How can I find the exponent of a log plot?

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Janna Hinchliff
Janna Hinchliff 2019 年 2 月 25 日
コメント済み: Harshita Gupta 2019 年 3 月 8 日
I'm plotting some data on a log scale using the code
for k = 1:length(dataeachcurrent)
temparray(k) = 1000./(T.TIcell{dataeachcurrent{k}{1}/10}(dataeachcurrent{k}{2})+273);
timeto50array(k) = dataeachcurrent{k}{3};
end
semilogy(fliplr(temparray),fliplr(timeto50array),'*','Color',[0.5 0 0.5])
fitEa = polyfit(fliplr(temparray),fliplr(timeto50array),1);
plotfit = polyval(fitEa,fliplr(temparray));
plot(fliplr(temparray),plotfit,'k--')
where temparray and timeto50array are taken from a set of cell arrays. This gives me a linear gradient on a log scale on the y axis, but I need the value of the exponent (which I think is given by the gradient) but I'm unsure how to get this value from the parameters I have. How can I extract this?
  2 件のコメント
Stephan
Stephan 2019 年 2 月 26 日
Harshita Gupta
Harshita Gupta 2019 年 3 月 8 日
I guess lscurvefit shoul work for your problem.

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