What does this plot mean? Can an equation be made of it?

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Blakeley Ficenec
Blakeley Ficenec 2023-2-2
评论： Image Analyst 2023-2-2
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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采纳的回答

Image Analyst
Image Analyst 2023-2-2
Ran in:
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
y = 20×30
1.0e+05 * 1.2372 0.4987 2.1508 0.6137 1.1839 1.1303 2.2412 0.5112 1.3909 0.5290 0.5288 1.0551 0.6578 0.5781 1.2955 0.9397 2.7138 0.6341 4.5755 0.9004 0.8226 0.8600 0.8630 0.9650 5.6663 0.9112 2.0831 1.7258 0.5786 0.9998 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000
plot(x, y, '.-', 'MarkerSize', 20);
grid on;
legend
whos x
Name Size Bytes Class Attributes x 1x20 160 double
whos y
Name Size Bytes Class Attributes y 20x30 4800 double
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 个评论
Blakeley Ficenec
Blakeley Ficenec 2023-2-2
I do currently get a plot with MATLAB, which is an energy fraction graph, but I wanted to understand what that line of code is doing to see if I can try to plot it in excel as well. I'm doing some research on Proper Orthogonal Decomposition.
Image Analyst
Image Analyst 2023-2-2
Well, like I said, it plots each column in your matrix as a separate curve.

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更多回答（2 个）

John D'Errico
John D'Errico 2023-2-2
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.
  1 个评论
Blakeley Ficenec
Blakeley Ficenec 2023-2-2
Thank you. I wanted to try and plot this in excel. I don't understand what the plot is currently doing in MATLAB.

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Cris LaPierre
Cris LaPierre 2023-2-2
编辑：Cris LaPierre 2023-2-2
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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