Using interp1 with semilog dataset

25 vues (au cours des 30 derniers jours)
Alex
Alex le 31 Août 2022
Modifié(e) : Alex le 31 Août 2022
I have 3 arrays:
  • 'Dose' contains 27 values ranging from 1x10^2 to 1x10^6
  • 'AlThick' contains 27 values ranging from 0.0001 to 20
  • 'Rays' contains 10000 values ranging from 0 to 20
The points in each data set are not equally spaced. Dose represents y values; AlThick and Rays both represent x values. I want to find the value of Dose for each non-zero value of Rays. I am doing the following and getting, as expected, 10000 values.
DoseMat = interp1(AlThick, Dose, rays(rays>0), 'linear')
However, I'm not sure if I need to be accounting for the fact that the y-axis data is plotted as a base-10 logarithm. I.e. it is a semilog plot with base-10 log values on the y-axis and linear values on the x-axis, as shown below.
Any advice is appreciated!
  2 commentaires
Torsten
Torsten le 31 Août 2022
Modifié(e) : Torsten le 31 Août 2022
And plotting "Dose" against "AlThick" in semilogy gives you a linear relationship ?
Could you include this plot ?
Alex
Alex le 31 Août 2022
I have updated the OP with the plot of Dose vs AlThick

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Réponse acceptée

Cris LaPierre
Cris LaPierre le 31 Août 2022
Modifié(e) : Cris LaPierre le 31 Août 2022
You do not need to take account of the fact that you are plotting your data in a semilog plot. That doesn't impact how you would interpolate the data. The 'linear' here is the interpolation method.
However, you are not using interp1 correctly. Interp1 is for (x,y) data. The syntax is
  • newY = interp1(knownX, knownY, newX, method).
You probably want to use interp2.
  • newZ = interp1(knownX, knownY, knownZ, newX, newY, method).
  4 commentaires
Image Analyst
Image Analyst le 31 Août 2022
@Alex, attaching data in a .mat file, and screenshots of it plotted, would help.
Alex
Alex le 31 Août 2022
I've updated the OP with the plot of Dose vs AlThick

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