Effacer les filtres
Effacer les filtres

How to plot minima and maxima of two matrices

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Vahid Esmaeili
Vahid Esmaeili le 12 Août 2020
Commenté : Vahid Esmaeili le 13 Août 2020
Hello
It was supposed the following lines find the maximum of the rows in RAJ_Mx1 taking into consideration only those values where the corresponding elements in RHMORx are greater and less than zero. When I plot RAJ_Mx1 and RHMORx together, I cannot find these values! Would you please assist me to find out why and how can I plot them properly to find the maxima that the program was supposed to provide with?
peakMOM_pos_R_Ank_pos_powerIny=mean(max(RAJ_In1.*(RAJMORy>0 ),[],2));
peakMOM_pos_R_Ank_neg_powerIny=mean(min(RAJ_In1.*(RAJMORy>0 ),[],2));
peakMOM_neg_R_Ank_pos_powerIny=mean(max(RAJ_In1.*(RAJMORy<0 ),[],2));
peakMOM_neg_R_Ank_neg_powerIny=mean(min(RAJ_In1.*(RAJMORy<0 ),[],2));
Thank you so much,

Réponse acceptée

Cris LaPierre
Cris LaPierre le 12 Août 2020
Modifié(e) : Cris LaPierre le 12 Août 2020
The four lines of code you shared each check a different condition.
  1. max value in each row of RAH_In1 but only considering values that correspond to positive values in RAJMORy.
  2. min value in each row of RAH_In1 but only considering values that correspond to positive values in RAJMORy.
  3. max value in each row of RAH_In1 but only considering values that correspond to negative values in RAJMORy.
  4. min value in each row of RAH_In1 but only considering values that correspond to negative values in RAJMORy.
Having found the corresponding value for each of the 9 rows, the last thing each line of code does is calculate the mean of the 9 values. The result is a single number. Because it is a mean, you will not find a corresponding value in either matrix.
peakMOM_pos_R_Ank_pos_powerIny = mean(max(RAJ_In1.*(RAJMORy>0),[],2));
  1 commentaire
Vahid Esmaeili
Vahid Esmaeili le 13 Août 2020
Dear Cris,
Thank you so much for your answer. I did not pay enough attention!
Vahid,

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