Optimization Toolbox. I haven't looked too much into command line optimization (though I will need to write it command line eventually) and I think I read somewhere that you can set the Upper Boundary and it does not have to be double?
Thank you so very much for the help, if someone is able. I apologize if this is a really nooby question.
What you are looking for are nonlinear constraints, fmincon can handle it (I only know the command, not the GUI) with the argument nonlcon. For more information look at this guide http://de.mathworks.com/help/optim/ug/fmincon.html
How would you implement this? First create a function
function [c, ceq] = mycondition(x)
c = -max(x)/min(x)/10;
ceq = 0;
I had to change the equation to match the correct formalism, i.e. c(x)<=0 is needed.
Maybe you could also create an anonymous function, I'm not sure (http://de.mathworks.com/help/matlab/matlab_prog/anonymous-functions.html).
Then use this function to feed the fmincon function using the # sign, i.e. at the specific location write
fmincon(...., #mycondition, ...)
Suppose that I have these Three variables in matlab Variables
I want to extract diverse values in NewGrayLevels and sum rows of OldHistogram that are in the same rows as one diverse value is.
For example you see in NewGrayLevels that the six first rows are equal to zero. It means that 0 in the NewGrayLevels has taken its value from (0 1 2 3 4 5) of OldGrayLevels. So the corresponding rows in OldHistogram should be summed.
So 0+2+12+38+113+163=328 would be the frequency of the gray level 0 in the equalized histogram and so on.
Those who are familiar with image processing know that it's part of the histogram equalization algorithm.
Note that I don't want to use built-in function "histeq" available in image processing toolbox and I want to implement it myself.
I know how to write the algorithm with for loops. I'm seeking if there is a faster way without using for loops.
The code using for loops:
for k=0:255
Condition = NewGrayLevels==k;
ConditionMultiplied = Condition.*OldHistogram;
NewHistogram(k+1,1) = sum(ConditionMultiplied);
end
I'm afraid if this code gets slow for high resolution big images.Because the variables that I have uploaded are for a small image downloaded from the internet but my code may be used for sattellite images.
I know you say you don't want to use histeq, but it might be worth your time to look at the MATLAB source file to see how the developers wrote it and copy the parts of their code that you would like to implement. Just do edit('histeq') or edit('histeq.m'), I forget which.
Usually the MATLAB code is vectorized where possible and runs pretty quick. This could save you from having to reinvent the entire wheel, just the parts you want to change.
I can't think a way to implement this without a for loop somewhere, but one optimisation you could make would be using indexing instead of multiplication:
for k=0:255
Condition = NewGrayLevels==k; % These act as logical indices to OldHistogram
NewHistogram(k+1,1) = sum(OldHistogram(Condition)); % Removes a vector multiplication, some additions, and an index-to-double conversion
end
Edit:
On rereading your initial post, I think that the way to do this without a for loop is to use accumarray (I find this a difficult function to understand, so read the documentation and search online and on here for examples to do so):
NewHistogram = accumarray(1+NewGrayLevels,OldHistogram);
This should work so long as your maximum value in NewGrayLevels (+1 because you are starting at zero) is equal to the length of OldHistogram.
Well I understood that there's no need to write the code that #Hugh Nolan suggested. See the explanation here:
%The green lines are because after writing the code, I understood that
%there's no need to calculate the equalized histogram in
%"HistogramEqualization" function and after gaining the equalized image
%matrix you can pass it to the "ExtractHistogram" function
% (which there's no loops in it) to acquire the
%equalized histogram.
%But I didn't delete those lines of code because I had tried a lot to
%understand the algorithm and write them.
For more information and studying the code, please see my next question.
well, I am new to matlab programming and I have been battling on the indexing issues. I am currently working on image processing which so far drive me crazy. anyways, lets jump to the questions.
I have the following code
perm=randperm(size(X,2));
CX=X(:,perm(1:nclus));
I tried to run the code but it triggers an error saying " Index exceeds the matrix dimensions. To my humble knowledge I think it is because the (:,perm(1:nclus)) is higher than the matrix dimensions. I would like to know how can i solve this problem.
Note that X: is the input points in the columns
nclus: number of clusters.
I highly appreciate if you guys clarify to me the error cause and the possible solution for it.
Thank you
Sami
Guessing that you just want to get nclus random columns from a 2 dimensional matrix X, try this:
perm=randperm(size(X,2));
CX=X(:,perm<=nclus);
The error that you experience should not come from X being called with too many dimensions, it is probably because the dimensions of perm are exceeded. Try running this line by line:
perm = randperm(size(X,2)); %Should be ok
idx = perm(1:nclus); %Probably fails
X(:,idx)
I just started working with Mathematica (5.0) for the first time, and while the manual has been helpful, I'm not entirely sure my technique has been correct using (Full)Simplify. I am using the program to check my work on a derived transform to change between reference frames, which consisted of multiplying a trio of relatively large square matrices.
A colleague and I each did the work by hand, separately, to make sure there were no mistakes. We hoped to get a third check from the program, which seemed that it would be simple enough to ask. The hand calculations took some time due to matrix size, but we came to the same conclusions. The fact that we had the same answer made me skeptical when the program produced different results.
I've checked and double checked my inputs.
I am definitely . (dot-multiplying) the matrices for correct multiplication.
FullSimplify made no difference.
Neither have combinations with TrigReduce / expanding algebraically before simplifying.
I've taken indices from the final matrix and tryed to simplify them while isolated, to no avail, so the problem isn't due to the use of matrices.
I've also tried to multiply the first two matrices, simplify, and then multiply that with the third matrix; however, this produced the same results as before.
I thought Simplify automatically crossed into all levels of Heads, so I didn't need to worry about mapping, but even where zeros would be expected as outputs in the matrix, there are terms, and where we would expect terms, there are close answers, plus a host of sin and cosine terms that do not reduce.
Does anyone frequent any type of technique with Simplify to get more preferable results, in contrast to solely using Simplify?
If there are assumptions on parameter ranges you will want to feed them to Simplify. The following simple examples will indicate why this might be useful.
In[218]:= Simplify[a*Sqrt[1 - x^2] - Sqrt[a^2 - a^2*x^2]]
Out[218]= a Sqrt[1 - x^2] - Sqrt[-a^2 (-1 + x^2)]
In[219]:= Simplify[a*Sqrt[1 - x^2] - Sqrt[a^2 - a^2*x^2],
Assumptions -> a > 0]
Out[219]= 0
Assuming this and other responses miss the mark, if you could provide an example that in some way shows the possibly bad behavior, that would be very helpful. Disguise it howsoever necessary in order to hide proprietary features: bleach out watermarks, file down registration numbers, maybe dress it in a moustache.
Daniel Lichtblau
Wolfram Research
As you didn't give much details to chew on I can only give you a few tips:
Mma5 is pretty old. The current version is 8. If you have access to someone with 8 you might ask him to try it to see whether that makes a difference. You could also try WolframAlpha online (http://www.wolframalpha.com/), which also understands some (all?) Mma syntax.
Have you tried comparing your own and Mma's result numerically? Generate a Table of differences for various parameter values or use Plot. If the differences are negligable (use Chop to cut off small residuals) the results are probably equivalent.
Cheers -- Sjoerd