MIchael -
Thanks for your insight. I think I see where you're going with this.
To make '==' comparisons for subsetting against an ordered factor, I've had
to create a lookup table for all possible values I'd ever want to compare
against (all dates covered by the quarters in question, in this
df - structure(c(106487, 495681, 1597442,
2452577, 2065141, 2271925, 4735484, 3555352,
8056040, 4321887, 2463194, 347566, 621147,
1325727, 1123492, 800368, 761550, 1359737,
1073726, 36, 53, 141, 41538, 64759, 124160,
69942, 74862, 323543, 247236, 112059, 16595,
Hi everyone / poLCA gurus,
I'm facing a problem regarding conditional dependence and poLCA.
Using a few data sets composed of only 1's and 2's, i'm able to do a latent
class analysis for independent data columns using poLCA.
The commands I use to do this are something like,
bcs =
Thanks, I got it! (And I think I should have googled what replicated
means!) However, then Bortz, Lienert, Boehnke are imprecise, if not
wrong: Der Friedman-Test setzt voraus, dass die N Individuen
wechselseitig unabhängig sind, dass also nicht etwa ein und dasselbe
Individuum zweimal oder
Dear people,
I created a plot which looks like this:
Ee1-matrix(c(88,86,74,62,41),ncol=5)
colnames(Ee1)-c(Lehrer,Lehrerinnen,Klassenkollegen,Klassenkolleginnen,Geschwister)
par(las=1)
par(mar=c(5,13,4,2))
barplot(Ee1,horiz=T,col=grey85,border=NA,xlim=c(0,100),axes=F)
axis(2,pos=10, tick=T,
Hi friends,
I have following data and would like to plot this with barchart() availble
with lattice package.
RsID Freqs Genotype
AAA 63.636 1/1
AAA 32.727 1/2
AAA 3.636 2/2
BBB 85.965 2/2
BBB 14.035 2/1
CCC 63.158 1/1
CCC 21.053 1/2
CCC 15.789 2/2
Hi Marion,
is all you want the white vertical lines? Then try
abline(v = seq(10, 90, by = 10), col = white)
instead of your axis commands.
Regards,
Enrico
Am 20.02.2012 11:04, schrieb Marion Wenty:
Dear people,
I created a plot which looks like this:
Ee1-matrix(c(88,86,74,62,41),ncol=5)
Hello,
I am new to R and currently have the following problem:
I have successfully loaded my data in R which consists of two numeric columns
(LI_F and female) and one character column (Strain). So far I can plot two
different set of boxplots for each of the numeric columns plotted by the groups
Hi all, I am having difficulties to understand how R sort strings:
If I do
R) sort(c(X.,X0B))
[1] X. X0B
So for me, as far as lexicographic order is concerned I can add whatever to
the end, the order will remain the same, but :
R) sort(c(X.Z,X0B.Z))
[1] X0B.Z X.Z
Can somebody give me a trick
Hi all,
I have a question concerning using several conditions in an ifelse function
used as the function in apply.
I want to create a new value with the function ifelse ‘ object which can be
coerced to logical mode “test[n,] 1 test[n-1,]==0”
With n I mean the row. I don’t know how I could
Hi Marion,
you can either use any of the *apply-functions or vectorize your
function (which internally uses mapply):
par(las=1)
par(mar=c(5,13,4,2))
barplot(Ee1,horiz=T,col=grey85,border=NA,xlim=c(0,100),axes=F)
#using sapply
invisible(sapply((1:9)*10,function(x)axis(2,pos=x,tick=T, tcl=F,
On Mon, Feb 20, 2012 at 02:18:42AM -0800, statquant2 wrote:
Hi all, I am having difficulties to understand how R sort strings:
If I do
R) sort(c(X.,X0B))
[1] X. X0B
So for me, as far as lexicographic order is concerned I can add whatever to
the end, the order will remain the same, but :
Hello,
I am trying to fit gamma, negative exponential and inverse power functions
to a dataset, and then test whether the fit of each curve is good. To do
this I have been advised to calculate predicted values for bins of data (I
have grouped a continuous range of distances into 1km bins), and
On 20.02.2012 01:54, David Winsemius wrote:
On Feb 19, 2012, at 7:45 PM, zheng wei wrote:
I did not know this before. I installed it as you suggested. what to
do next?
Read the Installation Manual?
And don't forget this is a source package for which no CRAN Windows
binary exists,
On 19.02.2012 18:58, Idielle Walters wrote:
Hi
I am struggling to install GTK+ for Windows 7. RGtk2 needs this
package to load. Does anybody know of a installation file that works?
See the ReadMe for Windows binary packages. For the current R-reelase
this is
Petr Savicky savi...@cs.cas.cz wrote in message
news:20120220105153.gc21...@cs.cas.cz...
On Mon, Feb 20, 2012 at 02:18:42AM -0800, statquant2 wrote:
Hi all, I am having difficulties to understand how R sort strings:
If I do
R) sort(c(X.,X0B))
[1] X. X0B
So for me, as far as
Hi everybody!
First of all, I would like to point that I am newebie in using R. The issue is
that I need to install lsa package in R. In theory, I have downloaded and
installed all necessary packages to run lsa library, but when I am going to
load it I get this message:
library(lsa)Loading
I am struggling to install GTK+ for Windows 7. RGtk2 needs this
package to load. Does anybody know of a installation file that works?
GTK+ is automatically installed when you install the RGtk2 package
(you'll be asked about it during installation). As of R-2.14.1, it is
installed under
On 20-02-2012, at 10:40, Nerak wrote:
Hi all,
I have a question concerning using several conditions in an ifelse function
used as the function in apply.
I want to create a new value with the function ifelse ‘ object which can be
coerced to logical mode “test[n,] 1 test[n-1,]==0”
See ?Comparison, which holds some warnings about what to expect when
sorting strings.
Am 20.02.2012 11:51, schrieb Petr Savicky:
On Mon, Feb 20, 2012 at 02:18:42AM -0800, statquant2 wrote:
Hi all, I am having difficulties to understand how R sort strings:
If I do
R) sort(c(X.,X0B))
[1] X.
Ok so it changed from 2.12.2 to 2.14.1 ??
Can somebody tell me how to modify my sort or whatever to get the save
resilt that I would get in 2.14.1 ?
Cheers
--
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Sent from the R help mailing list
Hi,
I need some neat ways of determing a subset of binary strings. For example,
x=c(0,0,1), y=c(0,1,1), z=c(0,1,0). So x is a subset of y and z is also a
subset of y, but x is not a subset of z.
I tried to search R functions and packages but no hits. Any ideas?
Best,
Jing
--
Jing Tang,
hi
i'm facing a problem
How to inset a cell into a matrix?
for example i have a matrix:
332 244 332244
665 332 665332
785 785 785665
i want to covert the matrix like this:
02440244
332 332
Hello R people,
I have created a '.csv' file of 100 rows by 20 columns whose each cell
contains a random numbers between 0 1, thru a Java program. Once that is
created a signal (just a letter) is send to the port of a socket
connection at localhost, which was earlier started by an R session.
Hi Enrico,
Yes, you were right, I just wanted to draw white vertical lines and with
your command it's much simpler. Thanks a lot!
Hi Eik,
I learned a lot from your tips about sapply and Vectorize!
I had only used apply, lapply and mapply but not yet sapply.
I find the Vectorize function very
I like it better. Thanks!
Ben
On Fri, Feb 17, 2012 at 11:38 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
On Sat, Feb 18, 2012 at 12:44 AM, Ben quant ccqu...@gmail.com wrote:
The code below works as expected but:
Using the proto package, is this the best way to 1) make a parameter
Hello,
I would like to perform triangular test for clinical trial with R.
can you help me please ?
Jan
[[alternative HTML version deleted]]
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the
On 20-02-2012, at 14:15, jing tang wrote:
Hi,
I need some neat ways of determing a subset of binary strings. For example,
x=c(0,0,1), y=c(0,1,1), z=c(0,1,0). So x is a subset of y and z is also a
subset of y, but x is not a subset of z.
I tried to search R functions and packages but no
Summary of the query: update does not work on a coxme object
I ran into this bug myself 2 days ago -- I rarely use update() so
hadn't encountered it before. The problem is that coxme breaks the
formula into fixed and random portions, and this confuses the default
method for formula.
Solution:
On Feb 20, 2012, at 1:45 AM, johnmark wrote:
MIchael -
Thanks for your insight. I think I see where you're going with this.
To make '==' comparisons for subsetting against an ordered factor,
I've had
to create a lookup table for all possible values I'd ever want to
compare
against (all
On Feb 20, 2012, at 5:34 AM, MLSC wrote:
Hi friends,
I have following data and would like to plot this with barchart()
availble
with lattice package.
RsID Freqs Genotype
AAA 63.636 1/1
AAA 32.727 1/2
AAA 3.636 2/2
BBB 85.965 2/2
BBB 14.035 2/1
CCC 63.158
Hi Jing,
I am not sure I got your definition of a subset right, but maybe this
helps.
Regards,
Enrico
x - c(0,0,1)
y - c(0,1,1)
z - c(0,1,0)
## is x a 'subset' of y?
isSubset - function(x, y) {
x - as.logical(x)
y - as.logical(y)
all(y[x] == TRUE)
}
isSubset(x, y)
isSubset(z,
On Feb 20, 2012, at 5:04 AM, Marion Wenty wrote:
Dear people,
I created a plot which looks like this:
Ee1-matrix(c(88,86,74,62,41),ncol=5)
colnames(Ee1)-
c
(Lehrer
,Lehrerinnen,Klassenkollegen,Klassenkolleginnen,Geschwister)
par(las=1)
par(mar=c(5,13,4,2))
On Feb 20, 2012, at 5:57 AM, Louise Mair wrote:
Hello,
I am trying to fit gamma, negative exponential and inverse power
functions
to a dataset, and then test whether the fit of each curve is good.
To do
this I have been advised to calculate predicted values for bins of
data (I
have
I did, but this does not give the answer to my question...
Anybody knows how to tweack the behaviour of sort or how to do ?
--
View this message in context:
http://r.789695.n4.nabble.com/Sorting-strings-tp4403696p4404091.html
Sent from the R help mailing list archive at Nabble.com.
The 10% rule does not provide a unique answer. Should it apply to the
cumulative probability, its logarithm, or log-log (log hazard scale)? Many
studies are too small to achieve 10% at any time point. I think it is more
traditional (but not without bias) to stop where fewer than 10 subjects are
I do not see the restricted range that you report.
There is probably some masking as David pointed out.
Try again in a fresh R session with --vanilla.
You probably want three additional arguments.
update(.Last.value, xlim=c(0, 100), between=list(x=1, y=1), origin = 0)
to the barchart function
I don't *think* it's version specific, but rather it depends on your
(still unstated) locale, as the documentation goes to great lengths to
point out. Change that and you might see different behaviors.
Michael
On Mon, Feb 20, 2012 at 8:55 AM, statquant2 statqu...@gmail.com wrote:
I did, but
This is copy paste from my session:
xyz-as.vector(c(ls(),as.matrix(lapply(ls(),class
dim(xyz)-c(length(xyz)/2,2)
allobj-function(){
+ xyz-as.vector(c(ls(),as.matrix(lapply(ls(),class;
+ dim(xyz)-c(length(xyz)/2,2);
+ return(xyz)
+ }
xyz
[,1] [,2]
[1,] a
Folks,
I'm trying to get stats from a matrix for each transition from one state to
another.
I have a matrix x as below.
structure(c(0, 2, 2, 2, 0, 0, 0, 1, 1, 1, 1, 2, 2, 1, 1, 1, 0,
0, 2, 2, 0.21, -0.57, -0.59, 0.16, -1.62, 0.18, -0.81, -0.19,
-0.76, 0.74, -1.51, 2.79, 0.41, 1.63, -0.86,
On Feb 20, 2012, at 10:07 AM, Ajay Askoolum wrote:
This is copy paste from my session:
xyz-as.vector(c(ls(),as.matrix(lapply(ls(),class
dim(xyz)-c(length(xyz)/2,2)
allobj-function(){
+ xyz-as.vector(c(ls(),as.matrix(lapply(ls(),class;
+ dim(xyz)-c(length(xyz)/2,2);
+ return(xyz)
+
Sorry, just checked it and you need to add .GlobalEnv to both ls() calls.
Michael
On Mon, Feb 20, 2012 at 10:17 AM, R. Michael Weylandt
michael.weyla...@gmail.com wrote:
Short answer, environments -- ls() looks (by default) in its current
environment, which is not the same as the global
Short answer, environments -- ls() looks (by default) in its current
environment, which is not the same as the global environment when
being called inside a function.
This would (I think) give the same answer but I haven't checked it. :
allobj-function(){
+
Sometimes you want to compute the physical size of a plot based on data.
In R itself this is no problem.
But is there a way to compute the values of height and width in S-weave, say:
graph,fig=TRUE,height=xx,width=yy=
where xx and yy are computed and not physically written in the document?
Hi
This is copy paste from my session:
xyz-as.vector(c(ls(),as.matrix(lapply(ls(),class
dim(xyz)-c(length(xyz)/2,2)
allobj-function(){
+ xyz-as.vector(c(ls(),as.matrix(lapply(ls(),class;
+ dim(xyz)-c(length(xyz)/2,2);
+ return(xyz)
+ }
xyz
[,1] [,2]
On Feb 20, 2012, at 10:11 AM, murali.me...@avivainvestors.com wrote:
Folks,
I'm trying to get stats from a matrix for each transition from one
state to another.
I have a matrix x as below.
structure(c(0, 2, 2, 2, 0, 0, 0, 1, 1, 1, 1, 2, 2, 1, 1, 1, 0,
0, 2, 2, 0.21, -0.57, -0.59, 0.16,
Thank you Chuck,
Here is the head of my data set (tjornres):
Fish.1 Fish.2 MORPHO DIET
1 1 20.03768 0.1559250
2 1 30.05609 0.7897060
3 1 40.03934 0.4638010
4 1 50.03363
On Mon, Feb 20, 2012 at 03:15:53PM +0200, jing tang wrote:
Hi,
I need some neat ways of determing a subset of binary strings. For example,
x=c(0,0,1), y=c(0,1,1), z=c(0,1,0). So x is a subset of y and z is also a
subset of y, but x is not a subset of z.
I tried to search R functions and
On Mon, Feb 20, 2012 at 05:55:30AM -0800, statquant2 wrote:
I did, but this does not give the answer to my question...
Anybody knows how to tweack the behaviour of sort or how to do ?
Hi.
Try this
Sys.setlocale(LC_COLLATE, C)
This comes from ?locale and reads there
Hello,
statquant2 wrote
Ok so it changed from 2.12.2 to 2.14.1 ??
Can somebody tell me how to modify my sort or whatever to get the save
resilt that I would get in 2.14.1 ?
Cheers
I don't know about 2.12.2 but for 2.12.0 I get:
R.version
_
Hello,
I am trying to do a meta-analysis where each study has two arms, similar to the
BCG data set. However, follow-up duration was different for each study arm, so
I would like to fit a model that uses the length of follow-up in each arm as a
moderator. Is this possible?
For example, if
Hi all,
I am trying to multiply each column of a matrix such to have a unique resulting
vector with length equal to the number of rows of the original matrix. In short
I would like to do what prod(.) function in Matlab does, i.e.
A -matrix(c(1:10),5,2)
V = A[,1]*A[,2]
Thank you
Graziano
Ok I have :
R) str(R.Version())
List of 13
$ platform : chr x86_64-unknown-linux-gnu
$ arch : chr x86_64
$ os: chr linux-gnu
$ system: chr x86_64, linux-gnu
$ status: chr
$ major : chr 2
$ minor : chr 12.2
$ year : chr
Dears,
I am a new R user and I am trying to analyze my data sets, R gives me a default
when I type in the regression formula as following:
fit1=gamlss(tot_remun_revenue$tot_remun.y~tot_remun_revenue$revenue.x,family=NO)
Fehler in model.frame.default(formula =
Try
apply(A, 1, prod)
I hope it helps.
Best,
Dimitris
On 2/20/2012 4:21 PM, Graziano Mirata wrote:
Hi all,
I am trying to multiply each column of a matrix such to have a unique resulting
vector with length equal to the number of rows of the original matrix. In short
I would like to do
On 20-02-2012, at 16:11, murali.me...@avivainvestors.com
murali.me...@avivainvestors.com wrote:
Folks,
I'm trying to get stats from a matrix for each transition from one state to
another.
I have a matrix x as below.
structure(c(0, 2, 2, 2, 0, 0, 0, 1, 1, 1, 1, 2, 2, 1, 1, 1, 0,
0,
It seems OS-dependent. I got different results when trying it on windows
xp and Redhat linux.
R.version
_
platform x86_64-unknown-linux-gnu
arch x86_64
os linux-gnu
system x86_64, linux-gnu
status
major 2
minor 9.1
year
On Mon, Feb 20, 2012 at 04:56:21PM +0100, Petr Savicky wrote:
On Mon, Feb 20, 2012 at 05:55:30AM -0800, statquant2 wrote:
I did, but this does not give the answer to my question...
Anybody knows how to tweack the behaviour of sort or how to do ?
Hi.
Try this
Sorry, just made a mistake. This is the result from windows xp.
sort(c(X.,X0B))
[1] X. X0B
sort(c(X.Z,X0B.Z))
[1] X.Z X0B.Z
R.version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major 2
minor
On Feb 20, 2012, at 10:26 AM, Thorsten Pöllinger wrote:
Dears,
I am a new R user and I am trying to analyze my data sets, R gives
me a default when I type in the regression formula as following:
fit1=gamlss(tot_remun_revenue$tot_remun.y~tot_remun_revenue
$revenue.x,family=NO)
Most
No, the authors are correct: the individuals (i.e. the 17 individuals) you have
need to be independent (i.e. no correlation between them, let alone any
individual running through your temporal experiment more than once, as
indicated in the citation), while the *observations* are of course
Many thanks for the suggestion.
I tried it already, but as I've never wrote a function, I had no luck.
If it's not asked too much, maybe you or somebody else could help me
getting the code for the function right.
Regards
Simon
Am 20/02/2012 14:52, schrieb Terry Therneau:
Summary of the
On 2012-2-20 23:15, Rui Barradas wrote:
Could it be OS related?
Yes, it seems. I tried it on my local windows xp and redhat linux
server, and got different results. Hope it will be fixed in the future
versions. Maybe we should keep alert to check whether the results are
consistent when
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
At a rough guess you may want to have a look at the mfrow in ?par but without
some sample data and a bit more information about what you need it is
On 20-Feb-2012 Petr Savicky wrote:
On Mon, Feb 20, 2012 at 05:55:30AM -0800, statquant2 wrote:
I did, but this does not give the answer to my question...
Anybody knows how to tweack the behaviour of sort or how to do ?
Hi.
Try this
Sys.setlocale(LC_COLLATE, C)
This comes from
Hi,
That is not what I get. After running your code I get
levels(df.m$Period)
[1] 1991-00 1901-10 1981-90 2001-06 1911-20 1881-90 1971-80
[8] 1921-30 1891-00 1961-70 1871-80 1851-60 1951-60 1861-70
[15] 1841-50 1941-50 1831-40 1931-40 1820-30
this is my version info:
sessionInfo()
R version
Simon,
Thanks alot for the further clarification. As i said some where in my
lengthy explanation - i don't what of the myriad steps were needed, only
that they were performed and i now have an up-to-date rJava.
Next time (actually coming up soon) i'll certainly be following the
couple
On Feb 20, 2012, at 11:41 AM, Simon Tragust wrote:
Many thanks for the suggestion.
I tried it already,
it means .. what exactly?
but as I've never wrote a function, I had no luck.
no luck means ... what?
If it's not asked too much, maybe you or somebody else could help me
getting the
I guess that is not possible with Sweave, but it is possible in the
knitr package (an alternative to Sweave). You can set
opts_knit$set(eval.opts = c('fig.height', 'fig.width'))
so that these two options will be evaluated as R expressions (e.g.
fig.height=x means it takes value from a variable
On 20-Feb-2012 Graziano Mirata wrote:
Hi all,
I am trying to multiply each column of a matrix such to have
a unique resulting vector with length equal to the number of
rows of the original matrix. In short I would like to do what
prod(.) function in Matlab does, i.e.
A -matrix(c(1:10),5,2)
Dear all,
I need to run a quantile regression but considering a different objective
function to be minimized: instead of finding the parameters that minimize
rho(t) = u*(t - I(u0)), I need to find the parameters beta that minimize a sum
of two rho functions rho(t1) = u1*(t1 - I(u10)) and
Hi! I'm deploying R behind a web-app on a linux-server and I don't want
to grant the users shell access through the system() function for
security reasons.
Is there any safe way to deny a user access to the function?
I tried
a) alter the function in the R-Sources before compiling them.
NICE DDE
It solves my problem !
Awesome stuff
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I have got two kinds of nomogram for the same data set analyzed by Cox
regression. My question is how to compare the concordance index between
these two nomograms with R?
--
View this message in context:
How can I had a trend line to my plot?
My data looks like this:
Date=seq(as.Date(1910/1/1), as.Date(1920/1/1), day)
Values=runif(length(Date), min=-5, max=5)
dataset=data.frame(Values,as.Date(Date))
I just want to add a linear trend line to this
plot(dataset,col=rgb(1,0,0,1/8),cex=0.5,pch=19)
Hello John,
Thanks for your fast answer. I will try to be clearer and more detailed this
time. At the moment I am importing a dataframe like below as a '.csv file'. I
want to generate a boxplot for M and F values grouped by X whereby boxplots for
M and F should be above or very close to each
On Feb 20, 2012, at 1:59 PM, Mirjam Appel wrote:
Hello John,
Thanks for your fast answer. I will try to be clearer and more
detailed this time. At the moment I am importing a dataframe like
below as a '.csv file'. I want to generate a boxplot for M and F
values grouped by X whereby
On Mon, Feb 20, 2012 at 06:17:23PM +0100, li...@mwoywod.de wrote:
Hi! I'm deploying R behind a web-app on a linux-server and I don't want
to grant the users shell access through the system() function for
security reasons.
Is there any safe way to deny a user access to the function?
Hi.
The http://mxcl.github.com/homebrew/ Homebrew package manager is an exelent
way to manage packages like R on osx. I highly recommend it to any
developer. I am hoping there are others in the R community that are
interested in getting this working too. Does anyone else run a Homebrew
install of
I want to use persp to graph my data and it keeps giving me the error
increasing 'x' and 'y' values expected, even though my data is in increasing
order with respect to x and y.
Here is the code I'm currently using:
bob= scan (SBA3dataTaxonB.txt,what=char)
labels = bob[1:3]
bob=bob[-c(1,2,3)]
Does something like the code below give you want you want?
It requires the ggplot2 package so you will likely have to install it.
John Kane
Kingston ON Canada
# sample data converted using dput
xx - structure(list(X = c(Exp1, Exp1, Exp1,
On Feb 20, 2012, at 2:15 PM, adick wrote:
I want to use persp to graph my data and it keeps giving me the error
increasing 'x' and 'y' values expected, even though my data is in
increasing
order with respect to x and y.
Do you have missing entries? (Running your data fragment through
On Feb 20, 2012, at 9:15 AM, BXC (Bendix Carstensen) wrote:
Sometimes you want to compute the physical size of a plot based on data.
In R itself this is no problem.
But is there a way to compute the values of height and width in S-weave, say:
graph,fig=TRUE,height=xx,width=yy=
where xx
[See at end]
On 20-Feb-2012 Ted Harding wrote:
On 20-Feb-2012 Graziano Mirata wrote:
Hi all,
I am trying to multiply each column of a matrix such to have
a unique resulting vector with length equal to the number of
rows of the original matrix. In short I would like to do what
prod(.)
This isn't being plotted with any special time series methods (because
it's not a time-based object here) so ?abline will work.
Michael
On Mon, Feb 20, 2012 at 12:14 PM, anaraster rrast...@gmail.com wrote:
How can I had a trend line to my plot?
My data looks like this:
I have three character strings represented below as seq1, seq2, and seq3. Each
string has a reference character different from the other. Thus, for seq1, the
reference character is U, seq2, S (3rd S from left where A is leftmost
character) and for seq3 Y.
seq1 = PQRTUWXYseq2 = AQSDSSDHRSseq3 =
Hello,
Is there any formula or way to compare slopes of different functions?
If we fit 2 functions in our data, and we have 2 slope parameters, how can we
compare these slopes? Plotting y=5x and y=exp(5x) in which slope is equal to 5
in both of them.. doesn't seem that it makes sense to
A nomogram is a graphical device for displaying a model. Please try again
with your question. If you also wrote me privately earlier today please
choose one mode of communication.
Thanks
Frank
lijundfgd wrote
I have got two kinds of nomogram for the same data set analyzed by Cox
regression.
Yes, with the old good cat() and results=tex, you can do anything. It
is just so unnatural. Why must a simple task like setting the size of
a plot involve with so much coding work?
% complete knitr code
setup, include=FALSE=
opts_knit$set(eval.opts = c('fig.height', 'fig.width'))
my.height = 6;
Hi John,
What do you mean by 'compare'? y = exp(5x) could equally be said: y =
exp(x)^5 so no, the slopes of those two lines are not the same or are
only equal when you take the natural logarithm of y (log_{e}(y) =
log_{e}(5x).
They specify _completely_ different relationships between x and y;
Hi, I am wondering if we can make prediction on a linear mixed model by lmer()
from lme4 package? Specifically I am fitting a very simple glmer() with
binomial family distribution, and want to see if I can get the predicted
probability like that in regular logistic regression?
Dear All,
I am new to calling R in C.
I want to call sample R function in C. i.e. I want to do a sample
equivalent to
n - c(1:10)
p - seq(0,10,length.out= 10)
sample(n,size = 1, prob = p, replace = FALSE)
how can I call this function directly in C?
any help would be great,
thanks,
nitin
Hi, all,
I have a really big matrix that I want to run k-means on.
I tried:
data -
read.big.memory('mydata.csv',type='double',backingfile='mydata.bin',descriptorfile='mydata.desc')
I'm using doMC to register multicore.
library(doMC)
registerDoMC(cores=8)
ans-bigkmeans(data,k)
In system monitor,
Hi,
On Mon, Feb 20, 2012 at 6:15 PM, nitin kumar nit...@gmail.com wrote:
Dear All,
I am new to calling R in C.
I want to call sample R function in C. i.e. I want to do a sample
equivalent to
n - c(1:10)
p - seq(0,10,length.out= 10)
sample(n,size = 1, prob = p, replace = FALSE)
how
Hi All,
I'm trying to label my plot axis with times (HH:MM) that correspond to a
numeric index (values 0:6) for my time variable. I'd like to plot 08:00,
12:00, and so on, instead of 0 through 6.
I have used the following line of code:
axis(1, 0:6, labels=c(08:00, 12:00, 16:00, 20:00,
This works for me:
plot(0:6, runif(7), xaxt=n)
axis(1, at=0:6, labels=c(08:00, 12:00, 16:00, 20:00, 24:00,
04:00, 08:00), cex=0.8)
You need the xaxt=n in the plot statement, and the correct form is at=0:6
Sarah
On Mon, Feb 20, 2012 at 6:39 PM, Gerard Smits g_sm...@verizon.net wrote:
Hi
Hello,
This is Elaine.
I am drawing a plot with x-axis label with km square as the unit.
Now I want to print km square in the form of km2 and output 2 as the
uppercase.
Please kindly help suggest command to show the uppercase.
Thank you.
Elaine
[[alternative HTML version deleted]]
Worked like a charm! Thanks for your help. Gerard
On Feb 20, 2012, at 3:52 PM, Sarah Goslee wrote:
This works for me:
plot(0:6, runif(7), xaxt=n)
axis(1, at=0:6, labels=c(08:00, 12:00, 16:00, 20:00, 24:00,
04:00, 08:00), cex=0.8)
You need the xaxt=n in the plot statement, and the
On 21/02/12 12:54, Elaine Kuo wrote:
Hello,
This is Elaine.
I am drawing a plot with x-axis label with km square as the unit.
Now I want to print km square in the form of km2 and output 2 as the
uppercase.
Please kindly help suggest command to show the uppercase.
When you say as [the]
Dear R-list,
I am wondering how to perform a bootstrap in R for the weighted time
dependent Cox model (Andersen–Gill format, with multiple observations
from each patients) to obtain the bootstrap standard error of the
treatment effect.
Below is an example dataset. Would 'censboot' be
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