Fit Plots to A Given Slope
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Hi, I have a set of data that I have plotted using MATLAB. The data I have is supposed to fit to a line with a slope of either 0.5, 1, or 2. Which line the data best fits to tells me what type of system the data is from. So if the data best fits to a line of slope 0.5 I know I am working with system A, if it best fits to a line of slope 1 I am working with system B, and if it best fits to a line of slope 2 I am working with system C. The intercept of the lines does not matter. What I need is a way to fit my data to a line of each of the above mentioned slopes and then to find the regression coefficient to tell which line best fits to the data. So in short I need to know how to fit my data to a line of a slope of 0.5, 1, or 2. Please note that I am working in a log-log plot and that I cannot have an intercept of less than zero. Thank you for your help.
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dpb
2013-8-21
Is this slope/fit in log-log space I presume?
Why not approach it the other way 'round--just do the fit and find the slope then compare it to the three choices for nearest neighbor --
Simple example...
>> y=rand(10,1); x=1:10; x=x'; % make up some data
>> c=polyfit(x,5*sort(y),1)
c =
0.4067 -0.0011
>> bnear=interp1([0.5 1 2]',[0.5 1 2]',c(1),'nearest','extrap')
bnear =
0.5000
>> c=polyfit(x,10*sort(y),1)
c =
0.8134 -0.0022
>> bnear=interp1([0.5 1 2]',[0.5 1 2]',c(1),'nearest','extrap')
bnear =
1
>> c=polyfit(x,20*sort(y),1)
c =
1.6268 -0.0045
>> bnear=interp1([0.5 1 2]',[0.5 1 2]',c(1),'nearest','extrap')
bnear =
2
>>
Easy enough to vectorize, of course...
采纳的回答
Walter Roberson
2013-8-21
x = [....]; %the vector of x values that "data" exist at
sd_half = std(data - 0.5 * x);
sd_one = std(data - x);
sd_two = std(data - 2 * x);
I would suggest, though, trying
coeffs = polyfit(x, data, 2);
slope = coeffs(1);
You might to use the three-output version of polyfit to have centering and scaling done.
4 个评论
dpb
2013-8-24
编辑:dpb
2013-8-25
What's wrong w/ my solution--just apply it in the log-log space if that's where the correlation is.
Walter's works, too, but seems simpler to me to do the match to the nearest computed slope rather than find the minimum variance his does (didn't think about the assumed slope of unity still leading to the smallest variance w/ choice of exponent was what got me initially) as only compute the one slope and a lookup as opposed to three variances and the comparison...overall, probably not a great deal of difference, though.
Pick one (or both and compare) and go with it... :)
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