How can I see if data fit a specified curve?

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Rachael Thompson 2019-11-20

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评论： Rachael Thompson 2019-11-21

I have expression data, some are correlated, but some follow a pattern of y = 1/x (only the positive values of x and y).

How do I rapidly tell if the data fit y = 1/x (I have a few thousand pairs of data to try, hence not wanting to plot each one individually). Is there an easy way of doing this with curve fitting functions?

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Adam Danz 2019-11-20

编辑：Adam Danz 2019-11-20

When you say that some are correlated, I'm not sure I understand what that means. Do you mean for some data, x approximately equals y? A screen shot of your plots may be helpful.

Rachael Thompson 2019-11-20

Yes! So some y=x and some y=1/x and some are just random clouds.

The y=x are easy to find with the spearman's correlation coefficient.

The ones where y=1/x are the ones I want to find.

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Answer 1

Adam Danz 2019-11-20

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https://ww2.mathworks.cn/matlabcentral/answers/492128-how-can-i-see-if-data-fit-a-specified-curve#answer_402455

编辑：Adam Danz 2019-11-20

在 MATLAB Online 中打开

There are several ways to classify your data into the three groups: 1) random, 2) y=x, and 3) y=1/x. I tend to favor low-level approaches whenever possible. The demo below creates a dataset that fits your description and then classifies each data point according to those three categories. It does so by computing the error between each coordinate and the two functions y=x and y=1/x. If the error is above a set (subjective) threshold for both functions, the coordinate is classified as random. Otherwise, it is classified according to the function with the smallest error. If your data contain a considerable amount of noise, first trying increasing the errThreshold. If that doesn't help, we may need to use a more sophistocated algorithm.

% Create x-data
% the +.2 is to avoid values near 0 which cause problems with plotting 1/x
x = rand(1,1000)*5 + .2;  
% Create 3 groups of y-data
%1) y = x
%2) y = 1/x
%3) random
randXIdx = randperm(numel(x)); 
sections = floor(numel(x).*[.33, .66]); 
y = rand(size(x))*5 + .2;                                     % random y values
y(randXIdx(1:sections(1)-1)) = x(randXIdx(1:sections(1)-1));  %y = x
y(randXIdx(sections(1):sections(2))) = 1./x(randXIdx(sections(1):sections(2)));  %y = 1/x
% Look at data
clf()
plot(x,y, 'o')
% Compute the error between the two function y=x and y=1/x.
err1 = abs(y-x); 
err2 = abs(y - 1./x); 
% Choose a threshold.  Plotting the error may be helpful. 
% Error greater than threshold will be assigned to random class
%     clf()
%     plot(err1,'ro')
%     hold on
%     plot(err2, 'bx')
errThreshold = 0.05;    % my subjective judgement
% Classify the coordinates based on minimum error.
% If the error for both is beyond threshold, classify as random. 
group = zeros(size(y)); 
group(err1 >= errThreshold & err2 >= errThreshold) = 1; % group 1 is random group
group(err1 < errThreshold) = 2; % group 2 is y=x group
group(err2 < errThreshold) = 3; % group 3 is y=1/x groups (and y=x if the point belongs to both groups)
% Check that all points are assigned to a group
if any(group==0)
    error('Point not assigned to group.')
end
% Plot results
clf()
plot(x,y, 'ko')
hold on
plot(x(group==1),y(group==1), 'r.', 'DisplayName', 'rand')
plot(x(group==2),y(group==2), 'b.', 'DisplayName', 'y=x')
plot(x(group==3),y(group==3), 'g.', 'DisplayName', 'y=1/x')
legend()