Search Results

Search found 1466 results on 59 pages for 'matlab toolbox'.

Page 24/59 | < Previous Page | 20 21 22 23 24 25 26 27 28 29 30 31  | Next Page >

  • « Pourquoi développer en MATLAB plutôt qu'en Java, Python ou C# ? », un séminaire gratuit de MathWorks le 13 décembre

    Pourquoi développer en MATLAB Plutôt qu'en Java, Python ou C# : un séminaire gratuit de MathWorks le 13 décembre MathWorks organise ce 13 décembre à La Défense un séminaire pour illustrer les atouts de MATLAB. « Lors de cette rencontre en matinée, vous découvrirez que MATLAB n'est pas simplement un langage de calcul scientifique mais un langage ouvert vous permettant de structurer vos développements d'applications d'entreprise », promet l'éditeur de MATLAB qui le compare aujourd'hui à Java, Python ou C#. La matinée illustrera l'utilisation d'interfaces externes (APIs) et la programmat...

    Read the article

  • Ubuntu 12.04 tilts when trying to open large excel file with libreoffice or matlab

    - by user1565754
    I have an xlsx-file of size 27.3MB and when I try to open it either in Libreoffice or Matlab the whole system slows down My processor is AMD Sempron(tm) 140 Processor (should be about 2.7Ghz) Memory I have about 1.7GB Any ideas? I opened this file in Windows no problem...of course it took a few seconds to load but Ubuntu freezes with this file completely...smaller files of size 3MB, 5MB etc open just fine... thnx for support =)

    Read the article

  • difference equations in MATLAB - why the need to switch signs?

    - by jefflovejapan
    Perhaps this is more of a math question than a MATLAB one, not really sure. I'm using MATLAB to compute an economic model - the New Hybrid ISLM model - and there's a confusing step where the author switches the sign of the solution. First, the author declares symbolic variables and sets up a system of difference equations. Note that the suffixes "a" and "2t" both mean "time t+1", "2a" means "time t+2" and "t" means "time t": %% --------------------------[2] MODEL proc-----------------------------%% % Define endogenous vars ('a' denotes t+1 values) syms y2a pi2a ya pia va y2t pi2t yt pit vt ; % Monetary policy rule ia = q1*ya+q2*pia; % ia = q1*(ya-yt)+q2*pia; %%option speed limit policy % Model equations IS = rho*y2a+(1-rho)yt-sigma(ia-pi2a)-ya; AS = beta*pi2a+(1-beta)*pit+alpha*ya-pia+va; dum1 = ya-y2t; dum2 = pia-pi2t; MPs = phi*vt-va; optcon = [IS ; AS ; dum1 ; dum2; MPs]; He then computes the matrix A: %% ------------------ [3] Linearization proc ------------------------%% % Differentiation xx = [y2a pi2a ya pia va y2t pi2t yt pit vt] ; % define vars jopt = jacobian(optcon,xx); % Define Linear Coefficients coef = eval(jopt); B = [ -coef(:,1:5) ] ; C = [ coef(:,6:10) ] ; % B[c(t+1) l(t+1) k(t+1) z(t+1)] = C[c(t) l(t) k(t) z(t)] A = inv(C)*B ; %(Linearized reduced form ) As far as I understand, this A is the solution to the system. It's the matrix that turns time t+1 and t+2 variables into t and t+1 variables (it's a forward-looking model). My question is essentially why is it necessary to reverse the signs of all the partial derivatives in B in order to get this solution? I'm talking about this step: B = [ -coef(:,1:5) ] ; Reversing the sign here obviously reverses the sign of every component of A, but I don't have a clear understanding of why it's necessary. My apologies if the question is unclear or if this isn't the best place to ask.

    Read the article

  • is there a faster version of fminbnd in matlab?

    - by kloop
    I am now using fminbnd in Matlab, and I find it relatively slow (I am using it inside a nested loop). The function itself, its interface and the values it returns are great, but when looking into the .m file I see it is not optimized. As a matter of fact, I was hoping for something like that to be written as a mex. Anyone knows of an alternative to fminbnd which works much faster and does not have much overhead?

    Read the article

  • How to access a matrix in a matlab struct's field from a mex function?

    - by B. Ruschill
    I'm trying to figure out how to access a matrix that is stored in a field in a matlab structure from a mex function. That's awfully long winded... Let me explain: I have a matlab struct that was defined like the following: matrixStruct = struct('matrix', {4, 4, 4; 5, 5, 5; 6, 6 ,6}) I have a mex function in which I would like to be able to receive a pointer to the first element in the matrix (matrix[0][0], in c terms), but I've been unable to figure out how to do that. I have tried the following: /* Pointer to the first element in the matrix (supposedly)... */ double *ptr = mxGetPr(mxGetField(prhs[0], 0, "matrix"); /* Incrementing the pointer to access all values in the matrix */ for(i = 0; i < 3; i++){ printf("%f\n", *(ptr + (i * 3))); printf("%f\n", *(ptr + 1 + (i * 3))); printf("%f\n", *(ptr + 2 + (i * 3))); } What this ends up printing is the following: 4.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 I have also tried variations of the following, thinking that perhaps it was something wonky with nested function calls, but to no avail: /* Pointer to the first location of the mxArray */ mxArray *fieldValuePtr = mxGetField(prhs[0], 0, "matrix"); /* Get the double pointer to the first location in the matrix */ double *ptr = mxGetPr(fieldValuePtr); /* Same for loop code here as written above */ Does anyone have an idea as to how I can achieve what I'm trying to, or what I am potentially doing wrong? Thanks! Edit: As per yuk's comment, I tried doing similar operations on a struct that has a field called array which is a one-dimensional array of doubles. The struct containing the array is defined as follows: arrayStruct = struct('array', {4.44, 5.55, 6.66}) I tried the following on the arrayStruct from within the mex function: mptr = mxGetPr(mxGetField(prhs[0], 0, "array")); printf("%f\n", *(mptr)); printf("%f\n", *(mptr + 1)); printf("%f\n", *(mptr + 2)); ...but the output followed what was printed earlier: 4.440000 0.000000 0.000000

    Read the article

  • Is there a better way to declare an empty, typed matrix in MATLAB?

    - by Arthur Ward
    Is there a way to "declare" a variable with a particular user-defined type in MATLAB? zeros() only works for built-in numeric types. The only solution I've come up with involves using repmat() to duplicate a dummy object zero times: arr = repmat(myClass(), [1 0]) Without declaring variables this way, any code which does "arr(end+1) = myClass()" has to include a special case for the default empty matrix which is of type double. Have I missed something a little more sensible?

    Read the article

  • Why is Matlab Stateflow 7.7 not throwing errors on undefined variables?

    - by Pyrolistical
    Previously in Matlab Stateflow 7.1 all variables and functions had to be included before they can be referred to in the state diagram or else it would throw an error when you tried to parse the diagram. But now in 7.7 it doesn't catch those kinds of errors. Its still compiling the diagram because it catches other syntactic errors. Am I missing an option somewhere? Can this be turned on?

    Read the article

  • How to download yahoo historical stock data into xls. format via matlab?

    - by Noob_1
    I have an xls sheet called Tickers (matrix 1 column 500 rows) with yahoo tickers. I want matlab to download the historical data for last 5 years for each stock ticker into a separate xls spreadsheet and save it in a given directory with title of the sheet = ticker. So that means i want a code that will create and save 500 tickers worth of data in 500 separate spreadhseets :) can anyone help or direct?

    Read the article

< Previous Page | 20 21 22 23 24 25 26 27 28 29 30 31  | Next Page >