if that is not specific (or not general) enough:
newDV <- dat$DV ## newDV is my DV & it is continuous.
newDV <- as.numeric(newDV)-5
str(newDV)
(i had to do a great deal of coding here so i am snipping down to the end)
tmp[which(dat$v1 == "stuff" & dat$v2 == "more stuff")] <- "lots of stuff"
hello. help with effects plots. here's the last bit of code before running
the model and then the effects, then the error.
nor.dem <- norway$v162 ## nor.dem is my DV & it is continuous.
nor.dem <- as.numeric(nor.dem)-5
str(nor.dem)
(i had to do a great deal of coding here so i am snipping dow
Michael Friendly yorku.ca> writes:
>
> For a paper dealing with generalized ellipsoids, I want to illustrate in
> 3D an ellipsoid that is unbounded
> in one dimension, having the shape of an infinite cylinder along, say,
> z, but whose cross-section in (x,y)
> is an ellipse, say, given by the
Inline.
On Fri, Mar 9, 2012 at 7:04 PM, RHelpPlease wrote:
> Hi there,
> I am having trouble subsetting a data frame by a conditional via one column
> (of many).
>
> I read the file into R through "read.fwf," where I specified column widths.
> Original data is .DAT. I then utilized "names" funct
Could you just add a log scale to the y dimension?
DAT <- data.frame(x = runif(1000, 0, 20), y = rnorm(1000))
plot(y ~ x, data = DAT, log = "y")
That lessens large dispersion (in some circumstances) but I'm not
really sure what that has to do with smoothingdo you mean
"smoothing" in the tech
Hmm, smooth the chart makes me think you are trying to find the trends:
require(ggplot2)
ggplot(mtcars, aes(mpg, hp)) +
geom_point() +
stat_smooth()
Try it out and see what you think---it adds a locally smoothed line
that does something like trace the means (that is a very over
simplificatio
The origin of this problem was that a plain scatter plot with too many
points with high dispersion generated too many points flying all over
places.
We are trying to smooth the charts a bit...
Any good recommendations?
Thanks a lot!
On Fri, Mar 9, 2012 at 8:59 PM, Michael wrote:
> Sorry for t
Sorry for the confusion Michael.
I myself am trying to figure out what my boss is requesting:
I am certain that I need to "plot the quantiles of each bin. " ...
But how are the quantiles plotted? Shall I specify 50% quantile, etc?
Being a diligent guy I am trying my hard to do some homework an
I understand that the NaN's mean that something can't be computed, I'm just
unsure as to what they refer to. Can anyone shed some light as to what the
beta and mix matrices are?
--
View this message in context:
http://r.789695.n4.nabble.com/poLCA-tp4458635p4461091.html
Sent from the R help mailin
On Fri, Mar 9, 2012 at 9:28 PM, Michael wrote:
> Thanks a lot Mike!
>
Michael if you don't mind. (Though admittedly it leads to some degree
of confusion in a conversation like this)
> Could you please explain your code a bit?
Which part?
>
> My imagination is that for each bin, I am plotting a
Thanks a lot Mike!
Could you please explain your code a bit?
My imagination is that for each bin, I am plotting a line which is the
quantile of the y-values in that bin?
I ran your program but couldn't figure out the meaning of the dots in your
plot?
Thanks again!
On Fri, Mar 9, 2012 at 7:07 P
Thanks Joshua.
The two criteria are separate...
So I have to produce two separate plots...
Thanks again!
On Fri, Mar 9, 2012 at 6:54 PM, Joshua Wiley wrote:
> Hi Michael,
>
> Although I do think ggplot2 does a superb job of elegant data
> visualization; I am not sure any graphics package will
i'll try to describe the data,
here [1] there is a subdatset (255 rows) 6 columns (a to f)
the last columns contains the Identification Number (ID) for a particular
species.
the ID in f are 20 different species and it should be my 'label':
16001
11012
25011
13011
11029
11027
10022
10024
20009
11
That doesn't really seem to make sense to me as a graphical
representation (transforming adjacent y values differently), but if
you really want to do so, here's what I'd do if I understand your goal
(the preprocessing is independent of the graphics engine):
DAT <- data.frame(x = runif(1000, 0, 20)
Hi Michael,
Although I do think ggplot2 does a superb job of elegant data
visualization; I am not sure any graphics package will do what you
want. I suspect you will first have to do some work binning your
data, and then plot in your package of choice.
In the situation that you have described, I
Dear Peter,
Thank you very much.
Best regards,
Ezhil
--- On Fri, 3/9/12, Petr Savicky wrote:
> From: Petr Savicky
> Subject: Re: [R] Correlation between 2 matrices but with subset of variables
> To: r-help@r-project.org
> Date: Friday, March 9, 2012, 1:18 PM
> On Thu, Mar 08, 2012 at 03:57:0
Dear All,
I am trying to install rgl on my mac notebook from the source file. I tried
using: /usr/bin/R64 CMD INSTALL rgl_0.92.798.tar.gz and get the following
error message:
checking for X... no
configure: error: X11 not found but required, configure aborted.
ERROR: configuration failed for pac
On 12-03-09 4:34 PM, Ruth Ripley wrote:
Dear all,
I have been running some tests of my package RSiena on different
platforms and trying to reconcile the results.
With Mac, the commands
options(digits=4)
round(1.81652, digits=4)
print 1.817
The value you're printing is 1.8165, so I believe W
Hi all,
I am trying hard to do the following and have already spent a few hours in
vain:
I wanted to do the scatter plot.
But given the high dispersion on those dots, I would like to bin the x-axis
and then for each bin of the x-axis, plot the quantiles of the y-values of
the data points in each
Just what null hypothesis are you trying to test or what question are
you trying to answer by comparing 2 matrices of different size?
I think you need to figure out what your real question is before
worrying about which test might work on it.
Trying to get your data to fit a given test rather tha
On Fri, Mar 9, 2012 at 1:50 PM, Massimo Di Stefano
wrote:
> Peter,
>
> really thanks for your answer.
>
>
>
> install.packages("flashClust")
> library(flashClust)
> data <- read.csv('/Users/epifanio/Desktop/cluster/x.txt')
> data <- na.omit(data)
> data <- scale(data)
>> mydata
>
Hi there,
I am having trouble subsetting a data frame by a conditional via one column
(of many).
I read the file into R through "read.fwf," where I specified column widths.
Original data is .DAT. I then utilized "names" function to read in column
headings.
For one column, PRVDR_NUM, I wish to f
Sarah Goslee gmail.com> writes:
>
> No, that was your answer, not a request for clarification.
> Type
> ?system
> at an R prompt and read the help file.
>
> Sarah
>
> On Fri, Mar 9, 2012 at 5:45 PM, Julio Sergio gmail.com>
wrote:
> > Sarah Goslee gmail.com> writes:
> >
> >>
> >> ?system
>
On Mar 9, 2012, at 4:36 PM, Max Kuhn wrote:
Can anyone recommend a good nonparametric density approach for data
bounded
(say between 0 and 1)?
I thought the "canonical" answer, at least the one that generally is
put forward whe people have difficulty with the stats::spline results
was t
It's here (search the sources for the appropriately named qbeta.c) but
I can't guarantee you'll understand it easily: Martin Maechler (and
others) works hard to make these the best in the business, but that
performance comes at a price, here paid in opacity:
http://svn.r-project.org/R/trunk/src/nm
Hello,
I don't know if it's the fastest but it's more natural to have an index
matrix with two columns only,
one for each coordinate. And it's fast.
fun <- function(valdata, inxdata){
nr <- nrow(inxdata)
nc <- ncol(inxdata)
mat <- matrix(NA, nrow=nr*nc, ncol=2)
i1
Hi David:
Thanks, I actually realized it was error on my part. I didnt think there
was a problem with R-code, just wanted to get a better understanding of the
function qbeta.
Anamika
On Fri, Mar 9, 2012 at 3:04 PM, David Winsemius wrote:
>
> On Mar 9, 2012, at 2:48 PM, Anamika Chaudhuri wrote:
On Fri, 09-Mar-2012 at 06:02AM -0800, RMSOPS wrote:
|> Hello
|>
|> Thanks for the reply. As yet I have not much experience in r, although I
|> make some mistakes.
|>Any tips to solve the problem of a more effective way.
When we know what problem you're trying to solve, we'll have various
On Mar 9, 2012, at 5:30 PM, Julio Sergio wrote:
Is there any way to issue operating system commands and geting back
the results,
in R?
I mean, for instance, in Linux, to execute from R the 'ls' command
and getting
back a list of files in the current directory, or, equivalently, in
Window
Rolf Turner xtra.co.nz> writes:
>
> On 10/03/12 11:30, Julio Sergio wrote:
> > Is there any way to issue operating system commands and geting back the
results,
> > in R?
> >
> > I mean, for instance, in Linux, to execute from R the 'ls' command and
getting
> > back a list of files in the curre
No, that was your answer, not a request for clarification.
Type
?system
at an R prompt and read the help file.
Sarah
On Fri, Mar 9, 2012 at 5:45 PM, Julio Sergio wrote:
> Sarah Goslee gmail.com> writes:
>
>>
>> ?system
>>
>> On Fri, Mar 9, 2012 at 5:30 PM, Julio Sergio gmail.com>
> wrote:
>> >
On 10/03/12 11:30, Julio Sergio wrote:
Is there any way to issue operating system commands and geting back the results,
in R?
I mean, for instance, in Linux, to execute from R the 'ls' command and getting
back a list of files in the current directory, or, equivalently, in Windows/DOS,
the 'dir'
Sarah Goslee gmail.com> writes:
>
> ?system
>
> On Fri, Mar 9, 2012 at 5:30 PM, Julio Sergio gmail.com>
wrote:
> > Is there any way to issue operating system commands and geting back the
results,
> > in R?
> >
> > I mean, for instance, in Linux, to execute from R the 'ls' command and
gettin
?system
On Fri, Mar 9, 2012 at 5:30 PM, Julio Sergio wrote:
> Is there any way to issue operating system commands and geting back the
> results,
> in R?
>
> I mean, for instance, in Linux, to execute from R the 'ls' command and getting
> back a list of files in the current directory, or, equival
Is there any way to issue operating system commands and geting back the
results,
in R?
I mean, for instance, in Linux, to execute from R the 'ls' command and getting
back a list of files in the current directory, or, equivalently, in
Windows/DOS,
the 'dir' command?
I'm not interested in the
With coordination with the code's author (Daniel),
The updated code has been uploaded to github here:
https://github.com/talgalili/R-code-snippets/blob/master/siegel.tukey.r
And also the following post was updated with the code:
http://www.r-statistics.com/2010/02/siegel-tukey-a-non-parametric-test
Peter,
really thanks for your answer.
install.packages("flashClust")
library(flashClust)
data <- read.csv('/Users/epifanio/Desktop/cluster/x.txt')
data <- na.omit(data)
data <- scale(data)
> mydata
a bc d e
1 -0.207709346 -6.61855
Can anyone recommend a good nonparametric density approach for data bounded
(say between 0 and 1)?
For example, using the basic Gaussian density approach doesn't generate a
very realistic shape (nor should it):
> set.seed(1)
> dat <- rbeta(100, 1, 2)
> plot(density(dat))
(note the area outside o
Dear all,
I have been running some tests of my package RSiena on different
platforms and trying to reconcile the results.
With Mac, the commands
options(digits=4)
round(1.81652, digits=4)
print 1.817
With Windows, the same commands print 1.816
I am not bothered which answer I get, but it w
Hi Chuck, thank you *very* much! That really helped! b
On 9 March 2012 17:15, wrote:
> Benilton Carvalho writes:
>
>> Hi,
>>
>> what is the proper of of "passing a missing value" so I can extract
>> the entire i-th row of a matrix (in a list of lists) without
>> pre-computing the number of cols
#The code of rank 1 in the previous post should have read
#rank1<-apply(iterator1,1,function(x) x+base1)
#corrected code below
siegel.tukey=function(x,y,id.col=TRUE,adjust.median=F,rnd=-1,alternative="two.sided",mu=0,paired=FALSE,exact=FALSE,correct=TRUE,conf.int=FALSE,conf.level=0.95){
if(id.col=
It's hard to help if you keep changing the framework of your problem,
first two matrices - now it's a data.frame and a list of subset row
names in a plotting method from whatever package "suprow" comes from.
Regardless, Michael's original answer already gave you a solution:
plot(table1,type='l',lw
On Thu, Mar 8, 2012 at 4:41 AM, Massimo Di Stefano
wrote:
>
> Hello All,
>
> i've a set of observations that is in the form :
>
> a, b, c, d, e, f
> 67.12, 4.28, 1.7825, 30, 3, 16001
> 67.12, 4.28, 1.7825, 30, 3, 16001
> 66.57, 4.28, 1.355, 30,
2012/3/9 Uwe Ligges :
> I think the main issue of the OP is that he geneartes a 55000x55000 distance
> matrix and has to calculate on it. Beside immense main memory consumption
> this may take ages to complete with hierarchical clustering.
Indeed. I missed that in the original email.
If a non-hie
Here is my latest. I kind of changed the problem (for speed). In real life
I have over 300 uadata type matrices, each having over 20 rows and over
11,000 columns. However the rddata file is valid for all of the uadata
matrices that I have (300). What I am doing now: I'm creating a matrix of
row ind
Take a look at n-x+1, the second parameter to the beta distribution:
> n <- c(10, 45, 38)
> x <- rbind(c( 7, 45, 31),
+c(10, 40, 35),
+c( 9, 44, 33),
+c( 8, 44, 31),
+c( 8, 45, 36))
> n - x + 1
[,1] [,2] [,3]
[1,]4 -6 15
[2,] 36 -29
On Mar 9, 2012, at 2:48 PM, Anamika Chaudhuri wrote:
HI All:
Does anyone know the code behind the qbeta function in R?
Well, yes, but don't you think it would be wise to question whether
your code might be the problem rather than the R code?
I am using it to calculate exact confidence i
HI All:
Does anyone know the code behind the qbeta function in R?
I am using it to calculate exact confidence intervals and I am getting
'NaN' at places I shouldnt be. Heres the simple code I am using:
k<-3
> x<-NULL
> p<-rbeta(k,3,3)# so that the mean nausea rate is alpha/(alpha+beta)
> min<-10
Seems to be fine for me.
Uwe Ligges
On 07.03.2012 03:24, Robert King wrote:
Where should I report mirror problems? There doesn't seem to be anywhere on
http://cran.r-project.org/mirrors.html listing contact emails for mirror
admins.
There is some problem with the debian binaries on
http://cran
I think the main issue of the OP is that he geneartes a 55000x55000
distance matrix and has to calculate on it. Beside immense main memory
consumption this may take ages to complete with hierarchical clustering.
Uwe Ligges
On 08.03.2012 15:02, Sarah Goslee wrote:
See inline:
On Thu, Mar 8,
Thanks, we will try to get it updated soon.
Uwe
On 08.03.2012 09:40, Caitlin wrote:
Hi all.
I just noticed that the release of version 2.14.2 was not announced on the
R home page.
Thanks,
~Caitlin
[[alternative HTML version deleted]]
__
For a paper dealing with generalized ellipsoids, I want to illustrate in
3D an ellipsoid that is unbounded
in one dimension, having the shape of an infinite cylinder along, say,
z, but whose cross-section in (x,y)
is an ellipse, say, given by the 2x2 matrix cov(x,y).
I've looked at rgl:::cylind
On 09-03-2012, at 20:00, Michael wrote:
> Hi all,
>
> In using "rlm" I've got a bunch of warnings... "failed to converge in 20
> steps", etc.
>
> My question is:
>
> what are the results then after the failure?
>
They haven't converged. So inaccurate. Maybe your model is badly formulated or
Benilton Carvalho writes:
> Hi,
>
> what is the proper of of "passing a missing value" so I can extract
> the entire i-th row of a matrix (in a list of lists) without
> pre-computing the number of cols?
>
> For example, if I know that the matrices have 2 columns, I can do the
> following:
>
> s
I'm accessing R via a socket connection. I set up a connection using
socketConnection and then use readLines inside of a while(TRUE) loop to
listen for activity. Is that the best way of doing this sort of activity?
It works, that's not the issue, I am just wondering if there's a better
way.
Tha
Hassan,
Others have provided you with better solutions, but I hope this allows you
to see why yours didn't work.
# first (going with your code) you needed a data.frame called "x"
# here is an example:
x <- structure(list(x1 = c(0.0986048226696643, -0.445652024980979,
0.0893989676314604, -3.02
The response much appreciated. They do match up, one is a small subset of
the other.
I have this:
> dput(table1)
structure(list(var1 = c(2L, 4L, 4L, 1L, 423L), var2 = c(3L, 5L,
6L, 342L, 3L)), .Names = c("var1", "var2"), class = "data.frame", row.names
= c("node1",
"node2", "node3", "node4", "node
Hi All,
s = Surv(outcome.[,1], outcome.[,2])
survplot= (survfit(s ~ person.list[,1]))
summary(survplot)
This prints a summary of all the curves at specified time intervals of
events. Is there a way to suppress this summary to only display a su
Hi all,
In using "rlm" I've got a bunch of warnings... "failed to converge in 20
steps", etc.
My question is:
what are the results then after the failure?
Will "rlm" automatically downgrade back to "lm" upon failure?
Thanks a lot!
[[alternative HTML version deleted]]
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of aoife doherty
>
> Thank you. Can the chi-squared test compare two matrices that
> are not the same size, eg if matrix 1 is a 2 X 4 table, and
> matrix 2 is a 3 X 5 matrix?
N
Do your matrices "match up" with each other in any meaningful way or
do you just want two independent plots on a single page?
You should probably provide the dput() output of each table object so
we can see what you've got.
Michael
On Fri, Mar 9, 2012 at 11:07 AM, aoife doherty wrote:
> Many th
You should probably read up on what the chi-squared test actually
tests: in one form, it asks whether some set of observations could
have come from a given multinomial distribution. Concretely, it asks
whether it is "reasonable" to get 3 blues, 4 reds, and 2 whites from a
uniform distribution over
On Mar 9, 2012, at 12:10 PM, David Winsemius wrote:
On Mar 8, 2012, at 8:02 AM, Mauricio Zambrano-Bigiarini wrote:
Dear list members,
Within a loop, I need to create an xyplot with only a legend, not
even
with the default external box drawn by lattice.
I already managed to remove the ax
On Mar 8, 2012, at 8:02 AM, Mauricio Zambrano-Bigiarini wrote:
Dear list members,
Within a loop, I need to create an xyplot with only a legend, not even
with the default external box drawn by lattice.
I already managed to remove the axis labels and tick marks, but I
couldn't find in the docum
R tends to see the ordering of factor levels as a property of the data
rather than a property of the table/graph. So it is generally best to
modify the data object (factor) to represent what you want rather than
look for an option in the table/plot function (this will also be more
efficient in the
Thank you. Can the chi-squared test compare two matrices that are not the
same size, eg if matrix 1 is a 2 X 4 table, and matrix 2 is a 3 X 5 matrix?
On Fri, Mar 9, 2012 at 4:37 PM, Greg Snow <538...@gmail.com> wrote:
> The chi-squared test is one option (and seems reasonable to me if it
> the t
Why do you want to do this? Lattice was not really designed to put
just part of the graph up, but rather to create the entire graph using
one command.
If you want to show a process, putting up part of a graph at a time,
it may be better to create the whole graph as a vector graphics file
(pdf, po
The chi-squared test is one option (and seems reasonable to me if it
the the proportions/patterns that you want to test). One way to do
the test is to combine your 2 matrices into a 3 dimensional array (the
abind package may help here) and test using the loglin function.
On Thu, Mar 8, 2012 at 5:
Thank you very much Michael!
Best,
Aurelie
On 2012-03-09, at 11:56 AM, R. Michael Weylandt wrote:
> You are overriding "b" at each loop iteration and consequently only
> keeping the last one.
>
> Perhaps
>
> b <- list()
>
> for(i in sort(unique(OT1$month))){
> a<-OT1[OT1$month==i,]
> b
Thank you Tal, useful link.
Best,
Aurelie
On 2012-03-09, at 11:53 AM, Tal Galili wrote:
> Hi Aurelie,
> Please give this a look:
> http://www.nealgroothuis.name/introduction-to-data-types-and-objects-in-r/
>
> And see if this resolves most, or all, of your questions...
>
>
> C
Many thanks for reply.
I have trouble understanding how to use response, i am sorry.
My question is i have two matrices. I then plot two matrices. Then I have 2
seperate plots. I can color the nodes in the plots in two different colors.
Then, how do i merge the two plots to view one overlapping the
You are overriding "b" at each loop iteration and consequently only
keeping the last one.
Perhaps
b <- list()
for(i in sort(unique(OT1$month))){
a<-OT1[OT1$month==i,]
b[[i]]<-ppp(a$longitude,a$latitude,marks=a$fTSUM,window=newW)
plot(b[[i]],main=i)
}
Generally it's bad practi
Hi Aurelie,
Please give this a look:
http://www.nealgroothuis.name/introduction-to-data-types-and-objects-in-r/
And see if this resolves most, or all, of your questions...
Contact
Details:---
Contact me: tal.gal...@gmail.com |
No idea what table1, table2 are
plot(1:5, type = "l")
points(5:1, col = 2)
should get you started.
Michael
On Fri, Mar 9, 2012 at 10:17 AM, aaral singh wrote:
> Hello.
>
> I have 2 plots.
>
>> plot1 <-plot(table1)
>> plot2 <-plot(table2)
>
> How may i plot these both on the same graph, i.e
On Mar 9, 2012, at 10:41 AM, David Winsemius wrote:
On Mar 9, 2012, at 5:31 AM, RMSOPS wrote:
Hello,
the idea is to copy the d for df, with new results.
x<-data.frame(name="x1",pos=4,age=20)
x<-rbind(x,data.frame(name="x2",pos=5,age=20))
x<-rbind(x,data.frame(name="x3",pos=6,age=21))
x<
Dear all,
I'm trying to create a list of point patterns ppp.object {spatstat} in a loop.
My dataset looks like this:
> names(OT1);head(OT1);dim(OT1)
[1] "EID" "latitude" "longitude" "month" "year" "CPUE"
"TSUM"
[8] "fTSUM"
EID latitud
I would like to read hdf5 or h5 files in R .
I found two packages hdf5 and h5r , but in both packages they have
information how to get attribute from data.
In my case I would like see what kind of parameters are saved in every file.
I have tried
h5 <- hdf5load("GOSATTFTS20090601_02C02SV0002R101
Hello.
I have 2 plots.
> plot1 <-plot(table1)
> plot2 <-plot(table2)
How may i plot these both on the same graph, i.e. layer one graph on top of
the other one.
The result should look similar to this the image below, where the black
lines indicate one plot, and the red dots indicate the second pl
On Mar 9, 2012, at 5:31 AM, RMSOPS wrote:
Hello,
the idea is to copy the d for df, with new results.
x<-data.frame(name="x1",pos=4,age=20)
x<-rbind(x,data.frame(name="x2",pos=5,age=20))
x<-rbind(x,data.frame(name="x3",pos=6,age=21))
x<-rbind(x,data.frame(name="x4",pos=7,age=24))
x<-rbind(
Dear R-users,
__
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.
Dear Federico,
I am using the r-plugin provided for VIM on unix. This has an option of
building tag files.
http://www.vim.org/scripts/script.php?script_id=2628
I 've just tried the command, and albeit some error messages showed up,
a tag file for the current dir was build. Not sure how the ca
Folks,
This is more of a stat question than an R question.
Apologies in advance!
Suppose I fit an AR(1) to a time-series and also fit an AR(1) to the logs of
the same time-series and then simulate future paths.
In my case I see a big difference in the resulting paths. If I simulate
thousands
Hi,
what is the proper of of "passing a missing value" so I can extract
the entire i-th row of a matrix (in a list of lists) without
pre-computing the number of cols?
For example, if I know that the matrices have 2 columns, I can do the following:
set.seed(1)
x0 <- lapply(1:10, function(i) repli
On Fri, Mar 09, 2012 at 05:37:42AM -0800, hubinho wrote:
> Hello.
>
> I'm looking for a method to simulate n different 2x2 tables having all the
> same odds ratio.
>
> For example.
>
> I need
> 100 tables with odds ratio 1
> 100 tables with odds ratio 2
> 100 tables with odds ratio 3
>
> and s
Thanks Pascal,
The good new is the code works, but it doesn't produce the result that I
expected, or at least it doesn't match what matlab does. I did a bit of search
myself and came across the following post on stackoverflow
http://stackoverflow.com/questions/7746529/smoothing-with-lowess
Whi
Hello.
I'm looking for a method to simulate n different 2x2 tables having all the
same odds ratio.
For example.
I need
100 tables with odds ratio 1
100 tables with odds ratio 2
100 tables with odds ratio 3
and so on.
All tables should have the same marginal frequencies.
Thank you
--
View th
On 2012-03-09 15:30, David Winsemius wrote:
On Mar 9, 2012, at 6:14 AM, Hans Ekbrand wrote:
I need confint() for glm() to supress the messages
I'm wondering if suppressMessages would be helpful? Which in turn
suggests that you do not know how to use "??", so firt you should get
in the habi
On Fri, Mar 09, 2012 at 01:24:00PM +0330, Hassan Eini Zinab wrote:
> Dear All,
>
> I have a data set with variables x1, x2, x3, ..., x20 and I want to
> create z1, z2, z3, ..., z20 with the following formula:
>
>
> z1 = 200 - x1
> z2 = 200 - x2
> z3 = 200 - x3
> .
> .
> .
> z20 = 200 - x20.
>
>
Hello everybody, I'm looking for someone who is able with MCA and would
like to gives some help.
If what I'm doing is not wrong, according to the purpose I have, I need to
understand how to create a dependence matrix, where I can analyze the
dependence between all my variables.
Till now this is wh
Hi Robert,
If you type
?loess
It pulls up the documentation. What about that function do you not like? As
you said, it needs two variables, but typically the second is just your
time index. Try this:
n <- 50
x <- rep(0,n)
for(i in 2:n){
x[i] <- rnorm(1,x[i-1])
}
loess(x ~ seq(1,n))
plot(1:n,x,t
On Mar 9, 2012, at 6:14 AM, Hans Ekbrand wrote:
I need confint() for glm() to supress the messages
I'm wondering if suppressMessages would be helpful? Which in turn
suggests that you do not know how to use "??", so firt you should get
in the habit of doing a helpSearch before posting.
?
You could also have a look at the LaF package which is written to
handle large text files:
http://cran.r-project.org/web/packages/LaF/index.html
Under the vignettes you'll find a manual.
Note: LaF does not help you to fit 9GB of data in 4GB of memory, but
it could help you reading your fi
You should send this to r-h...@stat.math.ethz.ch.
On 03/09/2012 09:21 AM, Andrea Sica wrote:
> Hello everybody, I'm looking for someone who is able with MCA and
> would like to gives some help.
>
> If what I'm doing is not wrong, according to the purpose I have, I
> need to understand how to cre
Hello
Thanks for the reply. As yet I have not much experience in r, although I
make some mistakes.
Any tips to solve the problem of a more effective way.
Regards
--
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Sent from
On Fri, 2012-03-09 at 12:00 +0100, r-help-requ...@r-project.org wrote:
> > A note on standard errors: ?S(t) +- std is a terrible confidence
> > interval. ?You will be much more accurate if you use log
> scale. ?(Some
> > argue for logit or log-log, in truth they all work well.) ? If n is
> large
>
On Mar 8, 2012, at 7:37 PM, Jeff Garcia wrote:
I have a simulated matrix of dates that I generated from a
probability function. Each column represents a single iteration.
I would like to bin each run _separately_ by decades and dump them
into a new matrix where each column is the length of
On 3/8/2012 1:08 PM, Gabriel Yospin wrote:
I would like to make a legible boxplot of tree growth rates for each of
seven tree species at each of seven different sites. It's a lot of data to
put on one figure, I know. I made a beautiful, interpretable figure using
color, but my target journal can'
First, be sure your R and the R associated with Rstudio are the same R
versions. In Rstudio, check Tools -> Options -> R version. It looks as
if your R-studio is running the 32-bit version of R.
Aayush Raman
Sent by: r-help-boun...@r-project.org
03/09/2012 07:11 AM
To
r-help@r-project.o
Perhaps ?polygon
Michael
On Fri, Mar 9, 2012 at 6:10 AM, aoife wrote:
> May I ask, is it possible using plotrix to shade a group of variables
> differentially from the rest of a graph, eg so the output looks similar to
> this, where the nodes of open circles are my nodes of interest:
>
> http://
On Fri, Mar 9, 2012 at 8:35 AM, carol white wrote:
> Hello,
> How is it possible to sort dates in R?
>
Your question has already been answered but note that if your data is
a time series and you represent it using zoo it will automatically be
sorted. Here dates is in reverse chronological order
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