For now, just change fun(x) to median(x) (or whatever) in your ci.fun()
below.
E.g.
lineplot.CI(x.factor = dose, response = len, data = ToothGrowth, ci.fun=
function(x) c(mean(x)-2*se(x), mean(x)+2*se(x)))
Otherwise, maybe the list members could help with a solution. An example
that illustrates
ror bars reflect this using the
> ci.fun argument. Can I use this argument or is some pre-processing
> necessary? If this information is located somewhere besides the
> ?bargraph.ci help I would greatly appreciate an indication of where it
> is.
>
> Thank you kindly,
> Mi
On Tue, 2009-11-03 at 03:51 -0600, Michael Just wrote:
> Hello,
> When using bargraph.CI in package sciplot can the bars for each group
> be different colors? How do I select the color for each group?
With the option err.col
bargraph.CI(dose, len, group = supp, data =ToothGrowth,
err.col=c("gray
Hi Tim,
I don't believe there is a satisfactory solution in R - at least yet -
for non-normal models. Ultimately, this should be possible using lmer()
but not in the near-term. One possibility is to use glmPQL as described
in:
Dormann, F. C., McPherson, J. M., Araújo, M. B., Bivand, R., Bolliger,
Hello list,
I'm trying to fit a model like beta[trt]/(1+alpha*x) where the data
include some grouping factor. The problem is that the estimate for alpha
is undefined for some of the treatments - any value greater than 20 is
equally good and a step function would suffice. Ignoring the grouping
stru
On Thu, 2009-06-11 at 16:10 -0400, Ben Bolker wrote:
> Manuel Morales wrote:
> > On Thu, 2009-06-11 at 12:08 -0700, Ben Bolker wrote:
> >>
> >> Manuel Morales wrote:
> >>> Hello list,
> >>>
> >>> I'm doing a bootstrap analysi
On Thu, 2009-06-11 at 12:08 -0700, Ben Bolker wrote:
>
>
> Manuel Morales wrote:
> >
> > Hello list,
> >
> > I'm doing a bootstrap analysis where some models occasionally fail to
> > converge. I'd like to automate the process of restricting AI
Hello list,
I'm doing a bootstrap analysis where some models occasionally fail to
converge. I'd like to automate the process of restricting AIC to the
models that do converge. A contrived example of what I'd like to do is
below:
resp <- c(1,1,2)
pred <- c(1,2,3)
m1 <- lm(resp~pred)
m2 <- lm(resp
>>>>> plotting functions associates closer to the confidence interval
> >>>>> ) error indication .
> >>>>> - Jarle Bjørgeengen
> >>>>> On May 24, 2009, at 3:02 , Manuel Morales wrote:
> >>>>>> You defin
You define your own function for the confidence intervals. The function
needs to return the two values representing the upper and lower CI
values. So:
qt.fun <- function(x) qt(p=.975,df=length(x)-1)*sd(x)/sqrt(length(x))
my.ci <- function(x) c(mean(x)-qt.fun(x), mean(x)+qt.fun(x))
lineplot.CI(x.f
On Wed, 2009-04-15 at 12:02 -0400, Michael A. Miller wrote:
> >>>>> "Manuel" == Manuel Morales writes:
>
> > nls(y ~ a[fac]*x^b, start=list(a=c(1,1), b=0.25))
>
> Did you mean a[f]?
>
> nls(y ~ a[f]*x^b, start=list(a=c(1,1), b=0.25))
>
Oops! I made a mistake. Corrected below.
On Wed, 2009-04-15 at 11:05 -0400, Manuel Morales wrote:
> A more compact way to code factors in nls is to use the syntax factor[].
>
> Here's an example using a simplified version of Ravi's example:
>
> n <- 200
> set.seed
A more compact way to code factors in nls is to use the syntax factor[].
Here's an example using a simplified version of Ravi's example:
n <- 200
set.seed(123)
x <- runif(n)
a <- gl(n=2, k=n/2) # a two-level factor
eps <- rnorm(n, sd=0.5)
y <- as.numeric(a) * x^.5 + eps
nls(y ~ a[]*x^b, start=l
On Sat, 2009-04-11 at 08:10 -0400, Manuel Morales wrote:
> On Fri, 2009-04-10 at 15:07 -0700, Metconnection wrote:
> > Hi there,
> > I wonder if anyone can help me. I'm trying to use bargraph.CI in the Sciplot
> > package when there is a missing combin
On Fri, 2009-04-10 at 15:07 -0700, Metconnection wrote:
> Hi there,
> I wonder if anyone can help me. I'm trying to use bargraph.CI in the Sciplot
> package when there is a missing combination of the factor levels.
> Unfortunately the standard errors on the plot do not appear to be correct.
>
>
On Mon, 2009-03-02 at 11:23 -0600, maxa0...@umn.edu wrote:
> I'm trying to create a bargraph of means with standard error bars using the
> function bargraph.CI (in the sciplot package). Like this:
>
> bargraph.CI(x.factor, response,data,xlab, ylab, par(family="serif"),font=11)
>
> However, an er
Hi list,
I'd like to install an archived version of lmer to compare results with
the current version. I guess one way to do this would be to download the
source, rename the package and then install it. Is there a better
alternative?
--
http://mutualism.williams.edu
signature.asc
Description: T
Hi all,
I'm trying to fit a model using the shorthand coeff[factor] instead of
coding dummy variables. Is there a way to keep this notation when
specifying constraints? See example below:
x = runif(200)
b0 = c(rep(0,100),runif(100))
b1 = 1
fac <- as.factor(rep(c(0,1), each=100))
y = b0+b1*x+rnorm
On Fri, 2008-10-10 at 18:15 -0300, Caio Azevedo wrote:
> Hi all,
>
> I am using the function "plotCI" with the following command:
>
> plotCI(m.residuos.p.2 [1:41],li=m.residuos.p.3 [1:41],ui=m.residuos.p.4
> [1:41],lty=1,ylab="")
>
> This generates exactly what I want except for the fact that I
On Wed, 2008-10-08 at 09:49 -0700, Dylan Beaudette wrote:
> On Wednesday 08 October 2008, Manuel Morales wrote:
> > Another option is bargraph.CI or lineplot.CI from the package sciplot.
> >
> > See http://mutualism.williams.edu/sciplot for examples.
> >
> >
On Wed, 2008-10-08 at 12:01 -0500, Michael Just wrote:
> Thank you all for you suggestions. They are all helpful. However, I have
> come to a more fundamental problem. Preparing my data to even make such a
> graph. I thought I was ready. I will obviously need to find the n, mean,
> and confidence
Another option is bargraph.CI or lineplot.CI from the package sciplot.
See http://mutualism.williams.edu/sciplot for examples.
On Tue, 2008-10-07 at 23:31 -0500, Michael Just wrote:
> Hello,
> I'd appreciate a suggestion on how to construct plots (barplots?) that use
> means on the Y axis instead
On Sat, 2008-03-08 at 08:07 -0600, Douglas Bates wrote:
> On Sat, Mar 8, 2008 at 2:57 AM, Alexandra Bremner
> <[EMAIL PROTECTED]> wrote:
> > I am attempting to model data with the following variables:
>
> > timepoint - n=48, monthly over 4 years
> > hospital - n=3
> > opsn1 - no of outcomes
Here's an updated version of lineplot.CI that will succeed even for
cases where data are not present in all factor combinations. Also, this
version has the option x.cont to specify that the x axis represents a
continuous variable with proportional spacing. A new version of sciplot
with these change
On Fri, 2008-02-15 at 15:18 +0100, Dieter Vanderelst wrote:
> Hi List,
>
> I have a problem plotting data using the lineplot.CI command in the sciplot
> package.
>
> I want to plot the data of 2 experimental cases using different lines
> (traces). Time is on the X-axis. The tricky thing is that
On Mon, 2008-02-11 at 09:31 -0800, questions? wrote:
> I have two distributions, represented by heights of several intervals.
> e.g. the distribution is partitioned into 10 segments, I have
> numbers(freq or counts) associated
> with each region in the format as:
>
> 0.2 0.3
> 0.1 0.1
> .
>
Hi Katherine,
If you have the original data, you can use the function barplot.CI from
the package sciplot which will calculate the means and CI's for you.
Examples are at: http://mutualism.williams.edu/sciplot
If you have the means and CI's already calculated, the following
function will do what
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