Hello there,
Can anyone point me to the code for logLik of an nls object? I found the
code for logLik of an lm but could not find exactly what function is used
for calculating the logLik of nls function?
I am using the nls to fit the following model to data -
Model 1: y ~ Ae^(-mx) + Be^(-nx) +c
Hi there,
I was wondering if there is any R package that one can use for plotting
that has more legend symbols - the standard pch has 18 symbols but I need
~30 for my application- and just using different colors is not an option.
Thank you in advance,
Diviya
[[alternative HTML version de
Hi there,
I am trying to fit the following model with a sum of exponentials -
y ~ Ae^(-md) + B e^(-nd) + c
the model has 5 parameters A, b, m, n, c
I am using nls to fit the data and I am using DEoptim package to pick the
most optimal start values -
fm4 <- function(x) x[1] + x[2]*exp(x[3] * -d
Hi there
I have an Rscript and I am looking for a way to install a package
non-interactively. In Rscript {Utils}, I saw an example which does
something like this, however this does not seem to work for my particular
example. I am trying to install the following package in an Rscript
(without switc
On Thu, Jul 26, 2012 at 5:16 PM, Diviya Smith wrote:
> Thank you for pointing me to the uniroot function?
>
> Is there a way to constrain this solution so that it only gives me values
> of 'a' between c(0,1)?
>
> I tried using nlminb and for some reason it always es
Sorry it is important for me to constrain the value of 'a' between c(0,1)
On Thu, Jul 26, 2012 at 4:48 PM, Diviya Smith wrote:
> Hi there,
>
> I would like to solve a simple equation in R
>
> a^2 - a = 8.313
>
> There is no real solution to this problem but I wou
Hi there,
I would like to solve a simple equation in R
a^2 - a = 8.313
There is no real solution to this problem but I would like to get an
approximate numerical solution. Can someone suggest how I can set this up?
Thanks in advance,
Diviya
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___
Hi there,
I would like to solve the following equation in R to estimate 'a'. I have
the amp, d, x and y.
amp*y^2 = 2*a*(1-a)*(-a*d+(1-a)*x)^2
test data:
amp = 0.2370 y=
0.0233 d=
0.002 x=
0.091
Can anyone suggest how I can set this up?
Thanks,
Diviya
[[alternative HTML version dele
n you can chew.
>
> -- Bert
>
> On Mon, Mar 19, 2012 at 11:29 AM, Diviya Smith
> wrote:
> > Hello there,
> >
> > I am new to using regression in R. I wanted to solve a simple regression
> > problem where I have 2 equations and 2 unknowns.
> >
> > So
Hello there,
I am new to using regression in R. I wanted to solve a simple regression
problem where I have 2 equations and 2 unknowns.
So lets say -
y1 = alpha1*A + beta1*B
y2 = alpha2*A + beta2*B
y1 <- runif(10, 0,1)
y2 <- runif(10,0,1)
alpha1 <- 0.6
alpha2 <- 0.75
beta1 <- 1-alpha1
b
Hi there,
I am trying to compute the autocorrelation in a dataset using R's acf
function. ACF automatically plots the results. This works well except in
some cases xlim doesnt work
data <- rnorm(2000,0,1)
acf(data,xlim=c(1,10)) # works - the plot starts at 1
acf(data,lag=100,xlim=c(1,100)) # t
Hello there,
I recently wrote some code to perform pairwise correlations between all
samples in a large dataset. So we are talking about performing pairwise
correlations between 400K vectors. Since R has a very rich library of
functions, it was very easy to code this in R. However, R was probably
wrote:
> > I'd so something like
> >
> > split(a, a$spending)
> >
> > and you can include a round(a$spending, -2) or something similar if
> > you want to group by the 100's.
> >
> > Michael
> >
> > On Mon, Dec 5, 2011 at 3:37 PM
Hello there,
I have a matrix with some data and I want to split this matrix based on the
values in one column. Is there a quick way of doing this? I have looked at
cut but I am not sure how to exactly use it?
for example:
I would like to split the matrix "a" based on the spending such that the
da
oes
> what you are talking about:
>
> barplot(sapply(unique(rownames(mdat)), function(n)
> colSums(mdat[n,,drop=F])), col = rainbow(2))
>
> Michael
>
> On Thu, Nov 10, 2011 at 9:21 PM, Diviya Smith
> wrote:
> > Hello there,
> >
> > I have a question re
Hello there,
I have a question regarding bar plots. I am trying to plot the data from
the following matrix as a barplot -
# input data
mdat <- matrix(c(0.1,0.9,0.9,0.1,0.5,0.5,0.45,1-0.45,0.6,0.4,0.8,0.2), nrow
= 6, ncol=2, byrow=TRUE,
+dimnames = list(c("Mon", "Mon", "Tues", "Tue
Hi there,
I am having some trouble with NLS convergence for my function. I was
wondering if there is a way to check the value of the indicator function for
all the iterations to look at the surface of the likelihood function.
Any help would be most appreciated.
Thanks,
Diviya
[[alternat
. Any
suggestions?
On Tue, Sep 20, 2011 at 5:59 PM, Diviya Smith wrote:
> I dont think *r* is related to the problem. I am not trying to estimate *r
> * and so basically I am giving the model the correct value of *r* and so
> log(1-r) should not go to infinity.
>
> For test data, I ge
following error -
Error in numericDeriv(form[[3L]], names(ind), env) :
Missing value or an infinity produced when evaluating the model
Any suggestions how to fix this?
Diviya
On Tue, Sep 20, 2011 at 2:37 PM, Jean V Adams wrote:
>
> Diviya Smith wrote on 09/20/2011 01:03:22 PM:
>
> >
&
Hello there,
I am using NLS for fitting a complex model to some data to estimate a couple
of the missing parameters. The model is -
y ~ (C+((log(1-r))*exp(-A*d)*(-1+r+exp(d*(A-B)))/(r*(-A*d+d*B+log(1-r)
where A, B and C are unknown.
In order to test the model, I generate data by setting value
Hi there,
I have a complex math equation which does not have a closed form solution.
It is -
y <- (p*exp(-a*d)*(1-exp((d-p)*(a-x[1]/((p-d)*(1-exp(-p*(a-x[1]
For this equation, I have all the values except for x[1]. So I need to solve
this problem numerically. Can anyone suggest an optimi
Hi there,
I am having a strange problem. I am running nls on some data.
#data
x <- -(1:100)/10
y <- 100 + 10 * (exp(-x / 2)
Using nls I fit an exponential model to this data and get a great fit
summary(fit)
Formula: wcorr ~ (Y0 + a * exp(m1 * -dist/100))
Parameters:
Estimate Std. E
1, 0.1, step = 0.01, initial = 0.04),
> b10=slider(0.01, 0.4, step = 0.01, initial = 0.3),
> b20 = slider(0.01, 0.1, step = 0.005, initial = 0.05),
> b00=slider(-0.01, -0.001, step=0.001,initial= -0.004))
>
> fit2 <- nls(y ~ k*(exp(-b1*x) + exp(-b2*x))
Hello there,
I am trying to use R function NLS to analyze my data and one of the examples
in the documentation is -
## the nls() internal cheap guess for starting values can be sufficient:
x <- -(1:100)/10
y <- 100 + 10 * exp(x / 2) + rnorm(x)/10
nlmod <- nls(y ~ Const + A * exp(B * x), trace=T
Hello there,
I want to perform a likelihood ratio test to check if a single exponential
or a sum of 2 exponentials provides the best fit to my data. I am new to R
programming and I am not sure if there is a direct function for doing this
and whats the best way to go about it?
#data
x <- c(1 ,10,
Hello there,
I am trying to fit an exponential fit using Least squares to some data.
#data
x <- c(1 ,10, 20, 30, 40, 50, 60, 70, 80, 90, 100)
y <- c(0.033823, 0.014779, 0.004698, 0.001584, -0.002017, -0.003436,
-0.06, -0.004626, -0.004626, -0.004626, -0.004626)
sub <- data.frame(x
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