R2 with a loglog plot
10 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Hi everyone.
I am quite new to matlab and I'd like to add a R2 to my loglog plot. I've seen some solution from few other posts, but none really does the job. My code is really simple:
bx= figure;
set(bx,'visible', 'on');
f = fit(x,y,'power1');
loglog(x,y,'O');
hold on;
plot(f);
[...]
the result look like that:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/155536/image.png)
So far I haven't find any way of determining the r2: That post (https://uk.mathworks.com/matlabcentral/answers/182998-r-squared-value-for-fitted-line) overestimates my r2, and this bit of code found somewhere on the ML forum as well:
ba = [x,ones(size(x))]\y;
ypred = ba(1)*x + ba(2);
SSE=sum((y-ypred).^2);
SST=sum((y-mean(y)).^2);
Rsq = 1 - (SSE/SST);
I also tried that way, but I think it works fine only for linear distributions:
X = [ones(length(x),1) x];
b = X\y;
yCalc = X*b;
r = 1 - sum((y - yCalc).^2)/sum((y - mean(y)).^2);
Thank you very much for your help :)
Flo
PS: the the r2 on excel is equal to 0.9597
0 Kommentare
Antworten (1)
dpb
am 8 Aug. 2016
I see nothing wrong with Star's Answer nor follow-up comment. To compute SSE from such a model requires evaluating the residual of the fit in original metric, not in log space.
But, fit returns a cfit for curves (your case) and some additional optional outputs, the second of which is a goodness-of-fit structure, gof in the documentation. Fields in gof include
sse - Error SSquares
R2 - Coefficient of determination ("raw" R-square)
adjustedR2 - DOF-adjusted R-square
stdError - RMS (or "standard") error
so to save yourself some effort, use it...
[f,gof] = fit(x,y,'power1');
7 Kommentare
dpb
am 12 Aug. 2016
Bearbeitet: dpb
am 12 Aug. 2016
Well, you need to look at the results obtained from the models (for the same, not disparate datasets) from Excel and Matlab to uncover where there's a difference. Clearly the results for the data you posted appear correct; if you got wildly different results from Excel, the most likely cause given the plot you posted is that it isn't the same dataset you're actually comparing to.
ADDENDUM
I attempted to read values off the above plot to see what kind of fit it actually provided; the Rsq was lower some but about 0.998 as 0.95 altho I could see guesstimating didn't work to get terribly close to the plot.
Agan, I can only suggest if you can reproduce the results you first quote in Excel, attach that set of data and model coefficients and results.
>> [f,gof]=fit(x,y,'power1')
f =
General model Power1:
f(x) = a*x^b
Coefficients (with 95% confidence bounds):
a = 1.026 (1.011, 1.04)
b = 1.013 (0.9852, 1.042)
gof =
sse: 1.6864e-04
rsquare: 0.9997
dfe: 13
adjrsquare: 0.9996
rmse: 0.0036
>> B % results from S-S:
B =
0.0254
1.0134
>> Rsq
Rsq =
0.9997
>>
BTW, I used fit in comparison to the fminsearch solution--results are essentially identical--
ADDENDUM 2
"*fit* in comparison to the fminsearch"
Actually, I just noticed there are two different solutions reached; the power term with fminsearch is about the magnitude of that of the fit solution less 1.0 -- 0.0254 vis a vis 1.026. How can that be???
Oh,
yfit = @(b,x) exp(b(1)) .* x.^b(2); % Power Function
has a definition problem; it's estimating log(B(1)) instead of B(1) directly...let's see what happens if redefine in same model space as fit uses--
>> yfit = @(b,x) b(1).*x.^b(2); % model A*x^B; B(1),B(2)-->A,B
>> SSECF = @(b) sum((yfit(b,x) - y).^2);
>> B = fminsearch(SSECF, [1; 1]);
>> B
B =
1.0256
1.0134
>>
Ah! As expected, now we agree...whew! :) Was worried there for a minute...
Siehe auch
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!