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

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Rabia Zulfiqar 2020-5-16

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评论： dpb 2020-5-20

For example I have this matrix:

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.

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Answer 1

dpb 2020-5-16

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编辑：dpb 2020-5-17

在 MATLAB Online 中打开

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.

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Rabia Zulfiqar 2020-5-19

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

dpb 2020-5-20

No problem...glad to help. :)

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Answer 2

Stephen23 2020-5-19

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编辑：Stephen23 2020-5-20

在 MATLAB Online 中打开

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.

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Stephen23 2020-5-20

在 MATLAB Online 中打开

@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 2020-5-20

" 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...

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