Find X of corresponding Y which is manipulated?
1 visualización (últimos 30 días)
Mostrar comentarios más antiguos
Sagar Dhage
el 16 de Jul. de 2014
Comentada: Andrei Bobrov
el 16 de Jul. de 2014
i need to extract the exact values where the increasing and decreasing trend starts.
I have Fx (column vector) and corresponding T values (column vector). step 1: the values which are less than 100, converts into 0
Fx(Fx <100)= 0;
step2: Fx(diff(Fx)==0) = []; % delete the element which is repeating
step3:B = [0; diff(Fx)]; %difference between element It will start with the first element in the matrix and then report the difference between that and the next element. There's no leading element before the first one so is just truncates the matrix by one element. We add a zero because there is no change there as it's the starting element.
step4: Result = find(B(1:end-1).*B(2:end)<0); This will return the index where you are on the cusp of the inflection. In this case it will be.
step5: New_Fx=Fx(Result); gives the Fx values where trends get changed
Now how to find corresponding T values? I tried New_T=T(New_Fx); but not it showing wrong T values. I think we have removed the elements in Fx which are repeating but not corresponding T values.
How to find these T values to corresponding New_Fx?
complete code is: Fx(Fx <100)= 0;
Fx(diff(Fx)==0) = [];
B = [0; diff(Fx)]; Result = find(B(1:end-1).*B(2:end)<0); New_Fx=Fx(Result);
0 comentarios
Respuesta aceptada
Andrei Bobrov
el 16 de Jul. de 2014
Editada: Andrei Bobrov
el 16 de Jul. de 2014
[~,i1] = findpeaks(Fx);
[~,i2] = findpeaks(-Fx);
out = T(sort([i1;i2]));
without findpeaks:
ll = find([true;diff(Fx)~=0]);
F = Fx(ll);
b = diff(F);
lll = [false;b(1:end-1).*b(2:end)<0;false];
out = T(ll(lll));
ADD
complete code:
x = xlsread('book3.xlsx');
xx = x(:,2);
xx(xx <100)= 0;
[~,i1] = findpeaks(xx);
[~,i2] = findpeaks(-xx);
out = x(sort([i1;i2]),1);
2 comentarios
Más respuestas (0)
Ver también
Categorías
Más información sobre Fourier Analysis and Filtering en Help Center y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!