Comments and Ratings

When I implement this function for noisy sine vawes, the amplitudes seem to be very far from reality. Often, amplitudes have even minus signs(?). Any ideas why ?

Very nice

This part of the submission is intended for Comments and Ratings. More over, I pleased anybody to write me by e-mail on errors and improvements, see the section Other requirements. This is the reason why I'll not communicate here anymore.

Hi

how can I give an interval for the results, i don't need negative values for answers or I need some answers higher
than some values,

you also helped me before Mira

thanks.

Professor Balda:
I truly appreciate you for helping me out to solve non-linear equations. Your code beautifully worked out for me. Also, thanks again to correct matrix in your weekend time.
Dr. Balda, Thanks again for your great help and prompt support.
D. Kim

txs a lot =)).
 i need it that mean

Most of this behavior can be handled by DEAL, for instance, a,b,c] = deal([])
(btw, the later example is rather strange as I would expect
"[E,b,c...] = arg2vars" to do the job ...)

The need for putting cell arrays into a cell to assign it to a single variable is in my opinion very contra-intuitive given the fact that a regular array is not split into separate variables.

To me, the only advantage of this submission seems to be that unassigned output arguments are set to empty, but a simple wrapper function around deal will do the same:

function [varargout] = mydeal(varargin) ;

varargout([1 2:nargout]) = {[]} ; % return [] when called without any in/output arguments
if nargin,
    N = max(min(nargin,nargout), 1) ;
    % how many things do we have to be DEALt with?
    [varargout{1:N}] = deal(varargin{1:N}) ;
end

Analysing the example in the introduction of the file (Rosenbrock) gives different results on my computer as given in the example (cnt=7; ssq much larger). Furthermore, the results using this M-File are identical as to the results obtained with fminsearch? What goes wrong on my computer?

Your solution is much simpler than that of mine, however, an operator doesn't know in advance, what he (she) is accepting as default.

This is a great bit of code that does exactly what it says. I used it to replace lsqcurvefit as I no longer have the optimisation toolbox.

I would suggest that the author also considers using additional examples, as the ones given are very complicated to understand if you come from a very different field. However, after a little search I found this by the author, which I found much more useful:
http://mathforum.org/kb/message.jspa?messageID=6283753&tstart=0

It tells you how to use the function to fit a curve to some data. This should be easily modified for fitting other functions/data etc.

As the previous submission - and what has changed, by the way? - this one just writes out a separator line. You can do anything that it does with a single line; in fact, you can do better, by writing out separated strings from a cell array with a single 'sprintf'. Btw, is 'inputparser' available in Matlab 2006b?

Works well, thanks.

For others who may be looking for the LMFtest dependencies, they are file ID numbers 9033, 9035 and 11725 (as documented in the pdf file appendix).

The solution does not work. I get an error whenever I copy and paste the code than run it on MATLAB

Wow!! I have been looking for LM algorithm (m file) for a long time.

The new extended description and help solves both notes.

Excuse me, sir. I can't catch the command:
R = @(x) sqrt(x'*x)-.5; % A distance from the radius r=0.5
ros= @(x) [10*(x(2)-x(1)^2); 1-x(1); (R(x)>0)*R(x)*1000];
I use MATLAB6.5.
Thanks.

.. or, you can use something like
> repmat('-',1,100)

I need your help Mr. Miroslav Balda.
This paper "Processing of Blade Monitoring System Data", complete document, data this proyect, information in general

Thanks for attention

I think that the title is misleading: the functions can be used to work with multivariate time series, but these are not date/time functions per se. I also wish the PDF offered more details on the functions: surprisingly, a chapter named 'An example' shows no Matlab code. Personally, I have not had trouble working with 3D array - how hard is it to code a loop, when 'x(i,j,:)' can't be used? - but other people may find your code useful, if you just explain upfront what it does. (For example, 'tsarr2mtx - ts array to ts matrix conversion' - what is that?). How about an HTML sample session?

Hi Miroslav!
Here is my comparison. It is known that the calc. speed depends on memory limitations. By me vectors 1e6 long are already problematic. Therefore I have selected 1e6, not 1e7. How many people need longer vectors?

Again, you have compared with MAXALL which DID INTERPOLATION. I compare both versions, with and without interpolation.

Yes, without interpolation your extr is apparently 0.5 sec faster for 1e6 vector. But this is again not completely correct, because both progs do different jobs. The user needs BOTH POSITIONS AND VALUES. You analize only y-data, and my prog. searches for both x and y values for extrema. But can a user write down several additional lines within 0.5 sec, in order to extract necessary positions and values?

I was very glad to see that the interpolation is extremely fast: only about 0.2 sec was additionally necessary. I disagree that interpolation is not important. It is frequently nesessary for real data. Typical and quite frequent case is std(diff(signal,2)) comparable with experimental noise amplitude.

As to plateaus: I could agree if you work mainly with double(int8 or int16, uint too). This is however a special niche. In a standard situation, if your sampled data obey a continuous distribution and are represented by double, the probability of such a plateau-extremum is close to zero.

Let users decide, which version they like better.
  
EXAMPLE: I have filtered somewhat randn, in order to create some similarity with possible real data.

x=savfilt(randn(1,1000000),10);
tic,L=extr(x);te=toc; %time extr
tic,[xp,yp]=maxall(x,0);tm=toc; %time maxall,no interp
tic,[xpi,ypi]=maxall(x);tmi=toc; %time maxall with interpolation

T=[te,tm,tmi] % all times
NUM=[sum(L{2}),length(xp),length(xpi)] %numb. of maxima

T = 0.406 0.906 1.125

NUM = 193980 193980 193980