How do i use least square method to fit a nonlinear curve to data set below ?

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maziar
maziar 2021 年 5 月 16 日
編集済み: Scott MacKenzie 2021 年 5 月 16 日
How do i use least square method to fit a nonlinear curve to data set below ?
p=[40 50 60 80 100]
r=[230.88 243.5 268.34 278.92 280]

採用された回答

Scott MacKenzie
Scott MacKenzie 2021 年 5 月 16 日
編集済み: Scott MacKenzie 2021 年 5 月 16 日
This solution uses polyfit and a log transformation, converting the power-law into a linear equation. There might by an easier way. Perhaps someone else will weigh in with an alternate solution.
p = [40 50 60 80 100];
r = [230.88 243.5 268.34 278.92 280];
% transform r = ap^n into log(r) = log(a) + n log(p)
pLog = log(p);
rLog = log(r);
% do the model fitting and get coefficients
pf = polyfit(pLog, rLog, 1);
b = pf(1);
a = exp(pf(2));
% get squared correlation coefficient
CM = corrcoef(pLog, rLog);
R2 = CM(1,2)^2;
% x/y data for power-law curve from x = 0 to x = 150
x = linspace(0,150);
y = a * x.^b;
% plot curve and scatter data
p1 = plot(x, y);
hold on;
s1 = scatter(p, r, 'filled');
ax = gca;
ax.XLim = [0 150];
ax.YLim = [0 350];
ax.XLabel.String = 'p';
ax.YLabel.String = 'r';
% print equation and line to curve
s = sprintf('%s = %5.3f%s^{%.4f}\n%s^2 = %.4f', '{\itr}', ...
a, '{\itp}', b, '{\itR}', R2);
text(50, 150, s);
line([40, 30], [175, a * 30^b], 'color', 'k');

その他の回答 (1 件)

Scott MacKenzie
Scott MacKenzie 2021 年 5 月 16 日
Which non-linear curve? Here are a few options for you, for polynomials with orders 1 through 6:
p = [40 50 60 80 100];
r = [230.88 243.5 268.34 278.92 280];
x = -1000:1000;
tiledlayout('flow');
for i=1:6
pf = polyfit(p, r, i); % assume p is x, r is y
y = polyval(pf, x);
nexttile;
plot(x,y);
end
  2 件のコメント
maziar
maziar 2021 年 5 月 16 日
thanks but i need to fit a curve like r=ap^n and get a and n .
Scott MacKenzie
Scott MacKenzie 2021 年 5 月 16 日
OK, got it. I've posted another answer for the power-law curve you are interested in.

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