Displaying 20 results from an estimated 400 matches similar to: "Obtain gradient at multiple values for exponetial decay model"
2018 Apr 05
0
Obtain gradient at multiple values for exponetial decay model
This smells like homework, which the Posting Guide indicates is off topic.
I am not aware of "the function" that will solve this, but if you know what a gradient is analytically then you should be able to put together a solution very similar to the code you already have with the addition of using the coef function.
--
Sent from my phone. Please excuse my brevity.
On April 5, 2018
2018 Apr 07
0
Obtain gradient at multiple values for exponential decay model
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
160,25
200,21
240,16
280,12
320,10
360,9
400,7
", header = TRUE )
nd <- c( 100, 250,
2018 Apr 06
3
Obtain gradient at multiple values for exponential decay model
> On Apr 6, 2018, at 8:03 AM, David Winsemius <dwinsemius at comcast.net> wrote:
>
>
>> On Apr 6, 2018, at 3:43 AM, g l <gnulinux at gmx.com> wrote:
>>
>>> Sent: Friday, April 06, 2018 at 5:55 AM
>>> From: "David Winsemius" <dwinsemius at comcast.net>
>>>
>>>
>>> Not correct. You already have
2018 Apr 05
4
Obtain gradient at multiple values for exponetial decay model
> Sent: Thursday, April 05, 2018 at 4:40 PM
> From: "Jeff Newmiller" <jdnewmil at dcn.davis.ca.us>
>
> the coef function.
>
For the benefit of other novices, used the following command to read the documentation:
?coef
Then tried and obtained:
> cvalue100<-coef(graphmodelp~100)
> cvalue100
NULL
Then looked at the model values which of course correspond
2018 Apr 06
0
Obtain gradient at multiple values for exponential decay model
> On Apr 6, 2018, at 3:43 AM, g l <gnulinux at gmx.com> wrote:
>
>> Sent: Friday, April 06, 2018 at 5:55 AM
>> From: "David Winsemius" <dwinsemius at comcast.net>
>>
>>
>> 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:
2018 Apr 06
0
Obtain gradient at multiple values for exponetial decay model
Try
coef( graphmodeld )
And you don't need to approximate if you use the newdata argument to the predict function.
I think reading the "Introduction to R" that comes with R would help.
--
Sent from my phone. Please excuse my brevity.
On April 5, 2018 2:00:45 PM PDT, g l <gnulinux at gmx.com> wrote:
>> Sent: Thursday, April 05, 2018 at 4:40 PM
>> From:
2018 Apr 06
2
Obtain gradient at multiple values for exponential decay model
> Sent: Friday, April 06, 2018 at 5:55 AM
> From: "David Winsemius" <dwinsemius at comcast.net>
>
>
> 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 interpolation explicity; thanks.
> The
2018 Apr 06
2
Obtain gradient at multiple values for exponential decay model
> Sent: Friday, April 06, 2018 at 4:53 AM
> From: "Jeff Newmiller" <jdnewmil at dcn.davis.ca.us>
> To: "g l" <gnulinux at gmx.com>
> 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 this
2018 Apr 06
0
Obtain gradient at multiple values for exponetial decay model
> On Apr 5, 2018, at 2:00 PM, g l <gnulinux at gmx.com> wrote:
>
>> Sent: Thursday, April 05, 2018 at 4:40 PM
>> From: "Jeff Newmiller" <jdnewmil at dcn.davis.ca.us>
>>
>> the coef function.
>>
>
> For the benefit of other novices, used the following command to read the documentation:
>
> ?coef
>
> Then tried and
2018 Apr 06
0
Obtain gradient at multiple values for exponential decay model
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 <gnulinux at gmx.com> wrote:
>> Sent: Friday, April 06, 2018 at 4:53 AM
>> From: "Jeff Newmiller" <jdnewmil at dcn.davis.ca.us>
2018 Apr 06
1
Obtain gradient at multiple values for exponential decay model
> Sent: Friday, April 06, 2018 at 1:44 PM
> From: "Jeff Newmiller" <jdnewmil at dcn.davis.ca.us>
>
> 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]
4.513544204 -0.006820623
This corresponds
2008 Aug 14
1
Graphing: plot 3rd variable based on color gradient
Hello,
I am searching for the best method to plot two variables with points
whose output color depends on the size of a third variable. For
example, the darkness of the x-y point would increase incrementally
based on the size of the z value, similar to the colramp parameter in
geneplotter. This would be analagous to symbols(), except changing the
selection from the color gradient rather than the
2010 Nov 04
1
Best Fit line trouble with rsruby
Hello, I am using R, through rsruby, to create a graph and best fit line for
a set of data points, regarding data collected in a Chemistry class. The
problem is that although the graph functions perfectly properly, the best
fit line will not work.
I initially used code I pretty much copied from a website with a tutorial on
this, which was:
graphData.png("/code/Beer's-Law
2007 Jul 10
6
Having trouble using data returned by Ajax.request
Hello everyone, I''m new here.
I''ve been working with prototype and plotr for about a month now, off
and on, and I have pretty much hit the wall on using the data returned
by Ajax.Request.
I''m using some php code to return a string:
{''foo'': [[0,0.0865334429075127], [1,0.0828179861705063],
[2,0.0828173042602942], [3,0.0841707718624196]]}
But I keep
2012 Feb 17
1
basic help: graph multivariate analysis.
Hey guys, I'd really appreciate any help.
I have a multivariate analysis done, the output of which is:
> GraphData <-read.table("eigen.coa")
> GraphData
V1 V2 V3 V4
1 1 0.371970 0.8552 0.8552
2 2 0.061785 0.1420 0.9972
3 3 0.001211 0.0028 1.0000
4 4 0.000000 0.0000 1.0000
> summary(GraphData)
V1 V2 V3
2010 Apr 24
1
help please: predict error code
Hello,
I am trying to calculate predicted values derived from one dataset into a hypothetical dataset. I tried this line of code:
graphdata$fmgpredvalues <- predict(Acs250.3.4, graphdata)
and received the following error message:
ERROR: ZXend[1], drop = FALSE] %*%lmeFit$beta
I have made sure all variable names are the same between the two datasets and all factors are appropriately
2009 Nov 09
0
Testing treatment effects on exponential decay models
Hello all:
I would like to test whether there are treatment effects on decomposition
rate, and I would like to inquire about the best, most appropriate means
using R.
I have plant decomposition data that is generally considered to follow an
exponential decay model as follows:
Wt = Wi * exp(-k * t)
Where Wt and Wi are the weights of the plant material at time t and 0,
respectively. k is a
2010 Jul 13
0
Neural Network package AMORE and a weight decay
Hi,
I want to use the neural network package AMORE and I don't find in the documentation the weight decay option.
Could someone tell if it is possible to add a regularization parameter (also known as a weight decay) to the training method.
Is it possible to alter the gradient descent rule for that?
Thanks,
Ron
2007 Mar 07
0
PeriodicalUpdater with Logarithmic decay
Greetings all,
So, I''ve finally found a place to play around with the
Ajax.PeriodicalUpdater. In looking at the API, I''m liking the decay
option -- not necessarily for my current purpose, but just to keep in
mind -- and I have a question: can the decay be a function which
returns an integer? Basically, why I''m looking for is a logarithmic
decay (where the system
2008 Dec 15
1
Population Decay in R
Hi,
I am new to R. I am trying to plot the decay of a population over time
(0-50yrs). I have the initial population value (5000) and the mortality
rate (0.26/yr) and I can't figure out how to apply this so I get a remaining
population value each year. In excel (ignoring headings) I would put 5000
in A1, in B2 I would enter the formula A1*0.26, and then in A2 (the next
years population) I