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
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] *
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
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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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 diviya.sm...@gmail.comwrote:
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
On Thu, Jul 26, 2012 at 5:16 PM, Diviya Smith diviya.sm...@gmail.comwrote:
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 estimates a = 0
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
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
beta2
, Diviya Smith diviya.sm...@gmail.com
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 lets say -
y1 = alpha1*A + beta1*B
y2 = alpha2*A + beta2*B
y1 - runif(10, 0,1)
y2 - runif
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)) #
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
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
)
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, Diviya Smith diviya.sm...@gmail.com
wrote:
Hello there,
I have a matrix with some data and I want to split this matrix based
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, Tues,
(rownames(mdat)), function(n)
colSums(mdat[n,,drop=F])), col = rainbow(2))
Michael
On Thu, Nov 10, 2011 at 9:21 PM, Diviya Smith diviya.sm...@gmail.com
wrote:
Hello there,
I have a question regarding bar plots. I am trying to plot the data from
the following matrix as a barplot
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
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
(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 jvad...@usgs.gov wrote:
Diviya Smith wrote on 09/20/2011 01:03:22 PM:
Hello there,
I am using NLS
suggestions?
On Tue, Sep 20, 2011 at 5:59 PM, Diviya Smith diviya.sm...@gmail.comwrote:
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 generate
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
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.
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 -
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 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),
(fit1, fit2)
HTH,
Dennis
On Sun, Jun 12, 2011 at 9:57 AM, Diviya Smith diviya.sm...@gmail.com
wrote:
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
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