Find X from Y in a graph

2 vues (au cours des 30 derniers jours)
msij
msij le 26 Avr 2016
Modifié(e) : msij le 27 Avr 2016
Hi,
I have plotted some x and y values to give me a hyperbola graph (I've attached the data in the excel file). I am trying to find what the x values are from a know y value but I am having some trouble.
An example:
From either the x/y values or the plotted graph; if y = -3.951, how can I find the corresponding x value?
I have looked around and most people recommend either using find (which I can't because the new y value won't definitely be in the given y values) or interp1. When I use interp1, it returns 'The grid vectors are not strictly monotonic increasing.'.
I used the line: x = interp1(y_values,x_values, m_A);
When I use,
x = interp1(x_values,y_values, -3.951);
I am returned with 'NaN'.
I'm not sure where to go from here. All help is appreciated.
Thanks!
EDIT:
Thanks for your help. I figured it out. I have decided to just separate out the x and y values so that they are unique. Then I can plot them oppositely.

Réponse acceptée

Azzi Abdelmalek
Azzi Abdelmalek le 26 Avr 2016
Modifié(e) : Azzi Abdelmalek le 26 Avr 2016
Try
x = interp1(x_values,y_values, -3.951,'linear','extrap');
  2 commentaires
msij
msij le 27 Avr 2016
I am not sure what this is doing. When I use the line above it returns an x-value of 1.6, but that isn't in the values of x on the graph.
I just need it to return the x-value for a given y, using the graph. Is this possible?
msij
msij le 27 Avr 2016
From what I've seen of interp, it can predict 'y' for a given 'x'. Because my graph is a kind of hyperbola, it won't let me plot them the other way around because interp1 returns 'The grid vectors are not strictly monotonic increasing.'

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