Want to fit linear curve on my data.

조회 수: 1 (최근 30일)
Phong Pham
Phong Pham 2013년 3월 18일
I have x, y data
x=[0;0.100000000000000;0.200000000000000;0.300000000000000;0.400000000000000;0.500000000000000;0.600000000000000];
y=[4.67178152947921e-06;4.67353333624452e-06;4.70560728038426e-06;4.74873086195845e-06;4.77333265701103e-06;4.84630647442201e-06;4.87015810633671e-06];
I want to plot x vs y and want to y-axis in log scale
plot(x,y)
set(gca,'YScale','log')
hold on
Note: x data starts from 0
Now I want to fit a line and show the slope of that curve fitting line + original curve
p=polyfit(x,(y),1);
q=polyval(p,x);
plot(x,q).
It seems to be not right because the fit line isn't straight ( it likes power fit or exponential) . Note log scale ( not log(data))
Please helps. Thanks
  댓글 수: 2
the cyclist
the cyclist 2013년 3월 18일
It would be tremendously helpful if you included a small sample of your data that exhibits the problem, so that we could run your code and see the results. Your statement that it "seems to be not right" is not quite enough.
Daniel Shub
Daniel Shub 2013년 3월 18일
Your x and y are not the same size ...

댓글을 달려면 로그인하십시오.

채택된 답변

Daniel Shub
Daniel Shub 2013년 3월 18일
편집: Daniel Shub 2013년 3월 20일
I don't like working on log scales, I would rather take the log transform of my data
x=[0;0.100000000000000;0.200000000000000;0.300000000000000;0.400000000000000;0.500000000000000;0.600000000000000];
y=[4.67178152947921e-06;4.67353333624452e-06;4.70560728038426e-06;4.74873086195845e-06;4.77333265701103e-06;4.84630647442201e-06;4.87015810633671e-06];
xx = x;
yy = log(y);
plot(xx, yy, '*');
lsline;
p = polyfit(xx, yy,1);
text(max(xlim), max(ylim), ['Slope: ', num2str(p(1))], 'HorizontalAlignment', 'Right', 'VerticalAlignment', 'Top')
set(gca, 'YTickLabel', 10^6*exp(get(gca, 'YTick')))
  댓글 수: 4
Phong Pham
Phong Pham 2013년 3월 19일
편집: Phong Pham 2013년 3월 19일
I dont know why but it has something weird to me. First, the y-scale has number likes 0.00344 not 10^-3 , etc .. ( log scale). Second, the slope is negative, the fitting line is straight up so it should give positive slope.
Thanks
Daniel Shub
Daniel Shub 2013년 3월 20일
I got the output of polyfit wrong (I thought the first coefficient was the intercept). I changed the code. I also changed the scaling of the ticks. You can set them to be whatever you want.

댓글을 달려면 로그인하십시오.

추가 답변 (1개)

Jan
Jan 2013년 3월 19일
When you want a line in the logspace diagram, you need an exponential fit on the data. Or build the log of the data at first and fit the line afterwards.

카테고리

Help CenterFile Exchange에서 Linear and Nonlinear Regression에 대해 자세히 알아보기

태그

Community Treasure Hunt

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

Start Hunting!

Translated by