Fit curve to nonstandard data

10 次查看（过去 30 天）
Pippa Williams
Pippa Williams 2018-6-15
回答： Star Strider 2018-6-16
I'm trying to fit a curve to 2 data sets, image attached. I want to end up with a vector (or function) which defines, for any given x value, what the most likely y value is. It should have the property y(x.1)<= y(x.2) i.e. for an increase in x you get an increase in y.
The data won't fit any traditional function - it's stepped and weirdly curved.
I started by defining a linspace for x, and then trying the median of values in a range around that x value:
XSpace = (0:3500)';
YSpace = 0*XSpace;
for index = 1:size(X)
YSpace(index,1) = median(Y(and(X>(XSpace(index)-100),X<(XSpace(index)+100))));
end
However, this is skewed as there aren't the same number of data points for any given X value (e.g. if there are more data points below the X value of interest than above, the median will be skewed low).
I've tried sorting the scatter data, and then applying a range of filters, but I struggle to maintain the shape. I thought a median filter would work, but it doesn't handle when the outliers become too frequent. I'm also struggling as there's lots of data for low x, and much less data for high x.
Has anyone got any bright ideas about how to fit a curve/space to this data? It seems very obvious looking at the data, how the curve should look, but I'm struggling to work out how to fit it with an algorithm.

请先登录，再回答此问题。

回答（2 个）

Stephan
Stephan 2018-6-15
编辑：Stephan 2018-6-15
Hi,
if you have the curve fitting toolbox, you could consider this for your purpose:
https://de.mathworks.com/help/curvefit/interpolation-methods.html
You also find an example which could meet your requirements here at about the half of the page:
Best regards
Stephan
Star Strider
Star Strider 2018-6-16
I am not certain what you want to do. The problem is that you have several sets of data in each vector, so taking the unique ‘x’ values and taking the mean of the corresponding ‘y’ values may be the only possible approach. You can then use interp1 to interpolate to get the ‘y’ corresponding to any ‘x’ within the limits of ‘x’ and ‘y’ for each data set.
The Code —
D = load('DataForForumQuestion.mat');
XData1 = D.XData1;
YData1 = D.YData1;
XData2 = D.XData2;
YData2 = D.YData2;
[UX1,~,ic] = unique(XData1); % Unique ‘x’ Values & Indices For Set #1
UY1 = accumarray(ic,YData1, [], @mean); % Use ‘bsxfun’ To Calculate ‘mean’ Of ‘y’ Values Corresponding To Unique ‘x’ Values
[UX2,~,ic] = unique(XData2); % Unique ‘x’ Values & Indices For Set #2
UY2 = accumarray(ic,YData2, [], @mean); % Use ‘bsxfun’ To Calculate ‘mean’ Of ‘y’ Values Corresponding To Unique ‘x’ Values
figure(1)
plot(XData1, YData1, '-b') % Display Raw Data
hold on
plot(XData2, YData2, '-r')
hold off
axis([xlim 0 350])
figure(2)
plot(UX1, UY1, '-b') % Display Processed Data
hold on
plot(UX2, UY2, '-r')
hold off
axis([xlim 0 350])
XD1i = 1502; % Choose ‘x’ To Interpolate
YD1i = interp1(UX1, UY1, XD1i); % Calculate Corresponding ‘y’
XD2i = 1502; % Choose ‘x’ To Interpolate
YD2i = interp1(UX2, UY2, XD2i); % Calculate Corresponding ‘y’
The Plot —
XD1i =
1502
YD1i =
157.2170
XD2i =
1502
YD2i =
68.2821

类别

在 Help Center 和 File Exchange 中查找有关 Get Started with Curve Fitting Toolbox 的更多信息

标签

产品


版本

R2017b

Community Treasure Hunt

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

Start Hunting!

Translated by