How to create a matrix whose elements are function of their indices?

I need to create a large square matrix, which I need to define elementwise. For that, each element of that matrix need to be a function of it's indices. I tried to do this, using two for loops, but that was taking too long. Is there a more faster way of doing this? If, there is some function like the 'matrix(m,n,f)' function of mupad, which returns a m×n matrix, whose element at the i'th row and j'th column is f(i,j). where f is a desired function, that would be the most apt one?

[Stephan]

Hi,

do you know how big your matrix will be from the beginning? Did you preallocate with

zeros(n)

or

ones(n)

? If you didnt it takes a lot of time changing the size of the matrix by every run through your loop.

Maybe there is such function you want - i do not know such a function. But if you show us your code it could help to give useful hints.

Best regards

Stephan

[Fangjun Jiang]

Can you give an example of matrix(m,n,f)?

[Saptarshi Biswas]

My matrix would be of dimension 1400×1400. And yes, I tried that too.

[Saptarshi Biswas]

f(x,y)=x^2+y^2 A=matrix(2,3,f) returns, A= 2 5 10 5 8 13

[Saptarshi Biswas]

https://in.mathworks.com/help/symbolic/mupad_ref/matrix.html I found that function in the above link.

[Stephan]

For my understanding i have to ask: tried with the

matrix (m,n,f)

this takes too long?

[Saptarshi Biswas]

I tried to use this function. But, when I try to execute it, matlab says undefined function.

[Fangjun Jiang]

It needs the Symbolic toolbox

[Stephan]

The code above is what you want to do i guess

tic
A=zeros(1400);
for m = 1:1400
    for n =1:1400
        A(m,n) = m.^2+n.^2;
    end
end
toc

takes 0.031684 sec on my mobile phone with matlab mobile...

Share your code please

Best regards

Stephan

[Saptarshi Biswas]

N=input('enter the number of rings=')

miu0=4*pi*10^(-7); l=.001; d=.0001; r=6; a=1; z=1;

syms i j

p(i)=i*r/N-r/(2*N)

R=[]

 for i=1:N
    R(i,1)=p(i)
 end
syms i j
k2(i,j)=4*p(i)*p(j)/(p(i)+p(j))^2
K(i,j)=ellipticK(k2(i,j))
E(i,j)=ellipticE(k2(i,j))
deff=l*tanh(d/l)
M=[]
for i=1:N
    for j=1:N
       if i==j
          M(i,j)=p(i)*(log(8*p(i)*2*pi*N/r)-2)
       else
           M(i,j)=(p(i)+p(j))*((1-k2(i,j)/2)*K(i,j)-E(i,j))
       end
    end
end

This was like the code I wrote and it didn't ever completed for N=1400

[Stephan]

Edit: comment moved to the answers area below.

[Image Analyst]

Stephan, could you carry on the discussion in an official "Answer" down there with the rest of us? That way you could even get reputation points for it. Since Saptarshi hasn't responded to any of the 4 other answers, I think he may only be seeing yours because he didn't scroll down far enough.

[Stephan]

Thanks for the hint Image Analyst. Made as suggested.

[Image Analyst]

You made a comment to Stephen Lord about his answer. Is that what you wanted to do? Or did you want your own independent answer?

[Stephan]

You are right again... ;-)

Has already been late yesterday...

Answer 1

[Image Analyst]

Use meshgrid(). Try this (uses your f(x,y)=x^2+y^2 function):

rows = 10;
columns = 20;
startTime = tic; % Start timer.
% Use meshgrid to get x and y
[x, y] = meshgrid(1:columns, 1:rows);
% Create our output matrix, f:
f = x .^ 2 + y .^ 2
fprintf('Elapsed time was %f seconds.\n', toc(startTime));

f =
     2     5    10    17    26    37    50    65    82   101   122   145   170   197   226   257   290   325   362   401
     5     8    13    20    29    40    53    68    85   104   125   148   173   200   229   260   293   328   365   404
    10    13    18    25    34    45    58    73    90   109   130   153   178   205   234   265   298   333   370   409
    17    20    25    32    41    52    65    80    97   116   137   160   185   212   241   272   305   340   377   416
    26    29    34    41    50    61    74    89   106   125   146   169   194   221   250   281   314   349   386   425
    37    40    45    52    61    72    85   100   117   136   157   180   205   232   261   292   325   360   397   436
    50    53    58    65    74    85    98   113   130   149   170   193   218   245   274   305   338   373   410   449
    65    68    73    80    89   100   113   128   145   164   185   208   233   260   289   320   353   388   425   464
    82    85    90    97   106   117   130   145   162   181   202   225   250   277   306   337   370   405   442   481
   101   104   109   116   125   136   149   164   181   200   221   244   269   296   325   356   389   424   461   500
Elapsed time was 0.001304 seconds.

Adapt the number of rows and columns as needed.

[Saptarshi Biswas]

Thank you very much for suggesting the mesh hrid technique. It is working real quick. Although I haven't yet completeted implementing it, since some minute details are still to be adjusted in my code. But, I highly hope, this suggestion can resolve my problem completely.

Answer 2

[Matt J]

There is no generically fast way to vectorize an arbitrary function f(). The speed you can achieve depends heavily on f() and how it is implemented.

Answer 3

[Fangjun Jiang]

I am not aware of such a built-in function in MATLAB without any special toolbox. A typical approach is to pre-allocate and then do a nested loop. You'll have to look at your desired function to see if there is any way to optimize it. Below is an example. The second approach takes less time because it re-uses the result and avoids duplicated calculation.

t=cputime;
f=@(x,y) x*x+y*y;
a=zeros(1300,1400);
for k=1:size(a,1)
    for j=1:size(a,2)
        a(k,j)=f(k,j);
    end
end
cputime-t
%
t=cputime;
a=zeros(1300,1400);
f1=(1:size(a,1)).^2;
f2=(1:size(a,2)).^2;
for k=1:size(a,1)
    for j=1:size(a,2)
        a(k,j)=f1(k)+f2(j);
    end
end
cputime-t

ans =

      1.1094

ans =

      0.0781

Answer 4

[Steven Lord]

If your generating function uses only functions and/or operations that support implicit expansion, and you're using a release that supports implicit expansion, you could generate vectors of your row and column indices and use those implicit expansion operations. If your release predates implicit expansion but all your functions/operations operate element-wise, use meshgrid or ndgrid on those vectors of indices.

As examples see the hilb function or the private function minij used by the gallery function.

type hilb.m

type private/minij.m

Answer 5

[Stephan]

For such calculations on this scale you should try without the symbolic toolbox.

If you look at the Wikipedia article for elliptic integral, you will find two potentially helpful hints:

The complete elliptic integral of the first kind is sometimes called the quarter period. It can be computed very efficiently in terms of the arithmetic–geometric mean:

The complete elliptic integral of the second kind can also be computed very efficiently using the arithmetic–geometric mean (Carlson 2010, 19.8).

Carlson, B. C. (2010), "Elliptic integral", in Olver, Frank W. J.; Lozier, Daniel M.; Boisvert, Ronald F.; Clark, Charles W., NIST Handbook of Mathematical Functions, Cambridge University Press, ISBN 978-0521192255, MR 2723248

Perhaps you can accelerate your calculations this way. I dont think you will get happy using symbolic toolbox in this case.

Also you could try

[K,E] = ellipke(m)

but i dont believe this will save much calculation time.

Best regards

Stephan

[Saptarshi Biswas]

Thank you very much for the answer.