Displaying 20 results from an estimated 10000 matches similar to: "fitting models with the subset argument"
2006 Feb 12
1
lme, nlsList, nlsList.selfStart
Dear listers,
I am trying to fit a model using nlsList() using alternately a SSfol()
selfstart function or its developped equivalent formulae.
This preliminary trial works well
mydata<-groupedData(Conc~Tps|Organ,data=mydata)
mymod1<-nls(Conc~SSfol(Dose,Tps,lKe,lKa,lCl),data=mydata)
as well as a developped form:
mymod2<-nls(Conc~Dose * exp(lKe+lKa-lCl) *
2024 Jul 26
1
Automatic Knot selection in Piecewise linear splines
dear all,
I apologize for my delay in replying you. Here my contribution, maybe
just for completeness:
Similar to "earth", "segmented" also fits piecewise linear relationships
with the number of breakpoints being selected by the AIC or BIC
(recommended).
#code (example and code from Martin Maechler previous email)
library(segmented)
o<-selgmented(y, ~x, Kmax=20,
2007 Feb 13
1
Missing variable in new dataframe for prediction
Hi,
I'm using a loop to evaluate several models by taking adjacent variables from my dataframe.
When i try to get predictions for new values, i get an error message about a missing variable in my new dataframe.
Below is an example adapted from ?gam in mgcv package
library(mgcv)
set.seed(0)
n<-400
sig<-2
x0 <- runif(n, 0, 1)
x1 <- runif(n, 0, 1)
x2 <- runif(n, 0, 1)
x3 <-
2024 Jul 16
2
Automatic Knot selection in Piecewise linear splines
>>>>> Anupam Tyagi
>>>>> on Tue, 9 Jul 2024 16:16:43 +0530 writes:
> How can I do automatic knot selection while fitting piecewise linear
> splines to two variables x and y? Which package to use to do it simply? I
> also want to visualize the splines (and the scatter plot) with a graph.
> Anupam
NB: linear splines, i.e. piecewise
2002 Aug 30
4
Intercept in model formulae.
Hi,
I'm trying to create a linear model for a dataset that has a breakpoint e.g.
# dummy dataset
x <- 1:20
y <- c(1:10,seq(10.5,15,0.5))
plot(x,y)
I've modelled this using the following formula:
temp <- lm(y ~ x*(x<=10)+x*(x>10))
I want to be able to omit the intercept (i.e. force the line through
zero) from the first of these segments (x<=10) so that I'm only
2003 Dec 14
3
Problem with data conversion
Hi All:
I came across the following problem while working with a dataset, and
wondered if there could be a solution I sought here.
My dataset consists of information on 402 individuals with the followng five
variables (age,sex, status = a binary variable with levels "case" or
"control", mma, dma).
During data check, I found that in the raw data, the data entry
2009 Sep 04
1
predicting from segmented regression
Hello
I'm having trouble figuring out how to use the output of "segmented()"
with a new set of predictor values.
Using the example of the help file:
??set.seed(12)
xx<-1:100
zz<-runif(100)
yy<-2+1.5*pmax(xx-35,0)-1.5*pmax(xx-70,0)+15*pmax(zz-.5,0)+rnorm(100,0,2)
dati<-data.frame(x=xx,y=yy,z=zz)
out.lm<-lm(y~x,data=dati)
o<-## S3
2003 Jul 10
1
The question is on Symmetry model for square table.
Please help,
I tried a program on S-plus, and it worked. Also I tried the same
program on R but not worked. Here is the programme. I put it in a
function form. The model and assumption are at the bottom.
where
counts<-c(22,2,2,0,5,7,14,0,0,2,36,0,0,1,17,10)
which is name.data, i is row size and j is the column size.
symmetry
function(i, j, name.data)
{
row <- (c(1:i))
col <-
2000 Mar 08
1
Coercing character to factor
I just downloaded version 1.0.0 and several binary libraries (VR, rpart,
norm, stataread) - WinNT version. I then converted a file from Stata 6.0
to R format by using the stataread library. The file converts perfectly
and I was able to use the VR function lda on the dataframe without
difficulty. I then tried to use the same dataframe with RPART. The model
statement:
2003 Apr 17
4
A function as argument of another function
Dear all,
I would like to write a function like:
myfun<-function(x,fn) {xx<-exp(x); x*fn(xx)}
where fn is a symbolic description of any function with its argument to be
specified. Therefore
myfun(5,"2+0.3*y^2")
should return 5*(2+0.3*exp(5)^2),
myfun(5,"log(y)") should return 5*log(exp(5)) and so on.
I tried with "expression" and others, but without success.
2008 Sep 04
1
pass data to log-likelihood function
Hi there,
When I do bootstrap on a maximum likelihood estimation, I try the
following code, however, I get error:
Error in minuslogl(alpha = 0, beta = 0) : object "x" not found
It seems that mle() only get data from workspace, other than the
boot.fun().
My question is how to pass the data to mle() in my case.
I really appreciated to any suggestions.
Best wishes,
Jinsong
2003 Dec 16
3
`bivariate apply'
dear all,
Given a matrix A, say, I would like to apply a bivariate function to each
combination of its colums. That is if
myfun<-function(x,y)cor(x,y) #computes simple correlation of two vectors x
and y
then the results should be something similar to cor(A).
I tried with mapply, outer,...but without success
Can anybody help me?
many thanks in advance,
vito
2002 Jan 04
1
taking variables from data.frame
Dear all,
Two of the arguments of my function fn(), say, are a glm-obj and a variable
in the data.frame where there are the variable of the obj too.
Is there a way to build a function that takes the variables form the same
data.frame? For instance, the update() function does so:
>obj<-glm(....,data=mydata)
>update(obj,.~.+x) #works becuse it searches x in mydata
But for my function:
2004 Oct 26
2
vcov method for 'coxph' objects
Dear all,
The help file for the generic function vcov states
"Classes with methods for this function include: 'lm', 'glm', 'nls', 'lme',
'gls', 'coxph' and 'survreg' (the last two in package 'survival')."
Since, I am not able to use vcov.coxph(), I am wondering whether I am
missing something (as I suspect..)
regards,
vito
2004 Oct 11
3
logistic regression
Hello,
I have a problem concerning logistic regressions. When I add a quadratic
term to my linear model, I cannot draw the line through my scatterplot
anymore, which is no problem without the quadratic term.
In this example my binary response variable is "incidence", the explanatory
variable is "sun":
> model0<-glm(incidence~1,binomial)
>
2008 May 16
1
xyplot: subscripts, groups and subset
I have stumbled across something in the Lattice package that is vexing me.
Consider the code below:
__________________________________________________________
library(lattice)
myData <- expand.grid(sub = factor(1:16), time = 1:10)
myData$observed <- rnorm(nrow(myData))
myData$fitted <- with(myData, ave(observed, sub, FUN = mean))
myData$event.time <- with(myData, ave(observed, sub,
2003 Mar 12
1
simulating 'non-standard' survival data
Dear all,
I'm looking for someone that help me to write an R function to simulate
survival data under complex situations, namely time-varying hazard ratio,
marginal distribution of survival times and covariates. The algorithm is
described in the reference below and it should be not very difficult to
implement it. However I tried but without success....;-(
Below there the code that I used; it
2002 Sep 18
3
problem in deparse(substitute())
Hi all,
I am experiencing the following quite strange (at least in my knowledge)
problem in a simple function like the following:
fn<-function(y,v=2){
n<-length(y)
y<-y[(v+1):(n-v)]
plot(y,type="l",lty=3,xlab="Time",ylab=deparse(substitute(y)))
}
fn(rnorm(50)) #look at the plot!!!
The plot appears with numbers on the left side!
If I delete the
2006 Sep 13
3
unexpected result in glm (family=poisson) for data with an only zero response in one factor
Dear members,
here is my trouble: My data consists of counts of trapped insects in different attractive traps. I usually use GLMs with a poisson error distribution to find out the differences between my traitments (and to look at other factor effects). But for some dataset where one traitment contains only zeros, GLM with poisson family fail to find any difference between this particular traitment
2007 Aug 26
3
subset using noncontiguous variables by name (not index)
Hi All,
I'm using the subset function to select a list of variables, some of
which are contiguous in the data frame, and others of which are not. It
works fine when I use the form:
subset(mydata,select=c(x1,x3:x5,x7) )
In reality, my list is far more complex. So I would like to store it in
a variable to substitute in for c(x1,x3:x5,x7) but cannot get it to
work. That use of the c function