[R] Problem on Matrix multiplication
Dear all r-users, I am getting a big problem with matrix multiplication suppose I have, weight Weight 1 1067640 2 8871500 3 42948778 4 127583735 5 2200 6 4400 7 5685 8 23805662 and, s a b c d e f g h a 402493.18 -133931.62 461483.3 -94042.86 674493.8 -1493713.2 - 714081.5 -1320551.6 b -133931.62 192766.63 -414674.4 -36700.74 -276595.3 798034.6 382661.6 883478.4 c 461483.25 -414674.43 1618660.4 660949.18 988526.5 -2795048.7 - 1483334.1 -1761234.9 d -94042.86 -36700.74 660949.2 1053310.43 -300291.4 -98814.9 - 263716.7 1130451.3 e 674493.80 -276595.33 988526.5 -300291.35 3090869.8 -3017747.3 - 1766040.6 -2295605.0 f -1493713.19 798034.60 -2795048.7 -98814.90 -3017747.3 10558734.1 4415620.8 6200897.7 g -714081.51 382661.62 -1483334.1 -263716.67 -1766040.6 4415620.8 2593947.5 2341455.0 h -1320551.58 883478.39 -1761234.9 1130451.26 -2295605.0 6200897.7 2341455.0 7756557.3 But when I want to get [transpose(weight)][s][weight] using following syntax, t(weight)%*%s%*%weight I got error: Error in t(weight) %*% s %*% weight : requires numeric matrix/vector arguments Can anyone please tell me that where my error is? Thanks and regards, [[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] Problem on Matrix multiplication
It looks like weight and s are data frames, not vectors/matrices as required (and as the error message tells you). Dan Daniel Nordlund Bothell, WA -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Arun Kumar Saha Sent: Wednesday, June 14, 2006 11:10 PM To: r-help@stat.math.ethz.ch Subject: [R] Problem on Matrix multiplication Dear all r-users, I am getting a big problem with matrix multiplication suppose I have, weight Weight 1 1067640 2 8871500 3 42948778 4 127583735 5 2200 6 4400 7 5685 8 23805662 and, s a b c d e f g h a 402493.18 -133931.62 461483.3 -94042.86 674493.8 -1493713.2 - 714081.5 -1320551.6 b -133931.62 192766.63 -414674.4 -36700.74 -276595.3 798034.6 382661.6 883478.4 c 461483.25 -414674.43 1618660.4 660949.18 988526.5 -2795048.7 - 1483334.1 -1761234.9 d -94042.86 -36700.74 660949.2 1053310.43 -300291.4 -98814.9 - 263716.7 1130451.3 e 674493.80 -276595.33 988526.5 -300291.35 3090869.8 -3017747.3 - 1766040.6 -2295605.0 f -1493713.19 798034.60 -2795048.7 -98814.90 -3017747.3 10558734.1 4415620.8 6200897.7 g -714081.51 382661.62 -1483334.1 -263716.67 -1766040.6 4415620.8 2593947.5 2341455.0 h -1320551.58 883478.39 -1761234.9 1130451.26 -2295605.0 6200897.7 2341455.0 7756557.3 But when I want to get [transpose(weight)][s][weight] using following syntax, t(weight)%*%s%*%weight I got error: Error in t(weight) %*% s %*% weight : requires numeric matrix/vector arguments Can anyone please tell me that where my error is? Thanks and regards, [[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] Latex-Style Characters in R
Dear All, I am starting to use R excellent graphical facilities to produce good-looking plots. However, I do not know yet how for use pedices/apices (e.g. when you write cm^3) and Greek letters (e.g. \sigma) which I really need now. Is there a special package to load? Could anyone post a simple example of such a plot so that I can re-use it? Cheers Lorenzo __ 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] Latex-Style Characters in R
Try ?plotmath Stefano On Thu, Jun 15, 2006 at 10:18:25AM +0200, Lorenzo Isella wrote: LorenzoDear All, LorenzoI am starting to use R excellent graphical facilities to produce Lorenzogood-looking plots. LorenzoHowever, I do not know yet how for use pedices/apices (e.g. when you Lorenzowrite cm^3) and Greek letters (e.g. \sigma) which I really need now. LorenzoIs there a special package to load? Could anyone post a simple example Lorenzoof such a plot so that I can re-use it? LorenzoCheers Lorenzo LorenzoLorenzo Lorenzo Lorenzo__ LorenzoR-help@stat.math.ethz.ch mailing list Lorenzohttps://stat.ethz.ch/mailman/listinfo/r-help LorenzoPLEASE 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] Latex-Style Characters in R
Hi, just take a look at demo(plotmath) and example(plotmath). Detlef On Thu, 15 Jun 2006 10:18:25 +0200 Lorenzo Isella [EMAIL PROTECTED] wrote: Dear All, I am starting to use R excellent graphical facilities to produce good-looking plots. However, I do not know yet how for use pedices/apices (e.g. when you write cm^3) and Greek letters (e.g. \sigma) which I really need now. Is there a special package to load? Could anyone post a simple example of such a plot so that I can re-use it? Cheers Lorenzo __ 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] MDS with missing data?
On Thu, 2006-06-15 at 07:13 +0300, Jari Oksanen wrote: 1) use nonmetric/gradient descent MDS which seems to allow missing data, or Not the isoMDS function in MASS. if N(N-1) is a problem, then nonmetric MDS may not be the solution. Sorry for the wrong information: isoMDS does handle NA. I remembered old times when I looked at the issue, but isoMDS changed since. Fine work! cheers, jari oksanen -- Jari Oksanen -- Dept Biology, Univ Oulu, 90014 Oulu, Finland email [EMAIL PROTECTED], homepage http://cc.oulu.fi/~jarioksa/ __ 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] survival probabilities with cph (counting process)
Hi, I have fitted a cox model with time-varying covariates (counting process style) using the cph function of the Design package. Now I want to know the survival probabilities at each time point given the history of a single individual. I know the survest function, but I am not sure how to interpretet its output when using time-varying covariates. Does it just give the probabilities as if it are independent individuals or can/does it take in consideration that it is the history of a single individual? Is this even possible? An example: Individual x has a history of 3 months and the cox model is fitted with two time-varying covariates: a b testcase - data.frame(a =[4 5 2], b = [1 0 1]) survest(coxmodel, testcase, time = c(1,2,3)) Is this the right way to compute the probabilities? Thank you in advance! Regards, Laurens Alberts [[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] Problem on Matrix multiplication
Hi, AKS But when I want to get [transpose(weight)][s][weight] using following AKS syntax, AKS t(weight)%*%s%*%weight I guess 'weight' and 's' are dataframes and you need matrices. Like here: x - rnorm(8) s - matrix(rnorm(64), ncol=8) t(x) %*% s %*% x # this is OK d - as.data.frame(x) t(d) %*% s %*% d # this is your error You can use data.matrix() to convert dataframes to matrices: t( data.matrix(d) ) %*% s %*% data.matrix(d) # again OK HTH, Michal ~,~`~,~`~,~`~,~`~,~`~,~`~,~`~,~`~,~ Michal Bojanowski ICS/Utrecht Utrecht University Heidelberglaan 2; 3584 CS Utrecht Room 1428 [EMAIL PROTECTED] __ 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 managment
If your df contains your data, try tmp - cbind( paste(df[ ,1], df[ ,2], sep=:), paste(df[ ,3], df[ ,4], sep=:) ) tmp - t( apply(tmp, 1, sort) ) out - data.frame( do.call(rbind, strsplit( tmp[,1], split=: )), do.call(rbind, strsplit( tmp[,2], split=: )) ) colnames(out) - colnames(df) out Regards, Adai On Wed, 2006-06-14 at 16:35 +0100, yohannes alazar wrote: First I would really like to thank the mailing list for help I got in the past, as a new to R I am really needing some support on hoe to code the following problem. I am trying to sort some data I have in a big file. The file has 4 columns and 19000 rows. An example of it looks like this:- G 0.892 A 0.108 G 0.883 T 0.117 T 0.5 C 0.5 A 0.617 G 0.383 G 0.925 A 0.075 A 0.967 G 0.033 C 0.883 T 0.117 C 0.633 T 0.367 G 0.95 A 0.05 C 0.742 G 0.258 G 0.875 T 0.125 T 0.167 C 0.833 C 0.792 A 0.208 Columns one and three are alphabets while three and four are their corresponding values. I wanted to sort this data so that my first and third columns are in alphabetic order. For example in the first row the order is G then A. This is not in alphabetic order therefore we swap them along with their values and it becomes: A0.108 G 0.892 Row two looks fine but row three needs the same rearrangement as row one. And the final out put looks like: A 0.108 G 0.892 G 0.883 T 0.117 C 0.5 T 0.5 A 0.617 G 0.383 A 0.075 G 0.925 A 0.967 G 0.033 C 0.883 T 0.117 C 0.633 T 0.367 A 0.05 G 0.95 C 0.742 G 0.258 G 0.875 T 0.125 C 0.833 T 0.167 A 0.208 C 0.792 Please some help with the relevant command names or a technique to code this task. Thank you in advance Regards Hannes [[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
Re: [R] appending
Hi Dave,
Does this do what you want?
Hank
comp.CAND.frm - read.table(comp.CAND.frm.frm, header=TRUE)
names(comp.CAND.frm) - make.names(gsub(_, ., names
(comp.CAND.frm)), unique=TRUE)
creatine.function.new = function(delta.0.Y.0=50, gamma=40,
dat=comp.CAND.frm) {
## function to calcuate the delta.i, i.e. the percent
## leftover ## gamma = rate of Cr going into bucket, e.g., mg/hr
## delta.0.Y.0 = product of delta.0 and Y.0 at baseline ##
##
unlist(
tapply(dat$Ucr, dat$Patient.no, function(Cs) {
n.obs - length(Cs)
Y=numeric(n.obs+1)
Y[1] = delta.0.Y.0 + gamma
delta = numeric(n.obs)
for (i in 1:n.obs) delta[i] - {
d - (Y[i] - Cs[i])/Y[i]
Y[i+1] = delta[i] * Y[i] + gamma
d
}
delta}
)
)
}
dim(comp.CAND.frm)
6*24
creatine.function.new()
### The names are the concatenation of the patient and the observation
## e.g. patient 24 observation 6 is labeled 246.
On Jun 14, 2006, at 2:35 PM, Afshartous, David wrote:
Hank,
Attached is the dataframe that can be supplied for the argument
comp.CAND.frm. The argument for comp.LIS.frm can be deleted since
it currently isn't used. The other arguments can be set as:
delta.0.Y.0 = 50
gamma = 40.
cheers,
dave
-Original Message-
From: Martin Henry H. Stevens [mailto:[EMAIL PROTECTED]
Sent: Wednesday, June 14, 2006 1:23 PM
To: Afshartous, David
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] appending
Hi David,
It would be helpful if you supply a little data, upon which this would
operate.
Hank
On Jun 14, 2006, at 12:31 PM, Afshartous, David wrote:
All,
In the function below I have 24 individuals and 6 calculations per
individual.
The 6 calculations are collected each time in a 1:24 loop when
calculating delta.
I'd like to collect all 144 = 24*6 calculations in one vector
(delta.patient.comb).
The function works as is via indexing, but is there an easier way to
collect the measurements via appendinng the 6 measurements each time
to the current set? I couldn't find anything in Intro to R on
appending.
cheers,
Dave
ps - please respond directly to [EMAIL PROTECTED]
creatine.function.new = function(delta.0.Y.0, gamma, comp.LIS.frm,
comp.CAND.frm) { ## function to calcuate the delta.i, i.e. the
percent
## leftover ## gamma = rate of Cr going into bucket, e.g., mg/hr ##
delta.0.Y.0 = product of delta.0 and Y.0 at baseline ##
Y.1 = delta.0.Y.0 + gamma
delta = numeric(6)
delta.patient = numeric(24)
delta.patient.comb = numeric(144)
##
for (k in 1:24) { ## each patient
patient.k.CAND = which(comp.CAND.frm$Patient_no == k)
Ucr.CAND.patient.k = comp.CAND.frm$Ucr[patient.k.CAND]
C = Ucr.CAND.patient.k ## 6 observed creatanine values
for each
patient
delta[1] = (Y.1 - C[1])/Y.1
Y.i = Y.1
delta.i = delta[1]
for (i in 1:5) {## six measurments per
patient
Y.i.plus.1 = delta.i * Y.i + gamma
delta.i.plus.1 = (Y.i.plus.1 -
C[i+1])/Y.i.plus.1
delta[i+1] = delta.i.plus.1
delta.i = delta[i+1]
Y.i = Y.i.plus.1
}
delta.patient[k] = list(delta)
delta.patient.comb[(6*(k-1)+1):(6*(k-1)+ 6)] = delta
}
list(delta.patient, delta.patient.comb) }
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Dr. M. Hank H. Stevens, Assistant Professor
338 Pearson Hall
Botany Department
Miami University
Oxford, OH 45056
Office: (513) 529-4206
Lab: (513) 529-4262
FAX: (513) 529-4243
http://www.cas.muohio.edu/~stevenmh/
http://www.muohio.edu/ecology/
http://www.muohio.edu/botany/
E Pluribus Unum
comp.CAND.frm.frm
Dr. M. Hank H. Stevens, Assistant Professor
338 Pearson Hall
Botany Department
Miami University
Oxford, OH 45056
Office: (513) 529-4206
Lab: (513) 529-4262
FAX: (513) 529-4243
http://www.cas.muohio.edu/~stevenmh/
http://www.muohio.edu/ecology/
http://www.muohio.edu/botany/
E Pluribus Unum
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Re: [R] A whine and a request
John Vokey wrote: Thanks. But, as is common with e-mail communication, nobody seems to have understood the issue I was whining about. Indeed, I was totally wrong. So, here is an actual case... Now, what I want is to regress the proportions of column(1)/(column(0) +column(1)) on x. I can accomplish the proportions as follows: props=tab[,2]/(tab[,1]+tab[,2]). So, I want to regress props on the *row labels* of tab. But I can't seem to get them to be anything *but* row labels. For example, tab[,0] ... extracts the row labels, *but still as labels*, not data. as.numeric(rownames(tab)) Jim __ 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] Problem with Julian function
Dear all, I have a problem with the function Julian, may be a bug in the function ? Here is a vector of character, which represents dates (May 18 to May 20 2000): amj - c(2000-05-18,2000-05-18,2000-05-18,2000-05-19,2000-05-19 ,2000-05-19, 2000-05-19, 2000-05-20, 2000-05-20, 2000-05-20) I load the date and chron libraries, I define the vector of character as a date variable using the dates function: ymd - dates(amj,format=c(dates=y-m-d),origin=c(month=1, day=1,year=2000)) Then when I apply the Julian function: julian(months(ymd),days(ymd),years(ymd),origin=c(month = 1, day = 1, year = as.numeric(years(ymd[1] I get the following result: [1] 137 137 137 138 138 138 138 139 139 139 However, when I check on the calendar, the dates I have correspond to the 139th, 140th and 141st days of the year (there is a two days shift because it is a leap year, for other years I have a one day shift). I tried the default origin when I use the dates function but I get the same results. Did anyone have this problem before ? I would appreciate some help, Thank you very much, Emmanuel [[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] Saving R graphics with version 2.3.1
[EMAIL PROTECTED] wrote: HI ! I´ve installed the latest R version (2.3.1) and I´ve had problems to save my graphics as JPEG, PNG and the others available formats. R saves the file with no extension. Then, I have to open this file using an image visualizer and save it as PNG, for instance. You can type the extension when you save it, if you're typing the filename. The default names generally include the extension. Previous R versions work without problems. Could you give a detailed description to reproduce the problem? Duncan Murdoch Has everyone had a similar problem? Is it a bug of R 2.3.1? Thanks in advance. Ilka Afonso Reis __ 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] select.list
Dear all,
I don't know, whether this is a bug or whether I', doing something wrong,
but since I installed the latest versions of R and ESS, R hangs within ESS,
when 'select.list' is the first command trying to open a window. Basically,
this hangs:
x - c('a', 'b', 'c', 'd')
y - select.list(x)
and this works fine:
winDialog('ok', 'Hello')
x - c('a', 'b', 'c', 'd')
y - select.list(x)
Here are my specs:
version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major 2
minor 3.1
year 2006
month 06
day01
svn rev38247
language R
version.string Version 2.3.1 (2006-06-01)
win.version()
[1] Windows XP Professional (build 2600) Service Pack 2.0
Furthermore, this is within GNU Emacs 21.3.1 (i386-mingw-nt5.1.2600) with
the latest ESS 5.3.1.
Any help is highly welcome.
Many thanks in advance and best regards,
Jan
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[R] help with table partition
Hi,
I have a test_table where the dim is 62220 by 73 (row by col)
I would like to partition the rows into 170 equal parts (170 tables
where each is of dim 366 by 73), and rearrange them horizontally. The
source codes I have:
for (i in 1:170) {
c = cbind(c,test_table[(367*i+1):(367*(i+1)),2:73]);
}
Unfortunately, using for loop and cbind for a table of this size causes
long running time. What is the most efficient way to get the table that
I want?
Thanks for any help.
K.
-
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[R] matrix selection return types
Dear Rusers, I would like some comments about the following results (under R-2.2.0) m = matrix(1:6 , 2 , 3) m [,1] [,2] [,3] [1,]135 [2,]246 z1 = m[(m[,1]==2),] z1 [1] 2 4 6 is.matrix(z1) [1] FALSE z2 = m[(m[,1]==0),] z2 [,1] [,2] [,3] is.matrix(z2) [1] TRUE Considered together, I'm a bit surprised about the returned types from z1 and z2. I would not have been surprised if z1 would still have been a matrix, or z2=NULL. There is certainly a logic behind this choice but it's not very clear for me, so any help/comment appreciated. Thanks Vincent __ 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] A question about stepwise procedures: step function
Hi, Step works for a Cox model. And I got the same error massage using stepAIC. Jia _ From: Ritwik Sinha [mailto:[EMAIL PROTECTED] Sent: Thursday, June 15, 2006 12:12 AM To: Li, Jia Subject: Re: [R] A question about stepwise procedures: step function Hi, The step documentation says object: an object representing a model of an appropriate class (mainly 'lm' and 'glm'). This is used as the initial model in the stepwise search. I wonder if it will work for a cox proportional hazard model. You could try stepAIC in MASS. On 6/14/06, Li, Jia [EMAIL PROTECTED] mailto:[EMAIL PROTECTED] wrote: Dear all, I tried to use step function to do model selection, but I got an error massage. What I don't understand is that data as data.frame worked well for my other programs, how come I cannot make it run this time. Could you please tell me how I can fix it? *** all-data.frame(z1,z2,z3) fit.model.all- coxph(Surv(t,cen) ~z1+z2+z3,data=all) reg.model.all-step(fit.model.all) Start: AIC= 689.1 Surv(t, cen) ~ z1 + z2 + z3 Error in as.data.frame.default(data) : cannot coerce class function into a data.frame *** Thanks a lot! Jia __ 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 -- Ritwik Sinha Graduate Student Epidemiology and Biostatistics Case Western Reserve University http://darwin.cwru.edu/~rsinha [[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] Access and assign list sub-elements using a string such as l$a$b
If I have a list I can set a sub-element as follows on the command line: people=list() people$tom$hair=brown people But what if I have a string containing the name of the sub-element that I want to access? subel= people$tom$hair get(subel) # returns error assign(subel,red) # silent but doesn't change list people The attempts above using assign/get won't do what I am trying to do [nor according to the help should they]. I would be very grateful for any suggestions. Many thanks, Greg. -- Gregory Jefferis, PhD and: Research Fellow Department of Zoology St John's College University of Cambridge Cambridge Downing Street CB2 1TP Cambridge, CB2 3EJ United Kingdom Lab Tel: +44 (0)1223 336683 Office: +44 (0)1223 339899 Lab Fax: +44 (0)1223 336676 http://www.zoo.cam.ac.uk/zoostaff/jefferis.html [EMAIL PROTECTED] __ 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 concerning mle
Hi I hope this is the right forum - if not, point me please to a better one. I am using R 2.3.0 on Linux, SuSE 10. I have a question concerning mle (method=BFGS). I have a few models which I am fitting to existing data points. I realised, that the likelihood is quite sensitive to the start values for one parameter. I am wondering: what is the best approach to identify the right initial values? Do I have to do it recursively, and if yes, how can I automate it? Or do I have to play with the system? I am quite confident that the resulting parameters are the optimal for my problem - but can I verify it? Thanks, Rainer -- Rainer M. Krug, Dipl. Phys. (Germany), MSc Conservation Biology (UCT) Department of Conservation Ecology and Entomology University of Stellenbosch Matieland 7602 South Africa __ 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] Problem with Julian function
This looks like a bug to me too although its one off rather than two off since julian uses origin 0, not 1, e.g. julian(1,1,2000, origin = c(1,1,2000)) [1] 0 Here is a workaround that seems to work ok and uses origin 1: as.numeric(format(as.Date(ymd), %j)) # ymd from your post [1] 139 139 139 140 140 140 140 141 141 141 On 6/15/06, Devred, Emmanuel [EMAIL PROTECTED] wrote: Dear all, I have a problem with the function Julian, may be a bug in the function ? Here is a vector of character, which represents dates (May 18 to May 20 2000): amj - c(2000-05-18,2000-05-18,2000-05-18,2000-05-19,2000-05-19 ,2000-05-19, 2000-05-19, 2000-05-20, 2000-05-20, 2000-05-20) I load the date and chron libraries, I define the vector of character as a date variable using the dates function: ymd - dates(amj,format=c(dates=y-m-d),origin=c(month=1, day=1,year=2000)) Then when I apply the Julian function: julian(months(ymd),days(ymd),years(ymd),origin=c(month = 1, day = 1, year = as.numeric(years(ymd[1] I get the following result: [1] 137 137 137 138 138 138 138 139 139 139 However, when I check on the calendar, the dates I have correspond to the 139th, 140th and 141st days of the year (there is a two days shift because it is a leap year, for other years I have a one day shift). I tried the default origin when I use the dates function but I get the same results. Did anyone have this problem before ? I would appreciate some help, Thank you very much, Emmanuel [[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
Re: [R] help with table partition
do.call(cbind, split(as.data.frame(test_table), rep(1:170,each=366)))
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---
Wong, Kim a écrit :
Hi,
I have a test_table where the dim is 62220 by 73 (row by col)
I would like to partition the rows into 170 equal parts (170 tables
where each is of dim 366 by 73), and rearrange them horizontally. The
source codes I have:
for (i in 1:170) {
c = cbind(c,test_table[(367*i+1):(367*(i+1)),2:73]);
}
Unfortunately, using for loop and cbind for a table of this size causes
long running time. What is the most efficient way to get the table that
I want?
Thanks for any help.
K.
-
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Re: [R] help with table partition
Hi
maybe ?split and ?t is what you want
mat-matrix(rnorm(1000), 100,10)
mat.s-split(data.frame(mat), rep(1:5, each=20))
#splits mat to list with 5 eqal submatrices
lapply(mat.s,t)
# transpose matrices in list
gives you a list of transposed tables, which is probably better than
separate tables. Just change rep(1:5,each=20) to rep (1:170,
each=366).
or
a quicker one without data frame
mat - matrix(rnorm(62220*73), 62220,73)
dim(mat) - c(366,73,170)
mat.i - array(0,dim=c(73,366,170))
for (i in 1:170) mat[ , , i] - t(mat[ , , i])
HTH
Petr
On 15 Jun 2006 at 9:38, Wong, Kim wrote:
Date sent: Thu, 15 Jun 2006 09:38:29 -0400
From: Wong, Kim [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Subject:[R] help with table partition
Hi,
I have a test_table where the dim is 62220 by 73 (row by col)
I would like to partition the rows into 170 equal parts (170 tables
where each is of dim 366 by 73), and rearrange them horizontally. The
source codes I have:
for (i in 1:170) {
c = cbind(c,test_table[(367*i+1):(367*(i+1)),2:73]);
}
Unfortunately, using for loop and cbind for a table of this size
causes long running time. What is the most efficient way to get the
table that I want?
Thanks for any help.
K.
-
CONFIDENTIALITY NOTICE: This message and any attachments
rel...{{dropped}}
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Petr Pikal
[EMAIL PROTECTED]
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[R] Array
Dear all R users, I am wondering if there is any way to define a 3 dimentional or 4 dimentional array in R. Sincerely yours, thanks in advance Send instant messages to your online friends http://in.messenger.yahoo.com Stay connected with your friends even when away from PC. Link: http://in.mobile.yahoo.com/new/messenger/ [[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] Array
Hi see ?array if you are interested array(0, c(5,5,5,5,5)) gives you 5 dimensional array. HTH Petr On 15 Jun 2006 at 15:34, stat stat wrote: Date sent: Thu, 15 Jun 2006 15:34:29 +0100 (BST) From: stat stat [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Subject:[R] Array Dear all R users, I am wondering if there is any way to define a 3 dimentional or 4 dimentional array in R. Sincerely yours, thanks in advance Send instant messages to your online friends http://in.messenger.yahoo.com Stay connected with your friends even when away from PC. Link: http://in.mobile.yahoo.com/new/messenger/ [[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 Petr Pikal [EMAIL PROTECTED] __ 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] matrix selection return types
Hi On 15 Jun 2006 at 15:41, [EMAIL PROTECTED] wrote: Date sent: Thu, 15 Jun 2006 15:41:14 +0200 From: [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Subject:[R] matrix selection return types Dear Rusers, I would like some comments about the following results (under R-2.2.0) m = matrix(1:6 , 2 , 3) m [,1] [,2] [,3] [1,]135 [2,]246 z1 = m[(m[,1]==2),] z1 [1] 2 4 6 is.matrix(z1) [1] FALSE z2 = m[(m[,1]==0),] z2 [,1] [,2] [,3] is.matrix(z2) [1] TRUE Considered together, I'm a bit surprised about the returned types from z1 and z2. I would not have been surprised if z1 would still have been a matrix, or z2=NULL. If you want z1 to be matrix see argument drop in ?[ z1 = m[(m[,1]==2),,drop=F] however why matrix is retained in second case i am not sure. Probably only if the result has exactly one dimension it is stripped from dim attribute by default. HTH Petr There is certainly a logic behind this choice but it's not very clear for me, so any help/comment appreciated. Thanks Vincent __ 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 Petr Pikal [EMAIL PROTECTED] __ 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] Standard Deviation Distribution
I'm having trouble with the standard deviation distribution
as shown on http://mathworld.wolfram.com/StandardDeviationDistribution.html .
(Eric Weisstein references Kenney and Keeping 1951, which I can't check.)
I believe the graphs they show, but when I code the function in R, according to
the listed formula,
I get very different graphs.
Would someone please point out my error or tell me where it's already
implemented in R?
Here is my version:
sddist
function(s,n) {
sig2 - n*s*s/(n-1)
2*(n/(2*sig2))^((n-1)/2) / gamma((n-1)/2) * exp(-n*s*s/(2*sig2)) * s^(n-2)
}
Version 2.3.1 (2006-06-01) on Windows XP SP2
Thanks for any help.
David L. Reiner
Rho Trading Securities, LLC
Chicago IL 60605
312-362-4963
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Re: [R] Access and assign list sub-elements using a string such as l$a$b
Hi very, very close On 15 Jun 2006 at 13:27, Gregory Jefferis wrote: Date sent: Thu, 15 Jun 2006 13:27:05 +0100 From: Gregory Jefferis [EMAIL PROTECTED] To: [EMAIL PROTECTED] [EMAIL PROTECTED] Forwarded to: r-help@stat.math.ethz.ch Forwarded by: Gregory Jefferis [EMAIL PROTECTED] Date forwarded: Thu, 15 Jun 2006 14:54:13 +0100 Subject:[R] Access and assign list sub-elements using a string such as l$a$b If I have a list I can set a sub-element as follows on the command line: people=list() people$tom$hair=brown people But what if I have a string containing the name of the sub-element that I want to access? subel= people$tom$hair get(subel) # returns error assign(subel,red) # silent but doesn't change list people See what happens when people-assign(subel, red) HTH Petr The attempts above using assign/get won't do what I am trying to do [nor according to the help should they]. I would be very grateful for any suggestions. Many thanks, Greg. -- Gregory Jefferis, PhD and: Research Fellow Department of Zoology St John's College University of Cambridge Cambridge Downing Street CB2 1TP Cambridge, CB2 3EJ United Kingdom Lab Tel: +44 (0)1223 336683 Office: +44 (0)1223 339899 Lab Fax: +44 (0)1223 336676 http://www.zoo.cam.ac.uk/zoostaff/jefferis.html [EMAIL PROTECTED] __ 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 Petr Pikal [EMAIL PROTECTED] __ 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] Install R 2.3.1 on SUSE Linux
Hello: I am new to Linux, and I am trying to install R 2.3.1 on SUSE Linux 10.0. The RPM installer, YAST, states that I need libgfortran.so.0. I have loaded Intel FORTRAN and GCC FORTRAN; but I still do not have the libgfortran.so.0 module YAST is asking for. Is there somewhere I can get this module for SUSE Linux 10.0. There is a post on the R-Help for this same problem; but I cannot seem to find the resolution. I would appreciate any help you could provide. Thanks Dennis This e-mail is intended only for the use of the individual or entity to which it is addressed and may contain information that is privileged and confidential. If the reader of this e-mail message is not the intended recipient, you are hereby notified that any dissemination, distribution or copying of this communication is prohibited. If you have received this e-mail in error, please notify the sender and destroy all copies of the transmittal. Thank you University of Chicago Hospitals [[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] Access and assign list sub-elements using a string suchas l$a$b
- Original Message - From: Petr Pikal [EMAIL PROTECTED] To: Gregory Jefferis [EMAIL PROTECTED] Cc: r-help@stat.math.ethz.ch Sent: Thursday, June 15, 2006 4:56 PM Subject: Re: [R] Access and assign list sub-elements using a string suchas l$a$b Hi very, very close On 15 Jun 2006 at 13:27, Gregory Jefferis wrote: Date sent: Thu, 15 Jun 2006 13:27:05 +0100 From: Gregory Jefferis [EMAIL PROTECTED] To: [EMAIL PROTECTED] [EMAIL PROTECTED] Forwarded to: r-help@stat.math.ethz.ch Forwarded by: Gregory Jefferis [EMAIL PROTECTED] Date forwarded: Thu, 15 Jun 2006 14:54:13 +0100 Subject:[R] Access and assign list sub-elements using a string such as l$a$b If I have a list I can set a sub-element as follows on the command line: people=list() people$tom$hair=brown people But what if I have a string containing the name of the sub-element that I want to access? subel= people$tom$hair get(subel) # returns error assign(subel,red) # silent but doesn't change list people See what happens when people-assign(subel, red) but I think this is not what Greg wanted; the above just assigns red to object 'people' (i.e., check `str(assign(subel, red))'). If I understood correctly, the following could be of help: people - list() people$tom$hair - brown people # subel - people$tom$hair eval(parse(text = subel)) eval(parse(text = paste(subel, - 'red'))) people Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax: +32/(0)16/337015 Web: http://med.kuleuven.be/biostat/ http://www.student.kuleuven.be/~m0390867/dimitris.htm HTH Petr The attempts above using assign/get won't do what I am trying to do [nor according to the help should they]. I would be very grateful for any suggestions. Many thanks, Greg. -- Gregory Jefferis, PhD and: Research Fellow Department of Zoology St John's College University of Cambridge Cambridge Downing Street CB2 1TP Cambridge, CB2 3EJ United Kingdom Lab Tel: +44 (0)1223 336683 Office: +44 (0)1223 339899 Lab Fax: +44 (0)1223 336676 http://www.zoo.cam.ac.uk/zoostaff/jefferis.html [EMAIL PROTECTED] __ 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 Petr Pikal [EMAIL PROTECTED] __ 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 Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.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
[R] xyplot problem
Dear all,
I have created the following data (that you can run) in order to explain my
problem:
y - rep(c(1,2), 8)
id - rep(1:8,each=2)
x1 - rep(c(-a,+a), each = 8)
x2 - rep(c(-b,+b), each = 2, times = 4)
x3 - rep(c(-c, +c), each = 4,2)
df - data.frame(cbind(id,y,x1,x2,x3))
If I do:
xyplot(y~ x3|x1*x2,data=df,groups=id,type = b)
then my id's are joined by lines which is what I want. However when I wanted
to add an horizontal line for each panel by using panel.abline the lines
disapears, i.e.
xyplot(y~ x3|x1*x2,data=df,groups=id,type = b,panel=function(x,y)
{
panel.xyplot(x,y)
panel.abline(h=1.5)
}
)
I would be grateful if someone can tell me how can I correct the second
statement in order to have horizontal lines for each panel and each id are
joined by lines.
Thank you,
Bernard,
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[R] y-axis label location relative to tick label width?
Dear R experts: I have an odd question: would it be better to calculate the location of the axis label relative to the furthest protruding tick label? the background is that I have my y-tick labels oriented horizontally rather than vertically. if R chooses to set a tick mark at 2, the y label should be relatively close to the axis. if R chooses to put 1.998, it should sit farther (because otherwise the 1 may touch the y label). possible? regards, /ivo __ 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] bubbleplot for matrix
Thanks Bogdan for the reply,
I almost got it working, but in my case, the rownames and colnames are
strings, not numbers, and I guess that this is a problem when using your
snippet:
a -
matrix(sample(1:5,100,replace=TRUE),nrow=10,dimnames=list(1:10,5*1:10))
rownames(a) =
c(aed,fde,fda,yxj,ijk,ddd,gcd,sbe,adc,asd)
colnames(a) =
c(aed,fde,fda,yxj,ijk,ddd,gcd,sbe,adc,asd)
x - y - z - vector()
for (i in 1:nrow(a)) {
x - c(x,rep(rownames(a)[i],ncol(a)))
y - c(y,colnames(a))
z - c(z,a[i,])
}
symbols(as.numeric(x),as.numeric(y),z,inches=0.2,bg=khaki)
text(as.numeric(x),as.numeric(y),labels=z)
symbols(as.numeric(x),as.numeric(y),z,inches=0.2,bg=khaki)
Error in plot.window(xlim, ylim, log, asp, ...) :
need finite 'xlim' values
In addition: Warning messages:
1: NAs introduced by coercion
2: NAs introduced by coercion
3: no finite arguments to min; returning Inf
4: no finite arguments to max; returning -Inf
5: no finite arguments to min; returning Inf
6: no finite arguments to max; returning -Inf
Any guess?
Thanks in advance,
Albert.
On Wed, 2006-06-14 at 16:47 -0400, bogdan romocea wrote:
Here's an example. By the way, I find that it's more convenient (where
applicable) to keep the data in 3 vectors/factors rather than one
matrix/data frame.
a - matrix(sample(1:5,100,replace=TRUE),nrow=10,dimnames=list(1:10,5*1:10))
x - y - z - vector()
for (i in 1:nrow(a)) {
x - c(x,rep(rownames(a)[i],ncol(a)))
y - c(y,colnames(a))
z - c(z,a[i,])
}
symbols(as.numeric(x),as.numeric(y),z,inches=0.2,bg=khaki)
text(as.numeric(x),as.numeric(y),labels=z)
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Albert Vilella
Sent: Tuesday, June 13, 2006 7:11 AM
To: r-help@stat.math.ethz.ch
Subject: [R] bubbleplot for matrix
Hi all,
I would like to ask if it is possible to use bubbleplot for a 20x20
matrix, instead of a dataframe with factors in columns.
The idea would be to get a tabular representation with bubbles like in
Rnews_2006_2 article, which look very nice.
Thanks in advance,
Albert.
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[R] how to set up parallel environment?
Dear list, My institution has a Beowulf Linux cluster with 128 nodes. I want to set up a parallel environment for R so that I can run some easily parallelizable jobs. I searched around and only found some packages (like snow, rpvm) with tutorials on how to run R functions in a already-setup parallel environment. But did not have success finding tutorials on how to setup parallel R. Should I compile R at each node? How should I install it in the main node? Is there any steo-by-step tutorial on R installation in Beowulf? Thanks a lot, mike [[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] Repost: Estimation when interaction is present: How do I get get the parameters from nlme?
Gday, This is a repost since I only had one direct reply and I remain mystified- This may be stupidity on my part but it may not be so simple. In brief, my problem is I'm not sure how to extract parameter values/effect sizes from a nonlinear regression model with a significant interaction term. My data sets are dose response curves (force and dose) for muscle that also have two treatments applied Treatment A (A- or A+) and Treatment B (B-/B+). A single muscle was used for each experiment - a full dose response curve and one treatment from the matrix A*B (A-/B-, A+/B-, A-/B+ and A+,B+). There are 8 replicates for each combination of treatments We fit a dose response curve to each experiment with parameters upper, ed50 and slope; we expect treatment A to change upper and ed50. We want to know if treatment B blocks the effect of treatment A and if so to what degree. This is similar to the Ludbrook example in Venables and Ripley, however they only had one treatment and I have two. my approach The dataframe is structured like this: expttreatA treatB doseforce. 1 - - 0.1 20 1 - - 0.2 40 ... 4 + + 0.1 20 4 + I used a groupedData object: mydata=groupedData(force ~ dose | expt) I used an nlme obect to model the data as follows (pseudocode): myfit.nlme - nlme(force ~ ss_tpl(dose, upper, ed50,slope), fixed=list(ed50~factor(treatA)*factor(treatB))) The function ss_tpl is a properly debugged and fully functional selfstarting three parameter logistic function that I wrote- no problem here. In my analysis I also included fixed terms for the other fit parameters; upper and slope, but my main problem is with the ed50 so that's all I've included here. Running an anova on the resulting object (anova(myfit.nlme) I found the A -/B- (control) to be significantly different from zero, treatment A was significantly different, treatment B had no significant effect and there was a significant interaction between treatment A and treatment B. The interaction term is likely to be real. The treatments are on sequential steps in a pathway and treatment B may be blocking the effect of treatment A, i.e. treatment B alone has no effect because it blocks a pathway that is not active, treatment A reduces force via this pathway and treament B therefore blocks the effect of treatment A when used together. From what I understand, please correct me if I'm wrong, the parameter estimates from summary(model.nlme) are not correct for main effects if a significant interaction is present. For example in my data treatment B alone has no signifcant effect in the anova but the interaction term A:B is significant. I believe The summary estimate for B is the estimate across all levels of A. What I want to do is pull out the estimate for B when A is not present. I suppose I can do it manually from the list of coefficients from nls or fit a oneway model with treatment levels A, B, AB. But I was kind of hoping there was some extractor function. The reason I need this is that the co-authors want to include a table of parameter values with std errs or confidence intervals ala: Treat upper ed50slope A-/B- x x x - shows value for comparison to control studies A+/B- x x x -Shows A is working0 A-/B+ x x x - Shows B has no effect alone A+/B+ x x x -shows B blocks A (not necessarily total) So back to my question,How do I extract estimates of the parameters from my model object for a specific combination of factors including the interaction term. i.e. what is the ed50 (and std err) for A-/B-, A+/B-, A-/B+, A+/B+ ? I think this is a fair question and one that many biomedical scientists would need. __ 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] xyplot problem
Marc Bernard wrote:
Dear all,
I have created the following data (that you can run) in order to explain
my problem:
y - rep(c(1,2), 8)
id - rep(1:8,each=2)
x1 - rep(c(-a,+a), each = 8)
x2 - rep(c(-b,+b), each = 2, times = 4)
x3 - rep(c(-c, +c), each = 4,2)
df - data.frame(cbind(id,y,x1,x2,x3))
If I do:
xyplot(y~ x3|x1*x2,data=df,groups=id,type = b)
then my id's are joined by lines which is what I want. However when I
wanted to add an horizontal line for each panel by using panel.abline the
lines disapears, i.e.
xyplot(y~ x3|x1*x2,data=df,groups=id,type = b,panel=function(x,y)
{
panel.xyplot(x,y)
panel.abline(h=1.5)
}
)
I would be grateful if someone can tell me how can I correct the second
statement in order to have horizontal lines for each panel and each id are
joined by lines.
Thank you,
Bernard,
This appears to work for me.
xyplot(y~ x3|x1*x2,data=df,groups=id,type = b,
panel=function(x,y,groups,...) {
panel.superpose(x,y,groups,...)
panel.abline(h=1.5)})
--
Kevin E. Thorpe
Biostatistician/Trialist, Knowledge Translation Program
Assistant Professor, Department of Public Health Sciences
Faculty of Medicine, University of Toronto
email: [EMAIL PROTECTED] Tel: 416.946.8081 Fax: 416.946.3297
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Re: [R] appending
Hi Hank,
Thanks so much for looking at this.
Yes, it does what I want and is much more efficient than my code.
I have a couple questions, please see my inserted comments after your
tapply statement and within your for loop. Also, how do you accomplish
what you mentioned RE the concatation in the naming?
cheers,
Dave
Ps - just noticed that a quick fix to my code would be to use unlist
towards my original
result delta.patient, then the creative indexing I used would not have
been needed (but your code still more efficient).
comp.CAND.frm - read.table(comp.CAND.frm.frm, header=TRUE)
names(comp.CAND.frm) - make.names(gsub(_, ., names
(comp.CAND.frm)), unique=TRUE)
#
creatine.function.new.R.help = function(delta.0.Y.0=50, gamma=40,
dat=comp.CAND.frm)
{
## function to calcuate the delta.i, i.e. the percent
## leftover ## gamma = rate of Cr going into bucket, e.g., mg/hr
## delta.0.Y.0 = product of delta.0 and Y.0 at baseline ##
unlist(
tapply(dat$Ucr, dat$Patient.no, function(Cs)
{ ## Cs is the argument to the function
defined
## for this tapply statement
## Cs will equal Ucr?
n.obs - length(Cs)
Y=numeric(n.obs+1)
Y[1] = delta.0.Y.0 + gamma
delta = numeric(n.obs)
for (i in 1:n.obs) delta[i] - ## why delta[i] outside
the {}?
{
d - (Y[i] - Cs[i])/Y[i]## why not just call
this d[i]?
Y[i+1] = delta[i] * Y[i] + gamma
d ## the output of each iteration; set equal to
d[i] each time?
}
delta # result of function(Cs) used in tapply
}
)
)
}
dim(comp.CAND.frm)
6*24
creatine.function.new()
### The names are the concatenation of the patient and the observation
## e.g. patient 24 observation 6 is labeled 246.
-Original Message-
From: Martin Henry H. Stevens [mailto:[EMAIL PROTECTED]
Sent: Thursday, June 15, 2006 7:00 AM
To: Afshartous, David
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] appending
Hi Dave,
Does this do what you want?
Hank
comp.CAND.frm - read.table(comp.CAND.frm.frm, header=TRUE)
names(comp.CAND.frm) - make.names(gsub(_, ., names
(comp.CAND.frm)), unique=TRUE)
creatine.function.new = function(delta.0.Y.0=50, gamma=40,
dat=comp.CAND.frm) {
## function to calcuate the delta.i, i.e. the percent
## leftover ## gamma = rate of Cr going into bucket, e.g., mg/hr
## delta.0.Y.0 = product of delta.0 and Y.0 at baseline ##
##
unlist(
tapply(dat$Ucr, dat$Patient.no, function(Cs) {
n.obs - length(Cs)
Y=numeric(n.obs+1)
Y[1] = delta.0.Y.0 + gamma
delta = numeric(n.obs)
for (i in 1:n.obs) delta[i] - {
d - (Y[i] - Cs[i])/Y[i]
Y[i+1] = delta[i] * Y[i] + gamma
d
}
delta}
)
)
}
dim(comp.CAND.frm)
6*24
creatine.function.new()
### The names are the concatenation of the patient and the observation
## e.g. patient 24 observation 6 is labeled 246.
On Jun 14, 2006, at 2:35 PM, Afshartous, David wrote:
Hank,
Attached is the dataframe that can be supplied for the argument
comp.CAND.frm. The argument for comp.LIS.frm can be deleted since
it currently isn't used. The other arguments can be set as:
delta.0.Y.0 = 50
gamma = 40.
cheers,
dave
-Original Message-
From: Martin Henry H. Stevens [mailto:[EMAIL PROTECTED]
Sent: Wednesday, June 14, 2006 1:23 PM
To: Afshartous, David
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] appending
Hi David,
It would be helpful if you supply a little data, upon which this would
operate.
Hank
On Jun 14, 2006, at 12:31 PM, Afshartous, David wrote:
All,
In the function below I have 24 individuals and 6 calculations per
individual.
The 6 calculations are collected each time in a 1:24 loop when
calculating delta.
I'd like to collect all 144 = 24*6 calculations in one vector
(delta.patient.comb).
The function works as is via indexing, but is there an easier way to
collect the measurements via appendinng the 6 measurements each time
to the current set? I couldn't find anything in Intro to R on
appending.
cheers,
Dave
ps - please respond directly to [EMAIL PROTECTED]
creatine.function.new = function(delta.0.Y.0, gamma, comp.LIS.frm,
comp.CAND.frm) { ## function to calcuate the delta.i, i.e. the
percent
## leftover ## gamma = rate of Cr going into bucket, e.g., mg/hr ##
delta.0.Y.0 = product of delta.0 and Y.0 at baseline ##
Y.1 = delta.0.Y.0 + gamma
delta = numeric(6)
delta.patient = numeric(24)
delta.patient.comb = numeric(144)
##
for (k in 1:24) { ## each patient
patient.k.CAND = which(comp.CAND.frm$Patient_no == k)
Ucr.CAND.patient.k = comp.CAND.frm$Ucr[patient.k.CAND]
C = Ucr.CAND.patient.k ## 6
Re: [R] A question about stepwise procedures: step function
Sorry, I still cannot find what's wrong with it. And it seems that
nothing is wrong with t.
Jia
z1-rnorm(N,0,1)
z2-rnorm(N,3,5)
z3-rbinom(N,1,0.6)
prop.cens-0.45
cen-rbinom(N,1,1-prop.cens) #-- censor indicator:45% censor in
the data.
t- rexp(N)
all-data.frame(z1,z2,z3)
fit.model.all- coxph(Surv(t,cen) ~z1+z2+z3,data=all)
fit.model.all
Call:
coxph(formula = Surv(t, cen) ~ z1 + z2 + z3, data = all)
coef exp(coef) se(coef) zp
z1 0.057466 1.059 0.1377 0.4173 0.68
z2 0.000907 1.001 0.0332 0.0273 0.98
z3 -0.349273 0.705 0.2867 -1.2184 0.22
Likelihood ratio test=1.81 on 3 df, p=0.613 n= 100
reg.model.all-step(fit.model.all)
Start: AIC= 376.39
Surv(t, cen) ~ z1 + z2 + z3
Error in as.data.frame.default(data) : cannot coerce class function
into a data.frame
_
From: Hong Ooi [mailto:[EMAIL PROTECTED]
Sent: Thursday, June 15, 2006 10:53 AM
To: Li, Jia
Subject: RE: [R] A question about stepwise procedures: step function
Note: This e-mail is subject to the disclaimer contained at the bottom
of this message.
_
t is the name of a function. If you have a variable called t in your
dataset, try renaming it.
_
From: [EMAIL PROTECTED] on behalf of Li, Jia
Sent: Thu 15/06/2006 11:52 PM
To: r-help@stat.math.ethz.ch; [EMAIL PROTECTED]
Subject: Re: [R] A question about stepwise procedures: step function
Hi,
Step works for a Cox model. And I got the same error massage using
stepAIC.
Jia
_
From: Ritwik Sinha [mailto:[EMAIL PROTECTED]
Sent: Thursday, June 15, 2006 12:12 AM
To: Li, Jia
Subject: Re: [R] A question about stepwise procedures: step function
Hi,
The step documentation says
object: an object representing a model of an appropriate class
(mainly 'lm' and 'glm'). This is used as the initial
model in the stepwise search.
I wonder if it will work for a cox proportional hazard model. You could
try stepAIC in MASS.
On 6/14/06, Li, Jia [EMAIL PROTECTED] mailto:[EMAIL PROTECTED]
wrote:
Dear all,
I tried to use step function to do model selection, but I got
an error massage. What I don't understand is that data as data.frame
worked well for my other programs, how come I cannot make it run this
time. Could you please tell me how I can fix it?
***
all-data.frame(z1,z2,z3)
fit.model.all- coxph(Surv(t,cen) ~z1+z2+z3,data=all)
reg.model.all-step(fit.model.all)
Start: AIC= 689.1
Surv(t, cen) ~ z1 + z2 + z3
Error in as.data.frame.default(data) : cannot coerce class
function into a data.frame
***
Thanks a lot!
Jia
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--
Ritwik Sinha
Graduate Student
Epidemiology and Biostatistics
Case Western Reserve University
http://darwin.cwru.edu/~rsinha
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Re: [R] matrix selection return types
m = matrix(1:6 , 2 , 3) ?'[' z1 - m[m[,1] == 2,] z1 [1] 2 4 6 is.matrix(z1) [1] FALSE z1 - m[m[,1] == 2,, drop=FALSE] z1 [,1] [,2] [,3] [1,]246 is.matrix(z1) [1] TRUE On 6/15/06, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote: Dear Rusers, I would like some comments about the following results (under R-2.2.0) m = matrix(1:6 , 2 , 3) m [,1] [,2] [,3] [1,]135 [2,]246 z1 = m[(m[,1]==2),] z1 [1] 2 4 6 is.matrix(z1) [1] FALSE z2 = m[(m[,1]==0),] z2 [,1] [,2] [,3] is.matrix(z2) [1] TRUE Considered together, I'm a bit surprised about the returned types from z1 and z2. I would not have been surprised if z1 would still have been a matrix, or z2=NULL. There is certainly a logic behind this choice but it's not very clear for me, so any help/comment appreciated. Thanks Vincent __ 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 -- Jim Holtman Cincinnati, OH +1 513 646 9390 (Cell) +1 513 247 0281 (Home) What is the problem you are trying to solve? [[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] Access and assign list sub-elements using a string suchas l$a$b
Hi yes you are correct, I remembered there is something with eval from older posts but did not find a connection to parse from eval help page. Shouldn't there be a link? Or even an example? Best regards Petr On 15 Jun 2006 at 17:21, Dimitris Rizopoulos wrote: From: Dimitris Rizopoulos [EMAIL PROTECTED] To: Petr Pikal [EMAIL PROTECTED], [EMAIL PROTECTED] Copies to: r-help@stat.math.ethz.ch Subject:Re: [R] Access and assign list sub-elements using a string suchas l$a$b Date sent: Thu, 15 Jun 2006 17:21:26 +0200 - Original Message - From: Petr Pikal [EMAIL PROTECTED] To: Gregory Jefferis [EMAIL PROTECTED] Cc: r-help@stat.math.ethz.ch Sent: Thursday, June 15, 2006 4:56 PM Subject: Re: [R] Access and assign list sub-elements using a string suchas l$a$b Hi very, very close On 15 Jun 2006 at 13:27, Gregory Jefferis wrote: Date sent: Thu, 15 Jun 2006 13:27:05 +0100 From: Gregory Jefferis [EMAIL PROTECTED] To: [EMAIL PROTECTED] [EMAIL PROTECTED] Forwarded to: r-help@stat.math.ethz.ch Forwarded by: Gregory Jefferis [EMAIL PROTECTED] Date forwarded: Thu, 15 Jun 2006 14:54:13 +0100 Subject:[R] Access and assign list sub-elements using a string such as l$a$b If I have a list I can set a sub-element as follows on the command line: people=list() people$tom$hair=brown people But what if I have a string containing the name of the sub-element that I want to access? subel= people$tom$hair get(subel) # returns error assign(subel,red) # silent but doesn't change list people See what happens when people-assign(subel, red) but I think this is not what Greg wanted; the above just assigns red to object 'people' (i.e., check `str(assign(subel, red))'). If I understood correctly, the following could be of help: people - list() people$tom$hair - brown people # subel - people$tom$hair eval(parse(text = subel)) eval(parse(text = paste(subel, - 'red'))) people Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax: +32/(0)16/337015 Web: http://med.kuleuven.be/biostat/ http://www.student.kuleuven.be/~m0390867/dimitris.htm HTH Petr The attempts above using assign/get won't do what I am trying to do [nor according to the help should they]. I would be very grateful for any suggestions. Many thanks, Greg. -- Gregory Jefferis, PhD and: Research Fellow Department of Zoology St John's College University of Cambridge Cambridge Downing Street CB2 1TP Cambridge, CB2 3EJ United Kingdom Lab Tel: +44 (0)1223 336683 Office: +44 (0)1223 339899 Lab Fax: +44 (0)1223 336676 http://www.zoo.cam.ac.uk/zoostaff/jefferis.html [EMAIL PROTECTED] __ 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 Petr Pikal [EMAIL PROTECTED] __ 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 Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm Petr Pikal [EMAIL PROTECTED] __ 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] Install R 2.3.1 on SUSE Linux
[EMAIL PROTECTED] I am new to Linux, and I am trying to install R 2.3.1 on SUSE Linux 10.0. The RPM installer, YAST, states that I need libgfortran.so.0. There is a SuSE 10.0 machine somewhere. Yes: I installed R on it, and it works well there: $ rpm -qf /usr/lib/libgfortran.so gcc-fortran-4.0.2_20050901-3 -- François Pinard http://pinard.progiciels-bpi.ca __ 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] Garch Warning
see inline Arun Kumar Saha wrote: Dear Spencer, Can you be more clear on your suggestion? On 6/15/06, *Spencer Graves* [EMAIL PROTECTED] mailto:[EMAIL PROTECTED] wrote: Your example is not reproducible. Can you find a simple example, preferably obtained either from a simple simulation or from a minor modification of an example included in some of the standard documentation. This listserve sometimes refuses to accept attachments, so I would not suggest you attach a data set unless it was quite small. You could, for example, try reducing the size of your data set, e.g., using a binary search to find the smallest number of observations that would still generate the same error message. Alternatively, the posting guide (www.R-project.org/posting-guide.html) includes the following: 'When providing examples, it is best to give an R command that constructs the data, as in the matrix() expression above. For more complicated data structures, dump(x, file=stdout()) will print an expression that will recreate the object x.' Have you tried walking through the code line by line after debug(garch)? I know I've discussed this issue on this listserve before. Therefore, I tried RSiteSearch(graves debug). This produced 77 hits, many of which were not relevant. The following seemed to be: http://finzi.psych.upenn.edu/R/Rhelp02a/archive/68215.html;. After you've tried that, if you'd still like help from this listserve, please submit a simple, self-contained / reproducible example, as suggested the posting guide! www.R-project.org/posting-guide.html http://www.R-project.org/posting-guide.html. hope this helps. spencer graves Arun Kumar Saha wrote: Dear all R-users, I wanted to fit a Garch(1,1) model to a dataset by: garch1 = garch( na.omit(dat)) But I got a warning message while executing, which is: Warning message: NaNs produced in: sqrt(pred$e) The garch parameters that I got are: garch1 Call: garch(x = na.omit(dat)) Coefficient(s): a0 a1 b1 1.212e-04 1.001e+00 1.111e-14 Can any one please tell me that why got this message? What is the remedy? Thanks and Regards [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailto:R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Repost: Estimation when interaction is present: How do I get get the parameters from nlme?
Hi John, I think a solution is to 1. recode A and B as a single factor, AB, with four levels, 2. define each fixed effect as a function of AB minus the intercept (e.g. ed50 ~ as.factor(AB)-1). 3. extract the tTable as a data.frame with summary(model)$tTable. I will be interested to see what other folks suggest. Cheers, Hank and then run the Probably not the best, nor the worst solution, might be to recode A and B On Jun 15, 2006, at 11:36 AM, John S. Walker wrote: Gday, This is a repost since I only had one direct reply and I remain mystified- This may be stupidity on my part but it may not be so simple. In brief, my problem is I'm not sure how to extract parameter values/effect sizes from a nonlinear regression model with a significant interaction term. My data sets are dose response curves (force and dose) for muscle that also have two treatments applied Treatment A (A- or A+) and Treatment B (B-/B+). A single muscle was used for each experiment - a full dose response curve and one treatment from the matrix A*B (A-/B-, A+/B-, A-/B+ and A+,B+). There are 8 replicates for each combination of treatments We fit a dose response curve to each experiment with parameters upper, ed50 and slope; we expect treatment A to change upper and ed50. We want to know if treatment B blocks the effect of treatment A and if so to what degree. This is similar to the Ludbrook example in Venables and Ripley, however they only had one treatment and I have two. my approach The dataframe is structured like this: expt treatA treatB doseforce. 1 - - 0.1 20 1 - - 0.2 40 ... 4 + + 0.1 20 4 + I used a groupedData object: mydata=groupedData(force ~ dose | expt) I used an nlme obect to model the data as follows (pseudocode): myfit.nlme - nlme(force ~ ss_tpl(dose, upper, ed50,slope), fixed=list(ed50~factor(treatA)*factor(treatB))) The function ss_tpl is a properly debugged and fully functional selfstarting three parameter logistic function that I wrote- no problem here. In my analysis I also included fixed terms for the other fit parameters; upper and slope, but my main problem is with the ed50 so that's all I've included here. Running an anova on the resulting object (anova(myfit.nlme) I found the A -/B- (control) to be significantly different from zero, treatment A was significantly different, treatment B had no significant effect and there was a significant interaction between treatment A and treatment B. The interaction term is likely to be real. The treatments are on sequential steps in a pathway and treatment B may be blocking the effect of treatment A, i.e. treatment B alone has no effect because it blocks a pathway that is not active, treatment A reduces force via this pathway and treament B therefore blocks the effect of treatment A when used together. From what I understand, please correct me if I'm wrong, the parameter estimates from summary(model.nlme) are not correct for main effects if a significant interaction is present. For example in my data treatment B alone has no signifcant effect in the anova but the interaction term A:B is significant. I believe The summary estimate for B is the estimate across all levels of A. What I want to do is pull out the estimate for B when A is not present. I suppose I can do it manually from the list of coefficients from nls or fit a oneway model with treatment levels A, B, AB. But I was kind of hoping there was some extractor function. The reason I need this is that the co-authors want to include a table of parameter values with std errs or confidence intervals ala: Treat upper ed50slope A-/B- x x x - shows value for comparison to control studies A+/B- x x x -Shows A is working0 A-/B+ x x x - Shows B has no effect alone A+/B + x x x -shows B blocks A (not necessarily total) So back to my question,How do I extract estimates of the parameters from my model object for a specific combination of factors including the interaction term. i.e. what is the ed50 (and std err) for A-/B-, A+/B-, A-/B+, A+/B + ? I think this is a fair question and one that many biomedical scientists would need. __ 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 Dr. M. Hank H. Stevens, Assistant Professor 338 Pearson Hall Botany Department Miami University Oxford, OH 45056 Office: (513) 529-4206 Lab: (513) 529-4262 FAX: (513) 529-4243 http://www.cas.muohio.edu/~stevenmh/
Re: [R] help with table partition
Hi, thank you all for the help.
The split function works very well.
I have an additional question. If I have a matrix of prices (row = 30,
col = 2) in matrix P
P:
30 40
31.5 42
32 43
What is the quickest way to get a new matrix, where each entry is the
ln(Pt/Pt-1)?
I have no prob doing this using a loop, but that might not be most
efficient if my table is huge. Moreover, I've read the apply/lapply
functions, but I could not get the right parameters to use.
Thank you all for help.
K.
-Original Message-
From: Petr Pikal [mailto:[EMAIL PROTECTED]
Sent: Thursday, June 15, 2006 10:35 AM
To: Wong, Kim
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] help with table partition
Hi
maybe ?split and ?t is what you want
mat-matrix(rnorm(1000), 100,10)
mat.s-split(data.frame(mat), rep(1:5, each=20))
#splits mat to list with 5 eqal submatrices
lapply(mat.s,t)
# transpose matrices in list
gives you a list of transposed tables, which is probably better than
separate tables. Just change rep(1:5,each=20) to rep (1:170,
each=366).
or
a quicker one without data frame
mat - matrix(rnorm(62220*73), 62220,73)
dim(mat) - c(366,73,170)
mat.i - array(0,dim=c(73,366,170))
for (i in 1:170) mat[ , , i] - t(mat[ , , i])
HTH
Petr
On 15 Jun 2006 at 9:38, Wong, Kim wrote:
Date sent: Thu, 15 Jun 2006 09:38:29 -0400
From: Wong, Kim [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Subject:[R] help with table partition
Hi,
I have a test_table where the dim is 62220 by 73 (row by col)
I would like to partition the rows into 170 equal parts (170 tables
where each is of dim 366 by 73), and rearrange them horizontally. The
source codes I have:
for (i in 1:170) {
c = cbind(c,test_table[(367*i+1):(367*(i+1)),2:73]);
}
Unfortunately, using for loop and cbind for a table of this size
causes long running time. What is the most efficient way to get the
table that I want?
Thanks for any help.
K.
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Re: [R] Saving R graphics with version 2.3.1
In response to the OP's query, we use numerous R scripts that save graphics as PNG and EPS through savePlot() commands. We have not experienced any problems with R 2.3.1. MHP on 6/15/2006 9:25 AM Duncan Murdoch said the following: [EMAIL PROTECTED] wrote: HI ! I´ve installed the latest R version (2.3.1) and I´ve had problems to save my graphics as JPEG, PNG and the others available formats. R saves the file with no extension. Then, I have to open this file using an image visualizer and save it as PNG, for instance. -- Michael Prager, Ph.D. Southeast Fisheries Science Center NOAA Center for Coastal Fisheries and Habitat Research Beaufort, North Carolina 28516 ** Opinions expressed are personal, not official. No ** official endorsement of any product is made or implied. __ 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] Name of a column
Hello, My problem is quite simply, but I didn't find any solution... I have a vector : truc longueur30 longueur40 longueur50 longueur60 longueur70 longueur80 longueur90 34 FALSE FALSE FALSE FALSE TRUE FALSE FALSE I would like to have the name of the column where there is TRUE. colnames(truc) [1] longueur30 longueur40 longueur50 longueur60 longueur70 [6] longueur80 longueur90 truc[truc == T] [1] TRUE colnames(truc[truc == T]) NULL How can I do it ? Thank you for your help. -- David [[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] Compact sums in functions definitions
I'm trying to make more compact the definition of a function as for example: f - function(x) 2/x+3/x by simply defining the array of coefficients arr = c(2,3) and setting: g - function(x) sum((arr/x)) Everything seems to work fine because the values returned by f and g result coincident for different values of their argument, but when I try to plot the function g using: x = seq(-1,1,.01) plot(x,g(x)) I receive the errors/warnings: Error in xy.coords(x, y, xlabel, ylabel, log) : 'x' and 'y' lengths differ In addition: Warning message: longer object length is not a multiple of shorter object length in: b/t Execution halted Any idea about that? Thanks. Alessandro -- Alessandro Antonucci Dalle Molle Institute for Artificial Intelligence (IDSIA) at Idsiae-mail: [EMAIL PROTECTED] Galleria 2 web: idsia.ch/~alessandro Via Cantonale mobile: +39 339-567-23-28 CH-6928 tel: +41 58-666-66-69 Manno - Lugano fax: +41 58-666-66-61 Switzerland skype: alessandro.antonucci __ 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] Access and assign list sub-elements using a string suchas l$a$b
Petr Pikal wrote: Hi yes you are correct, I remembered there is something with eval from older posts but did not find a connection to parse from eval help page. Shouldn't there be a link? Or even an example? would be a good thing to do (there only is a link from parse to eval). after all eval(parse(text = some_string)) is what many users (which might come from matlab/octave where this _would_ suffice) want when they try out eval(some_string)). and a standard eval(parse(text=...) example on the eval manpage (where most people probably would look, would be very good. Best regards Petr On 15 Jun 2006 at 17:21, Dimitris Rizopoulos wrote: From: Dimitris Rizopoulos [EMAIL PROTECTED] To: Petr Pikal [EMAIL PROTECTED], [EMAIL PROTECTED] Copies to:r-help@stat.math.ethz.ch Subject: Re: [R] Access and assign list sub-elements using a string suchas l$a$b Date sent:Thu, 15 Jun 2006 17:21:26 +0200 - Original Message - From: Petr Pikal [EMAIL PROTECTED] To: Gregory Jefferis [EMAIL PROTECTED] Cc: r-help@stat.math.ethz.ch Sent: Thursday, June 15, 2006 4:56 PM Subject: Re: [R] Access and assign list sub-elements using a string suchas l$a$b Hi very, very close On 15 Jun 2006 at 13:27, Gregory Jefferis wrote: Date sent: Thu, 15 Jun 2006 13:27:05 +0100 From: Gregory Jefferis [EMAIL PROTECTED] To: [EMAIL PROTECTED] [EMAIL PROTECTED] Forwarded to: r-help@stat.math.ethz.ch Forwarded by: Gregory Jefferis [EMAIL PROTECTED] Date forwarded: Thu, 15 Jun 2006 14:54:13 +0100 Subject:[R] Access and assign list sub-elements using a string such as l$a$b If I have a list I can set a sub-element as follows on the command line: people=list() people$tom$hair=brown people But what if I have a string containing the name of the sub-element that I want to access? subel= people$tom$hair get(subel) # returns error assign(subel,red) # silent but doesn't change list people See what happens when people-assign(subel, red) but I think this is not what Greg wanted; the above just assigns red to object 'people' (i.e., check `str(assign(subel, red))'). If I understood correctly, the following could be of help: people - list() people$tom$hair - brown people # subel - people$tom$hair eval(parse(text = subel)) eval(parse(text = paste(subel, - 'red'))) people Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax: +32/(0)16/337015 Web: http://med.kuleuven.be/biostat/ http://www.student.kuleuven.be/~m0390867/dimitris.htm HTH Petr The attempts above using assign/get won't do what I am trying to do [nor according to the help should they]. I would be very grateful for any suggestions. Many thanks, Greg. -- Gregory Jefferis, PhD and: Research Fellow Department of Zoology St John's College University of Cambridge Cambridge Downing Street CB2 1TP Cambridge, CB2 3EJ United Kingdom Lab Tel: +44 (0)1223 336683 Office: +44 (0)1223 339899 Lab Fax: +44 (0)1223 336676 http://www.zoo.cam.ac.uk/zoostaff/jefferis.html [EMAIL PROTECTED] __ 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 Petr Pikal [EMAIL PROTECTED] __ 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 Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm Petr Pikal [EMAIL PROTECTED] __ 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] Simulation / bayesian model
Hi, (In the bayesian model, this program makes estimate the value of theta (tn) with the empirical method.) I want to know how one can repeat this program 100 times, before to have 100 estimated values of tn. This is the program: a-1 theta-rexp(50,a) x-rpois(length(theta),theta) t-rexp(a) xn-rpois(1,t) tn-(1+xn)*mean(x)/(1+mean(x)) tn Thank you very much. __ [[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] R with tcl/tk 8.5
Hi List, Is it possible to tell R to use tcl/tk 8.5? My R package seems to depend on libtcl8.4.so. However, in Windows it seems to be possible to set TCL_LIBRARY and MY_TCLTK. Is there something similar possible in Linux? I have installed: $ rpm -q R R-2.3.1-1.fc5 $ locate libtcl8 /usr/lib/libtcl8.4.so /usr/local/lib/libtcl8.5.so __ 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] Name of a column
X - as.logical(rep(c(0,1), each=13)) names(X) - LETTERS which(X) N O P Q R S T U V W X Y Z 14 15 16 17 18 19 20 21 22 23 24 25 26 names(which(X)) [1] N O P Q R S T U V W X Y Z ?which David Hajage wrote: Hello, My problem is quite simply, but I didn't find any solution... I have a vector : truc longueur30 longueur40 longueur50 longueur60 longueur70 longueur80 longueur90 34 FALSE FALSE FALSE FALSE TRUE FALSE FALSE I would like to have the name of the column where there is TRUE. colnames(truc) [1] longueur30 longueur40 longueur50 longueur60 longueur70 [6] longueur80 longueur90 truc[truc == T] [1] TRUE colnames(truc[truc == T]) NULL How can I do it ? Thank you for your help. -- Chuck Cleland, Ph.D. NDRI, Inc. 71 West 23rd Street, 8th floor New York, NY 10010 tel: (212) 845-4495 (Tu, Th) tel: (732) 512-0171 (M, W, F) fax: (917) 438-0894 __ 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] flipping a plot vertically?
Hi, This seems like an obvious question but I can't find the answer in the par help document --- I'd like to make a plot where the 0,0 point is in the top left of the screen rather than bottom left... . This is just a regular plot that is flipped vertically -- i.e. the x-values will run along the top of the plot and the y-values will go from the top left (0,0) to the bottom left (0,ymax) -- I assume there is a simple way to do this but I can't find it. Any suggestions? Thanks! Tim __ 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] dpss.taper for spectral estimation
Hello R users, I'm running R version 2.2.1 with a fedora core 4 linux os. I want to use dpss (discrete prolate spheroidal sequences) tapers for consistent spectral estimation. Such tapers can be computed using the dpss.taper() function from the waveslim package (Brandon Whitcher 2004). The function computes the tapers, but I didn't manage to get the eigenvalues which are associated with. The documentation indicates 6 values for the output of this function and I only get one. How can I get these eigenvalues ? Thanks in advance for your help, Tristan __ 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] bubbleplot for matrix
This works, though I'm not sure why symbols() complains about
axes=FALSE while fulfilling the request.
a - matrix(sample(1:5,100,replace=TRUE),nrow=10)
rownames(a) - c(aed,fde,fda,yxj,ijk,ddd,gcd,sbe,adc,asd)
colnames(a) - c(aed,fde,fda,yxj,ijk,ddd,gcd,sbe,adc,asd)
x - y - z - vector()
for (i in 1:nrow(a)) {
x - c(x,rep(rownames(a)[i],ncol(a)))
y - c(y,colnames(a))
z - c(z,a[i,])
}
xp - as.numeric(as.factor(x))
yp - as.numeric(as.factor(y))
symbols(xp,yp,z,inches=0.2,bg=khaki,axes=FALSE)
axis(1,at=1:length(unique(x)),labels=sort(unique(x)))
axis(2,at=1:length(unique(y)),labels=sort(unique(y)))
box()
text(xp,yp,labels=z)
On 6/15/06, Albert Vilella [EMAIL PROTECTED] wrote:
Thanks Bogdan for the reply,
I almost got it working, but in my case, the rownames and colnames are
strings, not numbers, and I guess that this is a problem when using your
snippet:
a -
matrix(sample(1:5,100,replace=TRUE),nrow=10,dimnames=list(1:10,5*1:10))
rownames(a) =
c(aed,fde,fda,yxj,ijk,ddd,gcd,sbe,adc,asd)
colnames(a) =
c(aed,fde,fda,yxj,ijk,ddd,gcd,sbe,adc,asd)
x - y - z - vector()
for (i in 1:nrow(a)) {
x - c(x,rep(rownames(a)[i],ncol(a)))
y - c(y,colnames(a))
z - c(z,a[i,])
}
symbols(as.numeric(x),as.numeric(y),z,inches=0.2,bg=khaki)
text(as.numeric(x),as.numeric(y),labels=z)
symbols(as.numeric(x),as.numeric(y),z,inches=0.2,bg=khaki)
Error in plot.window(xlim, ylim, log, asp, ...) :
need finite 'xlim' values
In addition: Warning messages:
1: NAs introduced by coercion
2: NAs introduced by coercion
3: no finite arguments to min; returning Inf
4: no finite arguments to max; returning -Inf
5: no finite arguments to min; returning Inf
6: no finite arguments to max; returning -Inf
Any guess?
Thanks in advance,
Albert.
On Wed, 2006-06-14 at 16:47 -0400, bogdan romocea wrote:
Here's an example. By the way, I find that it's more convenient (where
applicable) to keep the data in 3 vectors/factors rather than one
matrix/data frame.
a - matrix(sample(1:5,100,replace=TRUE),nrow=10,dimnames=list(1:10,5*1:10))
x - y - z - vector()
for (i in 1:nrow(a)) {
x - c(x,rep(rownames(a)[i],ncol(a)))
y - c(y,colnames(a))
z - c(z,a[i,])
}
symbols(as.numeric(x),as.numeric(y),z,inches=0.2,bg=khaki)
text(as.numeric(x),as.numeric(y),labels=z)
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Albert Vilella
Sent: Tuesday, June 13, 2006 7:11 AM
To: r-help@stat.math.ethz.ch
Subject: [R] bubbleplot for matrix
Hi all,
I would like to ask if it is possible to use bubbleplot for a 20x20
matrix, instead of a dataframe with factors in columns.
The idea would be to get a tabular representation with bubbles like in
Rnews_2006_2 article, which look very nice.
Thanks in advance,
Albert.
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Re: [R] Simulation / bayesian model
Ahmed Elhabti wrote:
Hi,
(In the bayesian model, this program makes estimate the value of theta (tn)
with the empirical method.)
I want to know how one can repeat this program 100 times, before to have
100 estimated values of tn.
This is the program:
a-1
theta-rexp(50,a)
x-rpois(length(theta),theta)
t-rexp(a)
xn-rpois(1,t)
tn-(1+xn)*mean(x)/(1+mean(x))
tn
my.tn - function(){
a - 1
theta - rexp(50,a)
x - rpois(length(theta),theta)
t - rexp(a)
xn - rpois(1,t)
tn - (1+xn)*mean(x)/(1+mean(x))
return(tn)
}
replicate(100, my.tn())
[1] 0.5575221 0.5192308 1.0099010 1.2954545
[5] 0.4949495 1.0099010 0.5238095 0.4791667
[9] 1.8494624 1.4693878 0.5098039 0.8235294
[13] 0.5495495 0.4565217 0.5726496 1.4042553
[17] 0.5798319 2.2826087 1.0909091 1.000
[21] 0.5327103 0.4623656 1.0291262 0.7804878
[25] 1.4536082 0.9690722 0.9795918 0.889
[29] 0.5049505 0.5762712 0.4680851 0.4623656
[33] 0.5327103 1.3516484 0.9247312 0.4505495
[37] 0.889 0.5238095 1.7394958 0.5614035
[41] 0.5412844 0.5454545 1.0099010 0.4382022
[45] 0.444 0.500 0.9898990 0.444
[49] 0.9795918 1.778 0.5192308 1.0196078
[53] 0.4505495 0.5454545 1.4693878 2.2300885
[57] 0.5238095 0.5098039 1.8021978 1.9591837
[61] 0.4047619 1.0654206 0.5283019 0.4680851
[65] 0.5652174 2.5490196 1.1735537 1.0566038
[69] 2.1308411 2.5728155 0.4186047 1.0099010
[73] 0.4897959 1.3695652 0.4897959 1.0825688
[77] 0.5145631 0.4382022 1.0476190 0.5145631
[81] 0.5535714 0.4565217 0.5238095 0.4680851
[85] 0.4736842 0.5098039 0.5327103 1.9381443
[89] 0.5454545 0.5238095 0.5412844 0.4117647
[93] 0.444 1.1379310 0.5327103 1.0990991
[97] 0.4382022 0.444 1.1803279 0.5967742
?replicate
Thank you very much.
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--
Chuck Cleland, Ph.D.
NDRI, Inc.
71 West 23rd Street, 8th floor
New York, NY 10010
tel: (212) 845-4495 (Tu, Th)
tel: (732) 512-0171 (M, W, F)
fax: (917) 438-0894
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Re: [R] Compact sums in functions definitions
Hi Alessandro, The problem is that arr/x isn't quite doing what you thought it was. arr / x is something like c(arr[1] / x[1], arr[2] / x[2], arr[1] / x[3], ...) What I mean is, it divides arr[i] by x[i] and tries to lengthen arr by repeating itself (because it is shorter than x). It warns in case length(x) is not a multiple of length(arr). (or the other way around if arr was longer than x) I made a nice one for you: g - function(x) rowSums( t(t(1/x)) %*% arr ) It is likely someone else can do it much nicer (shorter), my R knowledge still has to be increased ... JeeBee. On Thu, 15 Jun 2006 18:55:15 +0200, Alessandro Antonucci wrote: I'm trying to make more compact the definition of a function as for example: f - function(x) 2/x+3/x by simply defining the array of coefficients arr = c(2,3) and setting: g - function(x) sum((arr/x)) Everything seems to work fine because the values returned by f and g result coincident for different values of their argument, but when I try to plot the function g using: x = seq(-1,1,.01) plot(x,g(x)) I receive the errors/warnings: Error in xy.coords(x, y, xlabel, ylabel, log) : 'x' and 'y' lengths differ In addition: Warning message: longer object length is not a multiple of shorter object length in: b/t Execution halted Any idea about that? Thanks. Alessandro __ 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] flipping a plot vertically?
[Tim Brown] This seems like an obvious question but I can't find the answer in the par help document --- I'd like to make a plot where the 0,0 point is in the top left of the screen rather than bottom left... . [...] Any suggestions? You might retry your plot, adding an ylim=c(HIGHEST, LOWEST) argument, that is, listing the maximum before the minimum. For example: plot(1:10, ylim=c(10, 1)) -- François Pinard http://pinard.progiciels-bpi.ca __ 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 with table partition
apply(log(P), 2, diff)
David L. Reiner
Rho Trading Securities, LLC
Chicago IL 60605
312-362-4963
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Wong, Kim
Sent: Thursday, June 15, 2006 11:30 AM
To: Petr Pikal; Jacques VESLOT
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] help with table partition
Hi, thank you all for the help.
The split function works very well.
I have an additional question. If I have a matrix of prices (row = 30,
col = 2) in matrix P
P:
30 40
31.5 42
32 43
What is the quickest way to get a new matrix, where each entry is the
ln(Pt/Pt-1)?
I have no prob doing this using a loop, but that might not be most
efficient if my table is huge. Moreover, I've read the apply/lapply
functions, but I could not get the right parameters to use.
Thank you all for help.
K.
-Original Message-
From: Petr Pikal [mailto:[EMAIL PROTECTED]
Sent: Thursday, June 15, 2006 10:35 AM
To: Wong, Kim
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] help with table partition
Hi
maybe ?split and ?t is what you want
mat-matrix(rnorm(1000), 100,10)
mat.s-split(data.frame(mat), rep(1:5, each=20))
#splits mat to list with 5 eqal submatrices
lapply(mat.s,t)
# transpose matrices in list
gives you a list of transposed tables, which is probably better than
separate tables. Just change rep(1:5,each=20) to rep (1:170,
each=366).
or
a quicker one without data frame
mat - matrix(rnorm(62220*73), 62220,73)
dim(mat) - c(366,73,170)
mat.i - array(0,dim=c(73,366,170))
for (i in 1:170) mat[ , , i] - t(mat[ , , i])
HTH
Petr
On 15 Jun 2006 at 9:38, Wong, Kim wrote:
Date sent: Thu, 15 Jun 2006 09:38:29 -0400
From: Wong, Kim [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Subject:[R] help with table partition
Hi,
I have a test_table where the dim is 62220 by 73 (row by col)
I would like to partition the rows into 170 equal parts (170 tables
where each is of dim 366 by 73), and rearrange them horizontally. The
source codes I have:
for (i in 1:170) {
c = cbind(c,test_table[(367*i+1):(367*(i+1)),2:73]);
}
Unfortunately, using for loop and cbind for a table of this size
causes long running time. What is the most efficient way to get the
table that I want?
Thanks for any help.
K.
-
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Re: [R] Compact sums in functions definitions
Slightly better: h - function(x) rowSums( as.matrix(1/x) %*% arr ) On Thu, 15 Jun 2006 19:41:12 +0200, JeeBee wrote: Hi Alessandro, The problem is that arr/x isn't quite doing what you thought it was. arr / x is something like c(arr[1] / x[1], arr[2] / x[2], arr[1] / x[3], ...) What I mean is, it divides arr[i] by x[i] and tries to lengthen arr by repeating itself (because it is shorter than x). It warns in case length(x) is not a multiple of length(arr). (or the other way around if arr was longer than x) I made a nice one for you: g - function(x) rowSums( t(t(1/x)) %*% arr ) It is likely someone else can do it much nicer (shorter), my R knowledge still has to be increased ... JeeBee. On Thu, 15 Jun 2006 18:55:15 +0200, Alessandro Antonucci wrote: I'm trying to make more compact the definition of a function as for example: f - function(x) 2/x+3/x by simply defining the array of coefficients arr = c(2,3) and setting: g - function(x) sum((arr/x)) Everything seems to work fine because the values returned by f and g result coincident for different values of their argument, but when I try to plot the function g using: x = seq(-1,1,.01) plot(x,g(x)) I receive the errors/warnings: Error in xy.coords(x, y, xlabel, ylabel, log) : 'x' and 'y' lengths differ In addition: Warning message: longer object length is not a multiple of shorter object length in: b/t Execution halted Any idea about that? Thanks. Alessandro __ 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] log returns (was: Re: help with table partition)
Please use a descriptive subject and not tag onto a prior thread
for new topics.
Assuming the first row is time 1 and the second row is time 2 and
so on try:
diff(log(P))
On 6/15/06, Wong, Kim [EMAIL PROTECTED] wrote:
Hi, thank you all for the help.
The split function works very well.
I have an additional question. If I have a matrix of prices (row = 30,
col = 2) in matrix P
P:
30 40
31.5 42
32 43
What is the quickest way to get a new matrix, where each entry is the
ln(Pt/Pt-1)?
I have no prob doing this using a loop, but that might not be most
efficient if my table is huge. Moreover, I've read the apply/lapply
functions, but I could not get the right parameters to use.
Thank you all for help.
K.
-Original Message-
From: Petr Pikal [mailto:[EMAIL PROTECTED]
Sent: Thursday, June 15, 2006 10:35 AM
To: Wong, Kim
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] help with table partition
Hi
maybe ?split and ?t is what you want
mat-matrix(rnorm(1000), 100,10)
mat.s-split(data.frame(mat), rep(1:5, each=20))
#splits mat to list with 5 eqal submatrices
lapply(mat.s,t)
# transpose matrices in list
gives you a list of transposed tables, which is probably better than
separate tables. Just change rep(1:5,each=20) to rep (1:170,
each=366).
or
a quicker one without data frame
mat - matrix(rnorm(62220*73), 62220,73)
dim(mat) - c(366,73,170)
mat.i - array(0,dim=c(73,366,170))
for (i in 1:170) mat[ , , i] - t(mat[ , , i])
HTH
Petr
On 15 Jun 2006 at 9:38, Wong, Kim wrote:
Date sent: Thu, 15 Jun 2006 09:38:29 -0400
From: Wong, Kim [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Subject:[R] help with table partition
Hi,
I have a test_table where the dim is 62220 by 73 (row by col)
I would like to partition the rows into 170 equal parts (170 tables
where each is of dim 366 by 73), and rearrange them horizontally. The
source codes I have:
for (i in 1:170) {
c = cbind(c,test_table[(367*i+1):(367*(i+1)),2:73]);
}
Unfortunately, using for loop and cbind for a table of this size
causes long running time. What is the most efficient way to get the
table that I want?
Thanks for any help.
K.
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Re: [R] flipping a plot vertically?
That works great. Thanks. Any idea how to get the x axis numbers to go along the top instead of the bottom? tim At 11:46 AM 6/15/2006, you wrote: [Tim Brown] This seems like an obvious question but I can't find the answer in the par help document --- I'd like to make a plot where the 0,0 point is in the top left of the screen rather than bottom left... . [...] Any suggestions? You might retry your plot, adding an ylim=c(HIGHEST, LOWEST) argument, that is, listing the maximum before the minimum. For example: plot(1:10, ylim=c(10, 1)) -- François Pinard http://pinard.progiciels-bpi.ca __ 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 managment
Thank you very much this does the work exactlly. regards Hannes On 6/15/06, Adaikalavan Ramasamy [EMAIL PROTECTED] wrote: If your df contains your data, try tmp - cbind( paste(df[ ,1], df[ ,2], sep=:), paste(df[ ,3], df[ ,4], sep=:) ) tmp - t( apply(tmp, 1, sort) ) out - data.frame( do.call(rbind, strsplit( tmp[,1], split=: )), do.call(rbind, strsplit( tmp[,2], split=: )) ) colnames(out) - colnames(df) out Regards, Adai On Wed, 2006-06-14 at 16:35 +0100, yohannes alazar wrote: First I would really like to thank the mailing list for help I got in the past, as a new to R I am really needing some support on hoe to code the following problem. I am trying to sort some data I have in a big file. The file has 4 columns and 19000 rows. An example of it looks like this:- G 0.892 A 0.108 G 0.883 T 0.117 T 0.5 C 0.5 A 0.617 G 0.383 G 0.925 A 0.075 A 0.967 G 0.033 C 0.883 T 0.117 C 0.633 T 0.367 G 0.95 A 0.05 C 0.742 G 0.258 G 0.875 T 0.125 T 0.167 C 0.833 C 0.792 A 0.208 Columns one and three are alphabets while three and four are their corresponding values. I wanted to sort this data so that my first and third columns are in alphabetic order. For example in the first row the order is G then A. This is not in alphabetic order therefore we swap them along with their values and it becomes: A0.108 G 0.892 Row two looks fine but row three needs the same rearrangement as row one. And the final out put looks like: A 0.108 G 0.892 G 0.883 T 0.117 C 0.5 T 0.5 A 0.617 G 0.383 A 0.075 G 0.925 A 0.967 G 0.033 C 0.883 T 0.117 C 0.633 T 0.367 A 0.05 G 0.95 C 0.742 G 0.258 G 0.875 T 0.125 C 0.833 T 0.167 A 0.208 C 0.792 Please some help with the relevant command names or a technique to code this task. Thank you in advance Regards Hannes [[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 [[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] flipping a plot vertically?
plot(runif(20), xaxt=n, ylim=c(1,0)) axis(side = 3) Tim Brown wrote: That works great. Thanks. Any idea how to get the x axis numbers to go along the top instead of the bottom? tim At 11:46 AM 6/15/2006, you wrote: [Tim Brown] This seems like an obvious question but I can't find the answer in the par help document --- I'd like to make a plot where the 0,0 point is in the top left of the screen rather than bottom left... . [...] Any suggestions? You might retry your plot, adding an ylim=c(HIGHEST, LOWEST) argument, that is, listing the maximum before the minimum. For example: plot(1:10, ylim=c(10, 1)) -- François Pinard http://pinard.progiciels-bpi.ca __ 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 -- Chuck Cleland, Ph.D. NDRI, Inc. 71 West 23rd Street, 8th floor New York, NY 10010 tel: (212) 845-4495 (Tu, Th) tel: (732) 512-0171 (M, W, F) fax: (917) 438-0894 __ 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] flipping a plot vertically?
Any idea how to get the x axis numbers to go along the top instead of the bottom? Use xaxt = 'n' in your plot call (?par for details) to suppress plotting of the axis and then add the axis via a call to axis(). If you do a lot of plotting, you may wish to purchase a copy of Murrell's R GRAPHICS or VR's MASS. At the very least, do read the relevant sections of an Introduction to R and the Reference Manual, as I believe this sort of thing is covered there. -- Bert Gunter Genentech Non-Clinical Statistics South San Francisco, CA The business of the statistician is to catalyze the scientific learning process. - George E. P. Box -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Tim Brown Sent: Thursday, June 15, 2006 11:18 AM To: r-help@stat.math.ethz.ch Subject: Re: [R] flipping a plot vertically? That works great. Thanks. Any idea how to get the x axis numbers to go along the top instead of the bottom? tim At 11:46 AM 6/15/2006, you wrote: [Tim Brown] This seems like an obvious question but I can't find the answer in the par help document --- I'd like to make a plot where the 0,0 point is in the top left of the screen rather than bottom left... . [...] Any suggestions? You might retry your plot, adding an ylim=c(HIGHEST, LOWEST) argument, that is, listing the maximum before the minimum. For example: plot(1:10, ylim=c(10, 1)) -- François Pinard http://pinard.progiciels-bpi.ca __ 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] SSPIR problem
Dear R-Users, I'm using SSPIR package for a spatio-temporal application. Is it possible to modify the structure of the involved matrixes (Fmat, Gmat, Vmat,Wmat)? I want to create a model like this #y(t)=k*theta(t)+epsilon(t) #theta(t)=h*theta(t-1)+eta(t) #epsilon(t) N(0,V) V=sigma2*I #eta(t) N(0,W)W=sigma2_eta where the state variable theta has dimension 1(p=1) and at each time time the observed variable y has dimension equal to d=3. Moreover I want to use white noise errors. But with the command SS(y) where y is my nXd data matrix (d=observations at each time, n=numbers of time points) I obtain a different model. Can you help me please? Thanks in advance Michela __ 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] Standard Deviation Distribution - solved
It turns out that I was confused by Weisstein's double use of the variable s.
sigma^2 = N*(observed s)/(N-1), hence constant in the function. So
sddist - function(s,s0,n) {
sig2 - n*s0*s0/(n-1)
2*(n/(2*sig2))^((n-1)/2) / gamma((n-1)/2) * exp(-n*s*s/(2*sig2)) * s^(n-2)
}
gives the plots on the web page I cited.
Thanks to Ed Pegg Jr. at MathWorld for clarifying.
David L. Reiner
Rho Trading Securities, LLC
Chicago IL 60605
312-362-4963
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[R] SSPIR problem
Dear R-Users, I'm using SSPIR package for a spatio-temporal application. Is it possible to modify the structure of the involved matrixes (Fmat, Gmat, Vmat,Wmat)? I want to create a model like this #y(t)=k*theta(t)+epsilon(t) #theta(t)=h*theta(t-1)+eta(t) #epsilon(t) N(0,V) V=sigma2*I #eta(t) N(0,W)W=sigma2_eta where the state variable theta has dimension 1(p=1) and at each time time the observed variable y has dimension equal to d=3. Moreover I want to use white noise errors. But with the command SS(y) where y is my nXd data matrix (d=observations at each time, n=numbers of time points) I obtain a different model. Can you help me please? Thanks in advance Michela __ 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] Problem with Fortran 95 code in R
Dear helpers: I am trying to call some Fortran 95 code with R-2.3.0 but had problem as below. Any pointers would be appreciated. Thanks, Zhu Wang On Linux, R CMD SHLIB rfh.f R dyn.load(rfh.so) Error in dyn.load(x, as.logical(local), as.logical(now)) : unable to load shared library '/home/zwang/rf/rfh.so': /home/zwang/rf/rfh.so: undefined symbol: _gfortran_allocate On Unix, R CMD SHLIB rfh.f R dyn.load(rfh.so) Error in dyn.load(x, as.logical(local), as.logical(now)) : unable to load shared library '/home/zwang/rf/rfh.so': ld.so.1: /usr/local/R-2.3.0/lib/R/bin/exec/R: fatal: relocation error: file /home/zwang/rf/rfh.so: symbol __f90_allocate2: referenced symbol not found __ 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] Histbackback and adding color
I am trying to plot two histograms using histbackback and then adding
color to each of the samples. The code I have seen on the R graph
gallery suggests the following method of using histbackback
require(Hmisc)
age - rnorm(1000,50,10)
sex - sample(c('female','male'),1000,TRUE)
out - histbackback(split(age, sex), probability=TRUE, xlim=c(-.06,.06),
main = 'Back to Back Histogram')
#! just adding color
barplot(-out$left, col=red , horiz=TRUE, space=0, add=TRUE,
axes=FALSE)
barplot(out$right, col=blue, horiz=TRUE, space=0, add=TRUE,
axes=FALSE)
I get the following message when I execute the snippet to add color
Warning messages:
1: parameter add could not be set in high-level plot() function
2: parameter add could not be set in high-level plot() function
Any help would be appreciated.
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Re: [R] flipping a plot vertically?
__ 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] more function in R?
I'm an R newbie and I just have a simple question. I'm using interactive R on a linux box and I'm trying to find out if there's a more or a less function. The data I want to look at is larger then my screen size and all I end up seeing is the last fews line of data. Is there a way that I can paginate through the data and the results of my summary functions? thanks, Heidi [[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] more function in R?
a - matrix(0, nr=1, nc=5) page(a, method=print) This works on Linux at least. Also, there might be a better way... Gabor On Thu, Jun 15, 2006 at 04:00:40PM -0400, Heidi Savin wrote: I'm an R newbie and I just have a simple question. I'm using interactive R on a linux box and I'm trying to find out if there's a more or a less function. The data I want to look at is larger then my screen size and all I end up seeing is the last fews line of data. Is there a way that I can paginate through the data and the results of my summary functions? thanks, Heidi [[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 -- Csardi Gabor [EMAIL PROTECTED]MTA RMKI, ELTE TTK __ 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] MDS with missing data? (change topic - isomap)
Thanks for the isoMDS pointer. I found one implementation of isomap at http://user.cs.tu-berlin.de/~astro/kmethods though the web page suggests the code may be immature. --- Jari Oksanen [EMAIL PROTECTED] wrote: On Thu, 2006-06-15 at 07:13 +0300, Jari Oksanen wrote: 1) use nonmetric/gradient descent MDS which seems to allow missing data, or Not the isoMDS function in MASS. if N(N-1) is a problem, then nonmetric MDS may not be the solution. Sorry for the wrong information: isoMDS does handle NA. I remembered old times when I looked at the issue, but isoMDS changed since. Fine work! cheers, jari oksanen -- Jari Oksanen -- Dept Biology, Univ Oulu, 90014 Oulu, Finland email [EMAIL PROTECTED], homepage http://cc.oulu.fi/~jarioksa/ __ 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] more function in R?
This works great! Thanks you're a lifesaver. On 6/15/06, Gabor Csardi [EMAIL PROTECTED] wrote: a - matrix(0, nr=1, nc=5) page(a, method=print) This works on Linux at least. Also, there might be a better way... Gabor On Thu, Jun 15, 2006 at 04:00:40PM -0400, Heidi Savin wrote: I'm an R newbie and I just have a simple question. I'm using interactive R on a linux box and I'm trying to find out if there's a more or a less function. The data I want to look at is larger then my screen size and all I end up seeing is the last fews line of data. Is there a way that I can paginate through the data and the results of my summary functions? thanks, Heidi [[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 -- Csardi Gabor [EMAIL PROTECTED]MTA RMKI, ELTE TTK [[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] individual scales in random subset of pairwise distance survey
Hello, I'm curious if anyone has encounted a version of this problem (and it's solution) involving finding a consistent set of scales for subsets of survey data. The goal is to obtain peoples' rankings of pairwise similarity of a large number of items, on a 1..5 scale for example, and average these across people to use as input to MDS: How similar is object A to Bon a 1..5 scale ___ How similar is object A to Con a 1..5 scale ___ etc. Because there are many items, there are N(N-1)/2 pairs, so it is not practical to show every pair to everyone. Showing people the pairs corresponding to random subsets of the objects seems desirable. THe problem is that, a particular random subset might by chance contain objects that would all be rated 5 if one were to see the entire dataset. When ranking pairs from this subset, the scale of 1..5 is different. If we ensure that each pair of people must see some data in common, then one can think about obtaining a set of scales, one for each person, that causes the data that is commonly ranked to have as similar scores as possible, summed across all pairs of people. Please let me know if you know of a standard procedure for this or any similar problems. 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] individual scales in random subset of pairwise distance survey
Perhaps you could try clustering the objects. # generate test data set.seed(1) n - 25 # number of items mat - matrix(0, n, n) # use different labelling scheme if 26 items rownames(mat) - colnames(mat) - letters[1:n] mat[lower.tri(mat)] - sample(5, n * (n-1)/2, TRUE) mat - mat + t(mat) + diag(1, n) # cluster and plot plot(hclust(as.dist(mat))) On 6/15/06, context grey [EMAIL PROTECTED] wrote: Hello, I'm curious if anyone has encounted a version of this problem (and it's solution) involving finding a consistent set of scales for subsets of survey data. The goal is to obtain peoples' rankings of pairwise similarity of a large number of items, on a 1..5 scale for example, and average these across people to use as input to MDS: How similar is object A to Bon a 1..5 scale ___ How similar is object A to Con a 1..5 scale ___ etc. Because there are many items, there are N(N-1)/2 pairs, so it is not practical to show every pair to everyone. Showing people the pairs corresponding to random subsets of the objects seems desirable. THe problem is that, a particular random subset might by chance contain objects that would all be rated 5 if one were to see the entire dataset. When ranking pairs from this subset, the scale of 1..5 is different. If we ensure that each pair of people must see some data in common, then one can think about obtaining a set of scales, one for each person, that causes the data that is commonly ranked to have as similar scores as possible, summed across all pairs of people. Please let me know if you know of a standard procedure for this or any similar problems. 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 __ 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] Touch options
Hi, Does anyone know a package which prices simple one-touch options, say using Rmetrics ? I want something which simply has a discontinuous payoff if an asset touches a single level. No knock-in or knock-out features. Thanks, Tolga __ 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] individual scales in random subset of pairwise distance survey
The diag(1,n) was not needed: set.seed(1) n - 25 mat - matrix(0, n, n) rownames(mat) - colnames(mat) - letters[1:n] mat[lower.tri(mat)] - sample(5, n * (n-1)/2, TRUE) mat - mat + t(mat) plot(hclust(as.dist(mat))) On 6/15/06, Gabor Grothendieck [EMAIL PROTECTED] wrote: Perhaps you could try clustering the objects. # generate test data set.seed(1) n - 25 # number of items mat - matrix(0, n, n) # use different labelling scheme if 26 items rownames(mat) - colnames(mat) - letters[1:n] mat[lower.tri(mat)] - sample(5, n * (n-1)/2, TRUE) mat - mat + t(mat) + diag(1, n) # cluster and plot plot(hclust(as.dist(mat))) On 6/15/06, context grey [EMAIL PROTECTED] wrote: Hello, I'm curious if anyone has encounted a version of this problem (and it's solution) involving finding a consistent set of scales for subsets of survey data. The goal is to obtain peoples' rankings of pairwise similarity of a large number of items, on a 1..5 scale for example, and average these across people to use as input to MDS: How similar is object A to Bon a 1..5 scale ___ How similar is object A to Con a 1..5 scale ___ etc. Because there are many items, there are N(N-1)/2 pairs, so it is not practical to show every pair to everyone. Showing people the pairs corresponding to random subsets of the objects seems desirable. THe problem is that, a particular random subset might by chance contain objects that would all be rated 5 if one were to see the entire dataset. When ranking pairs from this subset, the scale of 1..5 is different. If we ensure that each pair of people must see some data in common, then one can think about obtaining a set of scales, one for each person, that causes the data that is commonly ranked to have as similar scores as possible, summed across all pairs of people. Please let me know if you know of a standard procedure for this or any similar problems. 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 __ 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] plot two graphs with different length of data
thank you, Joerg. I am able to use points() to add the new data in the current plot. Eric On 6/13/06, Joerg van den Hoff [EMAIL PROTECTED] wrote: Eric Hu wrote: Hi I am trying to plot two data set in the same picture window without overlapping with each other. I am using the format plot(x1,y1,x2,y2) but get the following error message: plot(as.numeric(r0[,4]),as.numeric(r0[,7]),as.numeric(r0[,4]),as.numeric(r0[,7][ind[,1]])) Error in plot.window(xlim, ylim, log, asp, ...) : invalid 'ylim' value Can anyone tell me what went wrong? Thanks. Eric __ 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 matplot(cbind(x1,x2), cbind(y1,y2)) might be what you want. (if x1, x2 and y1,y2 are of equal length. otherwise pad the short ones with NAs or use `matplot' with type =n (to get the scaling of the plot right), followed by `plot(x1,y1), lines(x2,y2)') __ 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] running R in batch with stdin input
Hi I have a R script that needs to run a few times for different systems. I use R --no-save r.script for one system. I am trying with no luck to use R CMD BATCH to introduce an stdin input variable for the script. I wonder if anyone can provide the correct usage to put the variable in the command like R CMD BATCH r.script name_variable. Thanks. -Eric In the r.script I have name - readline(/dev/stdin) r0 - read.table(/usr/local/surface/$name/$name_c_r) ... I want to get at the end: name - 1BRS r0 - read.table(/usr/local/surface/1BRS/1BRS_c_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] Touch options
Tolga, You might be interested in the quantlib library: http://quantlib.org/. Currently, a wrapper for R can be found at http://dirk.eddelbuettel.com/code/rquantlib.html. Regards, Jeff Hammerbacher On 6/15/06, Tolga Uzuner [EMAIL PROTECTED] wrote: Hi, Does anyone know a package which prices simple one-touch options, say using Rmetrics ? I want something which simply has a discontinuous payoff if an asset touches a single level. No knock-in or knock-out features. Thanks, Tolga __ 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 [[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] running R in batch with stdin input
?commandArgs On Thu, 2006-06-15 at 16:05 -0700, Eric Hu wrote: Hi I have a R script that needs to run a few times for different systems. I use R --no-save r.script for one system. I am trying with no luck to use R CMD BATCH to introduce an stdin input variable for the script. I wonder if anyone can provide the correct usage to put the variable in the command like R CMD BATCH r.script name_variable. Thanks. -Eric In the r.script I have name - readline(/dev/stdin) r0 - read.table(/usr/local/surface/$name/$name_c_r) ... I want to get at the end: name - 1BRS r0 - read.table(/usr/local/surface/1BRS/1BRS_c_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 __ 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] Change the range of a 'ternaryplot'
Hi all Does someone know how can I alter the range of some variables in a 'ternaryplot' of the 'vcd' package, with the aim of make a zoom in some region of the plot? Thank you in advance Sincerely, Juan Carlos === Juan Carlos Martínez Ovando Banco de México Dirección General de Investigación Económica Cinco de Mayo No. 18, 4° Piso A Col. Centro, Del. Cuauhtémoc 06059, México D.F., MEXICO Tel.: +52 (55) 52.37.25.94 Fax: +52 (55) 52.37.27.03 e-mail: [EMAIL PROTECTED] === 2006, Año del Bicentenario del natalicio del Benemérito de las Américas, Don Benito Juárez García. __ 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
