I want to clean a decay by setting all values to zero after the first negative, please assist me on how to achieve that

B =

0.0374 0.0216 0.0185 -0.0320 -0.0012 0.0000 0.0063

0.0311 0.0157 0.0177 -0.0364 0.0029 -0.0022 0.0057

0.0244 0.0093 0.0170 -0.0400 -0.0044 -0.0044 0.0050

0.0178 0.0027 0.0164 -0.0428 -0.0056 -0.0065 0.0042

0.0114 -0.0039 0.0160 -0.0448 -0.0064 -0.0084 0.0034

for matrix B i want to set all values in a row after negative to zero.

e.g row 1 should look like;

0.0374 0.0216 0.0185 0.000 0.0000 0.0000 0.0060

OR row 2 to be;

0.0311 0.0157 0.0177 0.0000 0.0000 0.0000 0.0000

Answer 1

[Sameer]

Hi @Boitshepo Mpone

To transform the matrix "B", so that all values from the first negative value onwards in each row are set to zero, you can loop through each row, find the first negative value using the "find" function, and set that value and all subsequent values in the row to zero.

Here's how you can do it:

B = [
    0.0374    0.0216    0.0185   -0.0320   -0.0012    0.0000    0.0063;
    0.0311    0.0157    0.0177   -0.0364    0.0029   -0.0022    0.0057;
    0.0244    0.0093    0.0170   -0.0400   -0.0044   -0.0044    0.0050;
    0.0178    0.0027    0.0164   -0.0428   -0.0056   -0.0065    0.0042;
    0.0114   -0.0039    0.0160   -0.0448   -0.0064   -0.0084    0.0034
    ];
% Loop through each row
for i = 1:size(B, 1)
    % Find the index of the first negative value
    first_neg_idx = find(B(i, :) < 0, 1);
    
    % If a negative value is found, set it and all subsequent values to zero
    if ~isempty(first_neg_idx)
        B(i, first_neg_idx:end) = 0;
    end
end
disp(B);

    0.0374    0.0216    0.0185         0         0         0         0
    0.0311    0.0157    0.0177         0         0         0         0
    0.0244    0.0093    0.0170         0         0         0         0
    0.0178    0.0027    0.0164         0         0         0         0
    0.0114         0         0         0         0         0         0

Please refer to the below MathWorks documentation link:

https://www.mathworks.com/help/matlab/ref/find.html

Hope this helps!

[Boitshepo Mpone]

thanks Sameer this really helped me a lot much appreciated

Answer 2

[Star Strider]

Use logical operations and the cumprod function for this —

B = [0.0374    0.0216    0.0185   -0.0320   -0.0012    0.0000    0.0063
     0.0311    0.0157    0.0177   -0.0364   0.0029   -0.0022    0.0057
     0.0244    0.0093    0.0170   -0.0400   -0.0044   -0.0044    0.0050
     0.0178    0.0027    0.0164   -0.0428   -0.0056   -0.0065    0.0042
     0.0114   -0.0039    0.0160   -0.0448   -0.0064   -0.0084    0.0034];

multmtx = cumprod(B >= 0, 2)                                                % Logical Matrix Multiplied Cumulatively In The Column (2) Dimension

multmtx = 5×7

1 1 1 0 0 0 0 1 1 1 0 0 0 0 1 1 1 0 0 0 0 1 1 1 0 0 0 0 1 0 0 0 0 0 0

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B = B .* multmtx                                                            % Multiply By The Original ‘B’ Matrix

B = 5×7

0.0374 0.0216 0.0185 0 0 0 0 0.0311 0.0157 0.0177 0 0 0 0 0.0244 0.0093 0.0170 0 0 0 0 0.0178 0.0027 0.0164 0 0 0 0 0.0114 0 0 0 0 0 0

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.

[Image Analyst]

Clever. 🙂

[Star Strider]

Thank you!

[Boitshepo Mpone]

Wow thats cool

[Star Strider]

Thank you!

It’s also efficient, and while I kept the two statements separate to demonstrate how it works, they could be combined into a single statement —

B = [0.0374    0.0216    0.0185   -0.0320   -0.0012    0.0000    0.0063
     0.0311    0.0157    0.0177   -0.0364    0.0029   -0.0022    0.0057
     0.0244    0.0093    0.0170   -0.0400   -0.0044   -0.0044    0.0050
     0.0178    0.0027    0.0164   -0.0428   -0.0056   -0.0065    0.0042
     0.0114   -0.0039    0.0160   -0.0448   -0.0064   -0.0084    0.0034];

B = B .* cumprod(B >= 0, 2)                                                 % Multiply The Original ‘B’ Matrix By The Logical Matrix Multiplied Cumulatively In The Column (2) Dimension

B = 5×7

0.0374 0.0216 0.0185 0 0 0 0 0.0311 0.0157 0.0177 0 0 0 0 0.0244 0.0093 0.0170 0 0 0 0 0.0178 0.0027 0.0164 0 0 0 0 0.0114 0 0 0 0 0 0

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A vote for it would be appreciated!