Re: [R] how to solve the step halving factor problems in gnls and nls
Yimeng Lu yl2058 at columbia.edu writes: Could you give me some advice in solving the problem of such error message from gnls and nls? ## begin error message Problem in gnls(y1 ~ glogit4(b, c, m, t, x), data.frame(x..: Step halving factor reduced below minimum in NLS step Try to set nlsTol in the optional control parameter (gnlsControl) to a larger value, e.g. 0.1 instead of default 0.001, but be sure to check your results make sense. If this should help, you can try intermediate values. Dieter Menne __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] how to get dissimilarity matrix
Baoqiang == Baoqiang Cao [EMAIL PROTECTED]
on Mon, 18 Jul 2005 15:02:05 -0400 writes:
Baoqiang Hello All, I'm learning R. Just wonder, any
Baoqiang package or function that I can use to get the
Baoqiang dissimilarity matrix? Thanks.
Yes,
learn to use help.search() {also read the docu : ?help.search}
help.search(dissimilarity)
and find daisy() in recommended package 'cluster'.
There's also dist() in 'stats' which is a bit less versatile.
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Re: [R] New functions supporting GIF file format in R
JarekT == Tuszynski, Jaroslaw W [EMAIL PROTECTED] on Mon, 18 Jul 2005 16:00:43 -0400 writes: JarekT Hi, A minor announcement. I just added two functions JarekT for reading and writing GIF files to my caTools JarekT package. Input and output is in the form of standard JarekT R matrices or arrays, and standard R color-maps JarekT (palettes). The functions can read and write both JarekT regular GIF images, as well as, multi-frame animated JarekT GIFs. Most of the work is done in C level code JarekT (included), so functions do not use any external JarekT libraries. JarekT For more info and examples go to JarekT http://cran.r-project.org/doc/packages/caTools.pdf JarekT http://cran.r-project.org/doc/packages/caTools.pdf JarekT and click GIF. Wouldn't it make sense to donate these to the 'pixmap' package which is dedicated to such objects and has been in place for a very long time? Regards, Martin __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] listing datasets from all my packages
I did manage eventually to reproduce this via data(package=makecdfenv) which is a BioC package with a non-standard use of the data directory. (There are no objects in the single R file in the data directory, which from its name I suggest is a dummy and the data directory should be removed entirely. The indexing code fails to handle this correctly.) I had no difficulty on a system that has almost all CRAN packages installed. On Mon, 18 Jul 2005, Elizabeth Purdom wrote: Hi, I am using R 2.1.0 on Windows XP and when I type data() to list the datasets in R, there is a helpful hint to type 'data(package = .packages(all.available = TRUE))' to see the datasets in all of the packages -- not just the active ones. However, when I do this, I get the following message: data(package = .packages(all.available = TRUE)) Error in rbind(...) : number of columns of matrices must match (see arg 2) In addition: Warning messages: 1: datasets have been moved from package 'base' to package 'datasets' in: data(package = .packages(all.available = TRUE)) 2: datasets have been moved from package 'stats' to package 'datasets' in: data(package = .packages(all.available = TRUE)) I possibly have old libraries in my R libraries because I copy them forward and update them with new versions of R, rather than redownload them. Is there a way to fix this or do the same another way? (I saw something in archives about a problem similar to this with .packages(), but I got the impression it was fixed for 2.1.0) Thanks, Elizabeth __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] definition of index.array and boot.return in the code for boot
Or read the *sources* for package boot, and find the original code with all the comments. On Mon, 18 Jul 2005, Spencer Graves wrote: Excellent question. Try 'getAnywhere(index.array)'. It's hidden in namespace:boot. Ditto for boot.return. spencer graves Obrien, Josh wrote: Dear R friends, I am reading the code for the function boot in package:boot in an attempt to learn how and where it implements the random resampling used by the non-parametric bootstraps. The code contains two (apparent) functions - 'index.array' and 'boot.return' - for which I can find no documentation, and which don't even seem to exist anywhere on the search path. What are they? Also, if the meanings of those two don't answer my larger question, could you point me to the code that implements the random resampling? Thanks very much for your help, Josh O'Brien UC Davis __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Spencer Graves, PhD Senior Development Engineer PDF Solutions, Inc. 333 West San Carlos Street Suite 700 San Jose, CA 95110, USA [EMAIL PROTECTED] www.pdf.com http://www.pdf.com Tel: 408-938-4420 Fax: 408-280-7915 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] help: how to change the column name of data.frame
Just change the names! E.g. names(DF)[c(4,6)] - names(DF)[c(6,4)] Strictly a data frame has names, not column names, hence the use the names() and names-() functions here. On Tue, 19 Jul 2005, wu sz wrote: I have a data frame with 15 variables, and want to exchange the data of 4th column and 6th column. First I append a column in the data frame, copy the 4th column data there, then copy the 6th column data to 4th column, and copy the appended column data to 6th column, but the names of the 4th and 6th column are still unchanged. How can I exchange them? -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] how to change bar colours in plot.stl
My thanks to Achim Zeileis and Prof Ripley for their responses. In a very short time I not only had an answer and solved my problem, but also learned something about R that I can employ in other situations. Much appreciated, MT __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] colnames
Hi Adai,
Many Thanks for the examples.
I work for a financial institution. We are exploring R as a tool to implement
our portfolio optimization strategies. Hence, R is still a new language to us.
The script I wrote tried to make a returns matrix from the daily return indices
extracted from a SQL database. Please find below the output that produces the
'X' prefix in the colnames. The reason to preserve the column names is that
they are stock identifiers which are to be used by other sub systems rather
than R.
I would welcome any suggestion to improve the script.
Regards,
Gilbert
p.RIs2Returns -
+ function (RIm)
+ {
+ x-RIm[1:(nrow(RIm)-1), 1:ncol(RIm)]
+ y-RIm[2:nrow(RIm), 1:ncol(RIm)]
+ RReturns - (y/x -1)
+ RReturns
+ }
channel-odbcConnect(ourSQLDB)
result-sqlQuery(channel,paste(select * from equityRIs;))
odbcClose(channel)
result
stockidsdate dbPrice
1 899188 20050713 7.59500
2 899188 20050714 7.60500
3 899188 20050715 7.48000
4 899188 20050718 7.41500
5 902232 20050713 10.97000
6 902232 20050714 10.94000
7 902232 20050715 10.99000
8 902232 20050718 11.05000
9 901714 20050713 17.96999
10 901714 20050714 18.00999
11 901714 20050715 17.64999
12 901714 20050718 17.64000
13 28176U 20050713 5.19250
14 28176U 20050714 5.25000
15 28176U 20050715 5.25000
16 28176U 20050718 5.22500
17 15322M 20050713 11.44000
18 15322M 20050714 11.5
19 15322M 20050715 11.33000
20 15322M 20050718 11.27000
r1-reshape(result, timevar=stockid, idvar=sdate, direction=wide)
r1
sdate dbPrice.899188 dbPrice.902232 dbPrice.901714 dbPrice.28176U
dbPrice.15322M
1 20050713 7.595 10.97 17.96999 5.1925
11.44
2 20050714 7.605 10.94 18.00999 5.2500
11.50
3 20050715 7.480 10.99 17.64999 5.2500
11.33
4 20050718 7.415 11.05 17.64000 5.2250
11.27
#Set sdate as the rownames
rownames(r1) -as.character(r1[1:nrow(r1),1:1])
#Get rid of the first column
r1 - r1[1:nrow(r1),2:ncol(r1)]
r1
dbPrice.899188 dbPrice.902232 dbPrice.901714 dbPrice.28176U
dbPrice.15322M
20050713 7.595 10.97 17.96999 5.1925
11.44
20050714 7.605 10.94 18.00999 5.2500
11.50
20050715 7.480 10.99 17.64999 5.2500
11.33
20050718 7.415 11.05 17.64000 5.2250
11.27
colnames(r1) - as.character(sub([[:alnum:]]*\\.,, colnames(r1)))
r1
899188 902232 901714 28176U 15322M
20050713 7.595 10.97 17.96999 5.1925 11.44
20050714 7.605 10.94 18.00999 5.2500 11.50
20050715 7.480 10.99 17.64999 5.2500 11.33
20050718 7.415 11.05 17.64000 5.2250 11.27
RRs-p.RIs2Returns(r1)
RRs
X899188 X902232 X901714 X28176U X15322M
20050714 0.001316656 -0.002734731 0.002225933 0.011073664 0.005244755
20050715 -0.016436555 0.004570384 -0.019988906 0.0 -0.014782609
20050718 -0.008689840 0.005459509 -0.000566006 -0.004761905 -0.005295675
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Re: [R] help: how to change the column name of data.frame
On Tue, 19 Jul 2005, Prof Brian Ripley wrote: Just change the names! E.g. names(DF)[c(4,6)] - names(DF)[c(6,4)] Strictly a data frame has names, not column names, hence the use the names() and names-() functions here. That answered the subject line. If you want to exchange the columns (not just the data but also the names) you can just use DF[c(4,6)] - DF[c(6,4)] Please do try to be more precise as to what you want to do, for example by giving an example. On Tue, 19 Jul 2005, wu sz wrote: I have a data frame with 15 variables, and want to exchange the data of 4th column and 6th column. First I append a column in the data frame, copy the 4th column data there, then copy the 6th column data to 4th column, and copy the appended column data to 6th column, but the names of the 4th and 6th column are still unchanged. How can I exchange them? -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Michaelis-menten equation
Dear R users:
I encountered difficulties in michaelis-menten equation. I found
that when I use right model definiens, I got wrong Km vlaue,
and I got right Km value when i use wrong model definiens.
The value of Vd and Vmax are correct in these two models.
#-right model definiens
PKindex-data.frame(time=c(0,1,2,4,6,8,10,12,16,20,24),
conc=c(8.57,8.30,8.01,7.44,6.88,6.32,5.76,5.20,4.08,2.98,1.89))
mm.model - function(time, y, parms) {
dCpdt - -(parms[Vm]/parms[Vd])*y[1]/(parms[Km]+y[1])
list(dCpdt)}
Dose-300
modfun - function(time,Vm,Km,Vd) {
out - lsoda(Dose/Vd,time,mm.model,parms=c(Vm=Vm,Km=Km,Vd=Vd),
rtol=1e-8,atol=1e-8)
out[,2] }
objfun - function(par) {
out - modfun(PKindex$time,par[1],par[2],par[3])
sum((PKindex$conc-out)^2) }
fit - optim(c(10,1,80),objfun, method=Nelder-Mead)
print(fit$par)
[1] 10.0390733 0.1341544 34.9891829 #--Km=0.1341544,wrong value--
#-wrong model definiens
#-Km should not divided by Vd--
PKindex-data.frame(time=c(0,1,2,4,6,8,10,12,16,20,24),
conc=c(8.57,8.30,8.01,7.44,6.88,6.32,5.76,5.20,4.08,2.98,1.89))
mm.model - function(time, y, parms) {
dCpdt - -(parms[Vm]/parms[Vd])*y[1]/(parms[Km]/parms[Vd]+y[1])
list(dCpdt)}
Dose-300
modfun - function(time,Vm,Km,Vd) {
out - lsoda(Dose/Vd,time,mm.model,parms=c(Vm=Vm,Km=Km,Vd=Vd),
rtol=1e-8,atol=1e-8)
out[,2]
}
objfun - function(par) {
out - modfun(PKindex$time,par[1],par[2],par[3])
sum((PKindex$conc-out)^2)}
fit - optim(c(10,1,80),objfun, method=Nelder-Mead)
print(fit$par)
[1] 10.038821 4.690267 34.989239 #--Km=4.690267,right value--
What did I do wrong, and how to fix it?
Any suggestions would be greatly appreciated.
Thanks in advance!!
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Re: [R] Michaelis-menten equation
Chun-Ying Lee [EMAIL PROTECTED] writes:
Dear R users:
I encountered difficulties in michaelis-menten equation. I found
that when I use right model definiens, I got wrong Km vlaue,
and I got right Km value when i use wrong model definiens.
The value of Vd and Vmax are correct in these two models.
How do you know what the correct value is? Are you sure that the other
values are right?
I'm a bit rusty on MM, but are you sure your right model is right?
Try doing a dimensional analysis on the ODE. I kind of suspect that
Vd is entering in the wrong way. Since you're dealing in
concentrations, should it enter at all (except via the conc. at time
0, of course)?
Not knowing the context, I can't be quite sure, but generally, I'd
expect Vm*Km/(Km+y) to be the reaction rate, so that Vm is the maximum
rate, attained when y is zero and Km is the conc. at half-maximum
rate. This doesn't look quit like what you have.
#-right model definiens
PKindex-data.frame(time=c(0,1,2,4,6,8,10,12,16,20,24),
conc=c(8.57,8.30,8.01,7.44,6.88,6.32,5.76,5.20,4.08,2.98,1.89))
mm.model - function(time, y, parms) {
dCpdt - -(parms[Vm]/parms[Vd])*y[1]/(parms[Km]+y[1])
list(dCpdt)}
Dose-300
modfun - function(time,Vm,Km,Vd) {
out - lsoda(Dose/Vd,time,mm.model,parms=c(Vm=Vm,Km=Km,Vd=Vd),
rtol=1e-8,atol=1e-8)
out[,2] }
objfun - function(par) {
out - modfun(PKindex$time,par[1],par[2],par[3])
sum((PKindex$conc-out)^2) }
fit - optim(c(10,1,80),objfun, method=Nelder-Mead)
print(fit$par)
[1] 10.0390733 0.1341544 34.9891829 #--Km=0.1341544,wrong value--
#-wrong model definiens
#-Km should not divided by Vd--
PKindex-data.frame(time=c(0,1,2,4,6,8,10,12,16,20,24),
conc=c(8.57,8.30,8.01,7.44,6.88,6.32,5.76,5.20,4.08,2.98,1.89))
mm.model - function(time, y, parms) {
dCpdt - -(parms[Vm]/parms[Vd])*y[1]/(parms[Km]/parms[Vd]+y[1])
list(dCpdt)}
Dose-300
modfun - function(time,Vm,Km,Vd) {
out - lsoda(Dose/Vd,time,mm.model,parms=c(Vm=Vm,Km=Km,Vd=Vd),
rtol=1e-8,atol=1e-8)
out[,2]
}
objfun - function(par) {
out - modfun(PKindex$time,par[1],par[2],par[3])
sum((PKindex$conc-out)^2)}
fit - optim(c(10,1,80),objfun, method=Nelder-Mead)
print(fit$par)
[1] 10.038821 4.690267 34.989239 #--Km=4.690267,right value--
What did I do wrong, and how to fix it?
Any suggestions would be greatly appreciated.
Thanks in advance!!
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O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918
~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907
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Re: [R] help: how to change the column name of data.frame
Thanks a lot! But DF[c(4,6)] - DF[c(6,4)] seems to just exchange the data, not the names. If exchanging the columns(both data and names) needs two steps? DF[c(4,6)] - DF[c(6,4)] names(DF)[c(4,6)] - names(DF)[c(6,4)] Shengzhe __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] help: how to change the column name of data.frame
On Tue, 19 Jul 2005, wu sz wrote: Thanks a lot! But DF[c(4,6)] - DF[c(6,4)] seems to just exchange the data, not the names. If exchanging the columns(both data and names) needs two steps? DF[c(4,6)] - DF[c(6,4)] names(DF)[c(4,6)] - names(DF)[c(6,4)] Yes, it does. You did however say in your first message that you wanted to exchange the data, despite the subject line. As I asked before *PLEASE* do try to be precise in what you want to do. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Michaelis-menten equation
Chun-Ying Lee wrote:
Dear R users:
I encountered difficulties in michaelis-menten equation. I found
that when I use right model definiens, I got wrong Km vlaue,
and I got right Km value when i use wrong model definiens.
The value of Vd and Vmax are correct in these two models.
#-right model definiens
PKindex-data.frame(time=c(0,1,2,4,6,8,10,12,16,20,24),
conc=c(8.57,8.30,8.01,7.44,6.88,6.32,5.76,5.20,4.08,2.98,1.89))
mm.model - function(time, y, parms) {
dCpdt - -(parms[Vm]/parms[Vd])*y[1]/(parms[Km]+y[1])
list(dCpdt)}
Dose-300
modfun - function(time,Vm,Km,Vd) {
out - lsoda(Dose/Vd,time,mm.model,parms=c(Vm=Vm,Km=Km,Vd=Vd),
rtol=1e-8,atol=1e-8)
out[,2] }
objfun - function(par) {
out - modfun(PKindex$time,par[1],par[2],par[3])
sum((PKindex$conc-out)^2) }
fit - optim(c(10,1,80),objfun, method=Nelder-Mead)
print(fit$par)
[1] 10.0390733 0.1341544 34.9891829 #--Km=0.1341544,wrong value--
#-wrong model definiens
#-Km should not divided by Vd--
PKindex-data.frame(time=c(0,1,2,4,6,8,10,12,16,20,24),
conc=c(8.57,8.30,8.01,7.44,6.88,6.32,5.76,5.20,4.08,2.98,1.89))
mm.model - function(time, y, parms) {
dCpdt - -(parms[Vm]/parms[Vd])*y[1]/(parms[Km]/parms[Vd]+y[1])
list(dCpdt)}
Dose-300
modfun - function(time,Vm,Km,Vd) {
out - lsoda(Dose/Vd,time,mm.model,parms=c(Vm=Vm,Km=Km,Vd=Vd),
rtol=1e-8,atol=1e-8)
out[,2]
}
objfun - function(par) {
out - modfun(PKindex$time,par[1],par[2],par[3])
sum((PKindex$conc-out)^2)}
fit - optim(c(10,1,80),objfun, method=Nelder-Mead)
print(fit$par)
[1] 10.038821 4.690267 34.989239 #--Km=4.690267,right value--
What did I do wrong, and how to fix it?
Any suggestions would be greatly appreciated.
Thanks in advance!!
it is not clear to me what you are trying to do:
you seem to have a time-concentration-curve in PKindex and you seem to
set up a derivative of this time dependency according
to some model in dCpdt. AFAIKS this scenario is not directly related to
the situation described by the Michaelis-Menten-Equation which relates
some input concentration with some product concentration. If Vm and
Km are meant to be the canonical symbols,
what is Vd, a volume of distribution? it is impossible to see (at least
for me) what exactly you want to achieve.
(and in any case, I would prefer nls for a least squares fit instead
of 'optim').
joerg
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Re: [R] colnames
First, your problem could be boiled down to the following example. See
how the colnames of the two outputs vary.
df - cbind.data.frame( 100=1:2, 200=3:4 )
df/df
X100 X200
111
211
m - as.matrix( df ) # coerce to matrix class
m/m
100 200
1 1 1
2 1 1
It appears that whenever R has to create a new dataframe automatically,
it tries to get nice colnames. See help(data.frame). I am not exactly
sure why this behaviour is different when creating a matrix. But I do
not think this is a major problem for most people. If you coerce your
input to matrix, the problem goes away.
Next, note the following points :
a) mat[ 1:3, 1:ncol(mat) ] is equivalent to simply mat[ 1:3, ].
b) mat[ 2:nrow(mat), ] is equivalent to simply mat[ -1, ]
See help(subset) for more information.
Using the points above, we can simplify your function as
p.RIs2Returns - function (mat){
mat - as.matrix(mat)
x - mat[ -nrow(mat), ]
y - mat[ -1, ]
return( y/x -1 )
}
If your data contains only numerical data, it is probably good idea to
work with matrices as matrix operations are faster.
Finally, we can shorten your function. You can use the diff (which works
column-wise if input is a matrix) and apply function if you know that
y/x = exp(log(y/x)) = exp( log(y) - log(x) )
which could be coded in R as
exp( diff( log(r1) ) )
and then subtract 1 from above to get your returns.
Regards, Adai
On Tue, 2005-07-19 at 09:17 +0100, Gilbert Wu wrote:
Hi Adai,
Many Thanks for the examples.
I work for a financial institution. We are exploring R as a tool to implement
our portfolio optimization strategies. Hence, R is still a new language to us.
The script I wrote tried to make a returns matrix from the daily return
indices extracted from a SQL database. Please find below the output that
produces the 'X' prefix in the colnames. The reason to preserve the column
names is that they are stock identifiers which are to be used by other sub
systems rather than R.
I would welcome any suggestion to improve the script.
Regards,
Gilbert
p.RIs2Returns -
+ function (RIm)
+ {
+ x-RIm[1:(nrow(RIm)-1), 1:ncol(RIm)]
+ y-RIm[2:nrow(RIm), 1:ncol(RIm)]
+ RReturns - (y/x -1)
+ RReturns
+ }
channel-odbcConnect(ourSQLDB)
result-sqlQuery(channel,paste(select * from equityRIs;))
odbcClose(channel)
result
stockidsdate dbPrice
1 899188 20050713 7.59500
2 899188 20050714 7.60500
3 899188 20050715 7.48000
4 899188 20050718 7.41500
5 902232 20050713 10.97000
6 902232 20050714 10.94000
7 902232 20050715 10.99000
8 902232 20050718 11.05000
9 901714 20050713 17.96999
10 901714 20050714 18.00999
11 901714 20050715 17.64999
12 901714 20050718 17.64000
13 28176U 20050713 5.19250
14 28176U 20050714 5.25000
15 28176U 20050715 5.25000
16 28176U 20050718 5.22500
17 15322M 20050713 11.44000
18 15322M 20050714 11.5
19 15322M 20050715 11.33000
20 15322M 20050718 11.27000
r1-reshape(result, timevar=stockid, idvar=sdate, direction=wide)
r1
sdate dbPrice.899188 dbPrice.902232 dbPrice.901714 dbPrice.28176U
dbPrice.15322M
1 20050713 7.595 10.97 17.96999 5.1925
11.44
2 20050714 7.605 10.94 18.00999 5.2500
11.50
3 20050715 7.480 10.99 17.64999 5.2500
11.33
4 20050718 7.415 11.05 17.64000 5.2250
11.27
#Set sdate as the rownames
rownames(r1) -as.character(r1[1:nrow(r1),1:1])
#Get rid of the first column
r1 - r1[1:nrow(r1),2:ncol(r1)]
r1
dbPrice.899188 dbPrice.902232 dbPrice.901714 dbPrice.28176U
dbPrice.15322M
20050713 7.595 10.97 17.96999 5.1925
11.44
20050714 7.605 10.94 18.00999 5.2500
11.50
20050715 7.480 10.99 17.64999 5.2500
11.33
20050718 7.415 11.05 17.64000 5.2250
11.27
colnames(r1) - as.character(sub([[:alnum:]]*\\.,, colnames(r1)))
r1
899188 902232 901714 28176U 15322M
20050713 7.595 10.97 17.96999 5.1925 11.44
20050714 7.605 10.94 18.00999 5.2500 11.50
20050715 7.480 10.99 17.64999 5.2500 11.33
20050718 7.415 11.05 17.64000 5.2250 11.27
RRs-p.RIs2Returns(r1)
RRs
X899188 X902232 X901714 X28176U X15322M
20050714 0.001316656 -0.002734731 0.002225933 0.011073664 0.005244755
20050715 -0.016436555 0.004570384 -0.019988906 0.0 -0.014782609
20050718 -0.008689840 0.005459509 -0.000566006 -0.004761905 -0.005295675
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[R] Predict
When I callculate a linear model, then I can compute via confint the confidencial intervals. the interval level can be chosen. as result, I get the parameter of the model according to the interval level. On the other hand, I can compute the prediction-values for my model as well with predict(object, type=c(response) etc.). Here I have also the possibility to chose a level for the confidential intervals. the output are the calculatet values for the fit, the lower and upper level. the problem now is, that when I calculate the values through the linear model function with the parameter values I get from confint() an I compare them with the values I get from predict() these values differ extremely. Why is that so? Does the command predict() calculate the values through an other routine? That means the command predict() doesn't use the same parameters to calculate the prediction-values than the ones given by confint()? Greetings Matthias -- GMX DSL = Maximale Leistung zum minimalen Preis! __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Survival dummy variables and some questions
Stephen, This has nothing to do with your R but to do with your email settings. Are you sure you are sending mails in plain text ? Your email on the R-help mailing archive (see link below) appears to be unreadable https://stat.ethz.ch/pipermail/r-help/2005-July/074210.html Please try to use plain text. (See http://expita.com/nomime.html). Regards, Adai On Tue, 2005-07-19 at 08:59 +0200, Stephen wrote: Many thanks I follow you what you say You can request predicted values at any sequence of ages - I guess there are plenty of postings on how to do that Regards, Stephen - Original Message - From: Frank E Harrell Jr To: Stephen Cc: Prof Brian Ripley ; Sent: Monday, July 18, 2005 6:13 PM Subject: Re: [R] Survival dummy variables and some questions Stephen wrote: Hi 1. Right perhaps this should clarify. I would like to extract coefficeints for different levels of the IVs (covariate). So for instance, age of onset I would want Hazards etc for every 5 years and so on... The approach I took was to categorize the variables (e.g., age of onset) and then turn theu resultant categorical variable into a factor as opposed to a variable... that is when the problems began An alternative approach to pulling out different values at different levels of the variable is what I seek. 2. I looked for the link, but can't Your needs don't require categorization. You can request predicted values at any sequence of ages. If you want hazard ratios you can take differences in predicted log hazards and antilog them. Frank -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University ?? http://mail.nana.co.il [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] R: expression
hi all i am having a problem with the expression/paste command say we estimate a variable, named PHI it contains the value of say 2 and we want to display this value as hat(phi) = PHI onto a graphic i.e. hat(phi)=2 how does one do this? i've tried the following: 1. legend(-5,.3,expression(hat(phi)*=*PHI)) 2. legend(-5,.3,paste(expression(phi),=,PHI)) but they do not work. any help? / allan__ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] R: expression
Clark Allan wrote: hi all i am having a problem with the expression/paste command say we estimate a variable, named PHI it contains the value of say 2 and we want to display this value as hat(phi) = PHI onto a graphic i.e.hat(phi)=2 how does one do this? i've tried the following: 1.legend(-5,.3,expression(hat(phi)*=*PHI)) 2.legend(-5,.3,paste(expression(phi),=,PHI)) See ?plotmath or the Help Desk Article Automation of Mathematical Annotation in Plots in R News 2 (3), 32-34. legend(-5, .3, substitute(hat(phi) == PHI, list(PHI = PHI))) Uwe Ligges but they do not work. any help? / allan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] R: expression
Something like this : x - 0.5 plot( 1:10, main=substitute( hat(Phi) ~ = ~ x, list(x=x) ) ) Also see http://tolstoy.newcastle.edu.au/R/help/04/09/3371.html Regards, Adai On Tue, 2005-07-19 at 13:35 +0200, Clark Allan wrote: hi all i am having a problem with the expression/paste command say we estimate a variable, named PHI it contains the value of say 2 and we want to display this value as hat(phi) = PHI onto a graphic i.e.hat(phi)=2 how does one do this? i've tried the following: 1.legend(-5,.3,expression(hat(phi)*=*PHI)) 2.legend(-5,.3,paste(expression(phi),=,PHI)) but they do not work. any help? / allan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] dataframes of unequal size
Seems like no one had responded to this one yet, so I'll take a stab:
## Generate some bogus data:
set.seed(45)
dat - cbind(expand.grid(LETTERS[1:2], 1:3), round(runif(6), 2))
names(dat) - c(state, psu, weight)
dat2 - data.frame(state=sample(c(A, B), 100, replace=TRUE),
psu=sample(3, 100, replace=TRUE),
weight=rep(0, 100))
## The actual work:
split(dat2$weight, interaction(dat2$state, dat2$psu)) -
split(dat$weight, interaction(dat$state, dat$psu))
This, I think, will only work correctly if all state/psu combinations in
your C are also present in C1. If not, you can just augment C1 to
include them.
HTH,
Andy
From: Renuka Sane
I have two dataframes C and C1. Each has three columns viz. state, psu
and weight. The dataframes are of unequal size i.e. C1 could be
2/25/50 rows and C has 42000 rows. C1 is the master table i.e.
C1$state, C1$psu and C1$weight are never the same. ThisA. P., Urban, 0
is not so for C.
For example
C
state, psu,weight
A. P., Urban, 0
Mah., Rural, 0
W.B., Rural,0
Ass., Rural,0
M. P., Urban,0
A. P., Urban, 0
...
C1
state, psu, weight
A. P., Urban, 1.3
A. P., Rural, 1.2
M. P., Urban, 0.8
..
For every row of C, I want to check if C$state==C1$state and
C$psu==C1$psu. If it is, I want C$weight - C1$weight, else C$weight
should be zero.
I am doing the following
for( i in 1:length(C$weight)) {
C$w[C$state[i]==C1$state C$psu[i]==C1$psu] - C1$w[C$state[i] ==
C1$state C$psu[i] == C1$psu]
}
This gives me the correct replacements for the number of rows in C1
and then just repeats the same weights for the remaning rows in C.
Can someone point out the error in what I am doing or show the correct
way of doing this?
Thanks,
Renuka
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[R] data mining
Dear all, I'm looking for some material on data mining with R. I have something from Luis Torgo but I'd like to see something else. If anybody could help me I'll be thankful Adrián __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] reading data
Dear all, How can I read data from posgresql? Thanks Adrián __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] lme outer
Dear R users a question about outer explanatory variables in lme: I have measured size of a population of insects in fields. These fields were spread out over a large region. The fields are grouped (spatially) in pairs: one with fertiliser high, the other one low. I want to test effect of mean temperature and fertiliser on popsize. Meantemp was measured for each field, but measurements are correlated within pairs, and this should be taken into account to avoid pseudoreplication (in other words, I d like meantemp to be considered an outer variable). Do I need to replace the temperature measurements by the means for each pair? Or can I leave in the measurements per field pair? this is my data and model (with meantemp values for each field): popsize=c(8,19,13,28,30,29,45,41,21,30,20,32,44,52,65,45) meantemp=c(10,10.4,11.2,11.4,12,12.25,12.5,12.7,10.1,10.7,11.5,11.3,11.7,12.3,12.9,12.8) fertiliser=as.factor(rep(c(low,high),each=8)) pair=as.factor(rep(c(1:8),times=2)) model1=lme(popsize~meantemp+fertiliser, random=~1|pair) I now create a vector with the values of meantemp averaged per pair meantemp2=tapply(meantemp,pair,mean) meantemp2=meantemp2[pair] rerun a model with that explanatory variable: model2=lme(popsize~meantemp2+fertiliser, random=~1|pair) summary.lme and anova.lme suggest minute differences in the estimated parameters and DF (!) between model1 and model2. How do I explain these differences, especially in the DF? Is there a model to prefer? Sincerely, Yann -- Yann Clough Agroecology Georg-August University Waldweg 26 D-37073 Goettingen Tel: 0551/39-2358 email: [EMAIL PROTECTED] www: http://wwwuser.gwdg.de/~uaoe/mitarbeiter/y_clough_e.htm __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] data mining
On Tue, 19 Jul 2005 10:12:48 -0300 secretario academico FACEA wrote: Dear all, I'm looking for some material on data mining with R. I have something from Luis Torgo but I'd like to see something else. There have been several discussions on the list, so browsing the archives will probably bring up some helpful pointers. Also have a look at the MachineLearning view at http://CRAN.R-project.org/src/contrib/Views/MachineLearning.html Z If anybody could help me I'll be thankful Adrián __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] R: stats
for the stats gurus does anyone know if there exists a general formula relating the median of a continuous distribution to its moments. the distribution could be skewed or symmetric and is definitely not normal. the reason for asking is since the median of the particular distribution that i am interested in is difficult (probably impossible) to obtain. the median depends depends on an incomplete gamma distribution. the moments however can be obtained fairly easily. / allan__ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] R: stats
On 7/19/2005 9:28 AM, Clark Allan wrote: for the stats gurus does anyone know if there exists a general formula relating the median of a continuous distribution to its moments. the distribution could be skewed or symmetric and is definitely not normal. Not in general, and probably not any more practically useful than F^(-1)(0.5). the reason for asking is since the median of the particular distribution that i am interested in is difficult (probably impossible) to obtain. the median depends depends on an incomplete gamma distribution. the moments however can be obtained fairly easily. R can calculate the incomplete gamma function (see ?pgamma), so that's not necessarily a stumbling block. Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] R: expression
On Tue, 19 Jul 2005, Uwe Ligges wrote: Clark Allan wrote: hi all i am having a problem with the expression/paste command say we estimate a variable, named PHI it contains the value of say 2 and we want to display this value as hat(phi) = PHI onto a graphic i.e. hat(phi)=2 how does one do this? legend(-5, .3, substitute(hat(phi) == PHI, list(PHI = PHI))) or legend(-5, .3, bquote(hat(phi) == .(PHI))) -thomas Thomas Lumley Assoc. Professor, Biostatistics [EMAIL PROTECTED] University of Washington, Seattle __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] R Site Search is back up
The search site at http://finzi.psych.upenn.edu/ is back up. This allows you to search mailing list archives, R functions from most packages, and documents. You can also use the R function RSiteSearch() in R itself, which opens the results in your browser. I'm sorry the site was down for so long. I plan to work out a system that will allow faster restoration if this happens again (mutual complete backups with a colleague). Backups are useless if you don't have a machine to put them on, and it took 5 days to get one, then one day to restore. Some details that you don't need to read: The site was reconstructed from scratch. This means that links from one message in the archives to another message in the archives will not work anymore. They never worked very well, since the message numbers have been changed a few times over the years. (Threads work. Just links don't work.) In my rush to get this back up, I left out a few packages that would not install. I plan to install at least the help files from these in due course. I'm also not sure about Bioconductor packages that are not already on CRAN. I left out all of these. Let me know if you want them. Same with Jim Lindsey's packages. I also left out all the mail from r-sig-geo, which I had been including, although it seems hardly worth the effort since volume is so low now on that list. If you want it back, let me know. What happened is still not clear. There were several things wrong with the computer, still not fully diagnosed. The symptoms are consistent with overheating, but the fan was working fine, and so has the air conditioning in the building (which has been working very hard, global warming having finally hit Philadelphia). Recovery from the disk was possible but very expensive. What I learned from this is that backup systems do no good unless you have a computer to use. And also I could have backed up in a way that made it easier for myself to reconstruct, but that was not the major source of the delay. Jon -- Jonathan Baron, Professor of Psychology, University of Pennsylvania Home page: http://www.sas.upenn.edu/~baron __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] colnames
On Tue, 19 Jul 2005, Adaikalavan Ramasamy wrote: First, your problem could be boiled down to the following example. See how the colnames of the two outputs vary. df - cbind.data.frame( 100=1:2, 200=3:4 ) df/df X100 X200 111 211 That one is probably unintentional. m - as.matrix( df ) # coerce to matrix class m/m 100 200 1 1 1 2 1 1 It appears that whenever R has to create a new dataframe automatically, it tries to get nice colnames. See help(data.frame). I am not exactly sure why this behaviour is different when creating a matrix. But I do not think this is a major problem for most people. If you coerce your input to matrix, the problem goes away. A data frame is column-oriented, and can be used as a source of variables, e.g. by attach() and the data= argument of all the model-fitting functions. That is not the purpose of matrices, but is why data.frames are made to have syntactic names for columns by default. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] colnames
Hi Adai,
Thank you very much for your suggestions. Your optimized function would come in
very handy cause I will need to generate a matrix of size around 2250 * 1000.
Regards,
Gilbert
-Original Message-
From: Adaikalavan Ramasamy [mailto:[EMAIL PROTECTED]
Sent: 19 July 2005 12:20
To: Gilbert Wu
Cc: r-help@stat.math.ethz.ch
Subject: RE: [R] colnames
First, your problem could be boiled down to the following example. See
how the colnames of the two outputs vary.
df - cbind.data.frame( 100=1:2, 200=3:4 )
df/df
X100 X200
111
211
m - as.matrix( df ) # coerce to matrix class
m/m
100 200
1 1 1
2 1 1
It appears that whenever R has to create a new dataframe automatically,
it tries to get nice colnames. See help(data.frame). I am not exactly
sure why this behaviour is different when creating a matrix. But I do
not think this is a major problem for most people. If you coerce your
input to matrix, the problem goes away.
Next, note the following points :
a) mat[ 1:3, 1:ncol(mat) ] is equivalent to simply mat[ 1:3, ].
b) mat[ 2:nrow(mat), ] is equivalent to simply mat[ -1, ]
See help(subset) for more information.
Using the points above, we can simplify your function as
p.RIs2Returns - function (mat){
mat - as.matrix(mat)
x - mat[ -nrow(mat), ]
y - mat[ -1, ]
return( y/x -1 )
}
If your data contains only numerical data, it is probably good idea to
work with matrices as matrix operations are faster.
Finally, we can shorten your function. You can use the diff (which works
column-wise if input is a matrix) and apply function if you know that
y/x = exp(log(y/x)) = exp( log(y) - log(x) )
which could be coded in R as
exp( diff( log(r1) ) )
and then subtract 1 from above to get your returns.
Regards, Adai
On Tue, 2005-07-19 at 09:17 +0100, Gilbert Wu wrote:
Hi Adai,
Many Thanks for the examples.
I work for a financial institution. We are exploring R as a tool to implement
our portfolio optimization strategies. Hence, R is still a new language to us.
The script I wrote tried to make a returns matrix from the daily return
indices extracted from a SQL database. Please find below the output that
produces the 'X' prefix in the colnames. The reason to preserve the column
names is that they are stock identifiers which are to be used by other sub
systems rather than R.
I would welcome any suggestion to improve the script.
Regards,
Gilbert
p.RIs2Returns -
+ function (RIm)
+ {
+ x-RIm[1:(nrow(RIm)-1), 1:ncol(RIm)]
+ y-RIm[2:nrow(RIm), 1:ncol(RIm)]
+ RReturns - (y/x -1)
+ RReturns
+ }
channel-odbcConnect(ourSQLDB)
result-sqlQuery(channel,paste(select * from equityRIs;))
odbcClose(channel)
result
stockidsdate dbPrice
1 899188 20050713 7.59500
2 899188 20050714 7.60500
3 899188 20050715 7.48000
4 899188 20050718 7.41500
5 902232 20050713 10.97000
6 902232 20050714 10.94000
7 902232 20050715 10.99000
8 902232 20050718 11.05000
9 901714 20050713 17.96999
10 901714 20050714 18.00999
11 901714 20050715 17.64999
12 901714 20050718 17.64000
13 28176U 20050713 5.19250
14 28176U 20050714 5.25000
15 28176U 20050715 5.25000
16 28176U 20050718 5.22500
17 15322M 20050713 11.44000
18 15322M 20050714 11.5
19 15322M 20050715 11.33000
20 15322M 20050718 11.27000
r1-reshape(result, timevar=stockid, idvar=sdate, direction=wide)
r1
sdate dbPrice.899188 dbPrice.902232 dbPrice.901714 dbPrice.28176U
dbPrice.15322M
1 20050713 7.595 10.97 17.96999 5.1925
11.44
2 20050714 7.605 10.94 18.00999 5.2500
11.50
3 20050715 7.480 10.99 17.64999 5.2500
11.33
4 20050718 7.415 11.05 17.64000 5.2250
11.27
#Set sdate as the rownames
rownames(r1) -as.character(r1[1:nrow(r1),1:1])
#Get rid of the first column
r1 - r1[1:nrow(r1),2:ncol(r1)]
r1
dbPrice.899188 dbPrice.902232 dbPrice.901714 dbPrice.28176U
dbPrice.15322M
20050713 7.595 10.97 17.96999 5.1925
11.44
20050714 7.605 10.94 18.00999 5.2500
11.50
20050715 7.480 10.99 17.64999 5.2500
11.33
20050718 7.415 11.05 17.64000 5.2250
11.27
colnames(r1) - as.character(sub([[:alnum:]]*\\.,, colnames(r1)))
r1
899188 902232 901714 28176U 15322M
20050713 7.595 10.97 17.96999 5.1925 11.44
20050714 7.605 10.94 18.00999 5.2500 11.50
20050715 7.480 10.99 17.64999 5.2500 11.33
20050718 7.415 11.05 17.64000 5.2250 11.27
RRs-p.RIs2Returns(r1)
RRs
X899188 X902232 X901714 X28176U X15322M
20050714 0.001316656 -0.002734731 0.002225933 0.011073664 0.005244755
Re: [R] Survival dummy variables and some questions
Thanks For pointing that out. S - Original Message - From: Adaikalavan Ramasamy To: Stephen Cc: Frank E Harrell Jr ; Prof Brian Ripley ; Sent: Tuesday, July 19, 2005 1:30 PM Subject: Re: [R] Survival dummy variables and some questions Stephen, This has nothing to do with your R but to do with your email settings. Are you sure you are sending mails in plain text ? Your email on the R-help mailing archive (see link below) appears to be unreadable https://stat.ethz.ch/pipermail/r-help/2005-July/074210.html Please try to use plain text. (See http://expita.com/nomime.html). Regards, Adai On Tue, 2005-07-19 at 08:59 +0200, Stephen wrote: Many thanks I follow you what you say You can request predicted values at any sequence of ages - I guess there are plenty of postings on how to do that Regards, Stephen - Original Message - From: Frank E Harrell Jr To: Stephen Cc: Prof Brian Ripley ; Sent: Monday, July 18, 2005 6:13 PM Subject: Re: [R] Survival dummy variables and some questions Stephen wrote: Hi 1. Right perhaps this should clarify. I would like to extract coefficeints for different levels of the IVs (covariate). So for instance, age of onset I would want Hazards etc for every 5 years and so on... The approach I took was to categorize the variables (e.g., age of onset) and then turn theu resultant categorical variable into a factor as opposed to a variable... that is when the problems began An alternative approach to pulling out different values at different levels of the variable is what I seek. 2. I looked for the link, but can't Your needs don't require categorization. You can request predicted values at any sequence of ages. If you want hazard ratios you can take differences in predicted log hazards and antilog them. Frank -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University ?? http://mail.nana.co.il [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html ?? http://mail.nana.co.il [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] extracting row means from a list
Hello: I'm reading in a series of text files (100 files that are each 2000
rows by 6 columns). I wish to combine the columns (6) of each file (100) and
get the row mean. I'd like to end up with a data.frame of 2000 rows by 6
columns.
foo - list()
for(i in 1:10){
# The real data are read in from a series of numbered text files
foo[[i]] - data.frame(x1 = rnorm(100), x2 = rnorm(100), x3 =
rnorm(100),
x4 = rnorm(100), x5 = rnorm(100), x6 =
rnorm(100))
}
str(foo)
# by hand
mean.x1 -
rowMeans(cbind(foo[[1]][,1],foo[[2]][,1],foo[[3]][,1],foo[[4]][,1],foo[[5]][
,1]),
foo[[6]][,1],foo[[7]][,1],foo[[8]][,1],foo[[9]][,1
],foo[[10]][,1]))
mean.x2 -
rowMeans(cbind(foo[[1]][,2],foo[[2]][,2],foo[[3]][,2],foo[[4]][,2],foo[[5]][
,2]),
foo[[6]][,2],foo[[7]][,2],foo[[8]][,2],foo[[9]][,2
],foo[[10]][,2]))
# and so on to column 6
mean.x6 -
rowMeans(cbind(foo[[1]][,6],foo[[2]][,6],foo[[3]][,6],foo[[4]][,6],foo[[5]][
,6]),
foo[[6]][,6],foo[[7]][,6],foo[[8]][,6],foo[[9]][,6
],foo[[10]][,6]))
I've implemented this with nested loops that create temporary variables and
calc the mean, but the approach is clunky. E.g.,
# nested loops
for(i in 1:ncol(foo[[1]])){
for(j in 1:length(foo)){
# etc ...
}
}
Is there a way to build a better mouse trap?
TIA, Andy
Thanks, Andy
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[R] using argument names (of indeterminate number) within a function
Although I tried to find an answer in the manuals and archives, I cannot
solve this (please excuse that my English and/or R programming skills
are not good enough to state my problem more clearly):
I want to write a function with an indeterminate (not pre-defined)
number of arguments and think that I should use the ... construct and
the match.call() function. The goal is to write a function that (among
other things) uses cbind() to combine a not pre-defined number of
vectors specified in the function call. For example, if my vectors are
x1, x2, x3, ... xn, within the function I want to use cbind(x1, x2) or
cbind(x1, x3, x5) or ... depending on the vector names I use in the
funcion call. Additionally, the function has other arguments.
In the archives I found the following thread (followed by Marc Schwartz)
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/15186.html
[R] returning argument names from Peter Dalgaard BSA on 2003-04-10 (stdin)
that seems to contain the solution to my problem, but I am stuck because
sapply(match.call()[-1], deparse) gives me a vector of strings and I
don't know how to use the names in this vector in the cbind() function.
Up to now my (clearly deficit) function looks like:
test - function(..., mvalid=1)
{
args = sapply(match.call()[-1], deparse)
# and here, I don't know how the vector names in args
# can be used in the cbind() function to follow:
#
# temp - cbind( ???
if (mvalid 1)
{
# here it goes on
}
}
Ultimately, I want that the function can be called like
test(x1,x2,mvalid=1)
or
test(x1,x3,x5,mavlid=2)
and that within the function
cbind(x1,x2)
or cbind(x1,x3,x5)
will be used.
Can someone give and explain an example / a solution on how to proceed?
*
Dr. Dirk Enzmann
Institute of Criminal Sciences
Dept. of Criminology
Edmund-Siemers-Allee 1
D-20146 Hamburg
Germany
phone: +49-040-42838.7498 (office)
+49-040-42838.4591 (Billon)
fax: +49-040-42838.2344
email: [EMAIL PROTECTED]
www:
http://www2.jura.uni-hamburg.de/instkrim/kriminologie/Mitarbeiter/Enzmann/Enzmann.html
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Re: [R] colnames
Hi Adai,
When I tried the optimized routine, I got the following error message:
r1
899188 902232 901714 28176U 15322M
20050713 7.595 10.97 17.96999 5.1925 11.44
20050714 7.605 10.94 18.00999 5.2500 11.50
20050715 7.480 10.99 17.64999 5.2500 11.33
20050718 7.415 11.05 17.64000 5.2250 11.27
exp(diff(log(r1))) -1
Error in r[i1] - r[-length(r):-(length(r) - lag + 1)] :
non-numeric argument to binary operator
Any idea?
Many Thanks.
Gilbert
-Original Message-
From: Adaikalavan Ramasamy [mailto:[EMAIL PROTECTED]
Sent: 19 July 2005 12:20
To: Gilbert Wu
Cc: r-help@stat.math.ethz.ch
Subject: RE: [R] colnames
First, your problem could be boiled down to the following example. See
how the colnames of the two outputs vary.
df - cbind.data.frame( 100=1:2, 200=3:4 )
df/df
X100 X200
111
211
m - as.matrix( df ) # coerce to matrix class
m/m
100 200
1 1 1
2 1 1
It appears that whenever R has to create a new dataframe automatically,
it tries to get nice colnames. See help(data.frame). I am not exactly
sure why this behaviour is different when creating a matrix. But I do
not think this is a major problem for most people. If you coerce your
input to matrix, the problem goes away.
Next, note the following points :
a) mat[ 1:3, 1:ncol(mat) ] is equivalent to simply mat[ 1:3, ].
b) mat[ 2:nrow(mat), ] is equivalent to simply mat[ -1, ]
See help(subset) for more information.
Using the points above, we can simplify your function as
p.RIs2Returns - function (mat){
mat - as.matrix(mat)
x - mat[ -nrow(mat), ]
y - mat[ -1, ]
return( y/x -1 )
}
If your data contains only numerical data, it is probably good idea to
work with matrices as matrix operations are faster.
Finally, we can shorten your function. You can use the diff (which works
column-wise if input is a matrix) and apply function if you know that
y/x = exp(log(y/x)) = exp( log(y) - log(x) )
which could be coded in R as
exp( diff( log(r1) ) )
and then subtract 1 from above to get your returns.
Regards, Adai
On Tue, 2005-07-19 at 09:17 +0100, Gilbert Wu wrote:
Hi Adai,
Many Thanks for the examples.
I work for a financial institution. We are exploring R as a tool to implement
our portfolio optimization strategies. Hence, R is still a new language to us.
The script I wrote tried to make a returns matrix from the daily return
indices extracted from a SQL database. Please find below the output that
produces the 'X' prefix in the colnames. The reason to preserve the column
names is that they are stock identifiers which are to be used by other sub
systems rather than R.
I would welcome any suggestion to improve the script.
Regards,
Gilbert
p.RIs2Returns -
+ function (RIm)
+ {
+ x-RIm[1:(nrow(RIm)-1), 1:ncol(RIm)]
+ y-RIm[2:nrow(RIm), 1:ncol(RIm)]
+ RReturns - (y/x -1)
+ RReturns
+ }
channel-odbcConnect(ourSQLDB)
result-sqlQuery(channel,paste(select * from equityRIs;))
odbcClose(channel)
result
stockidsdate dbPrice
1 899188 20050713 7.59500
2 899188 20050714 7.60500
3 899188 20050715 7.48000
4 899188 20050718 7.41500
5 902232 20050713 10.97000
6 902232 20050714 10.94000
7 902232 20050715 10.99000
8 902232 20050718 11.05000
9 901714 20050713 17.96999
10 901714 20050714 18.00999
11 901714 20050715 17.64999
12 901714 20050718 17.64000
13 28176U 20050713 5.19250
14 28176U 20050714 5.25000
15 28176U 20050715 5.25000
16 28176U 20050718 5.22500
17 15322M 20050713 11.44000
18 15322M 20050714 11.5
19 15322M 20050715 11.33000
20 15322M 20050718 11.27000
r1-reshape(result, timevar=stockid, idvar=sdate, direction=wide)
r1
sdate dbPrice.899188 dbPrice.902232 dbPrice.901714 dbPrice.28176U
dbPrice.15322M
1 20050713 7.595 10.97 17.96999 5.1925
11.44
2 20050714 7.605 10.94 18.00999 5.2500
11.50
3 20050715 7.480 10.99 17.64999 5.2500
11.33
4 20050718 7.415 11.05 17.64000 5.2250
11.27
#Set sdate as the rownames
rownames(r1) -as.character(r1[1:nrow(r1),1:1])
#Get rid of the first column
r1 - r1[1:nrow(r1),2:ncol(r1)]
r1
dbPrice.899188 dbPrice.902232 dbPrice.901714 dbPrice.28176U
dbPrice.15322M
20050713 7.595 10.97 17.96999 5.1925
11.44
20050714 7.605 10.94 18.00999 5.2500
11.50
20050715 7.480 10.99 17.64999 5.2500
11.33
20050718 7.415 11.05 17.64000 5.2250
11.27
colnames(r1) - as.character(sub([[:alnum:]]*\\.,, colnames(r1)))
r1
899188 902232 901714 28176U 15322M
20050713 7.595 10.97 17.96999 5.1925 11.44
20050714 7.605 10.94 18.00999 5.2500
Re: [R] extracting row means from a list
Andy Bunn wrote:
Hello: I'm reading in a series of text files (100 files that are each 2000
rows by 6 columns). I wish to combine the columns (6) of each file (100) and
get the row mean. I'd like to end up with a data.frame of 2000 rows by 6
columns.
foo - list()
for(i in 1:10){
# The real data are read in from a series of numbered text files
foo[[i]] - data.frame(x1 = rnorm(100), x2 = rnorm(100), x3 =
rnorm(100),
x4 = rnorm(100), x5 = rnorm(100), x6 =
rnorm(100))
}
str(foo)
# by hand
mean.x1 -
rowMeans(cbind(foo[[1]][,1],foo[[2]][,1],foo[[3]][,1],foo[[4]][,1],foo[[5]][
,1]),
foo[[6]][,1],foo[[7]][,1],foo[[8]][,1],foo[[9]][,1
],foo[[10]][,1]))
mean.x2 -
rowMeans(cbind(foo[[1]][,2],foo[[2]][,2],foo[[3]][,2],foo[[4]][,2],foo[[5]][
,2]),
foo[[6]][,2],foo[[7]][,2],foo[[8]][,2],foo[[9]][,2
],foo[[10]][,2]))
# and so on to column 6
mean.x6 -
rowMeans(cbind(foo[[1]][,6],foo[[2]][,6],foo[[3]][,6],foo[[4]][,6],foo[[5]][
,6]),
foo[[6]][,6],foo[[7]][,6],foo[[8]][,6],foo[[9]][,6
],foo[[10]][,6]))
I've implemented this with nested loops that create temporary variables and
calc the mean, but the approach is clunky. E.g.,
# nested loops
for(i in 1:ncol(foo[[1]])){
for(j in 1:length(foo)){
# etc ...
}
}
Is there a way to build a better mouse trap?
TIA, Andy
I don't know of a way of getting around at least one for loop, but the
following might be want you need:
r - matrix(, NROW(foo[[1]]), length(foo))
for(i in 1:NCOL(foo[[1]]))
r[, i] - rowMeans(do.call(cbind, lapply(foo, [, i)))
dim(r)
This assumes each element of foo has identical dimensions. Otherwise
you'll get an error.
HTH,
--sundar
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Re: [R] using argument names (of indeterminate number) within a function
On Tue, 2005-07-19 at 17:29 +0200, Dirk Enzmann wrote:
Although I tried to find an answer in the manuals and archives, I cannot
solve this (please excuse that my English and/or R programming skills
are not good enough to state my problem more clearly):
I want to write a function with an indeterminate (not pre-defined)
number of arguments and think that I should use the ... construct and
the match.call() function. The goal is to write a function that (among
other things) uses cbind() to combine a not pre-defined number of
vectors specified in the function call. For example, if my vectors are
x1, x2, x3, ... xn, within the function I want to use cbind(x1, x2) or
cbind(x1, x3, x5) or ... depending on the vector names I use in the
funcion call. Additionally, the function has other arguments.
In the archives I found the following thread (followed by Marc Schwartz)
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/15186.html
[R] returning argument names from Peter Dalgaard BSA on 2003-04-10 (stdin)
that seems to contain the solution to my problem, but I am stuck because
sapply(match.call()[-1], deparse) gives me a vector of strings and I
don't know how to use the names in this vector in the cbind() function.
Up to now my (clearly deficit) function looks like:
test - function(..., mvalid=1)
{
args = sapply(match.call()[-1], deparse)
# and here, I don't know how the vector names in args
# can be used in the cbind() function to follow:
#
# temp - cbind( ???
if (mvalid 1)
{
# here it goes on
}
}
Ultimately, I want that the function can be called like
test(x1,x2,mvalid=1)
or
test(x1,x3,x5,mavlid=2)
and that within the function
cbind(x1,x2)
or cbind(x1,x3,x5)
will be used.
Can someone give and explain an example / a solution on how to proceed?
Hi Dirk,
How about this:
my.cbind - function(...)
{
do.call(cbind, list(...))
}
a - 1:10
b - 11:20
c - 21:30
d - 31:40
my.cbind(a, b)
[,1] [,2]
[1,]1 11
[2,]2 12
[3,]3 13
[4,]4 14
[5,]5 15
[6,]6 16
[7,]7 17
[8,]8 18
[9,]9 19
[10,] 10 20
my.cbind(b, c, d)
[,1] [,2] [,3]
[1,] 11 21 31
[2,] 12 22 32
[3,] 13 23 33
[4,] 14 24 34
[5,] 15 25 35
[6,] 16 26 36
[7,] 17 27 37
[8,] 18 28 38
[9,] 19 29 39
[10,] 20 30 40
my.cbind(a, b, c, d)
[,1] [,2] [,3] [,4]
[1,]1 11 21 31
[2,]2 12 22 32
[3,]3 13 23 33
[4,]4 14 24 34
[5,]5 15 25 35
[6,]6 16 26 36
[7,]7 17 27 37
[8,]8 18 28 38
[9,]9 19 29 39
[10,] 10 20 30 40
The use of list(...) in the function allows you to use list based
functions such as do.call() or lapply() against the argument objects
directly without having to deparse and re-parse the character names of
the arguments, which is the approach that Peter used in his response in
the thread you referenced.
The OP in that thread wanted the argument names as character vectors, as
opposed to the argument objects themselves, which is what you need here.
HTH,
Marc Schwartz
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[R] ROC curve with survival data
Hi everyone, I am doing 5 years mortality predictive index score with survival analysis using a Cox proportional hazard model where I have a continous predictive variable and a right censored response which is the mortality, and the individuals were followed a maximum of 7 years. I'd like to asses the discrimination ability of survival analysis Cox model by computing a ROC curve and area under the curve for a fixed time (5 years), taking into account that the response is not binary but right censored. So, is there a function that computes a ROC curve under right censored survival data? Thank you in advance. Isaac Subirana: [EMAIL PROTECTED] Institut Municipal d'Investigació Mèdica (IMIM), Barcelona (Spain) [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] .gct file
I have two files to compare, one is a regular txt file that I can read in no prob. The other is a .gct file (How do I read in this one?) I tried a simple read.table(data.gct, header = T) How do you suggest reading in this file?? thank you. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] .gct file
On 7/19/2005 12:10 PM, mark salsburg wrote: I have two files to compare, one is a regular txt file that I can read in no prob. The other is a .gct file (How do I read in this one?) I tried a simple read.table(data.gct, header = T) How do you suggest reading in this file?? .gct is not a standard filename extension. You need to know what is in that file. Where did you get it? What program created it? Chances are the easiest thing to do is to get the program that created it to export in a well known format, e.g. .csv. Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Predict
Dear Matthias Can you provide an example to demonstrate what you did? Two remarks to your email. Maybe that answers already your question. 1) Using predict() you will get the estimated value for each observation or for new data. You can reproduce this value by using the coefficients from your estimated model (see the example below). For the interval you can get a confidence interval for the expected value under fixed conditions of the explanatory variables or you can obtain a prediction interval for a single new observation. The latter is of course wider, since you try to catch a single observation and not the expected value. 2) Using confint() you will get the estimated parameters (which are random variables, too) and their confidence interval. You can use the estimated values to calculate the predicted values. But you can NOT use the upper values from confint to estimate the upper values from predict by just putting them into your regression model. Thats not the way how confidence intervals are constructed. (I am not sure if this was your intention. Maybe if you show a reproducible example you can correct me if you meant something different) ## R Code ## Creation of a dataframe set.seed(1) x1 - runif(40) f1 - rep(c(a, b, c,d), each = 10) y - 2*x1 + rep(c(0.5, 0.1, -0.6, 1.5), each = 10) + rnorm(40, 0, 2) dat - data.frame(y = y, f1 = f1, x1 = x1) ## regression model reg - lm(y~ x1 + f1, data = dat) summary(reg) confint(reg) predict(reg, type=c(response), interval = confidence) ## caluclation of predicted values using the estimated ## coefficients ## estimated coefficients co - summary(reg)$coefficients[,Estimate] ## Using the regression model with that coefficients ## for observation 11 co[(Intercept)] + dat[11,x1]*co[x1] + co[f1b] ## prediction of observation 11 predict(reg, type=c(response))[11] Regards, Christoph -- Christoph Buser [EMAIL PROTECTED] Seminar fuer Statistik, LEO C13 ETH (Federal Inst. Technology) 8092 Zurich SWITZERLAND phone: x-41-44-632-4673 fax: 632-1228 http://stat.ethz.ch/~buser/ -- Matthias Eggenberger writes: When I callculate a linear model, then I can compute via confint the confidencial intervals. the interval level can be chosen. as result, I get the parameter of the model according to the interval level. On the other hand, I can compute the prediction-values for my model as well with predict(object, type=c(response) etc.). Here I have also the possibility to chose a level for the confidential intervals. the output are the calculatet values for the fit, the lower and upper level. the problem now is, that when I calculate the values through the linear model function with the parameter values I get from confint() an I compare them with the values I get from predict() these values differ extremely. Why is that so? Does the command predict() calculate the values through an other routine? That means the command predict() doesn't use the same parameters to calculate the prediction-values than the ones given by confint()? Greetings Matthias -- GMX DSL = Maximale Leistung zum minimalen Preis! __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Padding in lattice plots
On 7/16/05, Federico Gherardini [EMAIL PROTECTED] wrote: On Friday 15 July 2005 17:00, Deepayan Sarkar wrote: On 7/15/05, Federico Gherardini [EMAIL PROTECTED] wrote: On Friday 15 July 2005 14:42, you wrote: Hi all, I've used the split argument to print four lattice plots on a single page. The problem now is that I need to reduce the amount of white space between the plots. I've read other mails in this list about the new trellis parameters layout.heights and layout.widhts but I haven't been able to use them properly. I've tried to input values between 0 and 1 as the padding value (both left and right and top and bottom) but nothing changed. It seems I can only increase the padding by using values 1. Any ideas? Thanks in advance for your help Federico Gherardini It seems like I've found an answer myself you have to use negative values to decrease the padding. I thought it was something like the cex parameter which acts like a multiplier I thought so too. but this is not the case. Could you post what you used? There are several different padding parameters you need to set to 0, did you change them all? Deepayan Hi Deepayan This is what I used I don't know if I did everything the proper way but at least I got the result I was seeking! :) trellis.par.set(list(layout.heights = list(top.padding = -1))) trellis.par.set(list(layout.heights = list(bottom.padding = -1, axis.xlab.padding = 1, xlab = -1.2))) trellis.par.set(list(layout.widths = list(left.padding = -1))) trellis.par.set(list(layout.widths = list(right.padding = -1, ylab.axis.padding = -0.5))) Do these settings make any sense? Yes, but there are other padding parameters. For example, at the top, there's $ top.padding : num 1 $ main.key.padding : num 1 $ key.axis.padding : num 1 The total amount of space left at the top (in the default case, with no main label and no key) is the sum of all three, so just setting one to 0 wouldn't be enough for what you want. (This is probably not very simple, but I couldn't think of anything that's simpler yet as flexible.) Deepayan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] initial points for arms in package HI
Dear R-users
I have a problem choosing initial points for the function arms()
in the package HI
I intend to implement a Gibbs sampler and one of my conditional
distributions is nonstandard and not logconcave.
Therefore I'd like to use arms.
But there seem to be a strong influence of the initial point
y.start. To show the effect I constructed a demonstration
example. It is reproducible without further information.
Please note that my target density is not logconcave.
Thanks for all comments or ideas.
Christoph Buser
## R Code:
library(HI)
## parameter for the distribution
para - 0.1
## logdensity
logDichteGam - function(x, u = para, v = para) {
-(u*x + v*1/x) - log(x)
}
## density except for the constant
propDichteGam - function(x, u = para, v = para) {
exp(-(u*x + v*1/x) - log(x))
}
## calculating the constant
(c - integrate(propDichteGam, 0, 1000, rel.tol = 10^(-12))$value)
## density
DichteGam - function(x, u = para, v = para) {
exp(-(u*x + v*1/x) - log(x))/c
}
## calculating 1000 values by repeating a call of arms (this would
## be the situation in an Gibbs Sample. Of course in a Gibbs sampler
## the distribution would change. This is only for demonstration
res1 - NULL
for(i in 1:1000)
res1[i] - arms(runif(1,0,100), logDichteGam, function(x) (x0)(x100), 1)
## Generating a sample of thousand observations with 1 call of arms
res2 - arms(runif(1,0,100), logDichteGam, function(x) (x0)(x100), 1000)
## Plot of the samples
mult.fig(4)
plot(res1, log = y)
plot(res2, log = y)
hist(res1, freq = FALSE, xlim = c(0,4), breaks = seq(0,100,by = 0.1),
ylim = c(0,1))
curve(DichteGam, 0,4, add = TRUE, col = 2)
hist(res2, freq = FALSE, xlim = c(0,4), breaks = seq(0,100,by = 0.1),
ylim = c(0,1))
curve(DichteGam, 0,4, add = TRUE, col = 2)
## If we repeat the procedure, using the fix intial value 1,
## the situation is even worse
res3 - NULL
for(i in 1:1000)
res3[i] - arms(1, logDichteGam, function(x) (x0)(x100), 1)
## Generating a sample of thousand observations with 1 call of arms
res4 - arms(1, logDichteGam, function(x) (x0)(x100), 1000)
## Plot of the samples
par(mfrow = c(2,2))
plot(res3, log = y)
plot(res4, log = y)
hist(res3, freq = FALSE, xlim = c(0,4), breaks = seq(0,100,by = 0.1),
ylim = c(0,1))
curve(DichteGam, 0,4, add = TRUE, col = 2)
hist(res4, freq = FALSE, xlim = c(0,4), breaks = seq(0,100,by = 0.1),
ylim = c(0,1))
curve(DichteGam, 0,4, add = TRUE, col = 2)
## If I generate the sample in a for-loop (one by one) I do not
## get the correct density. But this is exactly the situation in
## my Gibbs Sampler. Therfore I am concerned about the correct
## application of arms
--
Christoph Buser [EMAIL PROTECTED]
Seminar fuer Statistik, LEO C13
ETH (Federal Inst. Technology) 8092 Zurich SWITZERLAND
phone: x-41-44-632-4673 fax: 632-1228
http://stat.ethz.ch/~buser/
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Re: [R] .gct file
On Tue, 2005-07-19 at 12:28 -0400, Duncan Murdoch wrote: On 7/19/2005 12:10 PM, mark salsburg wrote: I have two files to compare, one is a regular txt file that I can read in no prob. The other is a .gct file (How do I read in this one?) I tried a simple read.table(data.gct, header = T) How do you suggest reading in this file?? .gct is not a standard filename extension. You need to know what is in that file. Where did you get it? What program created it? Chances are the easiest thing to do is to get the program that created it to export in a well known format, e.g. .csv. Duncan Murdoch A quick Google search would suggest Gene Cluster Text file: http://www.broad.mit.edu/cancer/software/genepattern/tutorial/gp_tutorial_fileformats.html#gct produced by Gene Pattern: http://www.broad.mit.edu/cancer/software/genepattern/ If correct, I would point Mark to the Bioconductor folks for more information and assistance: http://www.bioconductor.org/ HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] .gct file
ok so the gct file looks like this: #1.2 (version number) 7283 19 (matrix size) Name Description Values ... .. How can I tell R to disregard the first two lines and start reading the 3rd line in this gct file. I would just delete them, but I do not know how to open a gct. file thank you On 7/19/05, Duncan Murdoch [EMAIL PROTECTED] wrote: On 7/19/2005 12:10 PM, mark salsburg wrote: I have two files to compare, one is a regular txt file that I can read in no prob. The other is a .gct file (How do I read in this one?) I tried a simple read.table(data.gct, header = T) How do you suggest reading in this file?? .gct is not a standard filename extension. You need to know what is in that file. Where did you get it? What program created it? Chances are the easiest thing to do is to get the program that created it to export in a well known format, e.g. .csv. Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] .gct file
If it is a text file ?read.table should provide enough details to read the file into R. Based on the file format referenced below it shouldn't be too hard to get at the parts you want. Randy On 7/19/05 1:06 PM, Marc Schwartz (via MN) [EMAIL PROTECTED] wrote: On Tue, 2005-07-19 at 12:28 -0400, Duncan Murdoch wrote: On 7/19/2005 12:10 PM, mark salsburg wrote: I have two files to compare, one is a regular txt file that I can read in no prob. The other is a .gct file (How do I read in this one?) I tried a simple read.table(data.gct, header = T) How do you suggest reading in this file?? .gct is not a standard filename extension. You need to know what is in that file. Where did you get it? What program created it? Chances are the easiest thing to do is to get the program that created it to export in a well known format, e.g. .csv. Duncan Murdoch A quick Google search would suggest Gene Cluster Text file: http://www.broad.mit.edu/cancer/software/genepattern/tutorial/gp_tutorial_file formats.html#gct produced by Gene Pattern: http://www.broad.mit.edu/cancer/software/genepattern/ If correct, I would point Mark to the Bioconductor folks for more information and assistance: http://www.bioconductor.org/ HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html ~~ Randy Johnson Laboratory of Genomic Diversity NCI-Frederick Bldg 560, Rm 11-85 Frederick, MD 21702 (301)846-1304 ~~ __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] .gct file
On 7/19/2005 1:16 PM, mark salsburg wrote: ok so the gct file looks like this: #1.2 (version number) 7283 19 (matrix size) Name Description Values ... .. How can I tell R to disregard the first two lines and start reading the 3rd line in this gct file. I would just delete them, but I do not know how to open a gct. file Use skip=2. See ?read.table. Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Library mclust in 64bit compiled R
Hi, All;
I tried to use library mclust in 64-bit compiled R 2.0.1 but failed.
Installation went smoothly without any warning or error. However, when I
tried to use them with the following simple code, it crashed.
Library(mclust)
Dat - c(rnorm(20, mean=0, sd=0.2), rnorm(30, mean=1, sd=0.2))
Ind - Mclust(dat, 1, 5)$classification
cbind(Dat, Ind)
The error message was:
/usr/local/R-2.0.1_64bit/lib/R/bin/BATCH: line 55: 18097 Done
( echo invisible(options(echo = TRUE)); cat ${in}; echo proc.time() )
18099 Segmentation fault | ${R_HOME}/bin/R ${opts} ${out} 21
Can anybody help me with this?
Thanks in advance,
Tae-Hoon Chung
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Re: [R] .gct file
Try ?read.table or args(read.table). Might skip=2 do what you want? spencer graves p.s. I routinely readLines(File, n=11) to see how many headers there are AND identify the sep character. Then I quantile(count.fields(File, ...)) to see if all records have the same number of fields. Then I call something like read.table with the appropriate arguments. Then I print the first 2 or so rows of the result to make sure I read the file correctly. mark salsburg wrote: ok so the gct file looks like this: #1.2 (version number) 7283 19 (matrix size) Name Description Values ... .. How can I tell R to disregard the first two lines and start reading the 3rd line in this gct file. I would just delete them, but I do not know how to open a gct. file thank you On 7/19/05, Duncan Murdoch [EMAIL PROTECTED] wrote: On 7/19/2005 12:10 PM, mark salsburg wrote: I have two files to compare, one is a regular txt file that I can read in no prob. The other is a .gct file (How do I read in this one?) I tried a simple read.table(data.gct, header = T) How do you suggest reading in this file?? .gct is not a standard filename extension. You need to know what is in that file. Where did you get it? What program created it? Chances are the easiest thing to do is to get the program that created it to export in a well known format, e.g. .csv. Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Spencer Graves, PhD Senior Development Engineer PDF Solutions, Inc. 333 West San Carlos Street Suite 700 San Jose, CA 95110, USA [EMAIL PROTECTED] www.pdf.com http://www.pdf.com Tel: 408-938-4420 Fax: 408-280-7915 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Code Verification
Hi R Users
I have a code which I am running for my thesis work. Just want to make sure that
its ok. Its a t test I am conducting between two gamma distributions with
different shape parameters.
the code looks like:
sink(a1.txt);
for (i in 1:1000)
{
x-rgamma(40, 2.5, 10) # n = 40, shape = 2.5, Scale = 10
y-rgamma(40, 2.8, 10) # n = 40, shape = 2.8, Scale = 10
z-t.test(x, y)
print(z)
}
I will appreciate it if someone could tell me if its alrite or not.
thanks
-dev
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[R] integrate fails with errors
Hi all, i'm new to R, I need to modelize in R a statistic algorithm, This algo use Weibull, normal law, linear regression, normalisation, root mean square, to find eta and beta fitting the weibull model (to analyse few results) and further when we will get more information apply bayes model . the problem is when When i try to integrate it fails with errors. by the way i like to integrate something like that : beta0,eta0,n are initialized as single integer, temp is a 1 dimension array containing 9 integer. integrate(function(beta) ((beta/(eta0)^beta)^n)*prod(temp^(1-beta)*exp(-sum(temp^beta)/(eta^beta)))*(1/(sqrt(2*pi))*exp(((beta-beta0)^2,0,Inf) R says : The longuest object isn't a multiple of the shortest object in temp^beta , the same for temp^(1-beta). I don't understand why R fails to take calculate each temp^beta then sum it over again for each beta values. don't know if my post is explicit my head is burned out with these things today thks. // Webmail Oreka : http://www.oreka.com __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] .gct file
On Tue, 2005-07-19 at 13:16 -0400, mark salsburg wrote: ok so the gct file looks like this: #1.2 (version number) 7283 19 (matrix size) Name Description Values ... .. How can I tell R to disregard the first two lines and start reading the 3rd line in this gct file. I would just delete them, but I do not know how to open a gct. file thank you On 7/19/05, Duncan Murdoch [EMAIL PROTECTED] wrote: On 7/19/2005 12:10 PM, mark salsburg wrote: I have two files to compare, one is a regular txt file that I can read in no prob. The other is a .gct file (How do I read in this one?) I tried a simple read.table(data.gct, header = T) How do you suggest reading in this file?? .gct is not a standard filename extension. You need to know what is in that file. Where did you get it? What program created it? Chances are the easiest thing to do is to get the program that created it to export in a well known format, e.g. .csv. Duncan Murdoch The above would be consistent with the info in my reply. I guess if the format is consistent, as per Mark's example above, you can use: read.table(data.gct, skip = 2, header = TRUE) which will start by skipping the first two lines and then reading in the header row and then the data. See ?read.table HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] .gct file
This is all extremely helpful.
The data turns out is a little atypical, the columns are tab-delemited
except for the description columns
DATA1.gct looks like this
#1.2
23 3423
NAME DESCRIPTION VALUE
gene1 a protein inducer 1123
. . ..
How do I get R to read the data as tab delemited, but read in the 2nd
coloumn as one value based on the quotation marks..
thanks..
On 7/19/05, Marc Schwartz (via MN) [EMAIL PROTECTED] wrote:
On Tue, 2005-07-19 at 13:16 -0400, mark salsburg wrote:
ok so the gct file looks like this:
#1.2 (version number)
7283 19 (matrix size)
Name Description Values
... ..
How can I tell R to disregard the first two lines and start reading
the 3rd line in this gct file. I would just delete them, but I do not
know how to open a gct. file
thank you
On 7/19/05, Duncan Murdoch [EMAIL PROTECTED] wrote:
On 7/19/2005 12:10 PM, mark salsburg wrote:
I have two files to compare, one is a regular txt file that I can read
in no prob.
The other is a .gct file (How do I read in this one?)
I tried a simple
read.table(data.gct, header = T)
How do you suggest reading in this file??
.gct is not a standard filename extension. You need to know what is in
that file. Where did you get it? What program created it?
Chances are the easiest thing to do is to get the program that created
it to export in a well known format, e.g. .csv.
Duncan Murdoch
The above would be consistent with the info in my reply.
I guess if the format is consistent, as per Mark's example above, you
can use:
read.table(data.gct, skip = 2, header = TRUE)
which will start by skipping the first two lines and then reading in the
header row and then the data.
See ?read.table
HTH,
Marc Schwartz
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Re: [R] using argument names (of indeterminate number) within a function
On 7/19/05, Dirk Enzmann [EMAIL PROTECTED] wrote:
Although I tried to find an answer in the manuals and archives, I cannot
solve this (please excuse that my English and/or R programming skills
are not good enough to state my problem more clearly):
I want to write a function with an indeterminate (not pre-defined)
number of arguments and think that I should use the ... construct and
the match.call() function. The goal is to write a function that (among
other things) uses cbind() to combine a not pre-defined number of
vectors specified in the function call. For example, if my vectors are
x1, x2, x3, ... xn, within the function I want to use cbind(x1, x2) or
cbind(x1, x3, x5) or ... depending on the vector names I use in the
funcion call. Additionally, the function has other arguments.
In the archives I found the following thread (followed by Marc Schwartz)
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/15186.html
[R] returning argument names from Peter Dalgaard BSA on 2003-04-10 (stdin)
that seems to contain the solution to my problem, but I am stuck because
sapply(match.call()[-1], deparse) gives me a vector of strings and I
don't know how to use the names in this vector in the cbind() function.
Up to now my (clearly deficit) function looks like:
test - function(..., mvalid=1)
{
args = sapply(match.call()[-1], deparse)
# and here, I don't know how the vector names in args
# can be used in the cbind() function to follow:
#
# temp - cbind( ???
if (mvalid 1)
{
# here it goes on
}
}
Ultimately, I want that the function can be called like
test(x1,x2,mvalid=1)
or
test(x1,x3,x5,mavlid=2)
and that within the function
cbind(x1,x2)
or cbind(x1,x3,x5)
will be used.
If you just need to pass them to cbind then just use cbind(...), e.g.
test - function(..., m) if (m 1) cbind(...) else m
otherwise, use list(...) as shown by a previous answer or here
test2 - function(..., m) if (m 1) length(list(...)) else m
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Re: [R] .gct file
For the TAB delimited columns, adjust the 'sep' argument to:
read.table(data.gct, skip = 2, header = TRUE, sep = \t)
The 'quote' argument is by default:
quote = \'
which should take care of the quoted strings and bring them in as a
single value.
The above presumes that the header row is also TAB delimited. If not,
you may have to set 'skip = 3' to skip over the header row and manually
set the column names.
HTH,
Marc Schwartz
On Tue, 2005-07-19 at 13:52 -0400, mark salsburg wrote:
This is all extremely helpful.
The data turns out is a little atypical, the columns are tab-delemited
except for the description columns
DATA1.gct looks like this
#1.2
23 3423
NAME DESCRIPTION VALUE
gene1 a protein inducer 1123
. . ..
How do I get R to read the data as tab delemited, but read in the 2nd
coloumn as one value based on the quotation marks..
thanks..
On 7/19/05, Marc Schwartz (via MN) [EMAIL PROTECTED] wrote:
On Tue, 2005-07-19 at 13:16 -0400, mark salsburg wrote:
ok so the gct file looks like this:
#1.2 (version number)
7283 19 (matrix size)
Name Description Values
... ..
How can I tell R to disregard the first two lines and start reading
the 3rd line in this gct file. I would just delete them, but I do not
know how to open a gct. file
thank you
On 7/19/05, Duncan Murdoch [EMAIL PROTECTED] wrote:
On 7/19/2005 12:10 PM, mark salsburg wrote:
I have two files to compare, one is a regular txt file that I can read
in no prob.
The other is a .gct file (How do I read in this one?)
I tried a simple
read.table(data.gct, header = T)
How do you suggest reading in this file??
.gct is not a standard filename extension. You need to know what is in
that file. Where did you get it? What program created it?
Chances are the easiest thing to do is to get the program that created
it to export in a well known format, e.g. .csv.
Duncan Murdoch
The above would be consistent with the info in my reply.
I guess if the format is consistent, as per Mark's example above, you
can use:
read.table(data.gct, skip = 2, header = TRUE)
which will start by skipping the first two lines and then reading in the
header row and then the data.
See ?read.table
HTH,
Marc Schwartz
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Re: [R] Michaelis-menten equation
Peter Dalgaard [EMAIL PROTECTED] writes:
Chun-Ying Lee [EMAIL PROTECTED] writes:
Dear R users:
I encountered difficulties in michaelis-menten equation. I found
that when I use right model definiens, I got wrong Km vlaue,
and I got right Km value when i use wrong model definiens.
The value of Vd and Vmax are correct in these two models.
How do you know what the correct value is? Are you sure that the other
values are right?
I'm a bit rusty on MM, but are you sure your right model is right?
Try doing a dimensional analysis on the ODE. I kind of suspect that
Vd is entering in the wrong way. Since you're dealing in
concentrations, should it enter at all (except via the conc. at time
0, of course)?
Not knowing the context, I can't be quite sure, but generally, I'd
expect Vm*Km/(Km+y) to be the reaction rate, so that Vm is the maximum
rate, attained when y is zero and Km is the conc. at half-maximum
rate. This doesn't look quit like what you have.
Hmm, sorry, no. I'm talking through a hole in my head there.
Vm*y/(Km+y) makes OK sense. Vm is what you get for large y - passing
from 1st order to 0th order kinetics. However, looking at the data
plot(PKindex)
abline(lm(conc~time,data=PKindex))
shows that they are pretty much on a straight line, i.e. you are
in the domain of 0-order kinetics. So why are you expecting the rate
of decrease to have changed by roughly 3/4 (from 2/3*Vm/Vd at y=2*Km
to 1/2*Vm/Vd at y=Km when you reach 4.67)??
#-right model definiens
PKindex-data.frame(time=c(0,1,2,4,6,8,10,12,16,20,24),
conc=c(8.57,8.30,8.01,7.44,6.88,6.32,5.76,5.20,4.08,2.98,1.89))
mm.model - function(time, y, parms) {
dCpdt - -(parms[Vm]/parms[Vd])*y[1]/(parms[Km]+y[1])
list(dCpdt)}
Dose-300
modfun - function(time,Vm,Km,Vd) {
out - lsoda(Dose/Vd,time,mm.model,parms=c(Vm=Vm,Km=Km,Vd=Vd),
rtol=1e-8,atol=1e-8)
out[,2] }
objfun - function(par) {
out - modfun(PKindex$time,par[1],par[2],par[3])
sum((PKindex$conc-out)^2) }
fit - optim(c(10,1,80),objfun, method=Nelder-Mead)
print(fit$par)
[1] 10.0390733 0.1341544 34.9891829 #--Km=0.1341544,wrong value--
#-wrong model definiens
#-Km should not divided by Vd--
PKindex-data.frame(time=c(0,1,2,4,6,8,10,12,16,20,24),
conc=c(8.57,8.30,8.01,7.44,6.88,6.32,5.76,5.20,4.08,2.98,1.89))
mm.model - function(time, y, parms) {
dCpdt - -(parms[Vm]/parms[Vd])*y[1]/(parms[Km]/parms[Vd]+y[1])
list(dCpdt)}
Dose-300
modfun - function(time,Vm,Km,Vd) {
out - lsoda(Dose/Vd,time,mm.model,parms=c(Vm=Vm,Km=Km,Vd=Vd),
rtol=1e-8,atol=1e-8)
out[,2]
}
objfun - function(par) {
out - modfun(PKindex$time,par[1],par[2],par[3])
sum((PKindex$conc-out)^2)}
fit - optim(c(10,1,80),objfun, method=Nelder-Mead)
print(fit$par)
[1] 10.038821 4.690267 34.989239 #--Km=4.690267,right value--
What did I do wrong, and how to fix it?
Any suggestions would be greatly appreciated.
Thanks in advance!!
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--
O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918
~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907
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--
O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918
~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907
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Re: [R] .gct file
On Tue, 2005-07-19 at 13:08 -0500, Marc Schwartz (via MN) wrote: For the TAB delimited columns, adjust the 'sep' argument to: read.table(data.gct, skip = 2, header = TRUE, sep = \t) The 'quote' argument is by default: quote = \' which should take care of the quoted strings and bring them in as a single value. The above presumes that the header row is also TAB delimited. If not, you may have to set 'skip = 3' to skip over the header row and manually set the column names. One correction. If the final para applies and you need to use 'skip = 3', you would also need to leave out the 'header = TRUE' argument, which defaults to FALSE. Marc __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] .gct file
On Tue, 2005-07-19 at 13:08 -0500, Marc Schwartz (via MN) wrote: For the TAB delimited columns, adjust the 'sep' argument to: read.table(data.gct, skip = 2, header = TRUE, sep = \t) The 'quote' argument is by default: quote = \' which should take care of the quoted strings and bring them in as a single value. The above presumes that the header row is also TAB delimited. If not, you may have to set 'skip = 3' to skip over the header row and manually set the column names. One correction. If the final para applies and you need to use 'skip = 3', you would also need to leave out the 'header = TRUE' argument, which defaults to FALSE. Marc __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] CPU Usage with R 2.1.0 in Windows
Hi, I'm using a fairly simple HP Compaq desktop PC running Windows 2K. When running a large process in R, the process RGUI.exe will never exceed 50% of the CPU usage. The program used to be able to use more of the computer, but does not now. I don't believe this is a multiple processor machine. Can anyone give any advice on how to solve the problem? Thanks, Michael Greene Product Management Plymouth Rock Assurance Corp 617-951-1682 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] ROC curve with survival data
SUBIRANA CACHINERO, ISAAC wrote: Hi everyone, I am doing 5 years mortality predictive index score with survival analysis using a Cox proportional hazard model where I have a continous predictive variable and a right censored response which is the mortality, and the individuals were followed a maximum of 7 years. I'd like to asses the discrimination ability of survival analysis Cox model by computing a ROC curve and area under the curve for a fixed time (5 years), taking into account that the response is not binary but right censored. So, is there a function that computes a ROC curve under right censored survival data? Thank you in advance. I don't find ROC curves themselves very useful but the area under them is useful and is a simple translation of the Somers' Dxy rank correlation between predicted survival probability (or anything monotonically related to it just is log hazard) and observed survival time. Dxy = 2*(C-.5) where C is the concordance index, a generalization of ROC area not requiring choosing a specific time point. You can get this in the rcorr.cens function in the Hmisc package. Frank Isaac Subirana: [EMAIL PROTECTED] Institut Municipal d'Investigació Mèdica (IMIM), Barcelona (Spain) [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Taking the derivative of a quadratic B-spline
Hello, I have been trying to take the derivative of a quadratic B-spline obtained by using the COBS library. What I would like to do is similar to what one can do by using fit-smooth.spline(cdf) xx-seq(-10,10,.1) predict(fit, xx, deriv = 1) The goal is to fit the spline to data that is approximating a cumulative distribution function (e.g. in my example, cdf is a 2-column matrix with x values in column 1 and the estimate of the cdf evaluated at x in column 2) and then take the first derivative over a range of values to get density estimates. The reason I don't want to use smooth.spline is that there is no way to impose constraints (e.g. =0, =1, and monotonicity) as there is with COBS. However, since COBS doesn't have the 'deriv =' option, the only way I can think of doing it with COBS is to evaluate the derivatives numerically. Regards, Jim McDermott __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] CPU Usage with R 2.1.0 in Windows
Dear Michael: Why is it a problem that R is not using more CPU space than it seems to need? -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Greene, Michael Sent: Tuesday, July 19, 2005 2:29 PM To: '[EMAIL PROTECTED]' Subject: [R] CPU Usage with R 2.1.0 in Windows Hi, I'm using a fairly simple HP Compaq desktop PC running Windows 2K. When running a large process in R, the process RGUI.exe will never exceed 50% of the CPU usage. The program used to be able to use more of the computer, but does not now. I don't believe this is a multiple processor machine. Can anyone give any advice on how to solve the problem? Thanks, Michael Greene Product Management Plymouth Rock Assurance Corp 617-951-1682 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Taking the derivative of a quadratic B-spline
On 7/19/2005 2:53 PM, James McDermott wrote: Hello, I have been trying to take the derivative of a quadratic B-spline obtained by using the COBS library. What I would like to do is similar to what one can do by using fit-smooth.spline(cdf) xx-seq(-10,10,.1) predict(fit, xx, deriv = 1) The goal is to fit the spline to data that is approximating a cumulative distribution function (e.g. in my example, cdf is a 2-column matrix with x values in column 1 and the estimate of the cdf evaluated at x in column 2) and then take the first derivative over a range of values to get density estimates. The reason I don't want to use smooth.spline is that there is no way to impose constraints (e.g. =0, =1, and monotonicity) as there is with COBS. However, since COBS doesn't have the 'deriv =' option, the only way I can think of doing it with COBS is to evaluate the derivatives numerically. Numerical estimates of the derivatives of a quadratic should be easy to obtain accurately. For example, if the quadratic ax^2 + bx + c is defined on [-1, 1], then the derivative 2ax + b, has 2a = f(1) - f(0) + f(-1), and b = (f(1) - f(-1))/2. You should be able to generalize this to the case where the spline is quadratic between knots k1 and k2 pretty easily. Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Question about creating unique factor labels with the factor function
Hi guys, I ran into a problem of not being able to create unique labels when creating a factor. Consider an example below: hb - factor(c(1,1,1,2,2,2,3,3,3), levels=c(1,2,3),labels=c(1,1,2)) hb [1] 1 1 1 1 1 1 2 2 2 Levels: 1 1 2 unique(hb) [1] 1 1 2 Levels: 1 1 2 How come there are three unique levels, I thought this would only create one unique level? unique(as.ordered(hb)) [1] 1 2 Levels: 1 1 2 Is as.ordered the only solution? Thanks in advance, Greg __ [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] CPU Usage with R 2.1.0 in Windows
Dnia 2005-07-19 20:28, Użytkownik Greene, Michael napisał: Hi, I'm using a fairly simple HP Compaq desktop PC running Windows 2K. When running a large process in R, the process RGUI.exe will never exceed 50% of the CPU usage. If you have hyperthreading, R catches only one virtual processor (from two available), being not able to exceed half of total power (100% of one only). If you want to use full power, you should turn hyperthreading off, if your BIOS supports such option. Regards, -- Lukasz Komsta Department of Medicinal Chemistry Medical University of Lublin 6 Chodzki, 20-093 Lublin, Poland Fax +48 81 7425165 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Taking the derivative of a quadratic B-spline
I wish it were that simple (perhaps it is and I am just not seeing it). The output from cobs( ) includes the B-spline coefficients and the knots. These coefficients are not the same as the a, b, and c coefficients in a quadratic polynomial. Rather, they are the coefficients of the quadratic B-spline representation of the fitted curve. I need to evaluate a linear combination of basis functions and it is not clear to me how to accomplish this easily. I was hoping to find an alternative way of getting the derivatives. Jim McDermott On 7/19/05, Duncan Murdoch [EMAIL PROTECTED] wrote: On 7/19/2005 2:53 PM, James McDermott wrote: Hello, I have been trying to take the derivative of a quadratic B-spline obtained by using the COBS library. What I would like to do is similar to what one can do by using fit-smooth.spline(cdf) xx-seq(-10,10,.1) predict(fit, xx, deriv = 1) The goal is to fit the spline to data that is approximating a cumulative distribution function (e.g. in my example, cdf is a 2-column matrix with x values in column 1 and the estimate of the cdf evaluated at x in column 2) and then take the first derivative over a range of values to get density estimates. The reason I don't want to use smooth.spline is that there is no way to impose constraints (e.g. =0, =1, and monotonicity) as there is with COBS. However, since COBS doesn't have the 'deriv =' option, the only way I can think of doing it with COBS is to evaluate the derivatives numerically. Numerical estimates of the derivatives of a quadratic should be easy to obtain accurately. For example, if the quadratic ax^2 + bx + c is defined on [-1, 1], then the derivative 2ax + b, has 2a = f(1) - f(0) + f(-1), and b = (f(1) - f(-1))/2. You should be able to generalize this to the case where the spline is quadratic between knots k1 and k2 pretty easily. Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Taking the derivative of a quadratic B-spline
The derivative of a quadratic B-spline is the centered finite difference of a linear B-spline, so if you have access to the underlying coefficients of the B-spline expansion you can do this easily. I believe the coefficients are passed as the $coef component of the return value. Reid Huntsinger -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of James McDermott Sent: Tuesday, July 19, 2005 2:54 PM To: r-help@stat.math.ethz.ch Subject: [R] Taking the derivative of a quadratic B-spline Hello, I have been trying to take the derivative of a quadratic B-spline obtained by using the COBS library. What I would like to do is similar to what one can do by using fit-smooth.spline(cdf) xx-seq(-10,10,.1) predict(fit, xx, deriv = 1) The goal is to fit the spline to data that is approximating a cumulative distribution function (e.g. in my example, cdf is a 2-column matrix with x values in column 1 and the estimate of the cdf evaluated at x in column 2) and then take the first derivative over a range of values to get density estimates. The reason I don't want to use smooth.spline is that there is no way to impose constraints (e.g. =0, =1, and monotonicity) as there is with COBS. However, since COBS doesn't have the 'deriv =' option, the only way I can think of doing it with COBS is to evaluate the derivatives numerically. Regards, Jim McDermott __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Taking the derivative of a quadratic B-spline
On 7/19/2005 3:34 PM, James McDermott wrote: I wish it were that simple (perhaps it is and I am just not seeing it). The output from cobs( ) includes the B-spline coefficients and the knots. These coefficients are not the same as the a, b, and c coefficients in a quadratic polynomial. Rather, they are the coefficients of the quadratic B-spline representation of the fitted curve. I need to evaluate a linear combination of basis functions and it is not clear to me how to accomplish this easily. I was hoping to find an alternative way of getting the derivatives. I don't know COBS, but doesn't predict just evaluate the B-spline? The point of what I posted is that the particular basis doesn't matter if you can evaluate the quadratic at 3 points. Duncan Murdoch Jim McDermott On 7/19/05, Duncan Murdoch [EMAIL PROTECTED] wrote: On 7/19/2005 2:53 PM, James McDermott wrote: Hello, I have been trying to take the derivative of a quadratic B-spline obtained by using the COBS library. What I would like to do is similar to what one can do by using fit-smooth.spline(cdf) xx-seq(-10,10,.1) predict(fit, xx, deriv = 1) The goal is to fit the spline to data that is approximating a cumulative distribution function (e.g. in my example, cdf is a 2-column matrix with x values in column 1 and the estimate of the cdf evaluated at x in column 2) and then take the first derivative over a range of values to get density estimates. The reason I don't want to use smooth.spline is that there is no way to impose constraints (e.g. =0, =1, and monotonicity) as there is with COBS. However, since COBS doesn't have the 'deriv =' option, the only way I can think of doing it with COBS is to evaluate the derivatives numerically. Numerical estimates of the derivatives of a quadratic should be easy to obtain accurately. For example, if the quadratic ax^2 + bx + c is defined on [-1, 1], then the derivative 2ax + b, has 2a = f(1) - f(0) + f(-1), and b = (f(1) - f(-1))/2. You should be able to generalize this to the case where the spline is quadratic between knots k1 and k2 pretty easily. Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] extracting row means from a list
I think about half of my question in R can be solved with a judicious do.call. Thanks, Andy __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Using BRugs, in FUN: .C(..) 'type' not real
To All, I am using the BRugs package. In running the meta file BRugsFit() with a syntactically correct model .txt file, I see the message: Error in FUN(X[[1]], ...) : .C(..): 'type' must be real for this format I haven't found information on this kind of error either here or on the OpenBUGS/WinBUGS/BRugs bulletin boards. If there is a better place to post this message, please let me know. I am running Windows XP Professional (build 2600) Service Pack 2.0 on Intel Centrino, and using R version 2.1.1. Thank you, -- Seth Pruitt Department of Economics University of California, San Diego [EMAIL PROTECTED] http://dss.ucsd.edu/~sjpruitt [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] A warning message when using mix package.
Dear all,
I just start using package mix in R and am inexperienced with it. When I
ran my program several times, I sometimes got the warning message, and
sometimes not :
Warning message:
Loglik converged before variable 2 ; beta may be infinite. in: fitter(X,
Y, strats, offset, init, control, weights = weights,
I cannot find what is wrong with my program. Does anyone have an idea
with it and whether this would affect my results? Any help would be
appreciated.
Thanks!
By the way, I copy a part of my program and hope it would help,
beta - c(0, 0,0,0,0) # 'True' coefficients
#make a matrix of covariates---#
Y=cbind(z4,z5,z1,z2,z3)
s-prelim.mix(Y,2)
#run MI--#
MI-vector(list,10)
fit.model.mi-vector(list,10)
rngseed(1234567) # set random number generator seed
for (i in 1:10){
thetahat - em.mix(s)
newtheta - da.mix(s, thetahat, steps=2000, showits=TRUE)
MI[[i]] - imp.mix(s, newtheta)
}
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[R] sciviews installation
Hello 1. a few months ago, I had sciviews working fine with R (rw2001) under windows XP 2. now, upgrading to rw2011, the stuff seems fine (every package installed),but I find a conflict when launching sciviews: - it runs, apparently - but when I try to work (import/export In: text for instance), it asks for Rcmdr (Would you like to install it now?) 3. Rcmdr is already installed (with all dependencies) and works well when called directly in R gui 4. and it's impossible to make it reconized or to install it under sciviews I have all the latest packages, and I am going to get mad. what do you suggest to solve my problem ? Thanks Georges Moracchini __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] initial points for arms in package HI
Quoting Christoph Buser [EMAIL PROTECTED]:
Dear R-users
I have a problem choosing initial points for the function arms()
in the package HI
I intend to implement a Gibbs sampler and one of my conditional
distributions is nonstandard and not logconcave.
Therefore I'd like to use arms.
But there seem to be a strong influence of the initial point
y.start. To show the effect I constructed a demonstration
example. It is reproducible without further information.
Please note that my target density is not logconcave.
Thanks for all comments or ideas.
Christoph Buser
Dear Christoph,
There is a Metropolis step at each iteration of the ARMS sampler, in
which it may choose to reject the proposed move to a new point and stick
at the current point (This is what the M in ARMS stands for) If you
do repeated calls to arms with the same starting point, then the
iterations where the Metropolis step rejects a move will create a spike
in the sample density at your initial value. If you use a uniform random
starting point, then your sample density will be a mixture of the
target distribution (Metropolis accepts move) and a uniform distribution
(Metropolis rejects move).
You should be doing something like this:
res1 - arms(runif(1,0,100), logDichteGam, function(x) (x0)(x100), 1)
for(i in 2:1000)
res1[i] - arms(res1[i-1], logDichteGam, function(x) (x0)(x100), 1)
i.e., using each sampled point as the starting value for the next
iteration. The sequence of values in res1 will then be a correlated
sample from the given distribution:
acf(res1)
The bottom line is that you can't use ARMS to draw a single sample
from a non-log-concave density.
If you are still worried about using ARMS, you can verify your results
using the random walk Metropolis sampler (MCMCmetrop1R) in the package
MCMCpack.
Martyn
## R Code:
library(HI)
## parameter for the distribution
para - 0.1
## logdensity
logDichteGam - function(x, u = para, v = para) {
-(u*x + v*1/x) - log(x)
}
## density except for the constant
propDichteGam - function(x, u = para, v = para) {
exp(-(u*x + v*1/x) - log(x))
}
## calculating the constant
(c - integrate(propDichteGam, 0, 1000, rel.tol = 10^(-12))$value)
## density
DichteGam - function(x, u = para, v = para) {
exp(-(u*x + v*1/x) - log(x))/c
}
## calculating 1000 values by repeating a call of arms (this would
## be the situation in an Gibbs Sample. Of course in a Gibbs sampler
## the distribution would change. This is only for demonstration
res1 - NULL
for(i in 1:1000)
res1[i] - arms(runif(1,0,100), logDichteGam, function(x) (x0)(x100), 1)
## Generating a sample of thousand observations with 1 call of arms
res2 - arms(runif(1,0,100), logDichteGam, function(x) (x0)(x100), 1000)
## Plot of the samples
mult.fig(4)
plot(res1, log = y)
plot(res2, log = y)
hist(res1, freq = FALSE, xlim = c(0,4), breaks = seq(0,100,by = 0.1),
ylim = c(0,1))
curve(DichteGam, 0,4, add = TRUE, col = 2)
hist(res2, freq = FALSE, xlim = c(0,4), breaks = seq(0,100,by = 0.1),
ylim = c(0,1))
curve(DichteGam, 0,4, add = TRUE, col = 2)
## If we repeat the procedure, using the fix intial value 1,
## the situation is even worse
res3 - NULL
for(i in 1:1000)
res3[i] - arms(1, logDichteGam, function(x) (x0)(x100), 1)
## Generating a sample of thousand observations with 1 call of arms
res4 - arms(1, logDichteGam, function(x) (x0)(x100), 1000)
## Plot of the samples
par(mfrow = c(2,2))
plot(res3, log = y)
plot(res4, log = y)
hist(res3, freq = FALSE, xlim = c(0,4), breaks = seq(0,100,by = 0.1),
ylim = c(0,1))
curve(DichteGam, 0,4, add = TRUE, col = 2)
hist(res4, freq = FALSE, xlim = c(0,4), breaks = seq(0,100,by = 0.1),
ylim = c(0,1))
curve(DichteGam, 0,4, add = TRUE, col = 2)
## If I generate the sample in a for-loop (one by one) I do not
## get the correct density. But this is exactly the situation in
## my Gibbs Sampler. Therfore I am concerned about the correct
## application of arms
---
This message and its attachments are strictly confidential. ...{{dropped}}
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[R] Is it possible to create highly customized report in *.xls format by using R/S+?
I remember in one slide of Prof. Ripley's presentation overhead, he said the most popular data analysis software is excel. So is there any resource or tutorial on this topic? Thank you so much! __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Problems with date-format (R 2.1.1 + chron)
Hello, today I've updated on the newest R-Version. But sadly a function I needed didnt want to work: The input is e.g. days(as.Date(21-07-2005,%d-%m-%y)) the error is: Fehler in Math.Date(dts): floor nicht definiert für Date Objekte (Error in Math.Date(dts): floor not defined for date objects) Same for year. Only months gives me the correct output. In Version 2.01 it worked very well, with the same chron library. Whats wrong ? Carsten __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Indexing within scan
Dear R-help subscribers,
Can some one please help me figure out how to write code that will allow me to
use a for loop to scan a number of files one by one, and then save a summary of
each file as the for loop progresses. For example I have 24 files named a1
through a24, and I want to do something like:
results-numeric(24)
for (i in 1:24)
{
p-scan(ai.txt) # where the i is an index for each of the 24 files
results[i]-mean(p)
}
Thanks for any help,
Justin Rhodes
Behavioral Neuroscience
Oregon Health Science University
VA Medical Center (R D 12)
3710 SW US Veterans Hospital Rd
Portland, OR 97239
Phone: (503) 220-8262 extn 54392
Fax: (503) 721-1029
E-mail: [EMAIL PROTECTED]
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Re: [R] Indexing within scan
Justin Rhodes [EMAIL PROTECTED] writes:
Dear R-help subscribers,
Can some one please help me figure out how to write code that will allow me
to use a for loop to scan a number of files one by one, and then save a
summary of each file as the for loop progresses. For example I have 24 files
named a1 through a24, and I want to do something like:
results-numeric(24)
for (i in 1:24)
{
p-scan(ai.txt) # where the i is an index for each of the 24 files
results[i]-mean(p)
}
Thanks for any help,
paste(a, i, .txt, sep=)
should get you in the right direction. However, I don't think you
*really* want a for loop. Use vectorization and sapply instead:
names - paste(a, 1:24, .txt, sep=)
results - sapply(names, function(n) mean(scan(n)) )
--
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c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918
~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907
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Re: [R] extracting row means from a list
Justone more comment in addition to Sundar's solution: If these are all numeric matrices, I would read them into R as such, instead of data frames. Actually, I would read them all into a 3-dimensional array (2000 x 6 x # of files). Assuming you have such an array, then you can do something like: ## get your example into an array: need the abind package. bar - do.call(abind, c(lapply(foo, as.matrix), along=3)) str(bar) num [1:100, 1:6, 1:10] -0.78981 0.31939 -0.00819 1.59346 1.20498 ... - attr(*, dimnames)=List of 3 ..$ : chr [1:100] 1 2 3 4 ... ..$ : chr [1:6] x1 x2 x3 x4 ... ..$ : NULL str(m - rowMeans(bar, dims=2)) num [1:100, 1:6] -0.4401 0.5463 -0.0572 -0.1314 0.5177 ... - attr(*, dimnames)=List of 2 ..$ : chr [1:100] 1 2 3 4 ... ..$ : chr [1:6] x1 x2 x3 x4 ... Andy -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Andy Bunn Sent: Tuesday, July 19, 2005 3:59 PM To: Sundar Dorai-Raj Cc: R-Help Subject: Re: [R] extracting row means from a list I think about half of my question in R can be solved with a judicious do.call. Thanks, Andy __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] deriv - accessing numeric output for gradient
Hi,
I am interested in using the numeric output from the gradient attribute of
deriv's output in subsequent analyses.
But, I have so far been unable to determine how to do so.
I will use the example from the deriv help to illustrate.
## function with defaulted arguments:
(fx - deriv(y ~ b0 + b1 * 2^(-x/th), c(b0, b1, th),
function(b0, b1, th, x = 1:7){} ) )
fx(2,3,4)
This yields
[1] 4.522689 4.121320 3.783811 3.50 3.261345 3.060660 2.891905
attr(,gradient)
b0b1th
[1,] 1 0.8408964 0.1092872
[2,] 1 0.7071068 0.1837984
[3,] 1 0.5946036 0.2318331
[4,] 1 0.500 0.2599302
[5,] 1 0.4204482 0.2732180
[6,] 1 0.3535534 0.2756976
[7,] 1 0.2973018 0.2704720
I would greatly appreciate it if anyone could tell me how to convert the
numbers listed under b0, b1, and th into a matrix.
Thanks!
Jay Rotella
Ecology Department
Montana State University
Bozeman, MT 59717
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[R] deriv - accessing numeric output listed under gradient attribute
Hi,
I am interested in using the numeric output from the gradient attribute of
deriv's output in subsequent analyses.
But, I have so far been unable to determine how to do so.
I will use the example from the deriv help to illustrate.
## function with defaulted arguments:
(fx - deriv(y ~ b0 + b1 * 2^(-x/th), c(b0, b1, th),
function(b0, b1, th, x = 1:7){} ) )
fx(2,3,4)
This yields
[1] 4.522689 4.121320 3.783811 3.50 3.261345 3.060660 2.891905
attr(,gradient)
b0b1th
[1,] 1 0.8408964 0.1092872
[2,] 1 0.7071068 0.1837984
[3,] 1 0.5946036 0.2318331
[4,] 1 0.500 0.2599302
[5,] 1 0.4204482 0.2732180
[6,] 1 0.3535534 0.2756976
[7,] 1 0.2973018 0.2704720
I would greatly appreciate it if anyone could tell me how to convert the
numbers listed under b0, b1, and th into a matrix.
Thanks!
Jay Rotella
Ecology Department
Montana State University
Bozeman, MT 59717
__
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[R] Regression lines for differently-sized groups on the same plot
Hi there, I've looked through the very helpful advice about adding fitted lines to plots in the r-help archive, and can't find a post where someone has offered a solution for my specific problem. I need to plot logistic regression fits from three differently-sized data subsets on a plot of the entire dataset. A description and code are below: I have an unbalanced dataset consisting of three different species (hem, yb, and sm), with unequal numbers of wood pieces in each species group. I am trying to generate a plot that will show the size of the wood piece on the X axis, the probability of it having tree seedlings growing on it on the Y (a binomial yes or no variable), and three fitted curves showing how the probability of having tree seedlings changes with increasing wood piece size for each species. I have no problem generating fits using GLM, and no problem creating the plot. However, if I try to add a fitted curve based only on the hem data subset to a plot that shows the entire dataset, I get an error message that the lengths of those data sets differ. Error in xy.coords(x,y) : x and y lengths differ. I could see R's point -- you can't plot a regression line of babies born as a function of stork abundance on a graph of cherries produced (Y) versus rainfall (X), which for all the program knows, I'm trying to do. As a temporary fix, I added NAs to the end of the hem, yb, and sm subsets to make them the same length as the entire dataset. I can now add my fitted curves to the plot, but the lines are not connected. That is, if the hem group only contains wood pieces that are 1, 4, and 10 meters long, the plot has an X axis that ranges from 1 to 10, but line segments for the hem group regression line only appear above 1, 4, and 10. How can I fix this? An ideal solution would not require me to make the hem subset of my data the same length as the full dataset, either (although the summaries of regressions with the NAs (or zeroes) added and taken away are identical). I'd also settle for a work-around that would have R connect the pieces of the curve so that I get a solid line rather than small dots and dashes where actual data exist. Thanks so much for your help! Laura Marx Michigan State University, Dept. of Forestry #Note: hemdata has all the rows that are not hemlock species replaced with #NAs. hemhem=glm(hempresence~logarea, family=binomial(logit), data=hemdata) hemyb=glm(hempresence~logarea, family=binomial(logit), data=birchdata) hemsm=glm(hempresence~logarea, family=binomial(logit), data=mapledata) attach(logreg) #logreg is the full dataset plot(logarea, hempresence, xlab = Surface area of log (m2), ylab=Probability of hemlock seedling presence, type=n, font.lab=2, cex.lab=1.5, axes=TRUE) lines(logarea,fitted(hemhem), lty=1, lwd=2) lines(logarea,fitted(hemyb), lty=dashed, lwd=2) lines(logarea,fitted(hemsm), lty=dotted, lwd=2) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] deriv - accessing numeric output listed under gradient attribute
Jay Rotella wrote:
Hi,
I am interested in using the numeric output from the gradient attribute of
deriv's output in subsequent analyses.
But, I have so far been unable to determine how to do so.
I will use the example from the deriv help to illustrate.
## function with defaulted arguments:
(fx - deriv(y ~ b0 + b1 * 2^(-x/th), c(b0, b1, th),
function(b0, b1, th, x = 1:7){} ) )
fx(2,3,4)
This yields
[1] 4.522689 4.121320 3.783811 3.50 3.261345 3.060660 2.891905
attr(,gradient)
b0b1th
[1,] 1 0.8408964 0.1092872
[2,] 1 0.7071068 0.1837984
[3,] 1 0.5946036 0.2318331
[4,] 1 0.500 0.2599302
[5,] 1 0.4204482 0.2732180
[6,] 1 0.3535534 0.2756976
[7,] 1 0.2973018 0.2704720
I would greatly appreciate it if anyone could tell me how to convert the
numbers listed under b0, b1, and th into a matrix.
Thanks!
Jay Rotella
Ecology Department
Montana State University
Bozeman, MT 59717
Try:
attr(fx(2, 3, 4), gradient)
--sundar
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Re: [R] Memory limits using read.table on Windows XP Pro
Hello everyone, Would you please somebody explain me what my sin is (please see the code and timing bellow)? And how to improve myself and the following piece of R code? BTW, the code works. This is R 64-bit built by myself on Sun SPARC Solaris 9 with gcc-4.0.1 (64-bit) also built by yours truly. The machine is Sun Fire-V880 with 32GB memory and 8 cpu's. Nobody else was using it. Thank you very much Latchezar Dimitrov PS. Prof. Brian D. Ripley, thank you very much for not wasting your invaluable time responding to the above. I know your answer anyway. Sorry for making you read it though ... system.time( + haplo[i,2*j-1]-substr(as.character(geno[i,j]),1,1) + ,TRUE) [1] 66.27 13.67 80.02 0.00 0.00 ls() [1] P geno haplo i j lrmodel pheno i [1] 1 j [1] 1 dim(P) [1] 1 125000 str(P) num [1, 1:125000] 0.6188 0.0533 0.0893 0.8994 0.0316 ... str(pheno) `data.frame': 2500 obs. of 11 variables: $ pca: int 1 1 1 1 1 1 1 1 1 1 ... $ her: int 0 1 0 0 0 0 0 0 0 0 ... $ age: num 67.1 70.4 64.9 60.8 64.3 ... $ t : int 1 3 1 2 1 2 1 3 9 1 ... $ n : int 9 9 9 9 9 9 9 0 9 9 ... $ m : int 0 1 0 0 0 1 9 9 1 9 ... $ diffgrd: int 9 9 9 2 9 9 9 9 3 9 ... $ gs : int 6 7 6 7 7 7 5 6 NA 6 ... $ psa: num 13 75 11.3 13 9.1 51 4.3 10.7 NA 93 ... $ geo: int 2 2 1 1 1 1 1 1 1 1 ... $ ageg : int 6 7 5 5 5 5 3 5 4 5 ... str(geno) `data.frame': 2500 obs. of 125000 variables: ^C dim(geno) [1] 2500 125000 dim(haplo) [1] 2500 25 version() Error: attempt to apply non-function version _ platform sparc-sun-solaris2.9 arch sparc os solaris2.9 system sparc, solaris2.9 status Patched major2 minor1.1 year 2005 month07 day 09 language R __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Memory limits using read.table on Windows XP Pro
Really sorry for the wrong addressing. It was intended to the list only. I apologize. Latchezar -Original Message- From: Latchezar Dimitrov Sent: Tuesday, July 19, 2005 9:47 PM To: Latchezar Dimitrov; 'Prof Brian Ripley' Cc: 'r-help@stat.math.ethz.ch' Subject: RE: [R] Memory limits using read.table on Windows XP Pro Hello everyone, Would you please somebody explain me what my sin is (please see the code and timing bellow)? And how to improve myself and the following piece of R code? BTW, the code works. This is R 64-bit built by myself on Sun SPARC Solaris 9 with gcc-4.0.1 (64-bit) also built by yours truly. The machine is Sun Fire-V880 with 32GB memory and 8 cpu's. Nobody else was using it. Thank you very much Latchezar Dimitrov PS. Prof. Brian D. Ripley, thank you very much for not wasting your invaluable time responding to the above. I know your answer anyway. Sorry for making you read it though ... system.time( + haplo[i,2*j-1]-substr(as.character(geno[i,j]),1,1) + ,TRUE) [1] 66.27 13.67 80.02 0.00 0.00 ls() [1] P geno haplo i j lrmodel pheno i [1] 1 j [1] 1 dim(P) [1] 1 125000 str(P) num [1, 1:125000] 0.6188 0.0533 0.0893 0.8994 0.0316 ... str(pheno) `data.frame': 2500 obs. of 11 variables: $ pca: int 1 1 1 1 1 1 1 1 1 1 ... $ her: int 0 1 0 0 0 0 0 0 0 0 ... $ age: num 67.1 70.4 64.9 60.8 64.3 ... $ t : int 1 3 1 2 1 2 1 3 9 1 ... $ n : int 9 9 9 9 9 9 9 0 9 9 ... $ m : int 0 1 0 0 0 1 9 9 1 9 ... $ diffgrd: int 9 9 9 2 9 9 9 9 3 9 ... $ gs : int 6 7 6 7 7 7 5 6 NA 6 ... $ psa: num 13 75 11.3 13 9.1 51 4.3 10.7 NA 93 ... $ geo: int 2 2 1 1 1 1 1 1 1 1 ... $ ageg : int 6 7 5 5 5 5 3 5 4 5 ... str(geno) `data.frame': 2500 obs. of 125000 variables: ^C dim(geno) [1] 2500 125000 dim(haplo) [1] 2500 25 version() Error: attempt to apply non-function version _ platform sparc-sun-solaris2.9 arch sparc os solaris2.9 system sparc, solaris2.9 status Patched major2 minor1.1 year 2005 month07 day 09 language R __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Problems with date-format (R 2.1.1 + chron)
I think its likely that you are using different versions of chron. I noticed that version 2.2-33 of chron had the statement: tms - dts - trunc(dts) but version 2.2-35 seems to have replaced it with: tms - dts - floor(dts) and that seems to be causing the problem. As a workaround: floor.Date - floor.trunc days(as.Date(21-07-2005, %d-%m-%y)) On 7/19/05, Carsten Steinhoff [EMAIL PROTECTED] wrote: Hello, today I've updated on the newest R-Version. But sadly a function I needed didnt want to work: The input is e.g. days(as.Date(21-07-2005,%d-%m-%y)) the error is: Fehler in Math.Date(dts): floor nicht definiert für Date Objekte (Error in Math.Date(dts): floor not defined for date objects) Same for year. Only months gives me the correct output. In Version 2.01 it worked very well, with the same chron library. Whats wrong ? Carsten __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Taking the derivative of a quadratic B-spline
Would the unique quadratic defined by the three points be the same curve as the curve predicted by a quadratic B-spline (fit to all of the data) through those same three points? Jim On 7/19/05, Duncan Murdoch [EMAIL PROTECTED] wrote: On 7/19/2005 3:34 PM, James McDermott wrote: I wish it were that simple (perhaps it is and I am just not seeing it). The output from cobs( ) includes the B-spline coefficients and the knots. These coefficients are not the same as the a, b, and c coefficients in a quadratic polynomial. Rather, they are the coefficients of the quadratic B-spline representation of the fitted curve. I need to evaluate a linear combination of basis functions and it is not clear to me how to accomplish this easily. I was hoping to find an alternative way of getting the derivatives. I don't know COBS, but doesn't predict just evaluate the B-spline? The point of what I posted is that the particular basis doesn't matter if you can evaluate the quadratic at 3 points. Duncan Murdoch Jim McDermott On 7/19/05, Duncan Murdoch [EMAIL PROTECTED] wrote: On 7/19/2005 2:53 PM, James McDermott wrote: Hello, I have been trying to take the derivative of a quadratic B-spline obtained by using the COBS library. What I would like to do is similar to what one can do by using fit-smooth.spline(cdf) xx-seq(-10,10,.1) predict(fit, xx, deriv = 1) The goal is to fit the spline to data that is approximating a cumulative distribution function (e.g. in my example, cdf is a 2-column matrix with x values in column 1 and the estimate of the cdf evaluated at x in column 2) and then take the first derivative over a range of values to get density estimates. The reason I don't want to use smooth.spline is that there is no way to impose constraints (e.g. =0, =1, and monotonicity) as there is with COBS. However, since COBS doesn't have the 'deriv =' option, the only way I can think of doing it with COBS is to evaluate the derivatives numerically. Numerical estimates of the derivatives of a quadratic should be easy to obtain accurately. For example, if the quadratic ax^2 + bx + c is defined on [-1, 1], then the derivative 2ax + b, has 2a = f(1) - f(0) + f(-1), and b = (f(1) - f(-1))/2. You should be able to generalize this to the case where the spline is quadratic between knots k1 and k2 pretty easily. Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] nls
Dear R-helpers, I am trying to estimate a model that I am proposing, which consists of putting an extra hidden layer in the Markov switching models. In the simplest case the S(t) - Markov states - and w(t) - the extra hidden variables - are independent, and w(t) is constant. Formally the model looks like this: y(t)=c(1,y[t-1])%*%beta0*w+c(1,y[t-1])%*%beta1*(1-w). So I ran some simulations to obtain the y's, and I am putting it into the nls: res-nls(y~(a+b*x)*w+(c+d*x)*(1-w),start=list(a=1,b=0.3,c=-1,d=-0.2,w=0.5)) and the starting parameter values are similar to the ones I used for simulations, however I am getting Error in nlsModel(formula, mf, start) : singular gradient matrix at initial parameter estimates What am I doing wrong? I tried many different parameter values to no avail. Thank you so much in advance, Sincerely, Eugene McGill University __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] nls
On 7/19/05, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote: Dear R-helpers, I am trying to estimate a model that I am proposing, which consists of putting an extra hidden layer in the Markov switching models. In the simplest case the S(t) - Markov states - and w(t) - the extra hidden variables - are independent, and w(t) is constant. Formally the model looks like this: y(t)=c(1,y[t-1])%*%beta0*w+c(1,y[t-1])%*%beta1*(1-w). So I ran some simulations to obtain the y's, and I am putting it into the nls: res-nls(y~(a+b*x)*w+(c+d*x)*(1-w),start=list(a=1,b=0.3,c=-1,d=-0.2,w=0.5)) and the starting parameter values are similar to the ones I used for simulations, however I am getting Error in nlsModel(formula, mf, start) : singular gradient matrix at initial parameter estimates Your model is not identifiable. You are using 5 parameters to describe a two dimensional model -- in fact, y is linear in x so anything beyond the intercept and slope are redundant, viz. a singular gradient. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Michaelis-menten equation
Hi, We used known Vm and Km to simulate the data set (time, Cp) without adding random error in there. Yes, the line looks like very close to a straight line. But why can't we obtain the correct values with fitting process? We used optim first and then followed by using nls to fit the model. Thanks. regards, ---Chun-ying Lee Hmm, sorry, no. I'm talking through a hole in my head there. Vm*y/(Km+y) makes OK sense. Vm is what you get for large y - passing from 1st order to 0th order kinetics. However, looking at the data plot(PKindex) abline(lm(conc~time,data=PKindex)) shows that they are pretty much on a straight line, i.e. you are in the domain of 0-order kinetics. So why are you expecting the rate of decrease to have changed by roughly 3/4 (from 2/3*Vm/Vd at y=2*Km to 1/2*Vm/Vd at y=Km when you reach 4.67)?? __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Problems with date-format (R 2.1.1 + chron)
On 7/19/05, Gabor Grothendieck [EMAIL PROTECTED] wrote: I think its likely that you are using different versions of chron. I noticed that version 2.2-33 of chron had the statement: tms - dts - trunc(dts) but version 2.2-35 seems to have replaced it with: tms - dts - floor(dts) and that seems to be causing the problem. As a workaround: floor.Date - floor.trunc That should have been: floor.Date - trunc.Date days(as.Date(21-07-2005, %d-%m-%y)) On 7/19/05, Carsten Steinhoff [EMAIL PROTECTED] wrote: Hello, today I've updated on the newest R-Version. But sadly a function I needed didnt want to work: The input is e.g. days(as.Date(21-07-2005,%d-%m-%y)) the error is: Fehler in Math.Date(dts): floor nicht definiert für Date Objekte (Error in Math.Date(dts): floor not defined for date objects) Same for year. Only months gives me the correct output. In Version 2.01 it worked very well, with the same chron library. Whats wrong ? Carsten __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Regression lines for differently-sized groups on the same plot
Laura M Marx wrote: Hi there, I've looked through the very helpful advice about adding fitted lines to plots in the r-help archive, and can't find a post where someone has offered a solution for my specific problem. I need to plot logistic regression fits from three differently-sized data subsets on a plot of the entire dataset. A description and code are below: I have an unbalanced dataset consisting of three different species (hem, yb, and sm), with unequal numbers of wood pieces in each species group. I am trying to generate a plot that will show the size of the wood piece on the X axis, the probability of it having tree seedlings growing on it on the Y (a binomial yes or no variable), and three fitted curves showing how the probability of having tree seedlings changes with increasing wood piece size for each species. I have no problem generating fits using GLM, and no problem creating the plot. However, if I try to add a fitted curve based only on the hem data subset to a plot that shows the entire dataset, I get an error message that the lengths of those data sets differ. Error in xy.coords(x,y) : x and y lengths differ. I could see R's point -- you can't plot a regression line of babies born as a function of stork abundance on a graph of cherries produced (Y) versus rainfall (X), which for all the program knows, I'm trying to do. As a temporary fix, I added NAs to the end of the hem, yb, and sm subsets to make them the same length as the entire dataset. I can now add my fitted curves to the plot, but the lines are not connected. That is, if the hem group only contains wood pieces that are 1, 4, and 10 meters long, the plot has an X axis that ranges from 1 to 10, but line segments for the hem group regression line only appear above 1, 4, and 10. How can I fix this? An ideal solution would not require me to make the hem subset of my data the same length as the full dataset, either (although the summaries of regressions with the NAs (or zeroes) added and taken away are identical). I'd also settle for a work-around that would have R connect the pieces of the curve so that I get a solid line rather than small dots and dashes where actual data exist. Thanks so much for your help! Laura Marx Michigan State University, Dept. of Forestry #Note: hemdata has all the rows that are not hemlock species replaced with #NAs. hemhem=glm(hempresence~logarea, family=binomial(logit), data=hemdata) hemyb=glm(hempresence~logarea, family=binomial(logit), data=birchdata) hemsm=glm(hempresence~logarea, family=binomial(logit), data=mapledata) attach(logreg) #logreg is the full dataset plot(logarea, hempresence, xlab = Surface area of log (m2), ylab=Probability of hemlock seedling presence, type=n, font.lab=2, cex.lab=1.5, axes=TRUE) lines(logarea,fitted(hemhem), lty=1, lwd=2) lines(logarea,fitted(hemyb), lty=dashed, lwd=2) lines(logarea,fitted(hemsm), lty=dotted, lwd=2) Hi, Laura, Would ?predict.glm be better? plot(logarea, hempresence, xlab = Surface area of log (m2), ylab=Probability of hemlock seedling presence, type=n, font.lab=2, cex.lab=1.5, axes=TRUE) lines(logarea, predict(hemhem, logreg, response), lty=1, lwd=2) lines(logarea, predict(hemyb, logreg, response), lty=dashed, lwd=2) lines(logarea, predict(hemsm, logreg, response), lty=dotted, lwd=2) Without seeing more description of your data, this is still a guess. --sundar __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Turning off return warning messages.
Dear All, Is there a way I can turn off the following warning message for using multi-argument returns? multi-argument returns are deprecated in: return(p1, p2, p3, p4) Steve. ** Steve Su ([EMAIL PROTECTED]) Postdoctoral fellow Faculty of Business Queensland University of Technology Postal Address: Steve Su, School of Accountancy, QUT, PO Box 2434, Brisbane, Queensland, Australia, 4001. CRICOS No. 00213J Phone: +61 7 3864 4357 Fax:+61 7 3864 1812 Mobile: 0421 840 586 . _--_|\ / QUT \_.--._/ v ** [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Question about Installing SJava package
Dear all, I have an error message installing SJava package. So I searched web site(google) and R-mailing list to find a similar error message. But I couldn't find it. I installed R-2.1.1 like this on Fedora Core4 1) /configure --enable-R-shlib --with-libpng --with-jpeglib 2) make - make check - make install and then issuing on shell prompt (red lines are error messages) R CMD INSTALL -c /usr/local/src/R/SJava_0.68-0.tar.gz * Installing *source* package 'SJava' ... checking for java... /usr/java/jdk1.5.0_04//bin/java Java VM /usr/java/jdk1.5.0_04//bin/java checking for javah... /usr/java/jdk1.5.0_04//bin/javah Looking in /usr/java/jdk1.5.0_04/include Looking in /usr/java/jdk1.5.0_04/include/linux checking for g++... g++ checking for C++ compiler default output... a.out checking whether the C++ compiler works... yes checking whether we are cross compiling... no checking for suffix of executables... checking for suffix of object files... o checking whether we are using the GNU C++ compiler... yes checking whether g++ accepts -g... yes checking for gcc... gcc checking whether we are using the GNU C compiler... yes checking whether gcc accepts -g... yes checking for gcc option to accept ANSI C... none needed checking for Rf_initEmbeddedR in -lR... no No R shared library found configure: creating ./config.status config.status: creating Makevars config.status: creating src/Makevars config.status: creating src/RSJava/Makefile config.status: creating Makefile_rules config.status: creating inst/scripts/RJava.bsh config.status: creating inst/scripts/RJava.csh config.status: creating R/zzz.R config.status: creating cleanup config.status: creating inst/scripts/RJava Copying the cleanup script to the scripts/ directory Building libRSNativeJava.so in /tmp/R.INSTALL.tf2988/SJava/src/RSJava if test ! -d /usr/local/lib/R/library/SJava/libs ; then \ mkdir /usr/local/lib/R/library/SJava/libs ; \ fi gcc -g -O2 -D_R_ -I/usr/local/lib/R/include -I/usr/local/lib/R/include/R_ext -I/tmp/R.INSTALL.tf2988/SJava/src/RSJava -I. -I/tmp/R.INSTALL.tf2988/SJava/inst/include -I/usr/java/jdk1.5.0_04//include -I/usr/java/jdk1.5.0_04//include/linux -c CtoJava.c CtoJava.cweb:215: error: static declaration of 'std_env' follows non-static declaration CtoJava.cweb:195: error: previous declaration of 'std_env' was here make: *** [CtoJava.o] Error 1 Generating JNI header files from Java classes. RForeignReference, RManualFunctionActionListener, ROmegahatInterpreter REvaluator * Warning: At present, to use the library you must set the LD_LIBRARY_PATH environment variable to /usr/local/lib/R/library/SJava/libs:/usr/java/jdk1.5.0_04/jre/lib/i386/client:/usr/java/jdk1.5.0_04/jre/lib/i386:/usr/java/jdk1.5.0_04/jre/../lib/i386: or use one of the RJava.bsh or RJava.csh scripts * ** libs gcc -I/usr/local/lib/R/include -D_R_ -I/usr/local/lib/R/include -I/usr/local/lib/R/include/R_ext -I/tmp/R.INSTALL.tf2988/SJava/src/RSJava -I. -I/tmp/R.INSTALL.tf2988/SJava/inst/include -IRSJava -I/usr/java/jdk1.5.0_04//include -I/usr/java/jdk1.5.0_04//include/linux -I/usr/local/include -fPIC -g -O2 -c ConverterExamples.c -o ConverterExamples.o gcc -I/usr/local/lib/R/include -D_R_ -I/usr/local/lib/R/include -I/usr/local/lib/R/include/R_ext -I/tmp/R.INSTALL.tf2988/SJava/src/RSJava -I. -I/tmp/R.INSTALL.tf2988/SJava/inst/include -IRSJava -I/usr/java/jdk1.5.0_04//include -I/usr/java/jdk1.5.0_04//include/linux -I/usr/local/include -fPIC -g -O2 -c Converters.c -o Converters.o gcc -I/usr/local/lib/R/include -D_R_ -I/usr/local/lib/R/include -I/usr/local/lib/R/include/R_ext -I/tmp/R.INSTALL.tf2988/SJava/src/RSJava -I. -I/tmp/R.INSTALL.tf2988/SJava/inst/include -IRSJava -I/usr/java/jdk1.5.0_04//include -I/usr/java/jdk1.5.0_04//include/linux -I/usr/local/include -fPIC -g -O2 -c REmbed.c -o REmbed.o gcc -I/usr/local/lib/R/include -D_R_ -I/usr/local/lib/R/include -I/usr/local/lib/R/include/R_ext -I/tmp/R.INSTALL.tf2988/SJava/src/RSJava -I. -I/tmp/R.INSTALL.tf2988/SJava/inst/include -IRSJava -I/usr/java/jdk1.5.0_04//include -I/usr/java/jdk1.5.0_04//include/linux -I/usr/local/include -fPIC -g -O2 -c REmbedWin.c -o REmbedWin.o gcc -I/usr/local/lib/R/include -D_R_ -I/usr/local/lib/R/include -I/usr/local/lib/R/include/R_ext -I/tmp/R.INSTALL.tf2988/SJava/src/RSJava -I. -I/tmp/R.INSTALL.tf2988/SJava/inst/include -IRSJava -I/usr/java/jdk1.5.0_04//include -I/usr/java/jdk1.5.0_04//include/linux -I/usr/local/include -fPIC -g -O2 -c REval.c -o REval.o gcc -I/usr/local/lib/R/include -D_R_ -I/usr/local/lib/R/include -I/usr/local/lib/R/include/R_ext -I/tmp/R.INSTALL.tf2988/SJava/src/RSJava -I. -I/tmp/R.INSTALL.tf2988/SJava/inst/include -IRSJava -I/usr/java/jdk1.5.0_04//include -I/usr/java/jdk1.5.0_04//include/linux -I/usr/local/include -fPIC -g -O2 -c RFunctionListener.c -o RFunctionListener.o gcc -I/usr/local/lib/R/include -D_R_ -I/usr/local/lib/R/include -I/usr/local/lib/R/include/R_ext -I/tmp/R.INSTALL.tf2988/SJava/src/RSJava -I.
[R] class in apply
Numeric data that is part of a mixed type data frame is converted into
character. How can I tell apply to maintain the original class of a
column and not convert it into character. I would like to do this of
the vector and not inside the apply function individually over each
element. Consider the following two scenarios, in the second column
'a' maintained its class while it lost its numeric type in the first
case.
df = data.frame(a=c(1,2), b=c('A','B'))
df
a b
1 1 A
2 2 B
a=apply(df, 1, function(r) print(class(r['a'])))
[1] character
[1] character
a=apply(df, 1, function(r) print(class(r['b'])))
[1] character
[1] character
df = data.frame(a=c(1,2))
df
a
1 1
2 2
a=apply(df, 1, function(r) print(class(r['a'])))
[1] numeric
[1] numeric
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Re: [R] Turning off return warning messages.
On Wed, 20 Jul 2005, Steve Su wrote:
Is there a way I can turn off the following warning message
for using multi-argument returns?
multi-argument returns are deprecated in: return(p1, p2, p3, p4)
doubleEm - function(p1, p2, p3, p4) {
return(list(p1 = p1*p1,
p2 = p2*p2,
p3 = p3*p3,
p4 = p4*p4))
}
res - doubleEm(1, 2, 3, 4)
cat(p3 =, res$p3, \n)
--
SIGSIG -- signature too long (core dumped)
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