On Tue, Mar 17, 2009 at 12:41 PM, Greg Snow wrote:
> No, I meant the Combinations package, it is apparently an Omegahat package
> (http://www.omegahat.org/Combinations/). It looks similar to the permn
> function as far as the usage goes, but the documentation includes additional
> information
On Tue, Mar 17, 2009 at 5:24 AM, Gavin Simpson wrote:
> On Mon, 2009-03-16 at 18:43 -0400, Stavros Macrakis wrote:
>> ... no way to find relevant functions, and no way of knowing which one to
>> use if there is more than one.
> That is what the Task Views are meant to address, for discrete subje
Dear R-users,
I use glm() to do logistic regression and use stepAIC() to do stepwise model
selection.
The common AIC value comes out is about 100, a good fit is as low as around
70. But for some model, the AIC went to extreme values like 1000. When I
check the P-values, All the independent variab
Dear Andrew,
Try this:
data.frame(z=c(a$a,b$c))
HTH,
Jorge
On Tue, Mar 17, 2009 at 5:55 PM, Andrew McFadden <
andrew.mcfad...@maf.govt.nz> wrote:
> I all
>
> I am trying to combine columns from two dataframes to make a completely
> new dataframe consisting of one column of dates (ie a combinat
What about just:
data.frame(z=c(a$a, b$c))
?
--
David Winsemius
On Mar 17, 2009, at 10:44 PM, Steven McKinney wrote:
Not sure how well this gets at what you want,
but it seems close (if not kludgy!)
c <- data.frame(t(cbind(t(a), t(b))), stringsAsFactors = FALSE)[,
1, drop = FALSE]
c
Hi,
I thought I'd like to try out JGR, but after installing the package (and
dependencies) I receive the following when I try to load it:
> library(JGR)
Loading required package: rJava
Loading required package: JavaGD
Loading required package: iplots
Note: On Mac OS X we strongly recommend using
Hello,
In a human-subjects experiment collaborators (A & B) communicated
about how to categorize a number of multidimensional stimuli.
Following their conversation, each collaborator sorted the stimuli
into bins. This sort data was converted into square co-occurrence
matrices, where 1 indicated th
Not sure how well this gets at what you want,
but it seems close (if not kludgy!)
> c <- data.frame(t(cbind(t(a), t(b))), stringsAsFactors = FALSE)[, 1, drop =
> FALSE]
> c
a
1 2008-07-27
2 2008-10-01
3 2008-08-15
4 2008-08-14
5 2008-08-14
6 2008-09-20
7 2008-07-27
8 2008-10-01
> c$a <
Check help for tempfile() and tempdir()
You can set an environment variable
as discussed in the help
to point to D:\temp
>From ?tempfile:
By default, the filenames will be
in the directory given by tempdir().
This will be a subdirectory of the
temporary directory found by the
followin
?setwd()
I don't have a windows machine to work on, but this should work.
On Tue, Mar 17, 2009 at 10:14 PM, andrew wrote:
> setwd
>
> On Mar 18, 12:42 pm, Jonathan Greenberg wrote:
>> I'm trying to redirect where temporary files go under R to
>> D:\temp\somerandomname from its default (C:\Docume
setwd
On Mar 18, 12:42 pm, Jonathan Greenberg wrote:
> I'm trying to redirect where temporary files go under R to
> D:\temp\somerandomname from its default (C:\Documents and
> Settings\username\Temp\somerandomname) -- how do I go about doing this?
>
> --j
>
> --
>
> Jonathan A. Greenberg, PhD
> P
Paul,
I really appreciate all your help.
Here is what I ended up with:
http://n2.nabble.com/Can-R-produce-this-plot--td2489288.html#a2494992
This is really close. I might try to figure out a way to add the median to the
"boxplot", but this is really close and captures a ton of information on
I'm trying to redirect where temporary files go under R to
D:\temp\somerandomname from its default (C:\Documents and
Settings\username\Temp\somerandomname) -- how do I go about doing this?
--j
--
Jonathan A. Greenberg, PhD
Postdoctoral Scholar
Center for Spatial Technologies and Remote Sensin
Dear Thomas and John,
Thanks for your prompt reply and for looking at your code to explain
these differences. I see you replied very late at night, so I am sorry
if my little problem kept you awake!
As you pointed out, the model indeed converges (as reported in
model$converged) once I specif
Dear Thomas and John,
Thanks for your prompt reply and for looking at your code to explain
these differences. I see you replied very late at night, so I am sorry
if my little problem kept you awake!
As you pointed out, the model indeed converges (as reported in
model$converged) once I specif
readLines is doing exactly what you are asking:
Value
A character vector of length the number of lines read.
You still have to convert the character strings to numeric. Exactly
how large is "quite large"? What system are you running on? How much
memory do you have? What is the error message t
try something like this:
x <- rnorm(1)
# op <- par(mar=c(2,2,2,2), oma=c(0,0,0,0)+.8)
op <- par(mar=c(2, 2, 2, 2), oma=c(0,0,0,0)+.8)
layout(matrix(c(
4, 4, 4, 4, 4, 4,
2, 2, 2, 3, 3, 3,
1, 1, 1, 3, 3, 3,
1, 1, 1, 3, 3, 3,
> temp
type1 type2 a b c
1 male low 1 5 0
2 female med 2 NA 0
3 male high 4 5 1
4 female low NA 1 1
5 female med 3 2 NA
> temp[is.na(temp)] <- 0 ## the magic words...
> temp
type1 type2 a b c
1 male low 1 5 0
2 female med 2 0 0
3 male high 4 5 1
4 femal
Sorry, the data is:
temp <- data.frame(type1 = c("male", "female", "male", "female", "female"),
type2 = c("low", "med", "high", "low", "med"), a = c(1,2,4, NA, 3), b =
c(5,NA,5,1,2), c = c(0,0,1,1,NA))
On Tue, Mar 17, 2009 at 4:28 PM, Ferry wrote:
> Hi,
>
> I have newbie question. Suppose I hav
Erin Hodgess wrote:
Dear R People:
Here is a small data frame and two particular formulas:
test.df
y x
1 -0.9261650 1
2 1.5702700 2
3 0.1673920 3
4 0.7893085 4
5 0.3576875 5
6 -1.4620915 6
7 -0.5506215 7
8 -0.3480292 8
9 -1.2344036 9
10 0.8502660 10
summary
Hi,
I have newbie question. Suppose I have the following data:
temp <- data.frame(type1 = c("male", "female", "male", "female", "female"),
type2 = c("low", "med", "high", "low", "med"), a = c(1,2,4, NA, 3), b =
[TRUNCATED]
temp
type1 type2 a b c
1 male low 1 5 0
2 female med
Dear all,
I want to get the likelihood (or AIC or BIC) of a ridge regression model
using lm.ridge from the MASS library. Yet, I can't really find it. As
lm.ridge does not return a standard fit object, it doesn't work with
functions like e.g. BIC (nlme package). Is there a way around it? I would
ca
On 17-Mar-09 23:04:25, Erin Hodgess wrote:
> Dear R People:
> Here is a small data frame and two particular formulas:
>> test.df
> y x
> 1 -0.9261650 1
> 2 1.5702700 2
> 3 0.1673920 3
> 4 0.7893085 4
> 5 0.3576875 5
> 6 -1.4620915 6
> 7 -0.5506215 7
> 8 -0.3480292 8
Dear Neotropical Bat Risk Assessments,
Rounding is the problem. If you just want to lop off the seconds that is easy.
For example:
> d <- strptime(date(), "%a %b %d %H:%M:%S %Y")
> d
[1] "2009-03-18 08:52:41"
> format(d, "%Y-%m-%d %H:%M")
[1] "2009-03-18 08:52"
Here is a one way to do the rou
Dear R People:
Here is a small data frame and two particular formulas:
> test.df
y x
1 -0.9261650 1
2 1.5702700 2
3 0.1673920 3
4 0.7893085 4
5 0.3576875 5
6 -1.4620915 6
7 -0.5506215 7
8 -0.3480292 8
9 -1.2344036 9
10 0.8502660 10
> summary(lm(exp(y)~x))
Call:
On 18/03/2009, at 11:34 AM, Neotropical bat risk assessments wrote:
Hi all,
I need to compare between times and put all similar times in specific
1 minute bins.
Unfortunately the original data include seconds as well.
My data is in HH:MM:SS format but I need it rounded to only HH:MM and
tryin
I was initially going to say Uniform but you have too many parameters.
The Uniform distribution can only have a start point and an end point.
Perhaps the Mann-Whitney U? In R that could be qwilcox(p, n, m)
--
David Winsemius
On Mar 17, 2009, at 5:15 PM, MarcioRibeiro wrote:
Hi listers,
Does
Thanks Ray
On Thu, Mar 12, 2009 at 11:01 PM, Dr. Alireza Zolfaghari <
ali.zolfagh...@gmail.com> wrote:
> Hi list,
> I have a real problem with plotting US state map. When I try to plot the
> northern state, there will be some blank space in the top of graph (see case
> 1 example), and when I plot
Hi all,
I need to compare between times and put all similar times in specific
1 minute bins.
Unfortunately the original data include seconds as well.
My data is in HH:MM:SS format but I need it rounded to only HH:MM and
trying in Excel to display "unique" records only does not work since
the
Hi
Jason Rupert wrote:
> Paul,
>
> This is great! Like you said it is really close. I made a few
> changes and for some reason the y-axis label magically came back. I
> tried to remove it but it wouldn't stay away.
You have removed the 'axes=FALSE' from the qqplot() call.
Also, you have re
Hi listers,
Does anybody could tell me if there is a function for the DISTRIBUTION
FUNCTION U(p;m,n)!
Thanks,
Marcio
--
View this message in context:
http://www.nabble.com/Distribution-U-tp22567871p22567871.html
Sent from the R help mailing list archive at Nabble.com.
__
I actually pasted the wrong code! This attempts to replicate the
original request to replicate a JMP graphic:
library(ggplot2)
test_data<-rnorm(100,mean=10,sd=4)
a = data.frame(obs = test_data,condition = 'None')
p1 = ggplot(
data = a
,aes(
x = obs
)
)+geom_
I all
I am trying to combine columns from two dataframes to make a completely
new dataframe consisting of one column of dates (ie a combination of
dates from the two dataframes).
>From the following dataframes
a b
1 2008-07-27 1
2 2008-10-01 2
3 2008-08-15 3
4 2008-08-14 4
5 2008-08-14 5
6
Dear Luis,
You might be interested in ?title. Here is an example took from ?image:
require(grDevices) # for colours
x <- y <- seq(-4*pi, 4*pi, len=27)
r <- sqrt(outer(x^2, y^2, "+"))
image(z = z <- cos(r^2)*exp(-r/6), col=gray((0:32)/32))
# title added after plotting
title(main="Here you go!")
Nice work, Mike. Actually I think he was looking for it done this way.
library(ggplot2)
test_data<-rnorm(100)
a=data.frame(obs=test_data,condition='None')
p1=qplot(
data=a
,x=obs
,geom='histogram'
)+coord_flip()
p2=qplot(
data=a
,y=obs
,x=c
R doesn't keep track of when objects were created, so that's not possible.
If this is important to you, you could look at the 'trackObjs' package, which
does this and also stores individual objects in individual files (and writes
them to the file when they are changed in R).
-- Tony Plate
Ale
Using ggplot2:
library(ggplot2)
test_data<-rnorm(100)
a=data.frame(obs=test_data,condition='None')
p1=qplot(
data=a
,x=obs
,geom='histogram'
)+coord_flip()
p2=qplot(
data=a
,y=obs
,x=condition
,geom='boxplot'
)+opts(
On Mar 17, 2009, at 3:29 PM, steve_fried...@nps.gov wrote:
I'm moving my R applications to a Redhat OS and want to install ESS.
My sys admin has downloaded the rpm (emacs-common-
ess-5.3.8-1.fc8.src.rpm),
but when he tried to do the install he rec'd a number of warnings.
We are
not sure if
I'm moving my R applications to a Redhat OS and want to install ESS.
My sys admin has downloaded the rpm (emacs-common-ess-5.3.8-1.fc8.src.rpm),
but when he tried to do the install he rec'd a number of warnings. We are
not sure if the warnings are telling us that the installation did not occur
Paul,
This is great! Like you said it is really close. I made a few changes and for
some reason the y-axis label magically came back. I tried to remove it but it
wouldn't stay away.
Also, for some reason the title exceeds the margins of the layout. I am going
ot mess around with this a
In most biometric applications, those variances are treated as
nuisance parameters. They only need to be controlled for, while the
main purpose is to get the right point estimates and standard errors
for the fixed effects. In social science multilevel modeling (of which
education is probably the he
Hi
Jason Rupert wrote:
> I guess no reply means there is not an existing package to produce
> the plot?
>
> I will post the results of my script to hopefully help others who are
> trying to formulate the same plot.
>
> Thanks again.
I don't know of an existing function that does that particul
Ravi Varadhan wrote:
>
>
> require(BB)
>
> f2 <- function(x) {
> f <- rep(NA, length(x))
> f[1] <- 1 + 2 * x[1] * x[3] # x[3] is the Lagrangian multiplier
> f[2] <- 1 + 2 * x[2] * x[3]
> f[3] <- x[1]^2 + x[2]^2 - 1 # the equality constraint
> f
> }
>
>
You can also try my packag
When trying to remember what did I do in the session, especially after
coming back to it after a few days, I'd like to mimic Unix's ls -ltrh
-- does R retain the timing a certain variable is created? If not,
would it make a useful addition, to have ls with an option to sort by
creation tim
The code I'm using is shown below.
I would like to have a larger median at the top of the plot so that I can show
the entirity of "title_text".
Several times I tried messing with "par(mar", but that seemed to make matters
worse.
By any chance can anyone provide any insight as to the best
On Tue, Mar 17, 2009 at 7:10 PM, Ravi Varadhan wrote:
> Here is how you can implement the Lagrangian multiplier approach and solve
> the first-order KKT conditions to obtain the solution for Paul Smith's
> example:
>
> require(BB)
>
> f2 <- function(x) {
> f <- rep(NA, length(x))
> f[1] <-
Folks,
As reproduced in the code below, our data is being transformed during the
sqlFetch() operation. This has apparently been an issue since 2002. Are
there any ways for correctly reading data from MS SQL Server 2005? Why
does RODBC ignore the type information supplied by SQL Server?
req
Here is how you can implement the Lagrangian multiplier approach and solve
the first-order KKT conditions to obtain the solution for Paul Smith's
example:
require(BB)
f2 <- function(x) {
f <- rep(NA, length(x))
f[1] <- 1 + 2 * x[1] * x[3] # x[3] is the Lagrangian multiplier
f[2] <- 1 +
Lars,
Here is how you can solve the example given by Paul using spg() in "BB"
package.
f <- function(x, lambda) x[1] + x[2] - lambda * (x[1]^2 + x[2]^2 - 1)^2 #
look at how Penalized function is formulated
eps <- Inf
tol <- 1.e-05
p0 <- c(1, 1)
lambda <- 0.1 # start with a small value for
Can you reverse the x and y columns of the density object?
plot(density(c(-20,rep(0,98),20))$y, density(c(-20,rep(0,98),20))$x,
xlim = c(0,1), ylim=c(-4,4), type='l', lwd=3, col='red')
--
David Winsemius
On Mar 17, 2009, at 2:38 PM, Jason Rupert wrote:
Awesome.
This seems to
Hi,
In the package survival, using the function survreg for AFT model, I
only see 4 distributions for the response y: weibull, gaussian,
logistic, lognormal and log-logistic, which correspond to certain
distributions for the error terms. I'm wondering if there is a package
or how to obta
I've searched the help archives of both lists and apologize if I missed the
answer to my question:
Is there an update on developing mcmcsamp for glmer?
I'm using R v. 2.7.2 (on our Unix server - will hopefully be updated soon) and
2.8.1 on my PC and get the message for both:
gm1 <- glmer(cbind
Awesome.
This seems to produce the vertical histogram as needed, but is there then a way
to come back and add on a probability distribution line?
That is, add something like the following to the barplot:
points(density(x), type='l', lwd=3, col='red')
Thanks again.
--- On Tue, 3/17/09, Patr
You do not mention your OS, so further details may be needed.
Graphical output can be constrained by machine details. As a first
step, why not try a format that has multiple page possibilities such
as pdf ... this gives me 4 pages:
pdf(file ="outtest.pdf")
?lm
ctl <- c(4.17,5.58,5.18,6
Hi Lars,
Consider the following problem:
max x + y
subject to
x^2 + y^2 =1.
The solution is obviously (x,y) = (sqrt(2) / 2, sqrt(2) / 2).
Now, consider the unconstrained maximization problem on the variables
x, y and lambda:
max x + y + lambda * (x^2 + y^2 - 1)
(Notice that the objective fu
On Tue, Mar 17, 2009 at 12:54 PM, Bill Cunliffe wrote:
> Thanks for your help. Part of the problem was that I thought I was
> dealing with a data frame when it was actually an array. I forced all to be
> matrices for consistency. However, I see that
>
>
>
> avGain <- rbind(avGain[1:(j-1),],b,a
hello,
i use the vars package and would like to save the plots plotted with the
plot method for objects with class attribute varest etc.
so, if
library(vars)
data(Canada)
var.canada <- VAR(Canada)
then
plot(var.canada)
produces a plot for every of the four variable
On 3/17/2009 1:57 PM, kapo coulibaly wrote:
Is it possible to render a volume like the one attached, in R using packages
misc3d or rgl if I have the data for the individual surfaces (top and bottom
for each layer).
Your figure wasn't attached, but I'd guess the answer is yes. rgl
allows you
Is it possible to render a volume like the one attached, in R using packages
misc3d or rgl if I have the data for the individual surfaces (top and bottom
for each layer).
Thanks
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r
Yes, the slow convergence is easier to get with the log link. Overshooting the
correct coefficient vector has more dramatic effects on the fitted values and
weights, and in this example the starting value of (0,0) is a long way from the
truth. The same sort of thing happens in the Cox model
Dear List,
I recently got the chance to interview Jon Peck of SPSS Inc, a pioneering
technical statistician working since 1983 (when there were only two
substantial statistical software companies as per him ;) (not anymore ;)
and currently he is a Principal Software Engineer and Technical Advisor
Dear r helpers,
I'm trying to run the models of the BIOMOD package and got this message:
*
Error in nnet.default (x, y, w,...): **initial value in 'vmmin' is not finite*
What does this mean? and how can I correct it?
Thank you
*Andreia*
[[alternative HTML version deleted]]
In the case of 1:1 merging with distinct sets of non-ID variables in two
or more datasets, would the following code, which doesn't need to form
the larger merged data frame, be useful or faster? [A generalization of
with() would make this even better. I've often wondered about the
utility of
This question is better asked on the Bioconductor list
(bioconduc...@stat.math.ethz.ch).
James Mcininch wrote:
I would like to build a package for the HT HG-U133+ PM arrays from affy,
but I can't find any good documentation on how to go about it. Naively
using makecdfenv's make.cdf.package()
Hello,
I would be pleased if you could help me in determining the standard error of
the estimate for place2L1.N2 from the following table:
Call:
glm(formula = y ~ places2 + shellheight, family = quasibinomial)
Deviance Residuals:
Min1QMedian3Q Max
-3.41415 -
On Tue, Mar 17, 2009 at 12:56 PM, Patrick Burns wrote:
> Use 'do.call' as discussed in 'The R Inferno'
> and elsewhere.
Perfect, thanks.
--
Levi Waldron
post-doctoral fellow
Jurisica Lab, Ontario Cancer Institute
Division of Signaling Biology
IBM Life Sciences Discovery Centre
TMDT 9-304D
101
Thanks for your help. Part of the problem was that I thought I was dealing
with a data frame when it was actually an array. I forced all to be
matrices for consistency. However, I see that
avGain <- rbind(avGain[1:(j-1),],b,avGain[(j+1):nrow(avGain),])
would be a great solution as posit
The order() function allows you to specify multiple vectors, which are used
successively to break ties. If I want to use many vectors to break ties
(say, 25 or more), that are columns of a matrix or elements of a list, does
anyone know a shortcut to do this without passing 25 arguments to order()?
No, I meant the Combinations package, it is apparently an Omegahat package
(http://www.omegahat.org/Combinations/). It looks similar to the permn
function as far as the usage goes, but the documentation includes additional
information on other resources.
--
Gregory (Greg) L. Snow Ph.D.
Statis
OK, I can't see anything wrong with that.
a) post the results of ...
> sessionInfo()
... so we can check you're using a reasonably up to date system
b) send a copy of your data set to me at...
k.jewell at campden.co.uk
... and I'll see if I can reproduce your error. I can take most common da
try this:
create a file called mySweave.sty, containing the following code:
\NeedsTeXFormat{LaTeX2e}
\ProvidesPackage{mySweave}{}
\RequirePackage{ifthen}
\newboolean{swe...@gin}
\setboolean{swe...@gin}{true}
\newboolean{swe...@ae}
\setboolean{swe...@ae}{true}
\RequirePackage{ifthen}
\RequirePack
Jason,
be carefully to the order of intensities (counts or densities):
x=rnorm(1000)
par(mfrow=c(2,2))
h=hist(x,breaks=bk<-c(-5,-3,-2,-1,-.5,0,1,3,5))
barplot(rev(h$intensities),rev(bk[2:9]-bk[1:8]),space=0,horiz=T) # compare to
axis(2)
barplot(h$intensities,bk[2:9]-bk[1:8],space=0,horiz=T)
axis(2
I don't know if this will help, but look at the R graph gallery.
There may be something there, with code, that will work.
On Tue, Mar 17, 2009 at 10:39 AM, Jason Rupert wrote:
>
> I guess no reply means there is not an existing package to produce the plot?
>
> I will post the results of my script
Your first example that works would just repeat the atomic vector to the
length of the row. To replace a row with another data frame, you could use
rbind:
avGain <- rbind(avGain[-j,],b)
Or if the positioning is important,
2) avGain <- rbind(avGain[1:(j-1),],b,avGain[(j+1):nrow(avGain),])
I didn
That's by intent, by the way. The standard errors of the variance
components are only useful if the distribution is symmetric, and this is
not always true. If you were using lmer, and not lme, then you could use
the mcmcsamp function to look at the distribution of the random effects
to see if it is
not off the top of my head, but if you provide a reproducible example
(cut and paste into R) then I may be able to help you.
On Tue, Mar 17, 2009 at 10:44 AM, Bill Cunliffe wrote:
> I have the following data frames, avGain and retGain. They have the same
> dimensions.
>
>
>
> The following line
Dear all,
I read a file with all numbers with readLines function, as below,
> f <- file("data.txt")
> a <- readLines(f)
but all the values in a are in format "", and I cannot do the
calculation with them since they are not numeric. I wonder how should I skip
those quotes, thank you for help!
UNIX only (At least it works under Linux and Mac OS X):
(user <- system("users", intern=T))
Paolo
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide
Dear experts,
I use R to conduct multilevel modeling. However, I have a problem
about the interpretation of random effect. Unlike the variables in fixed
effects, the variables in random effects have not shown the standard error
(s.e.) and p-value, so I don't know whether the
Does anyone have a sample file with an example/template for Rdoc (as
implemented in R.oo)? The command I am using is Rdoc$compile(), but I
am having difficulties figuring out what the Macros look like.
Apparently, in the headers, I see that I should use one of:
# @RdocClass className
# @RdocMeth
I believe that hist will return a vector that could be passed to
barplot:
h.islands <- hist(islands)
> barplot(h.islands$intensities, horiz=TRUE) # or
> barplot(h.islands$counts, horiz=TRUE)
David Winsemius
On Mar 17, 2009, at 11:38 AM, Jason Rupert wrote:
Here is what I have so far:
I find Sweave very useful and I was trying to combine it with the
latex package fancyvrb. I was trying to get line numbering and labels
in order to reference the lines where particular commands occur.
Unfortunately, I haven't been able to figure out how to do it. Maybe
somebody can help me.
The
Here is what I have so far:
> test_data<-rnorm(100)
> par(mfrow=c(1,3)) # 3 rows by 1 columns layout of plots
> hist(test_data)
> boxplot(test_data)
> qqnorm(test_data)
I noticed that I can rotate a boxplot via "horizontal", but apparently "hist"
does not have that functionality.
I tried stack
My data frame is:
> str(act_data12)
'data.frame': 1687 obs. of 5 variables:
$ Date: chr "2006-02-22" "2006-02-22" "2006-02-22" "2006-02-22" ...
$ Dtime : chr "14:36:10" "14:36:11" "14:36:12" "14:37:38" ...
$ FocusApp_seq: chr "useractivity_act" "4" "3" "0" ...
$ App_duratio
I would like to build a package for the HT HG-U133+ PM arrays from affy,
but I can't find any good documentation on how to go about it. Naively
using makecdfenv's make.cdf.package() causes R to seg-fault.
I'm unfamiliar with the CDF format as such, but I'm guessing that it's
changed somewhat be
You can use something like this to make sure that h2 is a subset of h1
in consecutive order:
> h1 <- 1:50
> h2 <- 25:30
> hm <- match(h2, h1)
> all(diff(hm) == 1)
[1] TRUE
>
On Tue, Mar 17, 2009 at 6:42 AM, emj83 wrote:
>
> I have two numerical vectors; hun2 is a subset of hun1:
>
>> hun1
> [
In case it helps, this refers to the columns by position rather than name...
> aggregate(aDF[,4], by=aDF[,-4], sum)
It sums the fourth column (I'm guessing this is what Tammy meant by "merge")
for unique combinations of all columns except the fourth.
It doesn't need the column names inside the ag
Luis, take a look at the ?legend function, it can produce almost any kind of
legend you may need!
Hope It helps.
Rodrigo.
-Mensagem original-
De: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Em
nome de Luis Ridao Cruz
Enviada em: terça-feira, 17 de março de 2009 11:
I guess no reply means there is not an existing package to produce the plot?
I will post the results of my script to hopefully help others who are trying to
formulate the same plot.
Thanks again.
--- On Mon, 3/16/09, Jason Rupert wrote:
> From: Jason Rupert
> Subject: [R] R package to aut
R-help,
I'm sorry that I said " I don't have idea ".
What I mean t is that I just don't know how to set a legend
when "image" is called.
Thanks in advance
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PLEASE do
I have the following data frames, avGain and retGain. They have the same
dimensions.
The following line of code replaces row j of avGain as desired:
avGain[j, ] <- mean( retGain[jStart:j, ] )
However, the following line does not work:
avGain[j, ] <- ( avGain[j-1, ] * ( DAYS - 1 ) +
Looks as if you copied my code without modifying to suit your data frame.
My data frame (aDF) had column names V1, V2, V3, V4
> str(aDF)
'data.frame': 12 obs. of 4 variables:
$ V1: chr "2006-02-22 16:28:18" "2006-02-22 16:28:26" "2006-02-22
16:28:28" "2006-02-22 16:28:31" ...
$ V2: Factor w
R-help,
I have an image plot and I wish to set a legend to it
but I don't have an idea how to do it. Below it's the code line.
image(yyy, col = rev(heat.colors(10)), axes = T)
Thanks in advance.
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https://stat.ethz.c
Thanks.
Actually, I have been trying many time using method from KJ, There is always
the following information coming up or Rgui exit without reason.. I have no
idea about this?
gh<-with(act_data12, aggregate(V4, by=list(V1,V2,V3), sum))
Error in unlist(y, recursive = FALSE) :
promise alrea
Dear Francisco,
I was able to duplicate the problem that you reported, and in addition
discovered that the problem seems to be peculiar to the poisson family.
lm(y~ttment, data=dat) and biglm(y~ttment, data=dat) produce identical
results, as do glm(y~ttment, data=dat) and bigglm(y~ttment, data=dat
Dear R list members,
I'd like to make a graph of coefficients of the intercept, variable 1, and
variable 2 (and possibly the interaction between variable 1 and variable
2). When I use the lmList function as attached below, it shows a nice
coefficient graph.
> PACRP.lis <- lmList(PAffect ~ CRPC +
Tammy,
You won't get more help if you post this again every half an hour... I don't quite understand what you exactly want to do, but it seemed at
least to me that Keith's suggestion (see again below) pretty much does what you want. If not, could you please specify *EXACTLY* what you
want to do
Hi Lars,
The approach suggested by Paul Smith is one way to try and solve your problem.
Another aproach is to try and eliminate, if possible, one of the variables
using the equality constraint, so that the problem now become unconstrained in
(p-1) dimensions (where p is the original dimension
I still couldn't solve itany help would be preciated!
Tammy
> From: metal_lical...@live.com
> To: r-help@r-project.org
> Date: Tue, 17 Mar 2009 11:49:10 +0200
> Subject: [R] Merging
>
>
> Hi, All.
>
>
> I have a data frame with the part as :
>
> .
> 1422006-02-22 16:28
In thinking about this some more an SQL solution would be a
bit easier. Try this where species and temp are from
the prior post (i.e. after processing with chron).
The first statement finds the minimum distances for
each species and station_id combination. The
second finds all combinations of a
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