| T O P I C R E V I E W |
| lenaphys |
Posted - 08/17/2015 : 01:46:51 AM
Hi everyone!
Im using Hill function in Origin to fit my data.
I dont have much idea about how to make a good fit. Can anyone tell me how do it.
Thank you |
| 10 L A T E S T R E P L I E S (Newest First) |
| lenaphys |
Posted - 08/18/2015 : 08:39:03 AM
Dear Wojtek
I tried and the result is not so good.
thank you very much for kindness.
quote:
Originally posted by Wojtek.
Hi lenaphys,
did you try to set the bound for n parameter, for example n>2 and then play with k and A ?
I have no idea how to improve this fit if this does not help.
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| Wojtek. |
Posted - 08/18/2015 : 08:35:40 AM
Hi lenaphys,
did you try to set the bound for n parameter, for example n>2 and then play with k and A ?
I have no idea how to improve this fit if this does not help. |
| lenaphys |
Posted - 08/18/2015 : 08:26:26 AM
Dear Wojtek
Thank you very much for your help. The result looks very good.
But in my case I need the parameter n to be larger than 2. this makes the fitting not good, that's why I posted here to ask how to make the best fit in my case.
For your fitting, n is less than 2, it is meaningless to me.
For large value of n, the fitting curve looks like step function and does not fit well with the data plot.
quote:
Originally posted by Wojtek.
Hi lenaphys,
some time ago I read on this forum that is some cases the implict nonlinear fit works better. I perform two fitting procedures: first one with your Hill function as nonlinear fit (the red results in the figure bellow) and second with implicit nonlinear fitting function, where fitting formula is defined as: f=(A* (x^n)/(x^n+k^n))-y. For both fits I used a data selector, the results are as follow:
linear X scale:
log x scale:
In my opinion where you plot your data in linear scale the implicit fitting gives the better results - the green curve fits better to original data and the R^2 coefficient is little bit higher and the standard errors are smaller.
Regards,
Wojtek
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| Wojtek. |
Posted - 08/18/2015 : 08:06:49 AM
Hi lenaphys,
some time ago I read on this forum that is some cases the implict nonlinear fit works better. I perform two fitting procedures: first one with your Hill function as nonlinear fit (the red results in the figure bellow) and second with implicit nonlinear fitting function, where fitting formula is defined as: f=(A* (x^n)/(x^n+k^n))-y. For both fits I used a data selector, the results are as follow:
linear X scale:
log x scale:
In my opinion where you plot your data in linear scale the implicit fitting gives the better results - the green curve fits better to original data and the R^2 coefficient is little bit higher and the standard errors are smaller.
Regards,
Wojtek
|
| lenaphys |
Posted - 08/18/2015 : 07:27:46 AM
hi, this is my link for data which I want to fit with Hill function.
https://drive.google.com/file/d/0Bxy7qs3kEzEFakVodzN2ZlkwNGc/view?usp=sharing
thank you
quote:
Originally posted by Wojtek.
Hi lenaphys,
please attach some data which you fit with Hill function.
|
| Wojtek. |
Posted - 08/18/2015 : 07:18:14 AM
Hi lenaphys,
please attach some data which you fit with Hill function.
The residuals represents the difference between data and value of model for each point of your X, so you are able to observe the difference across the set of your X data.
Regards,
Wojtek |
| lenaphys |
Posted - 08/18/2015 : 06:59:48 AM
Dear Wojtek,
Thank you for your useful reply. I dont know what the residual tell us about the goodness of the fit. can you please explain me in detail?
this if the formula for Hill function: y=A* (x^n)/(x^n+k^n)
The data fit well when the parameter n is less than 2. But I want the parameter n to be larger than 2, in this case the goodness of the fit is not good. Do you know the way to utilize the fitting tool in origin to get a better fit?
thank you very much
quote:
Originally posted by Wojtek.
Hi lenaphys and JacquelineHe,
I think you should check the values of Residual parameters (in Origin there is a bookmarks in fitting window) as shown in figure bellow:
You can calculate manually the residuals and plot it versus your X.
For comparison of different fits I used also RMSE (root mean square error) value, which should be as low as possible.
Best regards,
Wojtek
|
| Wojtek. |
Posted - 08/17/2015 : 4:16:54 PM
Hi lenaphys and JacquelineHe,
I think you should check the values of Residual parameters (in Origin there is a bookmarks in fitting window) as shown in figure bellow:
You can calculate manually the residuals and plot it versus your X.
For comparison of different fits I used also RMSE (root mean square error) value, which should be as low as possible.
Best regards,
Wojtek |
| lenaphys |
Posted - 08/17/2015 : 08:54:06 AM
Thank you for your answer.
The only problem to me is that: I got good "Adj. R-square" value (around 0.98) eventhough the fitting curve does not fit well with the data plot.And sometime I DID NOT get good "Adj. R-square" value but the fitting curve looks so good, it fits the data plot well.
So, for two cases above which one is considered as the better one?
Thank you.
quote:
Originally posted by JacquelineHe
Hi,
If you do not get the good fitting result, you can consider change the suitable fitting function, or try to change the initialize parameters when you fit.
To change the initialize parameters:
After fitting, you can “Change Parameters” by clicking on the green lock to reopen the “NLFit” dialog.
The parameter values that will be used can be seen on the “Parameters” tab.
http://www.originlab.com/doc/Origin-Help/NLFit-Dialog-ParaTab
Thanks
Jacqueline
OriginLab
|
| JacquelineHe |
Posted - 08/17/2015 : 06:37:41 AM
Hi,
If you do not get the good fitting result, you can consider change the suitable fitting function, or try to change the initialize parameters when you fit.
To change the initialize parameters:
After fitting, you can “Change Parameters” by clicking on the green lock to reopen the “NLFit” dialog.
The parameter values that will be used can be seen on the “Parameters” tab.
http://www.originlab.com/doc/Origin-Help/NLFit-Dialog-ParaTab
Thanks
Jacqueline
OriginLab |
