I have never found the R symbolic differentiation helpful because my
functions are typically quite complicated, but was prompted by Steve
Ellison's suggestion to try it out in this case:
# reprex (see reprex package)
graphdta <- read.csv( text =
"t,c
0,100
40,78
80,59
120,38
160
You need to pay attention to the documentation more closely. If you don't
know what something means, that is usually a signal that you need to study
more... in this case about the difference between an input variable and a
design (model) matrix. This is a concept from the standard linear algebra
R-experts,
I have fitted many different lines. The fast-tau estimator (yellow line) seems
strange to me because this yellow line is not at all in agreement with the
other lines (reverse slope, I mean the yellow line has a positive slope and the
other ones have negative slope).
Is there somethi
> On Apr 6, 2018, at 8:15 AM, David Winsemius wrote:
>
>>
>> On Apr 6, 2018, at 8:03 AM, David Winsemius wrote:
>>
>>
>>> On Apr 6, 2018, at 3:43 AM, g l wrote:
>>>
Sent: Friday, April 06, 2018 at 5:55 AM
From: "David Winsemius"
Not correct. You already have `pr
> On Apr 6, 2018, at 8:03 AM, David Winsemius wrote:
>
>
>> On Apr 6, 2018, at 3:43 AM, g l wrote:
>>
>>> Sent: Friday, April 06, 2018 at 5:55 AM
>>> From: "David Winsemius"
>>>
>>>
>>> Not correct. You already have `predict`. It is capale of using the
>>> `newdata` values to do interpola
Since 2008, Microsoft staff and guests have written about R at the Revolutions
blog (http://blog.revolutionanalytics.com) and every month I post a summary of
articles from the previous month of particular interest to readers of r-help.
In case you missed them, here are some articles related to R f
> On Apr 6, 2018, at 3:43 AM, g l wrote:
>
>> Sent: Friday, April 06, 2018 at 5:55 AM
>> From: "David Winsemius"
>>
>>
>> Not correct. You already have `predict`. It is capale of using the `newdata`
>> values to do interpolation with the values of the coefficients in the model.
>> See:
>>
> coef(graphmodeld)
>(Intercept) graphdata[, 1]
>4.513544204 -0.006820623
>
> This corresponds to the linear model but it is still not understood how this
> helps
> to determine gradients at specific points for the exponential model.
You have fitted log(y) against x. Your fitted linear
> Sent: Friday, April 06, 2018 at 1:44 PM
> From: "Jeff Newmiller"
>
> You did not try my suggestion. You tried David's, which has a leftover
> mistake from your guesses about what the argument to coef should be.
Yes, sorry for the mistake.
coef(graphmodeld)
(Intercept) graphdata[, 1]
4.
You did not try my suggestion. You tried David's, which has a leftover mistake
from your guesses about what the argument to coef should be.
--
Sent from my phone. Please excuse my brevity.
On April 6, 2018 3:30:10 AM PDT, g l wrote:
>> Sent: Friday, April 06, 2018 at 4:53 AM
>> From: "Jeff Newm
> Sent: Friday, April 06, 2018 at 5:55 AM
> From: "David Winsemius"
>
>
> Not correct. You already have `predict`. It is capale of using the `newdata`
> values to do interpolation with the values of the coefficients in the model.
> See:
>
> ?predict
>
The § details did not mention interpola
> Sent: Friday, April 06, 2018 at 4:53 AM
> From: "Jeff Newmiller"
> To: "g l"
> coef( graphmodeld )
>
coef(graphmodelp)
Error: $ operator is invalid for atomic vectors
A quick search engine query revealed primarily references to the dollar sign
($) operator which does not seem relevant to th
12 matches
Mail list logo
