Hi fellow R users,
I'm trying to fit a model using nls with the following model definition:
y(t+1)=(th1*x1 + R1*x2) * exp(a1*x3) + (1-th1*x1 + R1*x2)*y(t)
y is the dependent variable (note on both sides of eq) and the x's represent
the regressors.
th1, R1 and a1 are parameters to be estimated.
smooth.spline is a handy function and does what you need:
Careful to not over fit though. Use different values of spar and df.
y<-c(467, 468, 460 ,460 ,450, 432, 419, 420, 423, 423)
x<-1:10
plot(x,y)
predict(smooth.spline(x,y,df=4),x=3.5)
lines(predict(smooth.spline(x,y,df=4),x=seq(1,10,length=1
Lookup:
?diag
?upper.tri
?lower.tri
Example:
my.mat<-matrix(1:16,ncol=4)
my.mat
my.mat[upper.tri(my.mat)]<-3
my.mat
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of kevinchang
Sent: 01 October 2007 17:37
To: r-help@r-project.org
Subject: [R] function for
When you call xyplot in a for loop you have to use the print command. For
instance modifying the xyplot example:
The following wont work:
Depth <- equal.count(quakes$depth, number=8, overlap=.1)
for(i in 1:3){xyplot(lat ~ long | Depth, data = quakes)}
But this will:
for(i in 1:3){print(x
Following on from your example:
write.table(M,file="Myfile.csv",sep=",",row.names=F)
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of Yves Moisan
Sent: 01 October 2007 14:45
To: r-help@r-project.org
Subject: Re: [R] export csv
Hello,
I wanna know how
x <- c("This ", "is ", "one ", "sentence.")
paste(x,collapse="")
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of Rainer M. Krug
Sent: 01 October 2007 13:23
To: r-help
Subject: [R] Concatenating one character vector into one string
Hi
I am sure this is si
Hi there, you could try:
library(scatterplot3d)
THese function are also quite handu for visualising 3d images in 2d by virtue
of contours, heat maps etc..
?image
?persp
?contour
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of John Sorkin
Sent: 01 October
Why dont you use the t.test within R? See help(t.test). It looks to have
everything you need:
here are the examples with different alternative hypothesese:
with(sleep, t.test(extra[group == 1], extra[group == 2],alternative =
"greater"))
with(sleep, t.test(extra[group == 1], extra[group == 2],
tapply is also very useful:
my.df<-data.frame(x=rnorm(20, 50, 10),group=factor(sort(rep(c("A", "B"), 10
tapply(my.df$x,my.df$group,function(x){(x-mean(x))/sd(x)})
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of Matthew Dubins
Sent: 26 September 2007
Not sure which of the questions yo want answered in your email.
However, if its the one regarding the boxplot try:
dd <- read.table("test.txt",header=T)
attach(dd)
boxplot(x)
outlier <- function(y){
out <- boxplot(y, range = 1)$out
outliers <- which(y %in% out)
return(list(out=out,outliers=
Hi there,
If the final predicted clusters vary according to a random starting cluster
then I suspect that your data is not clustering very well!!
A few reasons for this may be:
1) There are genuinely no clusters in the data!
2) You have chosen a poor distance measure.
3) You have picked an i
Hi there,
The tune function is extremely useful for quickly cross validating a model.
However, you often need to modify some of the predict functions.
Here is an example of tuning an svm and naiveBayes with the iris data set:
naiveBayes:
For some reason the naivebayes predict function
(g
Use table:
dat <- data.frame(c(60001,60001,60050,60050,60050),c(27,129,618,27,1579))
table(dat[,1],dat[,2])
27 129 618 1579
60001 1 1 00
60050 1 0 11
Regards
Wayne
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of F
look at the as.Date function
e.g.
as.Date("2007-04-21")- as.Date("2000-04-21")
also look at the chron library.
Regards
Wayne
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of Daniel Brewer
Sent: 24 September 2007 11:23
To: [EMAIL PROTECTED]
Subject: [R
The following link is an excellent reference which gives R code for making
different plots in R.
http://addictedtor.free.fr/graphiques/thumbs.php
Is it something like this you want:
http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=52
The code can be found to do this on the webp
I would suggest doing an F-test.A descrition is given here:
http://www.graphpad.com/curvefit/2_models__1_dataset.htm.
The method is valid becasue one of your models is a subset of another.
Correct use of the anova function does indeed perform this test.
For example:
data(airquality)
lm1<-l
Not sure what you mean by "group index" but try:
lapply(df.s,function(l){l$x})
or something like:
do.call("rbind",df.s)
to convert the result into a data.frame.
Regards
Wayne
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of Rick DeShon
Sent: 14 Sept
Google search "Logistic Regression using R"
There are loads of good links here. Basically you use a generalized linear
model.
Look up ?glm
Regards
Wayne
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of martin pareja
Sent: 13 September 2007 16:33
To: r-he
Testing if I can mail to the list! None of my emails seem to be getting through.
Apologies.
Wayne Jones
Statistical Consultant
Shell Global Solutions (UK)
Shell Technology Centre Thornton, P.O. Box 1, Cheste
Hi there,
Rather than cross validating or bootstrapping to prune a single tree you could
use random forest instead. Look at the overview in
http://www.stat.berkeley.edu/~breiman/RandomForests/cc_home.htm
THere is a package in R for doing this called library(randomForest). I have
found it to
?polygon
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of gallon li
Sent: 13 September 2007 11:52
To: r-help
Subject: [R] how to plot shaded area under a curve?
say, I am plotting
x=seq(0,5,len=100)
y=-(x-5)^2
plot(x,y)
how can I put some color or ver
?polygon
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of gallon li
Sent: 13 September 2007 11:52
To: r-help
Subject: [R] how to plot shaded area under a curve?
say, I am plotting
x=seq(0,5,len=100)
y=-(x-5)^2
plot(x,y)
how can I put some color or verti
?tapply
temp.data<-data[as.character(data$time %in% c("02:00","03:00","04:00"),]
# one quick and dirty method is to create a subset the data , not sure what
format your time data is so converted to character strings.
then something like:
tapply(temp.data$V1, factor(temp.data$time), mean)
W
Hi John,
It sounds like what you need is a mixed effects model. This will allow you to
model effects which apply to the whole population (fixed effects) and those
which are specific to an individual (random effects).The theory and practice of
mixed effects models is too large to discuss fully
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