T
hanks for your insights Rolf! The model I want to fit is y=x/a+x with no
intercept, so I transformed it to 1/y=1+a/x as they are the same. but i
will look up nls() and see how to fit the model without transformation.
On Tue, Aug 20, 2013 at 2:45 PM, Rolf Turner <rolf.tur...@xtra.co.nz> wrote:
>
> (1) It is not acceptable to use "wanna" in written English. You should say
> "I want to fit a model ....".
>
> (2) The model you have fitted is *not* equivalent to the model you first
> state.
>
> If you write "y ~ x/(a+x)" you are tacitly implying that
>
> y = x/(a+x) + E
>
> where the "errors" E are i.i.d. with mean 0.
>
> If this is the case then it will *not* be the case that
>
> 1/y = 1 + a/x + E
>
> with the E values being i.i.d. with mean 0.
>
> If the model "y ~ x/(a+x)" is really what you want to fit, then you should
> be using non-linear methods, e.g. by applying the function nls().
>
> cheers,
>
> Rolf Turner
>
>
>
> On 21/08/13 09:39, Ye Lin wrote:
>
>> Hey All,
>>
>> I wanna to fit a model y~x/(a+x) to my data, here is the code I use now:
>>
>> lm((1/y-1)~I(1/x)+0, data=b)
>>
>> and it will return the coefficient which is value of a
>>
>> however, if I use the code above, I am not able to draw a curve the
>> presents this equation. How can I do this?
>>
>
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