How to perform mathematical operation in columns on certain elements and make the other elements zero??

For example I have this matrix:

[0     0
0      0
1      0
2      7
5      8
3      11
4      12
2       9]

The min and max value of column one are (1,5) and column 2 are (7,12). I want to perform a specific operation a*b when the min value occurs in each column and perform a*(b-2) to its next value. similarly when max value occurs I want to perform a/b at max value and a/(b-2) to its next value. All the other values must remain zero.

Here

a=10;       a*b=10;   a*(b-2)=-10
b=1         a/b=10;   a/(b-2)=-10

The new matrix should be like:

Ans=[    0        0
         0        0
        10        0
       -10       10
        10      -10
       -10        0
         0       10
         0      -10]

Can someone please help????thankyou in advance.

Answer 1

[dpb]

Don't try to get too fancy...sometimes just deadahead is the simplest and best (and probably fastest, besides)...

a=10;                           % constants and compute insertion vectors
b=1;
vMIN=[a*b;a*(b-2)];
vMAX=[a/b;a/(b-2)];
A(A==0)=nan;                    % prelim fixup to exclude 0 from min/max
for i=1:size(A,2)               % engine for each column...
  [~,imn]=min(A(:,i));          % location min, max
  [~,imx]=max(A(:,i));
  A(imn:imn+1,i)=vMIN;          % replace element and next
  A(imx:imx+1,i)=vMAX;          
end
A(isnan(A)=0;                   % restore the zeros...

NB1: Must do search for locations and save before either substitution...

NB2: Presumes min/max is not in last row of A or will get out-of-bounds addressing error.

[Rabia Zulfiqar]

Thankyou so much for your help:)

[dpb]

If that solves the problem, please ACCEPT the Answer if for no other reason than to let others know it is.

[Rabia Zulfiqar]

Yes I have accepted your answer once again thank you so much your help:)

[dpb]

No problem...glad to help. :)

Answer 2

[Stephen23]

Something fancy...

>> M = [0,0;0,0;1,0;2,7;5,8;3,11;4,12;2,9]
M =
    0    0
    0    0
    1    0
    2    7
    5    8
    3   11
    4   12
    2    9
>> a = 10;
>> b = 1;
>> F = @(f,v) conv2(+(M==f(M./~~M,[],1)),[0;v],'same'); % requires >=R2016b
>> Q = F(@max,[a/b;a/(b-2)]) + F(@min,[a*b;a*(b-2)])
Q =
    0    0
    0    0
   10    0
  -10   10
   10  -10
  -10    0
    0   10
    0  -10

For earlier versions replace == with bsxfun.

[dpb]

Cute, Stephen... :)

[Stephen23]

PS: and not any slower either (1e4 iterations):

Elapsed time is 1.33313 seconds. % dpb's code
Elapsed time is 1.32813 seconds. % my code

[dpb]

Not surprised on speed w/ that implementation...I was thinking of convoluted accumarray -like solutions when writing the above comment.

Have to admit the convolution aspect here never crossed my mind...and doubt it ever would have.

[Stephen23]

@dpb: With so much data-duplication going on, to be honest I was quite surprised at the speed: most of the operators in that anonymous function create a new array with the same size as the input data. This does take its toll on larger input arrays: your loop tends to be faster on the larger arrays I tried.

The fastest approach I found was to split the vector indexing in your solution into two scalars, i.e.:

  A(imn,i)   = a*b;
  A(imn+1,i) = a*(b-2);
  etc.

No doubt all of these results are highly dependent on the MATLAB version anyway, and writing something readable/understandable/maintainable is of course much preferred.

[dpb]

" fastest approach I found was to split the vector indexing in your solution into two scalars"

The [;] operation has always been expensive. I get warnings about avoiding brackets all the time that I mostly ignore since not writing production code and I find them more legible...but if such code were buried in the bowels of a loop or iterative solution it then could be worth the less expressive form.

Interesting result...