Dear all,
I have tried to import a SPSS file in R, but always get the following
message:
--- cut here ---
Error in read.spss("spss-data.sav", :
error reading system-file header
In addition: Warning message:
In read.spss("spss-data.sav", :
spss-data.sav: Variable Y6B_A indicates variable la
This is not a graceful error, and it needs to be corrected by the package
(not 'library') maintainer. R's own devices which need X11 do fail
gracefully if the session is headless.
Meanwhile, there are other graphics devices which create SVGs and do not
do this. E.g. R-devel (to be 2.7.0) has
Hi everyone,
I plotted a chart this morning, which should have Chinese lables for the
points. I made the chart via:
coordinates:
angel77wq360.615019.278135
cngdsthuang1220 -278.74068 91.556843
c国际米兰c 125.47369 -66.589461
fszym2007 -174.86150 346.672588
john_azyb_lee
You need to tell us your OS: see the request in the posting guide for
'at a minimum' the result of sessionInfo().
We also need to know what graphics device(s) you are intending to use.
For postscript or PDF, see the article by Paul Murrell and myself in R
News 2006-2.
Note too that 'Chinese'
Thanks a lot Thierry and Frede!!! Thats great- I made it much more
complicated!!!
--
View this message in context:
http://www.nabble.com/Trellis-plots-with-two-regression-lines-tp15976467p16000637.html
Sent from the R help mailing list archive at Nabble.com.
Try
xyplot(abund ~ bif | experiment, data=graphs, groups=stage,
type = c("p", "r"))
See the help page for panel.xyplot().
Best regards
Frede Aakmann Tøgersen
Scientist
UNIVERSITY OF AARHUS
Faculty of Agricultural Sciences
Dept. of Genetics and Biotechnology
Blichers Allé 20, P.O. B
Hi,
I'm currently taking a look at possibilities to report my results, produced
by R.
After looking at LaTex, I'm now considering the odf format, as most people
at my
company work with MS Word. So using odfWeave would be my best bet probably.
But if I run some examples from the odfWeave package,
We need to be able to reproduce this, so can you please make the file
available? Please follow up to R-devel.
On Wed, 12 Mar 2008, K. Elo wrote:
> Dear all,
>
> I have tried to import a SPSS file in R, but always get the following
> message:
> --- cut here ---
> Error in read.spss("spss-data.sa
hello, ladyes and gentlemans.
check this:
means<-c(4,6,8)
stand.error<-c(0.1,0.3,0.5)
now i've strongly tryed to scatterplot the
means(y-axis),by showing their sd with the
arrow(..,code=3,angle=90) function.
The problem is that my x-axis has categorical values
(say, factor(x)), and the arrows()
Thanks Liviu, I am trying out what is proposed in the mentioned thread.
I start R by
"C:\Program Files\R\R-2.6.2\bin\Rgui.exe"
http_proxy="http://proxy.haifis.org:8080";
Then I do the following (selecting "Austria" as mirror)
> install.packages("package", method="wget")
--- Please select a CRAN
stephen sefick wrote:
station month bas
190 5 0.000
190 7 1.563
190 10 0.000
190 11 0.000
202 4 18.750
202 5 18.750
202 7 6.250
202 10 4.800
202 11 3.125
198 4 18.750
198 5 31.250
198 7
On 11/03/2008 2:40 PM, [EMAIL PROTECTED] wrote:
> someone sent in a question earlier about doing
> something in 3D so i took a stab at it purely
> for educational purposes ( i'm not even sure that I understood the question
> actually ).
>
> Unfortunately, persp gives me an error that I don't un
Hi,
i have to optimise a function f(a,b), with a, b vectors in R^d such that a and
b are orthogonal, that is a'b=0. Anybody has a suggestion?
Thanks, in advance, for your help,
Giovanna
_
[[elided Hotmail spam]]
[[alte
Dear Peter,
Actually, I'm aware of these distinctions. In my experience, identical
replicates are relatively rare, but do occur, e.g., when one inputs a
contingency table from a secondary source. On the other hand, I can't count the
times (including two days ago) that I've seen people do the fo
hi Giacomo,
consider this:
means<-c(4,6,8)
stand.error<-c(0.1,0.3,0.5)
lowlim<-means-stand.error
uplim<-means+stand.error
plot(as.factor(c("a","b","c")),means,ylim=c(min(lowlim),max(uplim)))
arrows(c(1:3),lowlim,c(1:3),uplim,code=3,angle=90)
Still, its just a workaround and I don't think you can
Don MacQueen llnl.gov> writes:
>
> The help page for readShapePoints says that the file name that you
> supply should not include the extension.
>
> The "system.file" part of the example in the help page is only to use
> the example file that came with the package. Since you want to open
> y
Hi,
I guess this problem is super-trivial but I dont even know where to
start looking.
After I installed R on my new computer, R has a strange new feature. It
shows any error messages not only on the promt (as usual), but also in a
little pop up window which I need to click away before I can pro
Hi Everyone,
I am doing a project based on "Spatially Lagged Predictor
Variable Models", I would like to know which package in R would execute this
model. Also, I am new to this field of spatial statistics. Any suggestions
for a good book on spatial regression analysis would be apprec
Hoping someone can help me with xtabs and ftable. I'm trying to get a pair
of ftables (possibly more) next to each other. For example:
> dunhill_lights_xtab<-ftable(xtabs(grossedupobs ~ gender+age_group +
dunhill_lights, data = ciggs))
> dunhill_lights_xtab
dunhill_lights
On Wed, Mar 12, 2008 at 5:37 AM, Giacomo Prodi <[EMAIL PROTECTED]> wrote:
> hello, ladyes and gentlemans.
>
> check this:
>
> means<-c(4,6,8)
> stand.error<-c(0.1,0.3,0.5)
>
> now i've strongly tryed to scatterplot the
> means(y-axis),by showing their sd with the
> arrow(..,code=3,angle=90) f
I don't think that you're actually creating a
scatterplot if the x-axis is a factor. You're getting
a boxplot.
If you want a scatterplot then something like this
might work.
==
means<-c(4,6,8)
stand.error<-c(0.1,0.3,0.5)
aa <- factor(c("A","B"
quick (and dirty) solution:
y.up<- means+stand.error
y.dwn<- means-stand.error
plot(means,ylim=c(3.5,10))
for (i in 1:length(means)) arrows(i,means[i],i,y.up[i],length=0.1)
for (i in 1:length(means)) arrows(i,means[i],i,y.dwn[i],length=0.1)
Giacomo Prodi wrote:
>
> hello, ladyes and gentlem
What is the simplest way to specify the location of text in a scatter plot
(created using the plot function) in relative terms rather than specific x-y
coordinates? For example, rather than putting text at (300,49) on a plot,
how do I put it 1/10 of the way over from the y axis and 1/2 of the way
Hi again!
Following up my previous posting below (to which no response
as yet), I have located a report which situates this type
of question in a longitudinal modelling context.
http://www4.stat.ncsu.edu/~dzhang2/paper/glm.ps
Generalized Linear Models with Longitudinal Covariates
Daowen Zhang & Xi
Please see the footer of email:
> 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.html
> and provide commented, minimal, self-contained, reproducible code.
try this
cbind(write.ftable(
Hi,
Can u please tell me which all packages do i need to install to
estimate the hurst parameter in R. I have tried installing all the possible
options but still it doesnt work.
basically i want to use 9 functions to estimate hurst parameter like
aggvarfit, rsfit, etc.
i will be very thankfu
Ted,
What you have can be rendered as a 2^5 (X1 by X2 by X3 by X4 by Y) table
of counts, right?
Why isn't this a vanilla log-linear modelling (as in loglin() ) problem?
It seems to me that the temporal aspect you describe suggests a sequence
of margins that could be studied, viz
lis
Try this:
plot(rnorm(100))
text(diff(par("usr")[1:2])/10, sum(par("usr")[3:4])/2, labels="Test")
On 12/03/2008, Tom La Bone <[EMAIL PROTECTED]> wrote:
>
> What is the simplest way to specify the location of text in a scatter plot
> (created using the plot function) in relative terms rather tha
See ?par, entry "usr". Either set a new coordinate system and use it, or
convert 1/10 to user coordinates.
In R-devel (2.7.0 to be) this is easier:
text(grconvertX(0.1,"npc"), grconvertY(0.5, "npc"), "some text")
cnvrt.coords() in package TeachingDemos does something similar (although
it d
Hello,
I am trying to call a FORTRAN subroutine from R. The Fortran code is @:
http://lib.stat.cmu.edu/apstat/206
It performs a bivariate isotonic regression on a rectangular grid (m X n)
matrix. I used the g77 compiler and successfully created a dll file and it also
loads successfully from R.
G'day Krishna,
On Wed, 12 Mar 2008 08:59:30 -0700 (PDT)
"[EMAIL PROTECTED]" <[EMAIL PROTECTED]> wrote:
> > output<-.C("smooth_",as.integer(NROW),as.integer(NCOL),as.integer(NDIM),as.real(X),as.real(W),as.real(A),as.real(B),as.integer(NCYCLE),as.integer(ICYCLE),as.real(G),as.real(EPS),as.integer(I
I had a similar problem when read one of my spss.sav with long variable
label. But when I read another spss.sav with short label the same way, I
don't have problem.
Seems like R only allows variable labels of lengths 1...255 as your
error message showed.
WZ
-Original Message-
From: [EMAI
How many trees are you growing in the forest? If you grow a 100-tree
forest, the possible predictions (for a two-class data) would simply be
{0, 0.01, 0.02, ..., 0.99, 1}
Then, of course, if you only grow three trees, the predictions can only
be
{0, 1/3, 2/3, 1}
You can ask for as many digits
Dear Group,
I have a data frame like the following:
x <- c("Mike","A",0.01)
x1 <- c("Carl","A",0.2)
x2 <- c("Gene","C",0.3)
x3 <- c("James","A",-0.3)
x4 <- c("Dough","B",0)
xx <- rbind(x,x1,x2,x3,x4)
colnames(xx)<-c("Name","Class","NES")
xx <-as.data.frame(xx)
> xx
Name Class NES
x Mike
Try str() on both objects and see what you get. It
does not look like you have two vectors of
length=1762.
See this for example
=
aa <- c("A", "B", "D", "D","B", "B","D", "A", "C",
"C", "A","D")
bb <- rnorm(length(aa))
cc <- matrix(rnorm(length(
Well,
i am not sure what do you want how to do the graphs you showed or how to
interpret them??
If you want to know how to do them, look at package ade4, especially functions
s.corcircle and scatter.dudi, but you can use the function plot as well and
choose the right variables since PCA
Tom La Bone asked:
>
> What is the simplest way to specify the location of text in a
> scatter plot
> (created using the plot function) in relative terms rather than
> specific x-y coordinates? For example, rather than putting text at
> (300,49) on a plot, how do I put it 1/10 of the way over f
I tested the main set of examples from cnvrt.coords using the new
grconvert functions and they worked perfectly. I have already noted in
the help page for cnvrt.coords (for the next version coming out soon)
that people should start using grconvertX and grconvertY when they
become available.
Are t
Perhaps in this case:
noquote(with(xx, tapply(NES, list(Class, Name), paste)))
On 12/03/2008, Srinivas Iyyer <[EMAIL PROTECTED]> wrote:
> Dear Group,
> I have a data frame like the following:
>
>
> x <- c("Mike","A",0.01)
> x1 <- c("Carl","A",0.2)
> x2 <- c("Gene","C",0.3)
> x3 <- c("James"
Dear R-users
I would like to integrate something like \int_k^\infty (1 - F(x)) dx, where
F(.) is a cumulative distribution function. As mentioned in the "integrate"
help-page: integrate(dnorm,0,2) ## fails on many systems. This does not
happen for an adaptive Simpson or Lobatto quadrature (
Hi Henrique,
Thanks for your tip.
how can I map xx onto xy (where xy is a matrix I
created).
> > > xy
Mike Carl Gene James Dough
A111 1 1
C111 1 1
B111 1 1
If I can map xx onto xy, I can have '1' without NAs.
Thanks
Srini
Try:
xy <- with(xx, tapply(NES, list(Class, Name), paste))
xy[is.na(xy)] <- 1
On 12/03/2008, Srinivas Iyyer <[EMAIL PROTECTED]> wrote:
> Hi Henrique,
> Thanks for your tip.
>
>
> how can I map xx onto xy (where xy is a matrix I
> created).
>
>
>
> > > > xy
> Mike Carl Gene James Dough
>
The way that you have set up the data.frame is rather
unusual. It makes NES a factor which I suspect you
don't want. Do str(xx) to see what I mean
Here is a way that I think does what you want but with
the data.frame constructed in a different manner.
Again do str(aa) to see the difference
==
Dear R users,
I wrote the following toy example to explain my problem :
a=0
f=function(x,y,z) {
if (a==0) x[1]+x[2]+y
if (a!=0) x[1]+x[2]+y+z
}
f(1:2,3)
I have not specified z and I get an error. Although a=0, R seems to want to
know z because it's in the expression x[1]+x[2]+y+z.
So I
Oops, I just realised that you asked for a matrix not
a data.frame Substituing for the last three lines of
code
final <- as.matrix(aba)
colnames(final) <- aa$Name
rownames(final) <- mm1$Class
should do it.
--- John Kane <[EMAIL PROTECTED]> wrote:
> The way that you have set up the data.frame
Hi,
Why do you need an extension of integrate()? integrate() is adaptive - it
uses an adaptive Gauss-Kronrod quadrature.
You can specify Inf and -Inf as upper and lower limits, resp., in
integrate(). In fact, this is what the help page recommends, and it also
discourages the use of a large numb
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Alberto Monteiro
> Sent: Wednesday, March 12, 2008 10:28 AM
> To: Tom La Bone; r-help@r-project.org
> Subject: Re: [R] Specifying relative position of text in a plot
>
>
> Tom La Bone asked:
> >
> > W
> mm1 <- cast(mm, Class~Name)
> aba <- mm1[,2:6]
> aba[is.na(aba)] <- 1
> final <- data.frame(mm1$Class, aba)
> names(final) <- names(mm1)
> final
You can abbreviate all this to:
final <- cast(mm, Class ~ Name, fill = 1)
Hadley
--
http://had.co.nz/
_
Note that integrate accepts upper=Inf.
All that is needed is to read the help and do as it asks.
On Wed, 12 Mar 2008, Lüthi David (luda) wrote:
Dear R-users
I would like to integrate something like \int_k^\infty (1 - F(x)) dx,
where F(.) is a cumulative distribution function. As mentioned in
In your particular case I don't think that censoring is an issue, at least
not
for the reason that you discuss. The basic censoring assumption in the Cox
model is that subjects who are censored have the same future risk as those who
were a. not censored and b. have the same covariates.
You just need to return the values computed when a==0 and when a!=0.
a=0
f=function(x,y,z=0) {
if (a==0) return(x[1]+x[2]+y)
if (a!=0) return(x[1]+x[2]+y+z)
}
> f(1:2,3)
[1] 6
>
Ravi.
---
Ravi Varadhan, Ph
Greetings,
I am interested in using a generalized linear mixed model with data that
best fits a beta distribution (i.e., the data is bounded between 0 and 1
but is not binomial). I noticed that the beta distribution is not
listed as an option in the "family objects" for glmmPQL or lmer. I
f
On Wed, 12 Mar 2008, Dwayne Blind wrote:
> Dear R users,
>
> I wrote the following toy example to explain my problem :
>
> a=0
> f=function(x,y,z) {
>if (a==0) x[1]+x[2]+y
>if (a!=0) x[1]+x[2]+y+z
> }
> f(1:2,3)
>
>
> I have not specified z and I get an error.
What was the error? It work
To left justify the text rather than center, use the adj argument, this
is easier and probably more reliable than trying to adjust the x
coordinate manually.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
> -Original Mess
Dear R-help-list,
The following is R function I wrote for computing multi-dimensional kernel
density. I am seeking R experts who can make the code to run faster, 50 times
faster ideally.
Specifically, for function
kernel.estimate = function(points, bw),
the argument points is a d by n matrix
Hi again,
many thanks for Your answers. So, if I understood Wei right:
Zhao, Wei (Cancer Center) wrote (12.3.2008):
> I had a similar problem when read one of my spss.sav with long
> variable label. But when I read another spss.sav with short label the
> same way, I don't have problem.
> Seems li
On Wed, 12 Mar 2008, Nordlund, Dan (DSHS/RDA) wrote:
>> -Original Message-
>> From: [EMAIL PROTECTED]
>> [mailto:[EMAIL PROTECTED] On Behalf Of Alberto Monteiro
>> Sent: Wednesday, March 12, 2008 10:28 AM
>> To: Tom La Bone; r-help@r-project.org
>> Subject: Re: [R] Specifying relative posi
On Wed, 12 Mar 2008, K. Elo wrote:
> Hi again,
>
> many thanks for Your answers. So, if I understood Wei right:
>
> Zhao, Wei (Cancer Center) wrote (12.3.2008):
>> I had a similar problem when read one of my spss.sav with long
>> variable label. But when I read another spss.sav with short label th
Hello, I have a general data analysis question. I recently visited a
lab where they are testing a new treatment and they had done the
experiment several times on different dates. They repeated the
experiment 3-5 times per day. And then for practical reasons they
repeated the whole procedure fo
glmmPQL can fit the same GLM families as glm() can -- it does not list
_any_ .
Howver, the beta distribution does not give a GLM family and hence your
subject line is strictly about a non-existent concept. I'm presuming that
you want to model the logit of the mean of a beta by a random effects
I am trying to generate a graphic with a matrix of 9 line graphs (3 rows, 3
columns), all with the same y-axis range, and only showing the y-axis
labels and title at the left edge of each row of the matrix. I have been
trying to use the widths argument in layout to specify a larger column
width in
Thanks Ted and Professor Ripley for the very helpful
answers! Now I know what the problem is in my case.
All the best,
Werner
--- [EMAIL PROTECTED] schrieb:
> On 11-Mar-08 08:58:55, Werner Wernersen wrote:
> > Hi,
> >
> > could anyone explain to me what this warning
> message
> > exactly me
The problem with your code is that you are passing widths to the matrix
function instead of the layout function. The matrix function does not
know what you want to do with the "widths" and gives the error. Move
that section to between the last 2 parentheses:
> layout( matrix( 1:9, ncol=3, byrow=
Hi,
My data was only a toy example that matched the real situation, with real data,
but i could not have posted the entire data.set and so i gave a self contained
example of what i thought was my problem. Of course you can see with the naked
eye that the data is unbalanced, (this was done inten
Hi Thomas
maybe try the following:
1) start R with option "C:\Program Files\R\R-2.6.2\bin\Rgui.exe" --
internet2
2) disable anything in your Rprofile.site and .Rprofile to have a
clean start of R
I recently upgraded from an earlier version of R to v 2.6.2, and was
using all my old libs, and
Hi all
I am trying to determine the distances between two datasets of x and y
points. The number of points in dataset One is very small i.e. perhaps
5-10. The number of points in dataset Two is likely to be very large
i.e. 20,000-30,000. My initial approach was to append the first dataset
to the
Hi,
I am trying to make use of "approximate" option to obtain null distribution
through Monte-Carlo resampling, as described in coin library documentation.
Unfortunately, this does not play well with my data -- permutation process
swallows astonishingly large amounts of RAM (4-5Gb) and runs far
Andrew McFadden said the following on 3/12/2008 1:47 PM:
> Hi all
>
> I am trying to determine the distances between two datasets of x and y
> points. The number of points in dataset One is very small i.e. perhaps
> 5-10. The number of points in dataset Two is likely to be very large
> i.e. 20,
I thought your question was well expressed and that you followed the
posting guide better than most.
I'm no expert on such issues, but I'd like to kick in a few opinions
(with which others may disagree).
(1) All of the anova stuff is based on the assumption of homogeneity
of variance. Howe
I have a very long list that I'd like to subset based on a logical value
within each element. Example below. I'd like to get just those list
elements for further study whose $sig.cor slot is TRUE. In this example,
I'd only want element [[2]].
Should be simple, I know. How can I do this? Thanks,
gene.pair.tf.lst[sapply(gene.pair.tf.lst, "[[", "sig.cor")]
b
On Mar 12, 2008, at 5:24 PM, Mark W Kimpel wrote:
I have a very long list that I'd like to subset based on a logical
value
within each element. Example below. I'd like to get just those list
elements for further study whose $sig.
I'm not sure exactly what you want, but try something like
lapply(gene.pair.tf.lst, function(x) if(x$sig.cor) x$gene.pair)
If you only want the non-NULL entries (and how would you know which they
are, given you have duplicates?)
gene.pair.tf.lst <- list(list(gene.pair="Lgals1:Pxmp2", sig.cor=FA
Hi,
i am sure, that this is a noob-question, but i have searched for
hours without any good result.
I want to draw a vertical line through the maximum of the first derivation.
Here is a small example.
--8<-[mydata.csv]
HM
115
222
323
4
Hi everybody, I really appeciate the help you have given me. I am getting
around R comfortably now and am able to import research project data, do
mathematical analysis of the data and produce graphs. My PhD project is
going really well in its early days, already collecting data and proposal
compl
Hi I have the smililar question.
I have a list:
my_list <- list(name="Fred", wife="Mary",
no.children=3, child.ages=c(4,7,9))
> my_list
$name
[1] "Fred"
$wife
[1] "Mary"
$no.children
[1] 3
$child.ages
[1] 4 7 9
Now I want to search "Fred" and get attribute of that
value which is 'name'.
Hello!
I have used read.csv to read in a data frame, and there are a few variables in
it, however, when I tried
is.list(data$V1)
>FALSE
In fact, I have tried, they are not vectors either.
I'm wondering:
1. What objects are these "lists" of data?
2. How could I find out about the type/inherite
class(data)
class(data$V1)
str(data)
str(data$V1)
dput(data)
dput(data$V1)
On Wed, Mar 12, 2008 at 7:17 PM, Chang Liu <[EMAIL PROTECTED]> wrote:
>
> Hello!
> I have used read.csv to read in a data frame, and there are a few variables
> in it, however, when I tried
> is.list(data$V1)
> >FALSE
> In
Try:
> names(which(my_list == 'Fred'))
[1] "name"
You can break down the two nested calls to figure out what is happening.
-Christos
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Srinivas Iyyer
> Sent: Wednesday, March 12, 2008 6:53 PM
> To: Be
Hello:
Using the built-in dataset aml as an example:
data(aml)
If I use instead dummy variables:
aml$x1 = (aml$x=="maintained")aml$x2 = (aml$x=="unmaintained")
and I want to plot the survival curve using x1, x2, and I just want the 2
levels, rather than 4 curves from:
fit <- survfit(Surv(
Here is one way.
...
usr <- par("usr") # get user coordinates
par(usr = c(0, 1, 0, 1)) # new relative user coordinates
text(0.1, 0.5, "Some text", adj = 0) # if that's what you want
par(usr = usr) # restore original user coordinates
...
If you were going to be d
Hi all:
How can I transpose this dataset from column to row
idweek value
1 5 51
2 6 73
3 7 41
4 8 22
5 9 83
6 10 55
7 11 42
to something like this...
id1 2 3 4 5 6 7
week 5 6 7 8 9 10 11
value 51 73 41
Here's what I would try. Suppose x1, y1 and x2, y2 are the two data
sets.
z1 <- complex(real = x1, imaginary = y1)
z2 <- complex(real = x2, imaginary = y2)
dMat <- outer(z1, z2, function(z1, z2) Mod(z1-z2))
Bill Venables
CSIRO Laboratories
PO Box 120, Cleveland, 4163
AUSTRALIA
Office Phone (emai
Hello,
Is there an easy function for switching list to matrix. My list is of
genetic distances between species pairs:
A A 0
A B .5
A C .25
B C .5
and I want a distance matrix such as:
A B C
A 0 .5 .25
B .5 0 .5
C .25 .5 0
for use in a mantel test.
Thank you for the help!
cheers,
charl
if x is your data: t(x)
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Im
Auftrag von Felipe Carrillo
Gesendet: Wednesday, March 12, 2008 8:19 PM
An: [EMAIL PROTECTED]
Betreff: [R]
is(data$V1)
Have a look at the refcard that you can download from cran.r-project.org or
please take a look at one of the many other manuals that are provided there
for free (many manuals are also available elsewhere on the internet, so you
can google them). Almost all of them answer these basic qu
Here is one way of doing it:
> x <- read.table(textConnection("A A 0
+ A B .5
+ A C .25
+ B C .5"), as.is=TRUE)
> closeAllConnections()
> # get the unique names
> x.names <- sort(unique(c(x[[1]], x[[2]])))
> # create a matrix of the right size and put names on it
> x.dist <- matrix(0, length(x.nam
Matlab has an aggregate variable called a structure, which consists of
component variables each of which can be of a different data type and
dimension. For example, I can have a single structure Model which
consists of Model.variable1, Model.variable2, and Model.variable3,
where variable1 i
Hello -
Scott Romans wrote:
> Matlab has an aggregate variable called a structure, which consists of
> component variables each of which can be of a different data type and
> dimension. For example, I can have a single structure Model which
> consists of Model.variable1, Model.variable2, and
?list
On 13/03/2008, at 3:50 PM, Scott Romans wrote:
> Matlab has an aggregate variable called a structure, which consists of
> component variables each of which can be of a different data type and
> dimension. For example, I can have a single structure Model which
> consists of Model.variable1,
Hello all,
Still fighting with regular expressions and such, I am again stuck:
Suppose I have a vector of character chains. In this vector, I wish to identify
which character chains start with a given pattern, and then replace everything
that comes after said pattern.
Here is a quick exampl
Try this one:
> gsub("^(plif)([a-z]*)", "\\1ONE", words)
[1] "plifONE" "plafboum" "ploufbang" "plifONE"
-Christos
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of GOUACHE David
> Sent: Wednesday, March 12, 2008 12:15 PM
> To: [EMAIL PROTECTED]
Hi,
I have what I think is a fairly straightforward problem. I've looked
through the FAQ's and mailing lists but have been unable to identify a
solution, probably because I don't understand the language well enough.
I have a set of data d, with 3 columns as shown,
I want to sort the data Group,
Stephen,
I am sure someone will have a more elegant solution, but the following
works. Mark
d.lst <- split(x = d, f = as.factor(d$Group), drop = FALSE)
d.lst.mn <- sapply(d.lst, FUN =
function(x){mean(as.numeric(as.character(x$Value)))})
o <- order(d.lst.mn, decreasing = TRUE)
d.lst.mn <- d.lst
Dear Prof. Therneau,
Many thanks for this,
On 3/13/08, Terry Therneau <[EMAIL PROTECTED]> wrote:
>
> In your particular case I don't think that censoring is an issue, at least
> not
> for the reason that you discuss. The basic censoring assumption in the Cox
> model is that subjects who are
Thanks Mark,
I see that I made an error in my original request for help. I got my
labels and groups mixed up (see below). Nonetheless, your code has been
a good pointer in the direction of a solution. I'll post it up when I
have it working.
Thanks again,
Stephen Barry.
I should have written i
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