What does this plot mean? Can an equation be made of it?
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Blakeley Ficenec
am 2 Feb. 2023
Kommentiert: Image Analyst
am 2 Feb. 2023
This is the plot I am trying to get an equation for:
plot((1:N_rel), [e_mean; mean(lambdaSO(1:end-1,:),2)]./e_total, 'ko');
This is what each value is:
N_rel = 20
e_mean = 4.6600e+04
lambdaSO = 20x30 double
e_total = 1x30 double
The first coordinate on the plot is (1, 0.996058).
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Image Analyst
am 2 Feb. 2023
With this:
N_rel = 20;
e_mean = 4.6600e+04;
lambdaSO = rand(20,30);
e_total = rand(1,30);
x = 1 : N_rel;
y = [e_mean; mean(lambdaSO(1:end-1,:),2)]./e_total
plot(x, y, '.-', 'MarkerSize', 20);
grid on;
legend
whos x
whos y
y is a matrix not a vector like x is. So it will plot columns. Each of the 30 columns of y is like a separate curve. What exactly are you hoping to see?
2 Kommentare
Image Analyst
am 2 Feb. 2023
Well, like I said, it plots each column in your matrix as a separate curve.
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John D'Errico
am 2 Feb. 2023
Can you get an "equation" for it? No. At least not any more than what you already have. Anyway, an equation involves an equality. You might hope to get an expression that approximates the relation you showed. But you aready have that. And there is no reason you have suggested that anything more meaningful exists, than what you already have.
Cris LaPierre
am 2 Feb. 2023
Bearbeitet: Cris LaPierre
am 2 Feb. 2023
There are likely many equations you could generate that could satisfy the one point you have shared. However, you must take all the points into consideration when coming up with the equation, and you have not shared that.
As John D'Errico said, that process most likely will be an approximation of the data. If that is what you want to do, then you might consider exploring the Curve Fitter app (this does require having the curve fitting toolbox installed).
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