inal Message
From: array chip
To: r-help@r-project.org
Sent: Tue, August 3, 2010 4:41:50 PM
Subject: [R] xlsx package
HI, I was trying to install xlsx package for reading in Excel 2007 files. The
installation went smoothly. But when I tried to load the library, I got the
following error me
Hi, I would like to a draw a scatterplot of x1 and x2 (plot (x1, x2)), and also
want to draw a sort of "regression" line across the data points. But x1 and x2
are just 2 independent variables, so in this case a regression of x1 over x2,
or
vice versa, is not appropriate per se. What would be an
Hi, I have a simple plot by plot(x,y, log='xy'). However, due to the large
range
of values of x, the x-axis annotation is printed as "2e+02 1e+03 5e+03 2e+04
1e+05" instead of "200 1000 5000 2 10". How can I make it printed
as in the later one?
Thanks
John
_
Thank you David!
- Original Message
From: David Winsemius
To: array chip
Cc: r-help@r-project.org
Sent: Mon, August 9, 2010 12:42:06 PM
Subject: Re: [R] x-axis annotation
On Aug 9, 2010, at 3:14 PM, array chip wrote:
> Hi, I have a simple plot by plot(x,y, log='xy').
Hi how can print x-axis labels in 45 degree in boxplot() (or plot in general)?
I
can use las=2 to print in 90 degree, but it looks ugly. Is there a simple
option
to do 45 degree easily?
Thanks
John
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https://stat.
Than you Marc.
John
- Original Message
From: Marc Schwartz
To: array chip
Cc: r-help@r-project.org
Sent: Thu, August 12, 2010 12:17:12 PM
Subject: Re: [R] x-axis label print in 45 degree
On Aug 12, 2010, at 2:14 PM, array chip wrote:
> Hi how can print x-axis labels in 45 deg
I searched with "print x-axis label in 45 degree" which didn't return useful
links. Apparently I used poor search keywords.
- Original Message
From: David Winsemius
To: Marc Schwartz
Cc: array chip ; r-help@r-project.org
Sent: Thu, August 12, 2010 12:34:16 PM
Sub
Hi,
When I plot, the axis ticks are printed as "50.00 25.00 10.00 1.00 0.05
0.01", is there any way to print them as "50 25 10 1 0.05 0.01" instead?
Thanks
John
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work - plot by plot. Any other suggestions?
Many thanks
John
- Original Message ----
From: David Winsemius
To: array chip
Cc: R-Help
Sent: Mon, August 16, 2010 1:19:57 PM
Subject: Re: [R] print numbers
On Aug 16, 2010, at 3:53 PM, array chip wrote:
> Hi,
>
> When I plot, the
Hi, I am really puzzled by this. hope someone can help me
I have a 2 small data frames "a" and "b" derived from a larger data frames.
They
look exactly the same to me, but identical() always returns FALSE.
> a
a b
2 10011048 L
4 10011048 R
6 10011049 L
8 10011049 R
> b
a b
1 1
Thank you all for the suggestions. They all worked. Now, those numbers on the
ticks are produced by default plot function "bxp()", i.e. they are different in
each plot, is there any functions I can retrieve them so I can use formatC() or
prettyNum() etc?
Thanks again,
John
- Original Me
Oops, I overlooked the row names. Sorry for my carelessness.
Thanks
- Original Message
From: Erik Iverson
To: array chip
Cc: r-h...@stat.math.ethz.ch
Sent: Mon, August 16, 2010 2:53:37 PM
Subject: Re: [R] identical()
Hello,
array chip wrote:
> Hi, I am really puzzled by t
Marc, this works perfectly!
Thanks
John
- Original Message
From: Marc Schwartz
To: William Dunlap
Cc: array chip ; r-help@r-project.org
Sent: Mon, August 16, 2010 3:23:33 PM
Subject: Re: [R] print numbers
Bill et al,
See ?axTicks
plot(3^(0:5), 0:5, log="x",
Hi all, let me give a simple example:
b<-20
I would like to print ylab as "P20" where "P" is printed in Italic font. When I
do the following:
plot(1, ylab=expression(paste(italic("P"),b,sep="")))
I got y axis label printed as "Pb" instead of "P20". What is the best solution
to print platmath s
Thanks, yes it worked!
What about if I want to print as "P2, A" where A is just letter A and 2 is from
variable b.
John
- Original Message
From: baptiste auguie
To: array chip
Cc: r-h...@stat.math.ethz.ch
Sent: Thu, August 19, 2010 11:08:10 AM
Subject: Re: [R] plotmat
Thanks David!
John
- Original Message
From: David Winsemius
To: array chip
Cc: baptiste auguie ; r-h...@stat.math.ethz.ch
Sent: Thu, August 19, 2010 11:34:07 AM
Subject: Re: [R] plotmath question
On Aug 19, 2010, at 2:24 PM, array chip wrote:
> Thanks, yes it worked!
>
David, yes, I now see how it worked.
Thanks again,
John
- Original Message
From: David Winsemius
To: array chip
Cc: baptiste auguie ; r-h...@stat.math.ethz.ch
Sent: Thu, August 19, 2010 12:12:46 PM
Subject: Re: [R] plotmath question
On Aug 19, 2010, at 2:46 PM, array chip wrote
Hi, I am a beginner of xyplot() (or lattice package). On one hand, I
immediately
realized it's a very powerful utility. On the other hand, there are too many
things for me to learn. Still haven't figure out a generalization of the syntax
and usage under many different circumstances.
Let me giv
What a simple way to do what I want to do! Thanks.
But if there is missing data in variable "y", then the averaged lines is broken
where the missing data is present. For example:
dat$y[c(10,185,200,400,450)]<-NA
then using type='a' option will result in broken lines.
xyplot(y~day|sex, groups=tr
Hi, is there anyway I can retrieve the user coordinates for the region of the
heatmap (only the heatmap, not include dendrogram, x- & y- axis annotation). I
found that par("usr") didn't give the user coordinates that I want. I want
those
user coordinates to add some additional information to th
Hi all,
I asked this before the holiday, didn't get any response. So would like to
resend the message, hope to get any fresh attention. Since this is not purely
lme technical question, so I also cc-ed R general mailing list, hope to get
some
suggestions from there as well.
I asked some ques
Hi, I am trying to run a simple logistic regression using lrm() to calculate a
odds ratio. I found a confusing output when I use summary() on the fit object
which gave some OR that is totally different from simply taking
exp(coefficient), see below:
> dat<-read.table("dat.txt",sep='\t',header=T
Dear R users, I am not asking questions specifically on R, but I know there are
many statistical experts here in the R community, so here it goes my questions:
Freedman (1982) propose an approximation of sample size/power calculation based
on log-rank test using the formula below (This is what n
p1=0.8 and p2=0.6 at 5 years at 5%
significance level with 80% power"
any comments are appreciated.
Â
John
--- On Thu, 5/6/10, Joris Meys wrote:
From: Joris Meys
Subject: Re: [R] sample size for survival curves
To: "array chip"
Date: Thursday, May 6, 2010, 8:12 PM
Thanks Kevin. I thought the time t is at the end of follow-up (length of
follow-up)?
John
--- On Thu, 5/6/10, Kevin E. Thorpe wrote:
> From: Kevin E. Thorpe
> Subject: Re: [R] sample size for survival curves
> To: "array chip"
> Cc: r-help@r-project.org
> Date: Th
Hi, I would like to use the curve() function to draw the predicted curve from
an nls() object. for example:
dd<-read.table("dd.txt",sep='\t',header=T,row.names=1)
obj<-nls(y~c+(d-c)/(1+(x/e)^b),data=dd,start=list(b=-1, c=0, d=100, e=150))
coef(obj)
b c d e
curve drawing on nls() object
> To: "array chip" , r-help@r-project.org
> Date: Tuesday, May 18, 2010, 7:42 PM
> I can't directly answer your question
> regarding 'expression', but can you just replace b, c,d, and
> e with coef(obj)[1], coef(obj)[2], ...
> etc.
010, 7:33 PM
> array chip yahoo.com> writes:
>
> > Hi, I would like to use the curve() function to draw
> the
> > predicted curve from an nls() object. for example:
> >
>
> Is there a reason you don't want to use plot() and
> predict() ???
>
I do know it, but wanted to spend a little more effort to make generalized
function for this type of plot.
Thanks
--- On Tue, 5/18/10, Shi, Tao wrote:
> From: Shi, Tao
> Subject: Re: [R] automate curve drawing on nls() object
> To: "array chip" , r-help@r-project.org
>
Hi, I am currently trying to do dose-response curves
using weighted 4-parameter model (4PL). The weighting was based on
1/(expected variance) derived from historical data. I tried both drm() from drc
package, and nls(), found very different
results derived from drm() vs. nls() using "weights=" arg
Hi, I am currently trying to do dose-response curves
using weighted 4-parameter model (4PL). The weighting was based on
1/(expected variance) derived from historical data. I tried both drm() from drc
package, and nls(), found very different
results derived from drm() vs. nls() using "weights=" arg
Hi, is there a way to create one image file (like using win.metafile(), bmp(),
etc) that contained multiple pages of plots, just like what postscript() does
in creating PDF file?
Thanks
John
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R-help@r-project.org mailing list
https://stat.ethz.ch/m
Hi, I am wondering if there is a way to extract studentized residues from a
nls() object? I searched archive, someone posted a similar question before
(http://tolstoy.newcastle.edu.au/R/e6/help/09/04/10845.html), but didn't get an
answer.
Thanks for any suggestions,
John
_
Hi, I am wondering how I can specify no intercept in a mixed model using
lmer().
Here is an example dataset attached ("test.txt"). There are 3 workers, in 5
days, measured a response variable "y" on independent variable "x". I want to
use a quadratic term (x2 in the dataset) to model the relat
Hi, is it possible to specify a constant intercept (based on prior knowledge)
in linear regression using lm()?
Thanks
John
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PLEASE do read the posting guide http://www.R-
Hi, I asked this before, but haven't got any response. So would like to have
another try. thanks for help. Also tried twice to join the model mailing list
so that I can ask question there, but still haven't got permission to join that
list yet.
===
Hi, I am wondering how I can spec
Hi, I have a dataset that has 2 groups of samples. For each sample, then
response measured is the number of success (no.success) obatined with the
number
of trials (no.trials). So a porportion of success (prpop.success) can be
computed as no.success/no.trials. Now the objective is to test if th
Hi Bert,
Thanks for your reply. If I understand correctly, prop.test() is not suitable
to
my situation. The input to prop.test() is 2 numbers for each group (# of
success
and # of trials, for example, groups 1 has 5 success out of 10 trials; group 2
has 3 success out of 7 trials; etc. prop.te
Hi, I initially posted this to the general R mailing list, but Bert Gunter
thought this may be a mixed model issue, so suggested me to post here.
I have a dataset that has 2 groups of subjects. For each subject in each group,
the response measured is the number of success (no.success) obatined w
e 0.15 to 0.85, so maybe
regular linear model may not be appropriate?
Thank you,
John
________
From: Robert A LaBudde
To: array chip
Cc: Bert Gunter ; r-h...@stat.math.ethz.ch
Sent: Thu, February 10, 2011 12:54:44 PM
Subject: Re: [R] comparing proportions
1.
Hi, I am wondering if there is a package for doing conditional logistic
regression for nested case-control study as described in "Estimation of
absolute
risk from nested case-control data" by Langholz and Borgan (1997) where
Horvitz-Thompson sampling weight (log of (number in the risk set divid
Thomas, thank you for pointing this out!
On another note, do you think coxph() will also take offset(logweight) as part
of the formula? Or just use the argument weight=logweight in coxph()?
Thanks!
John
From: Thomas Lumley
Sent: Sun, February 27, 2011 12:1
Terry, thanks very much!
Professor Langholz used a SAS software trick to estimate absolute risk by
creating a fake variable "entry_time" that is 0.001 less than the variable
"exit_time" (i.e. time to event), and then use both variables in Phreg. Is this
equivalent to your creating a dummy surv
Hi, I am experimenting with using glht() from multcomp package together with
coxph(), and glad to find that glht() can work on coph object, for example:
> (fit<-coxph(Surv(stop, status>0)~treatment,bladder1))
coxph(formula = Surv(stop, status > 0) ~ treatment, data = bladder1)
In case anyone is interested, I figure it out that when strata is used, I have
to specify the comparison matrix manually:
> (fit<-coxph(Surv(stop, status>0)~treatment+strata(enum),bladder1))
> coef(fit)
treatmentpyridoxine treatmentthiotepa
0.1877925 -0.2097894
> glht(fit,li
Hi, I am encountering a confusing problem when I tried to use read.ssd to read
SAS datasets. For one SAS dataset "a.sas7bdat", it did not work; while for
another SAS dataset "b.sas7bdat" it worked:
> tmp<-read.ssd("C:\\SASdata", "a",sascmd="C:/Program
>Files/SAS/SASFoundation/9.2/sas.exe")
SAS
2.40 seconds
cpu time0.59 seconds
Thank you!
John
________
From: peter dalgaard
Cc: r-help@r-project.org
Sent: Sun, March 6, 2011 1:51:51 AM
Subject: Re: [R] read.ssd() from foreign package
On Mar 6, 2011, at 09:34 , array chip wrote:
>
Hi, I am trying to run a conditional logistic model on a nested case-control
study using cph() and then estimate survival based on the model. The data came
from Prof Bryan Langholz website where he also has the SAS code to this, so I
am
trying to replicate the SAS results.
The data attached. B
Hi, let's say I have a simple ANOVA model with 2 factors A (level A1 and A2)
and
B (level B1 and B2) and their interaction:
aov(y~A*B, data=dat)
It turns out that the interaction term is not significant (e.g. P value = 0.2),
but if I used glht() to compare A1 vs. A2 within each level of B, I f
Hi Bert, thank you for your thoughtful and humorous comments, :-)
It is scientifically meaningful to do those comparisons, and the results of
these comparisons actually make sense to our hypothesis, i.e. one is
significant
at B2 level while the other is not at B1 level. Just unfortunately, the
e. If you have a priori interest in the comparison of A1 vs. A2 at
B2, then you can test it as a pre-planned contrast and not worry too
much about "protection" or multiplicity.
HTH,
Andy
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-bou
Hi, I have a character vector as below:
a<-c('10','3R','4','4R','5','5R','6','6R','7','8','9','7R','1','10R','11'
,'11R','12','12R','13','13R','14','14R','15','15R','1R','2','2R','3','8R'
,'9R')
Is there a clever way to sort this easily to return a vector of index that
would
produce a vector as
That's a smart way to do this. Thanks!
John
From: Berend Hasselman
To: r-help@r-project.org
Sent: Mon, March 14, 2011 10:49:37 AM
Subject: Re: [R] ideas on sorting
array chip wrote:
>
> Hi, I have a character vector as below:
>
> a<
Thank you Bill.
John
From: William Dunlap
Sent: Mon, March 14, 2011 10:47:39 AM
Subject: RE: [R] ideas on sorting
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of array ch
Thanks Martin!
From: Martin Morgan
Cc: r-help@r-project.org
Sent: Mon, March 14, 2011 12:33:40 PM
Subject: Re: [R] ideas on sorting
On 03/14/2011 10:32 AM, array chip wrote:
> Hi, I have a character vector as below:
>
> a<-c('10',
Hi, I have a microarray dataset of dimension 25000x30 and try to clustering
using hclust(). But the clustering on the rows failed due to the size:
> y<-hclust(dist(data),method='average')
Error: cannot allocate vector of size 1.9 Gb
I tried to increase the memory using memory.limit(size=3000), s
, TX 77030
Voice: (713) 798-6227 Fax: (713) 790-1275
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of array chip
Sent: Monday, March 14, 2011 4:03 PM
To: r-help@r-project.org
Subject: [R] hclust() memory issue
Hi, I have a microarray
Hi, I am stuck on this: how to specify a match pattern that means not to
include
"abc"?
I tried:
grep("^(abc)", "hello", value=T)
should return "hello".
while
grep("^(abc)", "hello abcd foo", value=T)
should return character(0).
But both returned character(0).
Thanks
John
[[alte
)
character(0)
I think both examples should return the text, but the 2nd example didn't.
What was wrong here?
Thanks
John
From: Bernd Weiss
Sent: Thu, March 31, 2011 5:32:25 PM
Subject: Re: [R] regular expression
Am 31.03.2011 19:31, schrieb array chip:
&
Ok then this code didn't do what I wanted. I want "not including 'arg' before
'.symptom'", not individual letters of "arg", but rather as a word.
Bill Dunlap suggested using invert=T, it works for single 1 condition, but not
for 2 conditions here: not including "arg" before ".", but at the same
Great. thank you Bernd! Learned a new thing here.
John
From: Bernd Weiss
Cc: r-help@r-project.org
Sent: Thu, March 31, 2011 6:19:25 PM
Subject: Re: [R] regular expression
Am 31.03.2011 21:06, schrieb array chip:
> Ok then this code didn't do what
Hi, I tried to pass the function dist() as an argument, but got an error
message. However, almost the same code with mean() as the function to be
passed,
it works ok.
foo<-function (x,
xfun = dist)
{
xfun(x)
}
foo(matrix(1:100,nrow=5))
Error in foo(matrix(1:100, nrow = 5)) : could not fi
OK, I figured it out, need to add stats::: before dist
foo<-function (x,
xfun = stats:::dist)
{
xfun(x)
}
John
To: r-help@r-project.org
Sent: Fri, April 1, 2011 2:56:06 PM
Subject: [R] function in argument
Hi, I tried to pass the function dist() as an ar
Dear all, this is not a pure R question, but really about how to set up a
multinomial logistic regression model to do a multi-class classification. I
would really appreciate if any of you would give me some of your thoughts and
recommendation.
Let's say we have 3-class classification problem: A
Hi, I am installing Excel using package "RExcelInstaller". When I tried to run
installRExcel()
I got this error message:
You don not have the R package rcom installed.
The (D)COM server installed which will aloow you to use the background server
in
RExcel.
Since rcom is not installed, foregro
s the room,
the thing itself have purpose? Or do we, what's the word... imbue it."
- Jubal Early, Firefly
r-help-boun...@r-project.org wrote on 04/11/2011 02:08:02 PM:
> [image removed]
>
> [R] RExcel
>
> array chip
>
> to:
>
> r-help
>
> 04
Hi, I am new to time-dependent Cox model to estimate time dependent hazard
ratios. Let me use aml dataset from survival package:
> aml3<-survSplit(aml2,cut=c(5,10,20),end="time",start="start",
event="status",episode="i")
If I want to esimate hazard ratio for each of the time intervals 0-5,
Sorry this is a re-post. I posted it last night, haven't heard from anyone,
hope
this moves the thread up a little and anyone can comment?
Thanks!
John
- Forwarded Message
From: array chip
To: r-help@r-project.org
Sent: Sun, April 17, 2011 11:33:32 PM
Subject: time depe
18, 2011, at 4:09 PM, array chip wrote:
> Sorry this is a re-post. I posted it last night, haven't heard from anyone,
>hope
> this moves the thread up a little and anyone can comment?
My observation is that coefficients of 19 in Cox (or other exponential models)
models gene
Hi, I was reading a paper published in JCO "Prediction of risk of distant
recurrence using 21-gene recurrence score in node-negative and node-positive
postmenopausal patients with breast cancer treated with anastrozole or
tamoxifen: a TransATAC study" (ICO 2010 28: 1829). The author uses a metho
some hints how to do this in R.
Thanks
John
- Original Message -
From: array chip
To: r-help
Cc:
Sent: Thursday, August 4, 2011 11:44 AM
Subject: [R] survival probability estimate method
Hi, I was reading a paper published in JCO "Prediction of risk of distant
recurrence usi
Dear Peter,
Thanks very much for the references. It seems the method is based on parametric
proportional hazard models by incorporating cubic spline of the baseline
hazards, not sure if tweaking survreg() would do?
Best,
John
- Original Message -
From: Peter Jepsen
To: 'array
Hi, I read on a paper the below statement, don't know how that was calculated.
Basically, there are 2 continuous variables x1 and x2, as independent variable
for predicting cancer recurrence. So this is a survival analysis. Now the
author try to check the correlation between x1 and x2. He calcul
Nice guess Duncan! Otherwise, I can't think of other ways to get that number.
Thanks
John
- Original Message -
From: Duncan Murdoch
To: array chip
Cc: r-help
Sent: Friday, August 5, 2011 11:21 AM
Subject: Re: [R] a question
On 05/08/2011 2:19 PM, array chip wrote:
> Hi, I
Hi al,
I have a dataset (see attached), which basically involves 4 treatments for a
chemotherapy drug. Samples were taken from 2 biopsy locations, and biopsy were
taken at 2 time points. So each subject has 4 data points (from 2 biopsy
locations and 2 time points). The objective is to study tre
+ time * trt, random=list(pid=pdIdent(~0 + biopsy.site)), data =
test, correlation=corAR1())
But they are not the same. What did I misunderstand here?
Many thanks
John
- Original Message -
From: "ONKELINX, Thierry"
To: array chip ; "r-sig-mixed-mod...@r-project.o
Hi, the calibrate.cph() function in rms package generate calibration curve for
Cox model on the same dataset where the model was derived using bootstrapping
or cross-validation. If I have the model built on dataset 1, and now I want to
produce a calibration curve for this model on an independent
is there a R function that produces calibration curve on an independetn data
automatically, just like what calibrate() does on the training data itself?
Thanks
John
From: Comcast
Cc: "r-help@r-project.org"
Sent: Monday, August 15, 2011 2:04 PM
Subject: Re:
; A combination of Predict (your newdata), cut2, and the plotting function
> of your choice ought to suffice. But thought that cross-validation was an
> option. Not at console at the moment (just off airplane.)
>
> Sent from my iPhone
>
> On Aug 15, 2011, at 5:26 PM, array chip &
Oops, thank for reminding. I found that an in-house package interfered with rms
package, which caused the error.
Thanks David!
John
- Original Message -
From: David Winsemius
To: array chip
Cc: Frank Harrell ; "r-help@r-project.org"
Sent: Tuesday, August 16, 201
Hi,
To use cuminc() from cmprsk package, if a subject has 2
events (both the event of interest and the event of competing risk),
should I create 2 observations for this subject in the dataset, one for
each event with different fstatus (1 and 2), or just 1 observation with
whatever event that
Hi Frank, it's true to one of your reply to my previous post, can only be seen
in Nabble.
- Original Message -
From: David Winsemius
To: Frank Harrell
Cc: r-help@r-project.org
Sent: Wednesday, August 17, 2011 3:08 PM
Subject: Re: [R] Labelling all variables at once (using Hmisc label)
Hi, in Design package, a plot of survival probability vs. a covariate can be
generated by survplot() on a cph object using the folliowing code:
n <- 1000
set.seed(731)
age <- 50 + 12*rnorm(n)
label(age) <- "Age"
sex <- factor(sample(c('male','female'), n, TRUE))
cens <- 15*runif(n)
h <- .02*exp(.
Thank you David. The plot from Design package will draw a plot of survival
probability at 5 years (in my example) versus different age. I will look into
Predict() in rms package to see how this can be done.
John
- Original Message -
From: David Winsemius
To: array chip
Cc: &q
Thanks Frank!
- Original Message -
From: Frank Harrell
To: r-help@r-project.org
Cc:
Sent: Thursday, August 25, 2011 4:33 PM
Subject: Re: [R] survplot() for cph(): Design vs rms
http://biostat.mc.vanderbilt.edu/Rrms shows differences between Design and
rms
Frank
array chip wrote
Hi,
basehaz() in survival package is said to estimate the baseline cumulative
hazard from coxph(), but it actually calculate cumulative hazard by
-log(survfit(coxph.object)). But survfit() calculate survival based on MEAN
covariate value, not covariate value of 0. I thought baseline cumulativ
Hi, I had a weird results from using apply(). Here is an simple example:
> y<-data.frame(list(a=c(1,NA),b=c('2k','0')))
> y
a b
1 1 2k
2 NA 0
> apply(y,1,function(x){x<-unlist(x); if (!is.na(x[2]) & x[2]=='2k' &
> !is.na(x[1]) & x[1]=='1') 1 else 0} )
This should print "1 0" as ou
Thanks Bill and David!
John
- Original Message -
From: William Dunlap
To: array chip ; "r-help@r-project.org"
Cc:
Sent: Monday, August 29, 2011 5:21 PM
Subject: RE: [R] weird apply() behavior
apply() should come with a big warning that it was
written for matrices and c
Hi, I am wondering if anyone can suggest how to test the equality of 2
proportions. The caveat here is that the 2 proportions were calculated from the
same number of samples using 2 different tests. So essentially we are comparing
2 accuracy rates from same, say 100, samples. I think this is lik
Hi all, thanks very much for sharing your thoughts. and sorry for my describing
the problem not clearly, my fault.
My data is paired, that is 2 different diagnostic tests were performed on the
same individuals. Each individual will have a test results from each of the 2
tests. Then in the end,
between the 2
tests.
Thanks again.
John
- Original Message -
From: Viechtbauer Wolfgang (STAT)
To: "r-help@r-project.org"
Cc: csrabak ; array chip
Sent: Thursday, September 8, 2011 1:24 AM
Subject: RE: [R] suggestion for proportions
I assume you mean Cohen's kappa. T
Hi, are the methods for multiple testing p value adjustment (Shaffer, Westfall,
free) implemented in the function adjusted() in multcomp package so called
closed testing procedure? what about those methods (holm, hochberg, hommel, BH,
BY) implemented in the p.adjust() in the stats package?
Tha
Hi, I have a dataset (see attached) with 2 variables "Y" is binary, "x" is a
continuous variable. I want to calculate area under the curve (AUC) for the ROC
curve, but I got different AUC values using ROC() from Epi package vs.
rcorr.cens() from rms package:
test<-read.table("test.txt",sep='\t'
Hi, I am wondering if anyone can explain to me if cumulative incidence (CI) is
just "1 minus kaplan-Meier survival"? Under what circumstance, you should use
cumulative incidence vs KM survival? If the relationship is just CI =
1-survival, then what difference it makes to use one vs. the other?
t.org
Sent: Mon, June 27, 2011 1:45:35 PM
Subject: Re: [R] cumulative incidence plot vs survival plot
On Jun 27, 2011, at 4:31 PM, array chip wrote:
> Hi, I am wondering if anyone can explain to me if cumulative incidence (CI) is
> just "1 minus kaplan-Meier survival"?
First tell us
Thank you David. Didn't realized someone posted related topic this morning.
John
- Original Message
From: David Winsemius
To: array chip
Cc: r-help@r-project.org
Sent: Mon, June 27, 2011 3:38:36 PM
Subject: Re: [R] cumulative incidence plot vs survival plot
On Jun 27, 2011,
chell)
To: array chip ; David Winsemius
Cc: r-help@r-project.org
Sent: Tue, June 28, 2011 9:20:22 AM
Subject: RE: [R] cumulative incidence plot vs survival plot
John,
Since death precludes recurrence, censoring deaths would violate the KM
estimator assumption that additional follow-up would event
Hi, how can I print "<=" (I mean the symbol of just one character) in the main
title of a plot?
for example:
plot(1:10, main=paste("x <=", x))
where variable x is some number generated on the fly.
Thanks
John
__
R-help@r-project.org mailing list
ht
this patient? I think I still code this patient as 0
(censoring) because lost-of-followup occurred before death, am I correct?
Thanks very much!
John
- Original Message ----
From: alanm (Alan Mitchell)
To: array chip ; David Winsemius
Cc: r-help@r-project.org
Sent: Tue, June 28, 2011 9:2
] cluster_1.13.3 grid_2.12.2 lattice_0.19-17 tools_2.12.2
- Original Message
From: Bert Gunter
To: array chip
Cc: R
Sent: Tue, June 28, 2011 10:32:29 AM
Subject: Re: [R] how to print "<=" in plot title
This is highly system dependent: what "character" d
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