Find X from Y in a graph

1 次查看（过去 30 天）
msij
msij 2016-4-26
编辑： msij 2016-4-27
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.

请先登录，再回答此问题。

采纳的回答

Azzi Abdelmalek
Azzi Abdelmalek 2016-4-26
编辑：Azzi Abdelmalek 2016-4-26
Try
x = interp1(x_values,y_values, -3.951,'linear','extrap');
  2 个评论
msij
msij 2016-4-27
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 2016-4-27
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.'

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Discrete Data Plots 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by