Zeros between Sign Change of Values in column
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Tommy Schumacher
el 14 de Abr. de 2021
Comentada: Jan
el 16 de Abr. de 2021
Hello,
i have a matrix A (1080x2). In the 2. Column i have values which contains sign changes
1.Column 2. Column
1 0
2 0
3 0.00047
4 -0.0177
5 -0.0328
6 0.0075
7 0
8 -0.0170
9 0
10 0
The values between sign changes in 2. column should be zero and keep the first value of sign change
1.Column 2. Column
1 0
2 0
3 0
4 -0.0177
5 0
6 0.0075
7 0
8 -0.0170
9 0
10 0
Maybe helpful is this Question/Answer:
Thank You
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Respuesta aceptada
Jan
el 15 de Abr. de 2021
data = [ ...
1 0; ...
2 0; ...
3 0.00047; ...
4 -0.0177; ...
5 -0.0328; ...
6 0.0075; ...
7 0; ...
8 -0.0170; ...
9 0; ...
10 0];
idx = [true; diff(data(:, 2) >= 0) == 0];
data(idx, 2) = 0
2 comentarios
Jan
el 16 de Abr. de 2021
Does this comment mean, that you want to treat the zero in another way?
Then please post an exhaustive set of input data to define uniquely, what you want to achieve. What is the wanted output for:
[-2, -1, 0, -1, -2]
[-2, -1, 0, 1, -2]
[1, 1, 0, 1, 1]
[1, 1, 0, -1, -1]
It is not automatically clear, how you want to treat the zeros or where the "sign change" occurs in [-1, 0, 1].
Más respuestas (2)
Mathieu NOE
el 14 de Abr. de 2021
hello
this is my suggestion below + see attachements
data = importdata('data.txt');
x = data(:,1);
y = data(:,2);
threshold = 0; %
[t0_pos,s0_pos,t0_neg,s0_neg]= crossing_V7(y,x,threshold,'linear'); % positive (pos) and negative (neg) slope crossing points
% t0 => corresponding time (x) values
% s0 => corresponding function (y) values , obviously they must be equal to "threshold"
figure(1)
plot(x,y,t0_pos,s0_pos,'+r',t0_neg,s0_neg,'+g','linewidth',2,'markersize',12);grid on
legend('signal','positive slope crossing points','negative slope crossing points');
% now let's force y values to be zero when x values are one step before positive and negative slope zero crossing points
dx = mean(diff(x)); % x axis spacing
indp_before= floor(t0_pos/dx); % index of values before positive slope zero crossing points
indn_before= floor(t0_neg/dx); % index of values before negative slope zero crossing points
yy = y;
yy(indp_before)= 0;
yy(indn_before)= 0;
figure(2)
plot(x,y,x,yy,t0_pos,s0_pos,'+r',t0_neg,s0_neg,'+g','linewidth',2,'markersize',12);grid on
legend('signal','signal modified','positive slope crossing points','negative slope crossing points');
1 comentario
Mathieu NOE
el 14 de Abr. de 2021
Forgot to mention my result :
yy =
0
0
0
-0.0177
0
0.0075
0
-0.0170
0
0
Tommy Schumacher
el 14 de Abr. de 2021
Editada: Tommy Schumacher
el 14 de Abr. de 2021
1 comentario
Mathieu NOE
el 15 de Abr. de 2021
hmm , I am not sure to understand the issue
I exported the original (first column) and modified signal (second column) in the attached excel file
can you add a 3rd column with the expected results ?
thanks
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