Generally nlsr package has better reliability in getting parameter estimates
because it tries to use automatic derivatives rather than a rather poor
numerical
estimate, and also uses a Levenberg-Marquardt stabilization of the linearized
model. However, nls() can sometimes be a bit more flexible.
Thanks Jeff
Bernard
Sent from my iPhone so please excuse the spelling!"
> On Apr 5, 2020, at 3:14 PM, Jeff Newmiller wrote:
>
> stats::nlm?
>
>> On April 5, 2020 11:53:10 AM PDT, Bernard Comcast
>> wrote:
>> Any recommendations on an R package to fit data to a nonlinear model
>> Y=f(x) with
err... stats::nls...
On April 5, 2020 12:14:15 PM PDT, Jeff Newmiller
wrote:
>stats::nlm?
>
>On April 5, 2020 11:53:10 AM PDT, Bernard Comcast
> wrote:
>>Any recommendations on an R package to fit data to a nonlinear model
>>Y=f(x) with a single x and y variable?
>>
>>I want to be able to gener
stats::nlm?
On April 5, 2020 11:53:10 AM PDT, Bernard Comcast
wrote:
>Any recommendations on an R package to fit data to a nonlinear model
>Y=f(x) with a single x and y variable?
>
>I want to be able to generate parameter uncertainty estimates and
>prediction uncertainties if possible.
>
>Berna
Any recommendations on an R package to fit data to a nonlinear model Y=f(x)
with a single x and y variable?
I want to be able to generate parameter uncertainty estimates and prediction
uncertainties if possible.
Bernard
Sent from my iPhone so please excuse the spelling!"
__
I think you are going to have to be more specific than "having some trouble".
Your plot used lka as the x-axis.
FWIW note that
lm(ruotsi.pist ~ mies + koulu + clka + koulu*clka, data=dta)
is the same as
lm(ruotsi.pist ~ mies + koulu*clka, data=dta)
--
Sent from my phone. Please excuse my brev
If your goal is to visualize the predicted curve from an lm fit (or
other model fit) then you may want to look at the Predict.Plot and
TkPredict functions from the TeachingDemos package.
On Sun, Sep 25, 2016 at 7:01 AM, Matti Viljamaa wrote:
> I’m trying to plot regression lines using curve()
> On 26 Sep 2016, at 19:41, Matti Viljamaa wrote:
>
> Thank you.
>
> However, I’m having some trouble converting your code to use clka, because
> the model I was using was:
>
> fit2 <- lm(ruotsi.pist ~ mies + koulu + clka + koulu*clka, data=dta)
I mean, not to use clka to replace lka. But to
Thank you.
However, I’m having some trouble converting your code to use clka, because the
model I was using was:
fit2 <- lm(ruotsi.pist ~ mies + koulu + clka + koulu*clka, data=dta)
> On 25 Sep 2016, at 21:23, Jeff Newmiller wrote:
>
> This illustrates why you need to post a reproducible exa
This illustrates why you need to post a reproducible example. You have a
number of confounding factors in your code.
First, "data" is a commonly-used function... avoid using it for variable
names.
Second, using the attach function this way leads to confusion... best to
forget this function u
> On 25 Sep 2016, at 19:37, Matti Viljamaa wrote:
>
> Okay here’s a pretty short code to reproduce it:
>
> data <-
> read.table("http://users.jyu.fi/~slahola/files/glm1_datoja/yoruotsi.txt";,
> header=TRUE)
data$clka <- I(data$lka - mean(data$lka))
> attach(data)
>
> fit2 <- lm(ruotsi.pist
Object clka not found.
Did you test run it in a fresh R environment?
--
Sent from my phone. Please excuse my brevity.
On September 25, 2016 9:37:56 AM PDT, Matti Viljamaa wrote:
>Okay here’s a pretty short code to reproduce it:
>
>data <-
>read.table("http://users.jyu.fi/~slahola/files/glm1_da
Okay here’s a pretty short code to reproduce it:
data <-
read.table("http://users.jyu.fi/~slahola/files/glm1_datoja/yoruotsi.txt";,
header=TRUE)
attach(data)
fit2 <- lm(ruotsi.pist ~ mies + koulu + lka + koulu*clka)
bs <- coef(fit2)
varitB <- c(data[koulu == 'B',]$mies)
varitB[varitB == 0] =
Go directly to C. Do not pass go, do not collect $200.
You think curve does something, but you are missing what it actually does.
Since you don't seem to be learning from reading ?curve or from our responses,
you need to give us an example you can learn from.
--
Sent from my phone. Please excu
On 2016-09-25 18:52, Jeff Newmiller wrote:
You seem to be confused about what curve is doing vs. what you are
doing.
But my x-range in curve()'s parameters from and to should be the entire
lka vector, since they are from=min(lka) and to=max(lka). Then why does
this not span the entire of lka?
You seem to be confused about what curve is doing vs. what you are doing.
A) Compute the points you want to plot and put them into 2 vectors. Then figure
out how to plot those vectors. Then (perhaps) consider putting that all into
one line of code again.
B) The predict function is the preferr
On 2016-09-25 18:30, Duncan Murdoch wrote:
On 25/09/2016 9:10 AM, Matti Viljamaa wrote:
Writing:
bs["(Intercept)"]+bs["mies"]*0+bs["kouluB"]+bs["lka"]*lka+bs["kouluB:clka"]*clka
i.e. without that being inside curve produces a vector of length 375.
So now it seems that curve() is really skippi
On 25/09/2016 9:10 AM, Matti Viljamaa wrote:
Writing:
bs["(Intercept)"]+bs["mies"]*0+bs["kouluB"]+bs["lka"]*lka+bs["kouluB:clka"]*clka
i.e. without that being inside curve produces a vector of length 375.
So now it seems that curve() is really skipping some lka-/x-values.
How could curve() k
Writing:
bs["(Intercept)"]+bs["mies"]*0+bs["kouluB"]+bs["lka"]*lka+bs["kouluB:clka"]*clka
i.e. without that being inside curve produces a vector of length 375.
So now it seems that curve() is really skipping some lka-/x-values.
> On 25 Sep 2016, at 16:01, Matti Viljamaa wrote:
>
> I’m trying
I’m trying to plot regression lines using curve()
The way I do it is:
bs <- coef(fit2)
and then for example:
curve(bs["(Intercept)"]+bs["mies"]*0+bs["kouluB"]+bs["lka"]*x+bs["kouluB:clka"]*clka,
from=min(lka), to=max(lka), add=TRUE, col='red')
This above code runs into error:
Error in curve(
Hi Waltenegus,
you should consider to show us your code and the data you used to fit the
curve. If you can't or if you prefer you can simply show the code and some
sample data on which the code can be run.
You'll find nice and useful tool in nlstools package.
Hope this help
Giuseppe
-
Gius
A.K
From: Luis Fernando García Hernández
To: arun
Sent: Monday, July 30, 2012 9:00 PM
Subject: Re: [R] curve comparison
Thanks both for your answer,
Will, do you have any reference which shows me how how to perform
this analysis? I´d like to know to find out how to do it. In my
case da
/geyer/old/5601/examp/kolmogorov.html).
I am not very familiar with non-parametric tests.
A.K.
From: Luis Fernando García Hernández
To: arun
Sent: Monday, July 30, 2012 9:00 PM
Subject: Re: [R] curve comparison
Thanks both for your answer,
Will, do you
: Monday, July 30, 2012 1:38 AM
Subject: [R] curve comparison
Dear R users,
I have seven regression lines I´d like to compare, in order to find out if
these are significatively different. The main problem is that these are
curves, non normal, non homogeneous data, I´ve tried to linearize them but
I'm very much a novice with R, and not a statistician. But as my group
has been working with generalized additive models (semi-parametric
regression), we have followed Wood's advice about using the R anova
function to do model comparison for different regressions. I would
imagine at least some
Dear R users,
I have seven regression lines I´d like to compare, in order to find out if
these are significatively different. The main problem is that these are
curves, non normal, non homogeneous data, I´ve tried to linearize them but
it has not worked. So I´d like to know if you know any comman
You could use the summation convention built into mgcv:gam for this. See
?linear.functional.terms
for details, but here is some example code, both for the exact match,
you describe, and a noisy version. best, Simon
library(mgcv)
f2 <- function(x) 0.2*x^11*(10*(1-x))^6+10*(10*x)^3*(1-x)^10 ## te
This sounds like possibly using logsplines may be what you want. See
the 'oldlogspline' function in the 'logspline' package.
On Thu, Apr 12, 2012 at 7:45 AM, Michael Haenlein
wrote:
> Dear all,
>
> This is probably more related to statistics than to [R] but I hope someone
> can give me an idea h
Dear all,
This is probably more related to statistics than to [R] but I hope someone
can give me an idea how to solve it nevertheless:
Assume I have a variable y that is a function of x: y=f(x). I know the
average value of y for different intervals of x. For example, I know that
in the interval[0
THanks so much Gene,
it worked
mintewab
Från: gley...@gmail.com [gley...@gmail.com] för Gene Leynes
[gleyne...@gmail.com]
Skickat: den 7 december 2011 18:43
Till: Mintewab Bezabih
Kopia: r-help@r-project.org
Ämne: Re: [R] curve fitted ... how to retreive
For the components:
result = predict(b, type="terms")
For the total fit:
result = predict(b)
result = b$fitted.values
On Wed, Dec 7, 2011 at 3:24 AM, Mintewab Bezabih <
mintewab.beza...@economics.gu.se> wrote:
> Dear R users,
>
> I have now managed to fit the curve using the thin plate spline a
Dear R users,
I have now managed to fit the curve using the thin plate spline as follows:
library(mgcv)
b <- gam(y~s(x1,x2,k=100),data =dat)
vis.gam(b)
What I want now is to get the fitted data for y and copy it so that I use it
for further analysis.
Many thanks in advance
mintewab
On 2010-12-14 08:43, Sarah Goslee wrote:
You are most likely to get reasonable help if you provide at a minimum
your OS, the version of R you're using, and the error message you get.
And also if you send your replies to the R-help list, not just me.
Sarah
Sarah's code works perfectly well on
You are most likely to get reasonable help if you provide at a minimum
your OS, the version of R you're using, and the error message you get.
And also if you send your replies to the R-help list, not just me.
Sarah
On Tue, Dec 14, 2010 at 10:58 AM, Val wrote:
> Thank you Sarah,
>
>
> It worked
Val,
Here's the complete console output. The graph produced is at:
http://www.functionaldiversity.org/temp/curve.png
R version 2.12.0 (2010-10-15)
Copyright (C) 2010 The R Foundation for Statistical Computing
ISBN 3-900051-07-0
Platform: x86_64-redhat-linux-gnu (64-bit)
R is free software and co
On Mon, Dec 13, 2010 at 2:12 PM, Ashta wrote:
> Thanks Sarah,
>
>> 1. to shade or color (blue) the curve using the criterion that any values
>> greater than 11,000
>
> I think I was not clear in the above point. I want shade not the line but
> the area under the curve,
Here's an example of how to
Thanks Sarah,
> 1. to shade or color (blue) the curve using the criterion that any values
greater than 11,000
I think I was not clear in the above point. I want shade not the line but
the area under the curve,
and
Your last line of code,
segments(x0=mean(test1), y0=0, y1=curveheight)
gave me th
Here's one way to do what I think you want:
test<- rnorm(5000,1000,100)
test1 <- subset(test, subset=(test > 1100))
d <- density(test)
plot(d, main="Density of production", xlab="")
lines(d$x[d$x > 1100], d$y[d$x > 1100], col="blue", lwd=2)
curveheight <- d$
Hi All,
I generated 5000 samples using the following script
test<- rnorm(5000,1000,100)
test1 <- subset(test, subset=(test > 1100))
d <- density(test)
plot(d, main="Density of production")
abline(v=mean(test1)
I wanted to do the following but faced diffic
ovided some technical and practical information which I could learn
>> from and be very thankful for.
>>
>> Regards,
>> Joseph
>>
>> On Fri, Apr 30, 2010 at 11:35 PM, Greg Snow
>> wrote:
>> >
>> >> -Original Message-----
>>
nal Message-
> From: Kyeong Soo (Joseph) Kim [mailto:kyeongsoo@gmail.com]
> Sent: Friday, April 30, 2010 5:24 PM
> To: Greg Snow
> Cc: r-help@r-project.org
> Subject: Re: [R] Curve Fitting/Regression with Multiple Observations
>
> I have already learned a lot from the
lto:r-help-boun...@r-
>> project.org] On Behalf Of Kyeong Soo (Joseph) Kim
>> Sent: Friday, April 30, 2010 4:10 AM
>> To: kMan
>> Cc: r-help@r-project.org
>> Subject: Re: [R] Curve Fitting/Regression with Multiple Observations
>
> [snip]
>
>> By the way, I won
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Kyeong Soo (Joseph) Kim
> Sent: Friday, April 30, 2010 4:10 AM
> To: kMan
> Cc: r-help@r-project.org
> Subject: Re: [R] Curve Fitting/Regression with
You can use nls2 to try many starting values. It works just like nls but:
- if you give it a two row data frame as the start value it will
create a grid between the upper and lower values of each parameter and
then run an optimization starting at each such point on the grid
returning the best
- i
ct.org
Date:
04/30/2010 03:33 PM
Subject:
[R] Curve Fitting
Sent by:
I am having troubles in fitting functions of the form
y~a*x^b+c
to data, for example
x<-c(0.1,0.36,0.63,0.90,1.166,1.43, 1.70, 1.96, 2.23)
y<-c(8.09,9.0,9.62,10.11,10.53,10.9, 11.25, 11.56, 11.86)
I tried for exam
I am having troubles in fitting functions of the form
y~a*x^b+c
to data, for example
x<-c(0.1,0.36,0.63,0.90,1.166,1.43, 1.70, 1.96, 2.23)
y<-c(8.09,9.0,9.62,10.11,10.53,10.9, 11.25, 11.56, 11.86)
I tried for example with nls, which did only work with really good initial
guessed values.
Any su
gsoo@gmail.com]
> Sent: Friday, April 30, 2010 4:10 AM
> To: kMan
> Cc: r-help@r-project.org
> Subject: Re: [R] Curve Fitting/Regression with Multiple Observations
>
> Dear Keith,
>
> Thanks for the suggestion and taking your time to respond to it.
>
> But, you misunde
n wrote:
>> > Dear Joseph,
>> >
>> > If you do not need to make any inferences, that is, you
>> just want it to look pretty, then drawing a curve by hand is
>> as good a solution as any. Plus, there is no reason for
>> expert testimony to say that the cu
eithC.
-Original Message-
From: Kyeong Soo (Joseph) Kim [mailto:kyeongsoo@gmail.com]
Sent: Friday, April 30, 2010 4:10 AM
To: kMan
Cc: r-help@r-project.org
Subject: Re: [R] Curve Fitting/Regression with Multiple Observations
Dear Keith,
Thanks for the suggestion and taking your time to re
rve does not mean anything.
> >
> > Sincerely,
> > KeithC.
> >
> > -Original Message-
> > From: Kyeong Soo (Joseph) Kim [mailto:kyeongsoo@gmail.com]
> > Sent: Tuesday, April 27, 2010 2:33 PM
> > To: Gabor Grothendieck
> > Cc: r
m: Kyeong Soo (Joseph) Kim [mailto:kyeongsoo@gmail.com]
> Sent: Tuesday, April 27, 2010 2:33 PM
> To: Gabor Grothendieck
> Cc: r-help@r-project.org
> Subject: Re: [R] Curve Fitting/Regression with Multiple Observations
>
> Frankly speaking, I am not looking for such a framework.
>
-
From: Kyeong Soo (Joseph) Kim [mailto:kyeongsoo@gmail.com]
Sent: Tuesday, April 27, 2010 2:33 PM
To: Gabor Grothendieck
Cc: r-help@r-project.org
Subject: Re: [R] Curve Fitting/Regression with Multiple Observations
Frankly speaking, I am not looking for such a framework.
The system I&#
Frankly speaking, I am not looking for such a framework.
The system I'm studying is a communication network (like M/M/1 queue,
but way too complicated to mathematically analyze it using classical
queueing theory) and the conclusion I want to make is qualitative
rather than quantatitive -- a high-l
If you are looking for a framework for statistical inference you could
look at additive models as in the mgcv package which has a book
associated with it if you need more info. e.g.
library(mgcv)
fm <- gam(dist ~ s(speed), data = cars)
summary(fm)
plot(dist ~ speed, cars, pch = 20)
fm.ci <- with(
Hello Gabor,
Many thanks for providing actual examples for the problem!
In fact I know how to apply and generate plots using various R
functions including loess, lowess, and smooth.spline procedures.
My question, however, is whether applying those procedures directly on
the data with multiple ob
This will compute a loess curve and plot it:
example(loess)
plot(dist ~ speed, cars, pch = 20)
lines(cars$speed, fitted(cars.lo))
Also this directly plots it but does not give you the values of the
curve separately:
library(lattice)
xyplot(dist ~ speed, cars, type = c("p", "smooth"))
On Tue,
-project.org] On
Behalf Of Kyeong Soo (Joseph) Kim
Sent: Tuesday, April 27, 2010 10:31 AM
To: r-help@r-project.org
Subject: [R] Curve Fitting/Regression with Multiple Observations
I recently came to realize the true power of R for statistical
analysis -- mainly for post-processing of data from large
I recently came to realize the true power of R for statistical
analysis -- mainly for post-processing of data from large-scale
simulations -- and have been converting many of existing Python(SciPy)
scripts to those based on R and/or Perl.
In the middle of this conversion, I revisited the problem o
.org
> 801.408.8111
>
>
> > -Original Message-
> > From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> > project.org] On Behalf Of Duncan Murdoch
> > Sent: Thursday, April 15, 2010 7:29 AM
> > To: Dwayne Blind
> > Cc: r-help@r-project.org
Behalf Of Duncan Murdoch
> Sent: Thursday, April 15, 2010 7:29 AM
> To: Dwayne Blind
> Cc: r-help@r-project.org
> Subject: Re: [R] curve
>
> On 14/04/2010 4:59 PM, Dwayne Blind wrote:
> > Dear R users,
> >
> > How can I use "curve" with a function of two va
On 14/04/2010 4:59 PM, Dwayne Blind wrote:
Dear R users,
How can I use "curve" with a function of two variables ?
See Ben Bolker's reply if you want to plot a surface. If you want to
plot a curve by holding one of the two variables fixed, just set it to a
constant value, and use "x" as t
Dwayne Blind gmail.com> writes:
> How can I use "curve" with a function of two variables ?
see "curve3d" in the emdbook package.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://
Dear R users,
How can I use "curve" with a function of two variables ?
Thank you very much,
Dwayne
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the postin
Pascale,
If you do want an nls fit with the associated error structure
assumptions, check ?SSlogis.
fm <- nls(y ~ SSlogis(x, Asy, xmid, scal))
summary(fm)
xx <- seq(123, 248, length = 101)
yy <- predict(fm, list(x = xx))
plot(x, y)
lines(xx, yy)
-Peter Ehlers
Gabor Grothendieck wrote:
A simple y vs log(x) fit seems to work pretty well here:
fit <- lm(y ~ log(x))
summary(fit)
plot(y ~ log(x))
abline(fit)
On Fri, Dec 4, 2009 at 9:06 AM, Pascale Weber wrote:
> Hi to all
>
> This is the first time I am quoting a question and I hope, my question is
> not too basic...
>
> For the
Hi to all
This is the first time I am quoting a question and I hope, my
question is not too basic...
For the following data, I wish to draw a fitted curve.
x <- c(123,129,141,144,144,145,149,150,158,159,163,174,183,187,242,248)
y <-
c(14.42,26.96,31.3,19.95,36.36,15.4,24.76,35.39,28.07,40.9
: r-help@r-project.org
Objet : Re: [R] curve from a formula with ggplot2
Hi Benoit,
Have a look at http://had.co.nz/ggplot2/stat_function.html. Does that help?
Hadley
On Tue, Jul 7, 2009 at 11:15 AM, Benoit
Boulinguiez wrote:
> Hi all,
>
> I'm smoothly transferring my lattice
Hi Benoit,
Have a look at http://had.co.nz/ggplot2/stat_function.html. Does that help?
Hadley
On Tue, Jul 7, 2009 at 11:15 AM, Benoit
Boulinguiez wrote:
> Hi all,
>
> I'm smoothly transferring my lattice graphs to ggplot2 graphs, but I'm stuck
> on representing a curve from a formula.
> I'm loo
Hi all,
I'm smoothly transferring my lattice graphs to ggplot2 graphs, but I'm stuck
on representing a curve from a formula.
I'm looking for the equivalent of curve() in ggplot2, Hadley Wickham
mentions geom_curve, but as far as I've seen in the help it doesn't exist.
My need is to plot a regul
Dear R Helper Community!
Thank you for all your help and suggestions.
For your reference and any future person searching the archives, here is the
solution that did what I wanted it to do.
As background, my goal was to find the coefficients for the following
equation form:
y ~ c1 * x1 *
or use nls.lm as in
install.packages("minpack.lm")
library(minpack.lm)
x <- c(2, 8, 14, 20, 26, 32, 38, 44, 50, 56, 62, 68, 74)
y <- c(100, 99, 99, 98, 97, 94, 82, 66, 48, 38, 22, 10, 1)
res <- function(p, x, y) y - ff(p,x)
ff <- function(p, x) 100*exp(p[1]*(1-exp(p[2]*x))/p[2])
aa <- nls.lm(par
Dear Dmitry,
Take a look at ?nls and its examples.
HTH,
Jorge
On Tue, May 12, 2009 at 5:44 PM, Dmitry Gospodaryov
wrote:
> I have the data:
> for x: 2, 8, 14, 20, 26, 32, 38, 44, 50, 56, 62, 68, 74,
> for y: 100, 99, 99, 98, 97, 94, 82, 66, 48, 38, 22, 10, 1.
> y depends on x by equation: y =
I have the data:
for x: 2, 8, 14, 20, 26, 32, 38, 44, 50, 56, 62, 68, 74,
for y: 100, 99, 99, 98, 97, 94, 82, 66, 48, 38, 22, 10, 1.
y depends on x by equation: y = 100*exp(b*(1-exp(c*x))/c),
where b and c are coefficients. I need to find coefficients in this
equation for given data. How can I do
Hi Thanks a lot,
I think you have covered the things I want to do for now so I will try to
implement them as soon I can.
<< A finite Fourier series could be the best tool IF the the multiple
periodicities are all integer fractions of a common scale.>>
This is certainly true for my repetitive
Dear Dr Gkikopoulos:
1. Have you looked at "bioconductor.org"? They have substantive
extensions to R specifically for "genomic data", which I assume would
include chromosome.
2. To "identify periodicities at different timescales", I agree
with Stephen that "spectrum" would l
There are a couple of different goals for this projects
*identify periodicities at different timescales (ie different dT)
*fit data into discrete number of curves, ie 6 different basic functions
should be enough to describe the basic repeating elements in this data (ie 6
different categories of
What is your end goal? If it is to try and account for the
variability of the "timeseries" you may want to look at ?spectrum
If it is to model the periodicity...
Stephen Sefick
On Fri, Apr 3, 2009 at 11:30 AM, trias wrote:
>
> Here is the gif that didn't come through earlier
> http://www.nabble
Here is the gif that didn't come through earlier
http://www.nabble.com/file/p22870832/signal.gif signal.gif
--
View this message in context:
http://www.nabble.com/Curve-fitting%2CFDA-for-biological-data-tp22868069p22870832.html
Sent from the R help mailing list archive at Nabble.com.
_
Dear all,
Another newbie just got attracted to this mailing list.
I am a biologist currently working my way through R, had sort play around with
python earlier this year.
I have some data exhibiting periodicity ** my data consists of peaks and
valleys, with peaks arising due to the presence
gregor rolshausen biologie.uni-freiburg.de> writes:
>
> ok. sorry for being blurry.
>
> I have x,y data, that probably fits a asymptotic curve (asymptote at N).
> now I want to fit a curve onto the data, that gives me the N. therefore
> I thought to fit an e-function, namely N(1-e^(y/x)) onto
ok. sorry for being blurry.
I have x,y data, that probably fits a asymptotic curve (asymptote at N).
now I want to fit a curve onto the data, that gives me the N. therefore
I thought to fit an e-function, namely N(1-e^(y/x)) onto the data and
get the N from the fitted curves' equation.
in the
gregor rolshausen wrote:
hello,
I want to fit a curve to a simple x,y dataset - my problem is, that I
want to fit it for the following term:
n(1-e^x/y) - so I get the n constant for my data...
Not an R problem in the first place, but the question arises what
"n(1-e^x/y)" means, its is jus
hello,
I want to fit a curve to a simple x,y dataset - my problem is, that I
want to fit it for the following term:
n(1-e^x/y) - so I get the n constant for my data...
can anyone help/comment on that?
cheers,
gregor
__
R-help@r-project.org mailing
Your model is singular. Varying m and log(l) have the same
effect: log(ir) = log(k) + m * log(l) * ox
Also with plinear you don't specify the linear coefficients but
rather an X matrix whose coefficients represent them:
If we use this model instead:
ir = k * exp(m * ox)
Then:
> mod0 <- lm(log(
I'm trying to fit a function y=k*l^(m*x) to some data points, with reasonable
starting value estimates (I think). I keep getting "singular matrix 'a' in
solve".
This is the code:
ox <- c(-600,-300,-200,1,100,200)
ir <- c(1,2.5,4,9,14,20)
model <- nls(ir ~ k*l^(m*ox),start=list(k=10,l=3,m=0.004
85 matches
Mail list logo
