Muchas gracias!! No me hab�a dado cuenta del guion bajo.
Saludos
Manuel Morales Ortiz
Profesor Titular Universidad
�rea de Metodolog�a de las Ciencias del Comportamiento
Tel�fono: +34 954 55 76 83
Web: http://personal.us.es/morales/
Solicitud de tutorias: https://calendly.com
Hola, estoy utilizando beamer dentro de TMarkdown y no consigo cambiar el
rótulo de 'Table' por el de 'Tabla'. En RMarkdown si lo consigo con la opción
'\renewcommand{\tablename}{Tabla}', pero si el documento es en beamer no lo
consigo. Tampoco consigo que en beamer me permita utilizar la
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
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,
the
?bargraph.ci help I would greatly appreciate an indication of where it
is.
Thank you kindly,
Michael Just
On Tue, Nov 3, 2009 at 8:53 AM, Manuel Morales
manuel.a.mora...@williams.edu wrote:
On Tue, 2009-11-03 at 03:51 -0600, Michael Just wrote:
Hello,
When using bargraph.CI in package sciplot
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.,
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
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 -
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 AIC to the
models that do converge. A contrived example of what
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 analysis where some models occasionally fail to
converge. I'd like to automate the process
for curiosity, any particular reason you chose standard error
, and not confidence interval as the default (the naming of the
plotting functions associates closer to the confidence interval
) error indication .
- Jarle Bjørgeengen
On May 24, 2009, at 3:02 , Manuel Morales wrote:
You
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))
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,
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(123)
x - runif(n)
a - gl(n=2, k=n/2
On Wed, 2009-04-15 at 12:02 -0400, Michael A. Miller wrote:
Manuel == Manuel Morales manuel.a.mora...@williams.edu 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))
Mike
Argh!!!
--
http
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 combination of the factor levels.
Unfortunately
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 error
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:
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 =
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 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
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 Tue, 2008-10-07 at 23:31 -0500, Michael Just
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
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 the
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
.
0.01
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