sounds like you downloaded corrupt files. This may be due to corruption on the
server, but more likely is due to whatever program you used to download the
installer. Try downloading from a different server and/or using a different
browser?
Thanks for your response! Yes, I uninstalled previous versions this time.
Earlier I used to run multiple versions of R without problems. I was
finally able to install XML in the Windows Vista R version 2.14.2.
I have not been able to install in R 2.15.x versions.
Santosh
On Fri, Jul 20, 2012 at
On 20/07/2012 22:06, Anthony Dick wrote:
Hello,
I am wondering if there is a robust estimation version of polr(),
similar to the lm() -> rlm(). If not, can anyone suggest a way to do
robust estimation with polr()?
Thanks,
Anthony
No: What would it mean to have a long-tailed distribution of a
Does not seem to be a good day for me! I am trying to install recent
versions of R in Windows Vista.
I got the earlier described errors in 2.14.1 and 2.15.1
I tried to install R2.15.1 I get this error now...
---
Error
---
C:\Program Files\R\R-2.1
Thanks for your response.
I tried R.2.15 also and just tried R.14.1 in 32-bit Windows Vista.
Below is the error message I received.
> install.packages("XML")
Warning message:
In getDependencies(pkgs, dependencies, available, lib) :
package XML is not available (for R version 2.14.1)
>
Santosh
Hello Rxperts..
I have a peculiar situation.. XML library is not available with R2.15.1
whereas I was able to install it with R version 2.13.1. Would highly
appreciate your suggestions. I am now trying to see if XML works with
previous versions of R.
Thanks,
Santosh
[[alternative HTML ve
Dear R People:
Could someone recommend a reference for R and Hadoop or Rhipe, please?
Thanks so much!
Sincerely,
Erin
--
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Sciences
University of Houston - Downtown
mailto: erinm.hodg...@gmail.com
_
Sorry about that, when I asked before I meant that I had tried your
proposed syntax. That has a problem, and it is that it uses the same set
of values for the character expansion *for both groups*, and it is not
what I wanted to do. If you plot the data (now a reduced set! sorry!)
you'll see that
Sorry, I had meant to include the following line:
Try: xyplot(y~x|group,cex=dd$z,data=dd)
Peter Ehlers
On 2012-07-20 18:00, "José M. Blanco Moreno" wrote:
Dear R-users,
I have tried, and I imagine it should be somewhere in the lines of passing
extra arguments to the panel function, but does
Using dput() is a good idea, but is the included dataset
really a _minimal_ set to illustrate your problem?
Peter Ehlers
On 2012-07-20 18:00, "José M. Blanco Moreno" wrote:
Dear R-users,
I have tried, and I imagine it should be somewhere in the lines of passing
extra arguments to the panel fu
On Saturday, July 21, 2012, olemissrebs1123 wrote:
> I am attempting to write what MATLAB does as : [S,A]= shaperead() which
> returns an N by 1 geographic array structure S containing geometric
> information, and a paralell N by 1 attribute structure array, A containing
> feature attribute inform
Dear R-users,
I have tried, and I imagine it should be somewhere in the lines of passing
extra arguments to the panel function, but does anyone know how to change the
character expansion factor that is affecting an individual point in each of the
panels of a lattice plot?
I have tried to pass a
Dear friends
i am trying to fit an Ornstein-Uhlenbeck process by MAXIMUM LIKELYHOOD
method.
i found these formulas on
http://www.sitmo.com/article/calibrating-the-ornstein-uhlenbeck-model/
this is the mean-reverting process
http://r.789695.n4.nabble.com/file/n4637271/process.txt process.txt
a
Hi,
I guess the ' +' strand was already ordered in the "mapping" set while '-' was
not ordered.
Try this:
test2<-mapping[with(mapping,rev(order(Chr.From))),]
rownames(test2)<-1:nrow(test2)
test2
Probe.Set.Name Chr Chr.Strand Chr.From Probe.X Probe.Y
1 ENSMUSG0047459_at 2
it is a typo
you sent
plot(x, y, type ="o", xlab="Panelist", ylab="T value",lwd=1.5,lty=1)
xlim=range(1:14),ylim=range(-8:8), las=1)
it should have been
plot(x, y, type ="o", xlab="Panelist", ylab="T value",lwd=1.5,lty=1,
xlim=range(1:14),ylim=range(-8:8), las=1)
On Fri, Jul 20, 2012 at 3:29
Hello!
I am trying to reproduce (for a publication) analyses that I ran several months
ago using lavaan, I'm not sure which version, probably 0.4-12. A sample model
is given below:
pathmod='mh30days.log.w2 ~ mh30days.log + joingroup + leavegroup + alwaysgroup
+ grp.partic.w2 + black + age + bi
I am attempting to write what MATLAB does as : [S,A]= shaperead() which
returns an N by 1 geographic array structure S containing geometric
information, and a paralell N by 1 attribute structure array, A containing
feature attribute information.
How can this be done with R code I tried using shp<
Hi,
I used GOstats to perform enrichment test on a set of genes (20).
There are 7 GO terms with pvalue less than cuttoff and therefore shown in
the result table.
How can I get the information that which gene in the input gene set belong
to which GO term of these enriched GO terms?
Thanks for a
whoops, backwards
new.x <- x[!remove.set,]
On Fri, Jul 20, 2012 at 2:51 PM, Richard M. Heiberger wrote:
> This works to multiply the ith row of a by the ith value of b.
> It might be what you can use
>
> a <- matrix(1:30, 6, 5)
> b <- 1:6
>
> a
> a*b
>
>
>
> To simplify your code, I think you c
I can't get the y axis to extend the full range that I need, which is -8 to 8
Here's my code. I tried using ylim, but it's still truncating at the
extremes in my data.
plot(x, y, type ="o", xlab="Panelist", ylab="T value",lwd=1.5,lty=1)
xlim=range(1:14),ylim=range(-8:8), las=1)
Any suggestions?
Hello,
Anyone know if R can read others format like the format.freeling
here the software: http://gramatica.usc.es/pln/tools/freeling.html
I want create a corpus with this format, or ¿ the best way is read from a
plain text?
thanks in advanced.
Greetings
Jenn.
[[alternative HTML versio
On Fri, Jul 20, 2012 at 12:38 PM, uday wrote:
> hi ,
>
> last time I did not mention the specific version of linux and it was problem
> with only syntax but with correction I got new error.
Nope -- you still have the same error Sarah responded to. It is
apt-get for Ubuntu (I believe) to get the n
Hello,
I am wondering if there is a robust estimation version of polr(),
similar to the lm() -> rlm(). If not, can anyone suggest a way to do
robust estimation with polr()?
Thanks,
Anthony
--
Anthony Steven Dick, Ph.D.
Assistant Professor
Director, Developmental Science Program
Department o
The nature of her inquiries suggests to me that Carol strongly needs
to consult a local statistician rather than fooling around with this
list.
-- Bert
On Fri, Jul 20, 2012 at 11:56 AM, John Fox wrote:
> Dear Carol,
>
>> -Original Message-
>> From: carol white [mailto:wht_...@yahoo.com]
Dear R users,
I am dealing with a data
set of aprox. 5 millions rows with data inconsistencies.
The data.frame is an
observation per claim with approximately 2 M unique ID's
Furthermore, one
individual could have one or more claims.
I have found that an
individual could have all his/h
Hi,
I was wondering what the best way is to create a new dataframe based on an
existing dataframe with only the unique available levels for each column (22
columns in total) in it.
If some columns have less unique values than others, then those columns can
be filled with blanks for the remaining
Dear Carol,
> -Original Message-
> From: carol white [mailto:wht_...@yahoo.com]
> Sent: July-20-12 2:45 PM
> To: John Fox
> Subject: Re: inverse normal transformation
>
> Thanks John for your quick reply.
>
> The purpose of applying inverse normal transformation is to reduce the
> impact
On Fri, 20 Jul 2012, Peter Ehlers wrote:
Well, You didn't say (in your original request) that you were using
the NADA package. The function is cenboxplot() and it's just a
wrapper for boxplot() and hence passes arguments to boxplot().
Thus the solution to your problem is just to add the 'names='
On 2012-07-20 12:09, Rui Barradas wrote:
Hello,
Still with aggregate, use length() not sum().
dtagroup <- aggregate(y ~ A + B + C, data=dtf, sum)
dtalength <- aggregate(y ~ A + B + C, data=dtf, length)
# Now merge the two
names(dtalength)[4] <- "count"
mm <- merge(dtagroup, dtalength)
# And
Hello,
Still with aggregate, use length() not sum().
dtagroup <- aggregate(y ~ A + B + C, data=dtf, sum)
dtalength <- aggregate(y ~ A + B + C, data=dtf, length)
# Now merge the two
names(dtalength)[4] <- "count"
mm <- merge(dtagroup, dtalength)
# And make it pretty
mm <- mm[, c("y", "A", "B",
Hi Michael, S Ellison,
I do not actually understand what you want to achieve with the M
estimates of rlm in MASS, but why you do not give a try of lmrob in
'robustbase'. Please have a llok in the references (?lmrob) about the
advantages of MM estimators over the M estimators.
Best regards,
Valent
As a follow up is there any way to also get the count for each combination?
For example
y A B C
0 11 2
0 12 1
1 11 2
0 11 2
1 11 2
1 12 1
0 12 2
Should become:
y A B C count
2 1 1 24
1 1 2
This works to multiply the ith row of a by the ith value of b.
It might be what you can use
a <- matrix(1:30, 6, 5)
b <- 1:6
a
a*b
To simplify your code, I think you can do this---one multiplication
xA <- x %*% A
Now you can do the tests on xA and not have any matrix multiplications
inside t
That is faster than what I was doing and reducing 15% of my iterations it
still very helpful.
Next question.
I need to multiply each row x[i,] of the matrix x by another matrix A.
Specifically
for(i in 1:n)
{
If (x[i,]%*%A[,1]<.5 || x[i,]%*%A[,2]<42 || x[i,]%*%A[,3]>150)
{
x<-x[-i,]
n<-n-1
}.
Hello,
Yes, Carol is comparing what can't be compared. Your code should do it,
I hope.
Rui Barradas
Em 20-07-2012 15:12, David L Carlson escreveu:
Rui's example included z-score data (drawn from rnorm). You converted your
data to z-scores so you need to compare your results to the z-scores n
On Fri, Jul 20, 2012 at 4:51 PM, Celine wrote:
> Hi,
>
> I am working with a SpatialPolygonsDataFrame of many islands. There are a
> lot of polygons (islands) composing my SpatialPolygonsDataFrame.
> I want to extract the elevation of each island.
Is each island a row in your SPDF? Or is it jus
On 2012-07-20 09:41, Rich Shepard wrote:
On Fri, 20 Jul 2012, MacQueen, Don wrote:
I don't have a cenbox() function that I can find, but your solution will
(probably? hopefully?) be along the lines of:
Don,
Well, I must have mis-typed that although I'm sure I read about it in the
NADA.pd
hi ,
last time I did not mention the specific version of linux and it was problem
with only syntax but with correction I got new error.
I found that some people has faced this problem before and some of them
have mentioned that it may be because of dependencies issue and
install.packages("Rcmd
Hi
Here is some code to illustrate how the correlations are calculated.
> data <- c("word1", "word1 word2","word1 word2 word3","word1 word2 word3
> word4","word1 word2 word3 word4 word5")
> frame <- data.frame(data)
> frame
data
1 word1
2
Hi,
Try this:
dat1<-read.table(text="
id date a b c y
1 1 8/6/2008 Red 15 B 22
2 1 8/6/2008 Green 15 B 22
",sep="",header=TRUE)
dat1$date<-with(dat1,as.factor(date))
dat1
id date a b c y
1 1 8/6/2008 Red
Hi. Is there any package available to calculate the systematic root mean
square error (RMSEs) and unsystematic RMSE (RMSEu)? I could not find such a
package. I am too lazy to calculate them step by step.
The calculation could be found in equation 3 and 4 in the Appendix A of the
following link:
ht
Thanks,
Y
On Fri, Jul 20, 2012 at 1:02 PM, arun wrote:
> Hi,
>
> Try this:
> dat1<-read.table(text="
>iddateab cy
> 1 1 8/6/2008Red15B 22
> 2 1 8/6/2008 Green 15B 22
> ",sep="",header=TRUE)
> dat1$date<-with(
On Fri, 20 Jul 2012, MacQueen, Don wrote:
I don't have a cenbox() function that I can find, but your solution will
(probably? hopefully?) be along the lines of:
Don,
Well, I must have mis-typed that although I'm sure I read about it in the
NADA.pdf or Dennis' book. I'll look again. I don't
Thanks,
Y
On Fri, Jul 20, 2012 at 11:47 AM, R. Michael Weylandt <
michael.weyla...@gmail.com> wrote:
> d$date <- factor(d$date)
>
> Best,
> Michael
>
> On Fri, Jul 20, 2012 at 9:44 AM, Yolande Tra
> wrote:
> > Hello,
> >
> > I would like to convert date as a factor to represent time in a repeate
Hi Rich,
I don't have a cenbox() function that I can find, but your solution will
(probably? hopefully?) be along the lines of:
foo <- boxplot( y ~ x, data=sdf, plot=FALSE)
foo$names <- ifelse(foo$names, "Label for TRUE", "Label for FALSE")
bxp(foo)
where sdf is a dataframe containing your d
Yes,
that's this.
Thank you very much.
2012/7/20 Peter Ehlers [via R]
> On 2012-07-20 04:05, carla moreira wrote:
>
> >
> > Hi,
> >
> > I would like to evaluate a function, with 3 arguments, for instance,
> >
> > myfunc<-function(a,b,c) { sqrt(a)-exp(b)+4*c
> >
HI,
I don't have much experience with laercio package. If it works for you, it is
good.
A.K.
- Original Message -
From: Michael Eisenring
To: arun
Cc:
Sent: Friday, July 20, 2012 8:38 AM
Subject: Re: [R] TukeyHSD not working
Hi
Thanks for the quick response. I installed the laerc
Hi,
Try this:
dat1<-data.frame(x,y,z)
xtabs(z~x+y,aggregate(z~x+y,dat1,mean))
y
x 10 20 30
1 100 0 0
2 0 200 0
3 0 0 300
table(dat1$z,dat1$y)
10 20 30
100 2 0 0
200 0 3 0
300 0 0 3
or,
table(dat1$x,dat1$z)
100 200 300
1 2 0 0
Contact me with your full names, phone numbers and residential address if you
own this email r-h...@stat.math.ethz.ch.
I have a proposal for you
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz
Hi,
Check this link.
http://stackoverflow.com/questions/5904513/tukeyhsd-after-within-factors-anova
A.K.
- Original Message -
From: Michael Eisenring
To: r-help@r-project.org
Cc:
Sent: Friday, July 20, 2012 6:15 AM
Subject: [R] TukeyHSD not working
Dear r-help members.
I would like t
Hi,
Just a doubt regarding the dataset.
Suppose, I include two more patients F and G with different missing values as
in this new dataset and run the code.
dat1<-read.table(text="
Patient Cycle V1 V2 V3 V4 V5
A 1 0.4 0.1 0.5 1.5 NA
A 2 0.3 0.2 0.5 1.6 NA
A 3 0.3 NA 0.6 1.7
Hi,
I am working with a SpatialPolygonsDataFrame of many islands. There are a
lot of polygons (islands) composing my SpatialPolygonsDataFrame.
I want to extract the elevation of each island.
I need to separate the different polygons (like dissolve function in
arcgis), to have the elevation of eac
Dear R help list,
I have done a lot of searching but have not been able to find an answer to
my problem. I apologize in advance if this has been asked before.
I am applying a mixed model to my data using lmer. I will use sample data
to illustrate my question:
>library(lme4)
>library(arm)
>data
d$date <- factor(d$date)
Best,
Michael
On Fri, Jul 20, 2012 at 9:44 AM, Yolande Tra wrote:
> Hello,
>
> I would like to convert date as a factor to represent time in a repeated
> measure situation in the following code. How would I do that?
>> d <- read.csv(file.path(dataDir,"data.csv"), as.is=
Thanks the aggregate() command is what I was looking for.
Chris
On Thu, Jul 19, 2012 at 9:03 PM, David L Carlson wrote:
> > dtf <- read.table(text="y A B C
> + 0 11 2
> + 0 12 1
> + 1 11 2
> + 0 11 2
> + 1 11 2
> + 1 12 1
> + 0
"Reith, William [USA]" wrote on 07/20/2012
09:52:02 AM:
> Would this matrix eat up memory making the rest of my program
> slower? Each x needs to be multiplied by a matrix and the results
> checked against a set of thresholds. Doing them one at a time takes
> at least 24 hours right now.
>
"Reith, William [USA]" wrote on 07/20/2012
09:52:02 AM:
> Would this matrix eat up memory making the rest of my program
> slower? Each x needs to be multiplied by a matrix and the results
> checked against a set of thresholds. Doing them one at a time takes
> at least 24 hours right now.
>
Michael Eisenring wrote on 07/20/2012 09:35:03 AM:
> Dear Jean,
> thanks for this email. I'm grateful for this instruction Just to
> make sure that I understood you properly: something like this?:
Michael,
Yes, this is perfect. Very helpful.
> My data:
>
> dput(geo_anova_nested_input):
>
Duncan Murdoch-2 wrote
>
> On 12-07-20 2:50 AM, Martin Ivanov wrote:
>> Hello,
>>
>> I want to create and save objects in a loop, but this is precluded by the
>> following obstacle:
>> this part of the script fails to work:
>>
>> assign(x=paste("a", 1, sep=""), value=1);
>> save(paste("a", 1, s
> -Original Message-
> Subject: [RsR] How does "rlm" in R decide its "w" weights for
> each IRLS iteration?
> I am also confused about the manual:
>
>a. The input arguments:
>
> wt.method are the weights case weights (giving the relative
> importance of case, so a weight
Bert,
The only thing wrong is that I'm still 75% asleep! Yikes!!
Thanks for the heads-up.
Carla: See Bert's solution.
Peter Ehlers
On 2012-07-20 07:10, Bert Gunter wrote:
Inline.
-- Bert
On Fri, Jul 20, 2012 at 6:59 AM, Peter Ehlers wrote:
On 2012-07-20 04:05, carla moreira wrote:
Hi,
On Fri, Jul 20, 2012 at 04:26:34PM +0200, Petr Savicky wrote:
> On Fri, Jul 20, 2012 at 05:45:30AM -0700, wwreith wrote:
> > General problem: I have 20 projects that can be invested in and I need to
> > decide which combinations meet a certain set of standards. The total
> > possible combinations c
Petr,
This is great.
MUCH faster than the code I provided.
And much more elegant code.
Thanks for posting!
Jean
Petr Savicky wrote on 07/20/2012 09:26:34 AM:
> On Fri, Jul 20, 2012 at 05:45:30AM -0700, wwreith wrote:
> > General problem: I have 20 projects that can be invested in and I need
Hello,
I would like to convert date as a factor to represent time in a repeated
measure situation in the following code. How would I do that?
> d <- read.csv(file.path(dataDir,"data.csv"), as.is=T,stringsAsFactors =
FALSE)
> d[1:2,]
id date ab c y
1
Dear Prof Ripley,
Many thanks for your response. In fact, latticeExtra and ggplot2 both state
they work with "layers",
but essentially this is nothing more than sequentially adding graphical output
to an existing device.
But given that PDF can include layers since version 1.5 it would be intere
On Fri, Jul 20, 2012 at 05:45:30AM -0700, wwreith wrote:
> General problem: I have 20 projects that can be invested in and I need to
> decide which combinations meet a certain set of standards. The total
> possible combinations comes out to 2^20. However I know for a fact that the
> number of proje
Michael,
Use dput() to output your data (or perhaps a small subset). Then paste
the result of that call and your R code (just those lines of code that are
needed to reproduce the problem) right in your message to R-help. That
makes it easier for the R-help list readers to help you troubleshoo
On 20/07/2012 8:28 AM, carol white wrote:
Thanks Rui.
I changed my scripts to the followings and I think that it still is not correct.
You haven't told us clearly what you are trying to achieve, and you
don't tell us what is wrong with what you have. How do you expect
anyone to help?
Dunca
Rui's example included z-score data (drawn from rnorm). You converted your
data to z-scores so you need to compare your results to the z-scores not
the original data.
Change these lines:
tmp.qnorm = qnorm(tmp.p/2,lower.tail=FALSE)*sign(scale(tmp))
# sign is of scale(tmp) not tmp
equal(scale(tmp),
Inline.
-- Bert
On Fri, Jul 20, 2012 at 6:59 AM, Peter Ehlers wrote:
> On 2012-07-20 04:05, carla moreira wrote:
>>
>>
>> Hi,
>>
>> I would like to evaluate a function, with 3 arguments, for instance,
>>
>> myfunc<-function(a,b,c) { sqrt(a)-exp(b)+4*c
>>
I've had to do something similar, so I wrote a small function to help.
This runs in about 1/4 the time of your code on my machine.
Others may have a more efficient approach.
all.combs <- function(num, from=0, to=num) {
# create a matrix of all possible combinations of num items
# r
On 2012-07-20 04:05, carla moreira wrote:
Hi,
I would like to evaluate a function, with 3 arguments, for instance,
myfunc<-function(a,b,c) { sqrt(a)-exp(b)+4*c
}
How to execute myfunc(x,y,z), for all x, all y and all z, where x,y,z are
vectors
It would help if you supplied some code for us to see what you have tried.
However something like this would presumably give you the sample data sets
--note only n=100 in the example.
mymat <- matrix(NA, ncol= 100, nrow= 100)
for (n in 1: 100) {
mymat[,n] <- sample(c(0,1), 100, replace =
There are lots of possibilities. Here's one using only xtabs():
dframe <- na.omit(data.frame(x, y, z))
zsum <- xtabs(z~x+y, dframe)
zcount <- xtabs(~x+y, dframe)
zmean <- ifelse(is.nan(zsum/zcount), 0, zsum/zcount)
--
David L Carlson
Associate Professor
On Jul 20, 2012, at 5:30 AM, vioravis wrote:
> I have the following data:
>
> x <- as.factor(c(1,1,1,2,2,2,3,3,3))
> y <- as.factor(c(10,10,10,20,20,20,30,30,30))
> z <- c(100,100,NA,200,200,200,300,300,300)
>
> I could create the cross tab of x and y with Sum of z as its elements using
> the xt
Dear R users,
I have the following problem. I plot a SpatialPixelsDataFrame object with
spplot and I need to add a contourplot
of another spatial variable. To be more precise, I plot the
SpatialPixelsDataFrame of the mean precipitation over Germany with coloured
regions and I want to overlay
Not quite what you asked for but would this do?
library(plyr)
x <- as.factor(c(1,1,1,2,2,2,3,3,3))
y <- as.factor(c(10,10,10,20,20,20,30,30,30))
z <- c(100,100,NA,200,200,200,300,300,300)
df1 <- data.frame(x, y, z)
(dsum <- ddply(df1, .(x, y), summarise, sum = sum(z, na.rm = TRUE),
Well, what do you want to control there?
Need a subset? Need an ordering?
On 20.07.2012, at 15:00, Carla Moreira wrote:
> Yes, I do.
>
> But I need to control how the permutations are done.
>
> Thank you.
>
> 2012/7/20 Jessica Streicher
> You mean executing the function for all combinatio
I think we need some raw data. Have a look at ?dput for a way to supply it.
It might help if you supplied some sample code of what you have tried.
"I want a line plot" is not particularly helpful.
John Kane
Kingston ON Canada
> -Original Message-
> From: anamika...@gmail.com
> Sent:
Yes, I do.
But I need to control how the permutations are done.
Thank you.
2012/7/20 Jessica Streicher
> You mean executing the function for all combinations of values?
> For example, if you have a<-b<-c<-1:2
> you would get back the values of
>
> myfunc(1,1,1)
> myfunc(1,1,2)
> myfunc(1,2,1)
General problem: I have 20 projects that can be invested in and I need to
decide which combinations meet a certain set of standards. The total
possible combinations comes out to 2^20. However I know for a fact that the
number of projects must be greater than 5 and less than 13. So far the the
code
Thanks Rui.
I changed my scripts to the followings and I think that it still is not
correct. See also the attached file.
Thanks for your help,
tmp
[1] 2.502519 1.828576 3.755778 17.415000 3.779296 2.956850 2.379663
[8] 1.103559 8.920316 2.744500 2.938480 7.522174 10.629200 8.5522
See below for the complete mail to which I reply which was not sent to rhelp.
==
emptyexpandlist2<-list(ne=0,l=array(NA, dim=c(1, 1000L)),len=1000L)
addexpandlist2<-function(x,prev){
if(prev$len==prev$ne){
n2<-prev$len*2
prev <- list(ne=prev$ne, l=array(prev$l, dim=c(1, n2))
Hi Tjun Kat,
you can define variables outside the ode function, but normally NOT
state variables, because their values need to be updated by the solver
during the simulation process.
But, if you want to block this for any debugging purposes and want to
e.g. fix a derivative to a certain valu
Not quite sure what you are aiming at, but looking at ?mapply or ?expand.grid
could be helpful
Benno
On Jul 20, 2012, at 1:05 PM, carla moreira wrote:
>
> Hi,
>
> I would like to evaluate a function, with 3 arguments, for instance,
>
> myfunc<-function(a,b,c) { sqrt(a)-exp(b)+4*c
>
You mean executing the function for all combinations of values?
For example, if you have a<-b<-c<-1:2
you would get back the values of
myfunc(1,1,1)
myfunc(1,1,2)
myfunc(1,2,1)
myfunc(1,2,2)
myfunc(2,1,1)
myfunc(2,1,2)
myfunc(2,2,1)
myfunc(2,2,2)
?
On 20.07.2012, at 13:05, carla moreira wrote:
Hello,
No it's not correct, you are computing a what seems to be a two-tailed
probabiity, so the inverse should account for it. Look closely: you take
the absolute value, then the upper tail probability, then multiply 2
into it. Reverse these steps to get the correct value.
# Helper function
e
I have a continuous data. So to calculate the inverse normal transformation, I
thought that I should first calculate the Z-score normalized data and then,
calculate the p-value et the quantile transformation. Does this seem to be more
sensible
my_data.p =2*pnorm(abs(scale(my_data)),lower.tail=
HI,
Probably this might be helpful for you.
http://rgm2.lab.nig.ac.jp/RGM2/func.php?rd_id=fBasics:dist-nigFit
A.K.
- Original Message -
From: carol white
To: "r-h...@stat.math.ethz.ch"
Cc:
Sent: Friday, July 20, 2012 6:21 AM
Subject: [R] function for inverse normal transformation
H
Hello,
I have a file like this (just a snapshot) where I have numerical
values for various genes, I want a line plot with shading (may be
using smooth ?) using qplot or ggplot :
Gene1 10 14 12 23 11 11 33 1 ..(multiple columns)
Gene2 4 2 1 1 3 4 1 2 .
Gene3 2 5 7 5 6 89 7 3 ..
Gene4 1
I have the following data:
x <- as.factor(c(1,1,1,2,2,2,3,3,3))
y <- as.factor(c(10,10,10,20,20,20,30,30,30))
z <- c(100,100,NA,200,200,200,300,300,300)
I could create the cross tab of x and y with Sum of z as its elements using
the xtabs function as follows:
# X Vs. Y with Sum Z
xtabs(z ~ x +
Dear r-help members.
I would like to compare species numbers of moths between eight different
forests (each sampled for six nights). I would like to do a nested anova to
compare species numbers between forests and nights.
For more site specific details I wanted to do a Tukey test (TukeyHSD).
Unf
Thanks David for the solution, I paste here the code with a few changes:
par(xpd = NA, oma = c(5, 0, 0, 0))
matplot(t(buffersump[,1:6]), type="l", xaxt="n", lwd=3, ylab="Porcentajes de
respuestas afirmativas")
axis(1, 1:6, colnames(buffersump2))
legend("bottomright", legend=rownames(buffersump2),
On 20/07/12 12:07, Christoph Scherber wrote:
Dear all,
Is it possible to create a pdf file with layers using the pdf() device in R?
No. Is it possible to specify layers in the R graphics language or any
device? (From what I understand by 'layers', no.)
The author of pdf()
Many thanks f
Hi,
I would like to evaluate a function, with 3 arguments, for instance,
myfunc<-function(a,b,c) { sqrt(a)-exp(b)+4*c
}
How to execute myfunc(x,y,z), for all x, all y and all z, where x,y,z are
vectors?
Thank you very much in advance
--
View
Dear prof. Harrell,
I'm not able to use the force option with fastbw, here an example of the error
I've got (dataset stagec rpart package):
> fitstc <- cph(Surv(stagec$pgtime,stagec$pgstat) ~ age + eet + g2 + grade +
> gleason + ploidy, data=stagec)
> fbwstc <- fastbw(fitstc,rule="aic",type="in
Dear Peter,
This indeed resolves the problem.
Many thanks. My apologies for not starting a new thread. I am a new R user
and not yet fully integrated into the R community.
Kind regards,
Bart Ferket
--
View this message in context:
http://r.789695.n4.nabble.com/rcspline-problem-tp3501627p46
On 12-07-20 7:36 AM, carol white wrote:
Thanks for your reply.
So to derive it from a given data set, is the following correct to do?
my_data.p =2*pnorm(abs(my_data),lower.tail=FALSE)
my_data.q = qnorm(my_data.p)
I don't know what you're trying to do, but that doesn't look like it
does some
Hi List
I have a dataframe (~1,200,000 rows deep) and I'd like to conditionally reorder
groups of rows in this dataframe.
I would like to reorder any rows where the Chr.Strand column contains a '-' but
reorder within subsets delineated by the Probe.Set.Name column.
# toy example
library
Thanks for your reply.
So to derive it from a given data set, is the following correct to do?
my_data.p =2*pnorm(abs(my_data),lower.tail=FALSE)
my_data.q = qnorm(my_data.p)
Cheers,
From: Duncan Murdoch
Cc: "r-h...@stat.math.ethz.ch"
Sent: Friday, July 20,
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