Re: [R] Longitudinal categorical response data
lmer is not designed for ordered categorical data as yours are. You could take a look at the ordinal package which is designed for this type of data including mixed models (function clmm) which you probably want to use. Best, Rune Den 24/03/2011 21.03 skrev "Rasanga Ruwanthi" : > > Dear List, > > I have some longitudinal data, each patient was followed at times 0, 12, 16, 24 weeks and measure severity of a illness (0-worse, 1-same, 2-better). So, longitudinal response is categorical. I was wondering whether lmer in R can fit a model for this type of data. If so, how we code? Or any other function in R that can fit this type of longitudinal data? Any suggestion would be greatly appreciated. > > Thanks > Ruwanthi > > > > >[[alternative HTML version deleted]] > > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Matching package - Match function
I have figured out the solution to this. It appears that "distance" isn't defined as the difference in propensity scores. Instead, this difference is then scaled by the standard deviation of the predicted dependent-variable values of the corresponding logistic regression, and then squared. Therefore, instead of checking for: > summary(abs(logit.reg$fitted[match$index.treated]-logit.reg$fitted[match$index.control])) > Min. 1st Qu.Median Mean 3rd Qu. Max. 7.453e-13 2.959e-07 5.849e-07 5.842e-07 8.741e-07 1.167e-06 the correct form is: > summary(((logit.reg$fitted[match$index.treated]-logit.reg$fitted[match$index.control])/sd(logit.reg$fitted))^2) Min. 1st Qu.Median Mean 3rd Qu. Max. 4.078e-18 6.426e-07 2.512e-06 3.337e-06 5.609e-06 1.000e-05 which then gives the expected maximum value of 1e-5, as set in the distance.tolerance option. sunny wrote: > > Hi. > > I am using the Matching package for propensity score matching. For each > treated unit, I want to find all control units whose propensity scores lie > within a certain distance from the treated unit. The sample code is as > follows: > >> library(Matching) > >> x <- rnorm(10) >> y <- rnorm(10) >> z <- rbinom(10,1,0.002) > >> logit.reg <- glm(z~x+y,family=binomial(link='logit')) > >> match <- >> Match(Y=NULL,Tr=z,X=logit.reg$fitted,version='fast',ties=TRUE,M=1,distance.tolerance=1e-5) > > According to the function definition > (http://sekhon.berkeley.edu/matching/Match.html): > > "distance.tolerance: This is a scalar which is used to determine if > distances between two observations are different from zero. Values less > than distance.tolerance are deemed to be equal to zero. This option can be > used to perform a type of optimal matching" > > Thus, for each treated unit I should get all control units whose > difference in propensity scores from the treated unit is less than 1e-5. > However, the actual difference between the treated unit's and the control > units' propensity is distributed as follows: > >> summary(abs(logit.reg$fitted[match$index.treated]-logit.reg$fitted[match$index.control])) > Min. 1st Qu.Median Mean 3rd Qu. Max. > 7.453e-13 2.959e-07 5.849e-07 5.842e-07 8.741e-07 1.167e-06 > > -- View this message in context: http://r.789695.n4.nabble.com/Matching-package-Match-function-tp3406144p3406919.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] bwplot [lattice]: how to get different y-axis scales for each row?
Dear expeRts,
How can I get ...
(1) different y-axis scales for each row
(2) while having the same y-axis scales for different columns?
I coulnd't manage to do this with relation="free" [which gives (1) but not (2)].
I also tried relation="sliced", but it did not give the same y-axis scales
within each row (see the fourth row). Further, it "separates" the panels.
Cheers,
Marius
## minimal example:
library(lattice)
## build example data set
dim <- c(100, 6, 2, 3) # n, groups, methods, attributes
dimnames <- list(n=paste("n=", seq_len(100), sep=""),
groups=paste("group=", seq_len(6), sep=""),
methods=paste("method=", seq_len(2), sep=""),
attr=paste("attribute=", seq_len(3), sep=""))
set.seed(1)
data <- rexp(prod(dim))
arr <- array(data=data, dim=dim, dimnames=dimnames)
arr[,2,,] <- arr[,2,,]*10
arr[,4,2,2] <- arr[,4,2,2]*10
z <- abs(sweep(arr, 3, 1))
df <- as.data.frame.table(z, responseName="error")
## box plot
bwplot(error ~ methods | attr * groups, data=df,
as.table=TRUE, notch=TRUE,
scales=list(alternating=c(1,1), tck=c(1,0)))
## with relation="sliced"
bwplot(error ~ methods | attr * groups, data=df,
as.table=TRUE, notch=TRUE,
scales=list(alternating=c(1,1), tck=c(1,0), relation="sliced"))
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[R] bwplot: how to get plotmath labels?
Dear expeRts,
How can I get plotmath-labels in the bwplot below?
As you can see, I couldn't manage to pass the expressions through the dimnames
argument.
Cheers,
Marius
library(lattice)
## data
dim <- c(100, 6, 2, 3)
dimnames <- list(n=paste("n=", seq_len(100), sep=""),
groups=paste("group=", seq_len(6), sep=""),
methods=c(expression(method==1), expression(alpha=1)),
attr=paste("attribute=", seq_len(3), sep=""))
data <- rexp(prod(dim))
arr <- array(data=data, dim=dim, dimnames=dimnames)
z <- abs(sweep(arr, 3, 1))
df <- as.data.frame.table(z, responseName="error")
## box plot
bwplot(error ~ methods | attr * groups, data=df,
as.table=TRUE, outer=FALSE, strip=TRUE, notch=TRUE,
scales=list(alternating=c(1,1), tck=c(1,0)))
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Re: [R] Error: cannot allocate vector of size
Without more detailed information I would say that R runs out of memory...and furthermore: PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. cheers, Paul On 03/25/2011 11:32 PM, mipplor wrote: i run a model ,but i turn out to be like this. but i have run this model days ago and it works well whats going on here? any suggestion. model1<‐siarmcmcdirichletv4(data,sources,tef,concdep=0,50,5) Error: cannot allocate vector of size 2.2 Gb In addition: Warning messages: 1: In matrix(1, ncol = (numsources + numiso) * numgroups, nrow = (siardata$iterations - : Reached total allocation of 895Mb: see help(memory.size) 2: In matrix(1, ncol = (numsources + numiso) * numgroups, nrow = (siardata$iterations - : Reached total allocation of 895Mb: see help(memory.size) 3: In matrix(1, ncol = (numsources + numiso) * numgroups, nrow = (siardata$iterations - : Reached total allocation of 895Mb: see help(memory.size) 4: In matrix(1, ncol = (numsources + numiso) * numgroups, nrow = (siardata$iterations - : Reached total allocation of 895Mb: see help(memory.size) -- View this message in context: http://r.789695.n4.nabble.com/Error-cannot-allocate-vector-of-size-tp3406597p3406597.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Paul Hiemstra, MSc Department of Physical Geography Faculty of Geosciences University of Utrecht Heidelberglaan 2 P.O. Box 80.115 3508 TC Utrecht Phone: +3130 253 5773 http://intamap.geo.uu.nl/~paul http://nl.linkedin.com/pub/paul-hiemstra/20/30b/770 currently @ KNMI paul.hiemstra_AT_knmi.nl __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] linear constrained optimization in R
sammyny caa.columbia.edu> writes:
>
> I am trying to use
> http://rss.acs.unt.edu/Rdoc/library/stats/html/constrOptim.html in R to do
> optimization in R with some given linear constraints but not able to figure
> out how to set up the problem.
>
> For example, I need to maximize $f(x,y) = log(x) + \frac{x^2}{y^2}$ subject
> to constraints $g_1(x,y) = x+y < 1$, $g_2(x,y) = x > 0$ and $g_3(x,y) = y >
> 0$. How do I do this in R? This is just a hypothetical example. Do not worry
> about its structure, instead I am interested to know how to set this up in
> R.
>
> thanks!
>
To get a reasonable solution, avoid coming near to x=0 and y=0. With
x >= 0.0001 and y >= 0.0001 constrOptim() can be called as follows:
constrOptim(c(0.25, 0.25), function(x) -log(x[1])-x[1]^2/x[2]^2, NULL,
matrix(c(-1,1,0, -1,0,1), 3, 2), c(-1, 0.0001, 0.0001))
--Hans Werner
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[R] Effect size in multiple regression
Dear all, is there a convenient way to determine the effect size for a regression coefficient in a multiple regression model? I have a model of the form lm(y ~ A*B*C*D) and would like to determine Cohen's f2 (http://en.wikipedia.org/wiki/Effect_size) for each predictor without having to do it manually. Thanks, Michael Michael Haenlein Associate Professor of Marketing ESCP Europe Paris, France [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Effect size in multiple regression
Hi Michael, You can just fit your model, and then use anova() to get the Sum of Squares. ## fit and store model m <- lm(mpg ~ hp * wt * vs, data = mtcars) ## store ANOVA from model msum <- anova(m) ## divide all Sums of Squares by the sum() of the Sums of Squares msum[["Sum Sq"]]/sum(msum[["Sum Sq"]]) do note that this will be order dependent. HTH, Josh On Sat, Mar 26, 2011 at 4:03 AM, Michael Haenlein wrote: > Dear all, > > is there a convenient way to determine the effect size for a regression > coefficient in a multiple regression model? > I have a model of the form lm(y ~ A*B*C*D) and would like to determine > Cohen's f2 (http://en.wikipedia.org/wiki/Effect_size) for each predictor > without having to do it manually. > > Thanks, > > Michael > > > > Michael Haenlein > Associate Professor of Marketing > ESCP Europe > Paris, France > > [[alternative HTML version deleted]] > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > -- Joshua Wiley Ph.D. Student, Health Psychology University of California, Los Angeles http://www.joshuawiley.com/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] NetCDF - rolling means and StdDev
Hello,
I am trying to come up with a routine to calculate a rolling mean and
standard deviation for three parameters stored in a NetCDF file. The NetCDF
file contains 10 years of daily records for each of the three parameters,
and I need the statistics for each of the three parameters.
Below I include some code illustrating what I've done so far. Obviously this
approach does not get at the problem. It loops thru the data, to get to the
daily data, but stops short of trying to get the running mean/standard
deviation, an assignment that came after I developed the approach shown
here.
Where should the rolling mean and standard deviation functions be integrated
in the script, and what recommendations - approach can you offer to
accomplish this task.
I am running this on a Windows 7 machine with R 2.12.2
Thanks for taking the time to help.
Steve
setwd("C:\\Woodstorks")
getwd()
library(tis) # includes function for leap year
library(ncdf)
woodstorks <- open.ncdf("WoodStork_Eden_2000_2009.nc")
print(woodstorks)
print(v2)
# next two lines create index values for temporal sequences.
DaysToMonth365 <- c(0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334,
365)
DaysToMonth366 <- c(0, 31, 60, 91, 121, 152, 182, 213, 244, 274, 305, 335,
366)
Lyear <- isLeapYear(2000:2009) # returns boolean TRUE FALSE vector for
years in sequence
# next function is used latter in script see line 50
processMonth<-function(x){
print(x)
}
start_year<- 1999
# initiates 1 year before to capture December of correct year, next line
uses the for loop to run thru
# the number of years. Here 3 is used to limit the number used in a testing
mode.
for(t in 1:3) {
currentYear = start_year + t
leapYear <- isLeapYear(currentYear:currentYear) # returns boolean vector
TRUE FALSE for years in sequence
#print(leapYear)
Days<-DaysToMonth365
if(leapYear){
Days<-DaysToMonth366
}
print(Days)
data <- array(data, dim = c(Days[1], 405, 265))
for(d in 1: 226) {
for(y in 1: 405) {
for(x in 1:287) {
Day = 1
counter <- 2 # first element in array
#print(paste("Day = ", d))
if(d >= Days[counter]) {
counter <- counter + 1
#print(paste("here = ", t))
processMonth(data)
data <- array(data, dim = c(Days[1], 405, 265))
Day = 1
print(Day)
}
else
{
index<-c(x,y,d,t)
count<-c(1,1,1,1)
temp<- get.var.ncdf(woodstorks, v2, start=index, count =
count)
i <- (d*405*287) + (y*287) + x #map 3d to 1dimension
data[i] = temp[1]
#print(data[i])
Day <- Day + 1
print(Day)
}
}
}
}
}
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Re: [R] A question on glmnet analysis
Hi Nick, Thanks a lot for your quick response. Sorry for delayed response becasue of time difference between us. (11/03/25 22:40), Nick Sabbe wrote: > > I haven't read all of your code, but at first read, it seems right. > > > > With regard to your questions: > > 1. Am I doing it correctly or not? > > Seems OK, as I said. You could use some more standard code to convert your > > data to a matrix, but essentially the results should be the same. > > Also, lambda.min may be a tad to optimistic: to correct for the reuse of > > data in crossvalidation, one normally uses the minus one se trick (I think > > this is described in the helpfile for glmnet.cv, and that is also present in > > the glmnet.cv return value (lambda.1se if I'm not mistaken)) Actually, fit.1cv$lambda.min [1] 0.036 fit.1$lambda.1se [1] 0.121 coef(fit.1cv, s=fit.1cv$lambda.1se) 16 x 1 sparse Matrix of class "dgCMatrix" 1 (Intercept) -1.1707519 V1 0.000 V2 0.000 V3 0.000 V4 0.000 V5 0.000 V6 0.000 V7 0.000 V8 0.2865651 V9 0.000 V10 0.000 V11 0.000 V12 0.000 V13 0.000 V14 0.000 V15 0.000 Do you mean that I should select one parameter model (only V8 included) described above? > > 2. Which model, I mean lasso or elastic net, should be selected? and > > why? Both models chose the same variables but different coefficient values. > > You may want to read 'the elements of statistical learning' to find some > > info on the advantages of ridge/lasso/elnet compared. Lasso should work fine > > in this relatively low-dimensional setting, although it depends on the > > correlation structure of your covariates. > > Depending on your goals, you may want to refit a standard logistic > > regression with only the variables selected by the lasso: this avoids the > > downward bias that is in (just about) every penalized regression. fit1.glm <- glm(outcome ~ x1+x2+x3+x4, family="binomial", data=MyData) summary(fit1.glm) Call: glm(formula = outcome ~ x1+x2+x3+x4, family = "binomial", data = MyData) Deviance Residuals: Min 1Q Median 3Q Max -1.806 -0.626 -0.438 -0.191 2.304 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -1.855 0.393 -4.72 2.3e-06 *** x11.037 0.5581.86 0.0630 . x21.031 0.3263.16 0.0016 ** x3 -0.591 0.323 -1.83 0.0678 . x40.347 0.2931.18 0.2368 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 114.717 on 103 degrees of freedom Residual deviance: 87.178 on 99 degrees of freedom AIC: 97.18 Number of Fisher Scoring iterations: 5 library(epicalc) logistic.display(fit1.glm) Logistic regression predicting postopDWI_HI2 crude OR(95%CI) adj. OR(95%CI)P(Wald's test) P(LR-test) x1 1 vs 0 2.45 (0.98,6.13) 2.82 (0.95,8.42) 0.063 0.059 x2 2.76 (1.6,4.77) 2.8 (1.48,5.31) 0.002 < 0.001 x3 0.64 (0.36,1.14) 0.55 (0.29,1.04) 0.068 0.051 x4 2.07 (1.28,3.34) 1.41 (0.8,2.51) 0.237 0.234 Log-likelihood = -43.5889 No. of observations = 104 AIC value = 97.1777 The AUC of this glm model is 0.81, which means good predictive ability. My next question, however, is which coefficients value should I select, glmnet model or glm model for presentation? Is it O.K. to select variables by glmnet and present coefficients (odds ratio) by glm? > > > > 3. Is it O.K. to calculate odds ratio by exp(coefficients)? And how can > > you calculate 95% confidence interval of odds ratio? > > Or 95%CI is meaningless in this kind of analysis? > > At this time, confidence intervals for lasso/elnet in GLM settings is an > > open problem (the reason being that the L1 penalty is not differentiable). > > Some 'solutions' exist (bootstrap, for one), but they have all been shown to > > have (statistical) properties that make them - at the least - doubtful. I > > know, because I'm working on this. Short answer: there is no way to do this > > (at this time). Well, thank you for a clear-cut answer. :-) > > > > HTH (and hang on there in Japan), > > Nick Sabbe Your answers helped me very much. Actually, I live in the west part of Japan and fortunately had no damage around neighborhodd by the earthquake. Now we are worried about the nuclear power plant... Best regards, KH __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] A question on glmnet analysis
(11/03/26 0:12), Mike Marchywka wrote: Would you post your data or if you did I missed it? Sorry, I did not. This is patients' data and I have to obtain some permission from my office to send it, so ... I don't have an answer for you but curious as I was interested in similar analyses with mostly continuous prognostic factors and continuous outcome ( survival). My interest was making a blind test to see if analysis could separate drug and treatment patients in a "triple blind" test. The end result would create preidctions ( drug or treeatment for each patient) and that would need to be compared to real assignments and checked for significance. I want to divide my data into training and test data. But I only have 104 observations. That's a problem. Best regards, KH __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] grofit package
Does anyone know if it is possible to get the length of the log phase (the main growth phase) from a grofit logistic model? I can get the lag phase length, maximum growth and max rate of growth, but I also want to know the length of the log phase (ie, for how long do the bacteria grow exponentially?) Thank you for any help in advance [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Error: cannot allocate vector of size
On 25.03.2011 23:32, mipplor wrote: i run a model ,but i turn out to be like this. but i have run this model days ago and it works well whats going on here? any suggestion. If it worked exactly the way before on the same machine, you probably have too huge objects in your workspace. Uwe Ligges model1<‐siarmcmcdirichletv4(data,sources,tef,concdep=0,50,5) Error: cannot allocate vector of size 2.2 Gb In addition: Warning messages: 1: In matrix(1, ncol = (numsources + numiso) * numgroups, nrow = (siardata$iterations - : Reached total allocation of 895Mb: see help(memory.size) 2: In matrix(1, ncol = (numsources + numiso) * numgroups, nrow = (siardata$iterations - : Reached total allocation of 895Mb: see help(memory.size) 3: In matrix(1, ncol = (numsources + numiso) * numgroups, nrow = (siardata$iterations - : Reached total allocation of 895Mb: see help(memory.size) 4: In matrix(1, ncol = (numsources + numiso) * numgroups, nrow = (siardata$iterations - : Reached total allocation of 895Mb: see help(memory.size) -- View this message in context: http://r.789695.n4.nabble.com/Error-cannot-allocate-vector-of-size-tp3406597p3406597.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] bwplot: how to get plotmath labels?
On Mar 26, 2011, at 4:47 AM, Marius Hofert wrote: Dear expeRts, How can I get plotmath-labels in the bwplot below? As you can see, I couldn't manage to pass the expressions through the dimnames argument. It would have been helpful, and may still be necessary, for you to explain more fully what you were attempting to do. See if this effort was informative. Marius library(lattice) ## data dim <- c(100, 6, 2, 3) # did not find the dimnames object useful # at least to the extent I understood the question data <- rexp(prod(dim)) arr <- array(data=data, dim=dim, dimnames=dimnames) z <- abs(sweep(arr, 3, 1)) df <- as.data.frame.table(z, responseName="error") ## box plot bwplot(error ~ methods | attr * groups, data=df, as.table=TRUE, outer=FALSE, strip=TRUE, notch=TRUE, scales=list(alternating=c(1,1), tck=c(1,0))) bwplot(error ~ methods | attr * groups, data=df, as.table=TRUE, outer=FALSE, strip=TRUE, notch=TRUE, scales=list(y=list(alternating=TRUE, tck=c(1,0)), x=list(labels=c( expression(method == 1), expression(alpha == 1)) ) ) ) David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Free license for xlsReadWritePro
Some years ago I travelled through Libya and it's with much emotion that I follow the news. The license below is dedicated to Mohammed Nabbous and will let you run xlsReadWritePro for free (Windows, R 32-bit): library( "xlsReadWritePro" ) xls.lic( action = "register", miscData = list( User = "R.I.P. Mohammed Nabbous", Company = "Let's hope that Libya will become free", Product = "xlsReadWritePro", Date = "2011-03-19", LicenseType = "SingleUser", Count = 1, Key = "76991137f5f554a288e8cee343a24722", WhereToStore = "ProgramFolder" ) ) Info/download at http://swissr.org. Direct download links at http://dl.dropbox.com/u/2602516/swissrpkg/index.html. Support (tracker) at http://dev.swissr.org/. Let's hope that Libya will become free, Hans-Peter __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] bwplot: how to get plotmath labels?
Dear David, many thanks, that was it. Cheers, Marius On 2011-03-26, at 18:04 , David Winsemius wrote: > > On Mar 26, 2011, at 4:47 AM, Marius Hofert wrote: > >> Dear expeRts, >> >> How can I get plotmath-labels in the bwplot below? >> As you can see, I couldn't manage to pass the expressions through the >> dimnames >> argument. > > It would have been helpful, and may still be necessary, for you to explain > more fully what you were attempting to do. See if this effort was informative. > >> Marius >> >> library(lattice) >> >> ## data >> dim <- c(100, 6, 2, 3) > > # did not find the dimnames object useful > # at least to the extent I understood the question > >> data <- rexp(prod(dim)) >> arr <- array(data=data, dim=dim, dimnames=dimnames) >> z <- abs(sweep(arr, 3, 1)) >> df <- as.data.frame.table(z, responseName="error") >> >> ## box plot >> bwplot(error ~ methods | attr * groups, data=df, >> as.table=TRUE, outer=FALSE, strip=TRUE, notch=TRUE, >> scales=list(alternating=c(1,1), tck=c(1,0))) > > bwplot(error ~ methods | attr * groups, data=df, > as.table=TRUE, outer=FALSE, strip=TRUE, notch=TRUE, > scales=list(y=list(alternating=TRUE, > tck=c(1,0)), > x=list(labels=c( expression(method == 1), expression(alpha > == 1)) ) >) ) > > > > David Winsemius, MD > West Hartford, CT > __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Building a matrix so that matrix(r, c)<-matrix(c, r) with No For Loops
Hello,
I would like to take advantage of the upper.tri() function here but I don't
know exactly. Here is some working code...
i<-5
fi<-matrix(0,nrow=i,ncol=i)
for(r in 1:i){
for(c in 1:i){
if(r==c){
fi[r,c]<-1
}else if(rhttps://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Tinn-R looses connection to R (Windows Vista)
Nutter, Benjamin ccf.org> writes: I've noticed it, but I haven't looked into it much since I rarely work on Vista. I have found that opening R before I open Tinn-R tends to work better than using Tinn-R to open the preferred GUI. Benjamin -Original Message- From: r-help-bounces r-project.org [mailto:r-help-bounces r- project.org] On Behalf Of Agi Richard Sent: Wednesday, November 17, 2010 4:43 AM To: R-help r-project.org Subject: [R] Tinn-R looses connection to R (Windows Vista) Hi everyone, I wonder, if anyone can help me with this annoying issue, although it is related to Tinn-R and Windows Vista... I am running R 2.11.1 and Tinn-R 2.3.5.2 together. I saved 2 R-hotkeys: ALT+A for sending the whole content of one file and ALT+S for a ALT+selection from one file from Tinn-R to R. Everything works fine with other windows versions, also with Vista at the beginning of every session, but after a certain time (not always the same, neither the number of times, that I used the hotkeys is the same, so it occurs randomly, but quite fast, lets say on average within half an hour - 2hs) ALT+S does not work anymore, but instead this hotkey opens the start menue of Vista!! (The same function the windows key has!) I would be very happy, if anyone has an idea why this could be!! Is it related to R, to Tinn-R or to Vista? Thank you very very much in advance! Carol Hi: Did you fix the problem? I just recently started experiencing the same thing with my laptop(windows vista). I don't have that problem with my desktop which it still has windows XP. Thanks __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Sampling Weights in HB Choice Modelling (e.g., rhierMnlRwMixture)
Is anyone familiar with a way to account for sampling weights (e.g., in order to cope with selection bias) for individual respondents using the bayesm package (e.g., rhierMnlRwMixture)? In the regular MNL this can easily be done in STATA using the mlogit function with pweights option. However, I am unfamiliar with a way to do it in HB estimation. Any help or hints are appreciated. Best, Klaus -- View this message in context: http://r.789695.n4.nabble.com/Sampling-Weights-in-HB-Choice-Modelling-e-g-rhierMnlRwMixture-tp3407486p3407486.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Sweave and Textwrangler
Hi, I am trying to get TextWrangler to work with LaTeX and Sweave. Ideally I would call a script from TextWrangler that would run Sweave on a document, then LaTeX (using SyncTeX), and finally open the corresponding pdf in Skim. Of course I don't always need to run Sweave and would be looking for the option of LaTeX -> Open. I found this http://www.xs4all.nl/~msneep/latex/ but it doesn't work for me as it gives me an error. Something similar to do that would be ideal. Additionally the option to use BibTeX would be great too. Additionally, I tried this http://www.stat.umn.edu/~arendahl/computing/index.html and Sweave.sh script and trying to alter that (and the *.engine files) to work with TextWrangler but to no avail. Does anyone do this and have a tip they could give me? Sorry if this is off-topic (if so please reply to me off the list). Thanks! Chris [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Building a matrix so that matrix(r, c)<-matrix(c, r) with No For Loops
On Mar 26, 2011, at 9:44 AM, Brian Pellerin wrote:
Hello,
I would like to take advantage of the upper.tri() function here but
I don't
know exactly. Here is some working code...
i<-5
fi<-matrix(0,nrow=i,ncol=i)
for(r in 1:i){
for(c in 1:i){
if(r==c){
fi[r,c]<-1
}else if(rfi[lower.tri(fi)]<-fi[upper.tri(fi)]#This entry is not correct.
fi[r,c] ! ==
fi[c,r]
I've always found using the upper.tri and lower.tri functions error
prone in my hands, because they are really logical matrices for
selection rather than returning values as I naively expect. Try this:
i<-5
fi<-diag(1,i,i)
fi[upper.tri(fi)]<-1-runif(length(fi[upper.tri(fi)]))^.5
fi[lower.tri(fi)]<-t(fi)[lower.tri(fi)]
fi
It may seem odd to use lower.tri(fi) inside `[ ]` since the values of
`fi` in the lower triangle are all zero, but you are really just using
it to extract from `t(fi)`.
--
David.
Any suggestions?
Sincerely,
Brian
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Re: [R] Sweave and Textwrangler
On 11-03-26 10:58 AM, Christopher Desjardins wrote: Hi, I am trying to get TextWrangler to work with LaTeX and Sweave. Ideally I would call a script from TextWrangler that would run Sweave on a document, then LaTeX (using SyncTeX), and finally open the corresponding pdf in Skim. Of course I don't always need to run Sweave and would be looking for the option of LaTeX -> Open. I found this http://www.xs4all.nl/~msneep/latex/ but it doesn't work for me as it gives me an error. Something similar to do that would be ideal. Additionally the option to use BibTeX would be great too. Additionally, I tried this http://www.stat.umn.edu/~arendahl/computing/index.html and Sweave.sh script and trying to alter that (and the *.engine files) to work with TextWrangler but to no avail. Does anyone do this and have a tip they could give me? Sorry if this is off-topic (if so please reply to me off the list). I don't use TextWrangler, but I got TeXShop working pretty nicely with Sweave. The "engine" file simply does this: Rscript -e "patchDVI::SweavePDF( '$1' )" The patchDVI package is on R-forge here: https://r-forge.r-project.org/R/?group_id=233 This needs a very recent version of Sweave.sty if you have a recent pdflatex. You say you don't want to run Sweave every time, and want to sometimes run only Latex. I don't see how this would be desirable, because your edits to the .tex file are lost the next time you run Sweave, but the above engine skips Sweave if it is given a .tex file instead of something it recognizes as Sweave input. I gave a little tutorial on this recently; it's online here: http://www.umanitoba.ca/statistics/seminars/2011/3/4/duncan-murdoch-using-sweave-R/ The last section of the sample document shows how to set up TeXShop. I suspect setting up TextWrangler would be somewhat similar. Duncan Murdoch __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] 2 questions about probplot in package e1071
The contributed package e1071 does exactly what I want except that I need to have (1) the abscissa and ordinate axes swapped, with the probability scale on the bottom and the quantiles scale on the LHS. Using the following example: library(e1071) x <- rnorm(100, mean=5) probplot(x, line=FALSE) and (2) I need to have lines connecting the plotted symbols, as you get with: x<-log(seq(1:20)) plot(x,type='b') How can I do these two things; I've done a bunch of searching, but have not come across anything yet. Regards, Tom -- Thomas E Adams National Weather Service Ohio River Forecast Center 1901 South State Route 134 Wilmington, OH 45177 EMAIL: thomas.ad...@noaa.gov VOICE: 937-383-0528 FAX:937-383-0033 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] index of sort
Hi How can I return the index of sort, when I use R function sort? or any other sorting functions in R For example, I sorted a vector, but R just return the sorted value without giving me the original index of these data. Thanks a lot. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] round number
Hi, > a <- 4 > a*0.2 [1] 0.8 ok!! Is there a method to obtain this: > a*0.2 [1] 0.80 I need to round the number also with the zero. Thanks in advance, Alfredo __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] 2 questions about probplot in package e1071
On Mar 26, 2011, at 4:06 PM, Thomas Adams wrote: The contributed package e1071 does exactly what I want except that I need to have (1) the abscissa and ordinate axes swapped, with the probability scale on the bottom and the quantiles scale on the LHS. Using the following example: library(e1071) x <- rnorm(100, mean=5) probplot(x, line=FALSE) and (2) I need to have lines connecting the plotted symbols, as you get with: x<-log(seq(1:20)) plot(x,type='b') How can I do these two things; I've done a bunch of searching, but have not come across anything yet. The code is all there. Just rework it. Type: probplot then... probplot2 <- function( and paste in your re-worked code that swaps labels, the x and y vectors, and change axis(1) to axis 2 and axis(2, ...) to axis(1, ...) a few other swaps like abline(h=...) to abline(v=...) -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] index of sort
On Mar 26, 2011, at 2:03 PM, Quan Zhou wrote: Hi How can I return the index of sort, when I use R function sort? or any other sorting functions in R For example, I sorted a vector, but R just return the sorted value without giving me the original index of these data. You seem to be looking for: ?order -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] 2 questions about probplot in package e1071
David, Thanks! This is very helpful! I'm still very much a novice… Tom On 3/26/11 4:11 PM, David Winsemius wrote: On Mar 26, 2011, at 4:06 PM, Thomas Adams wrote: The contributed package e1071 does exactly what I want except that I need to have (1) the abscissa and ordinate axes swapped, with the probability scale on the bottom and the quantiles scale on the LHS. Using the following example: library(e1071) x <- rnorm(100, mean=5) probplot(x, line=FALSE) and (2) I need to have lines connecting the plotted symbols, as you get with: x<-log(seq(1:20)) plot(x,type='b') How can I do these two things; I've done a bunch of searching, but have not come across anything yet. The code is all there. Just rework it. Type: probplot then... probplot2 <- function( and paste in your re-worked code that swaps labels, the x and y vectors, and change axis(1) to axis 2 and axis(2, ...) to axis(1, ...) a few other swaps like abline(h=...) to abline(v=...) -- Thomas E Adams National Weather Service Ohio River Forecast Center 1901 South State Route 134 Wilmington, OH 45177 EMAIL: thomas.ad...@noaa.gov VOICE: 937-383-0528 FAX:937-383-0033 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] round number
On 11-03-26 2:05 PM, Alfredo Alessandrini wrote:
Hi,
a<- 4
a*0.2
[1] 0.8
ok!!
Is there a method to obtain this:
a*0.2
[1] 0.80
It is easy to print values with trailing zeros, but I think it's hard to
make it the default. The sprintf() function is the one I'd use for the
formatting:
sprintf("%.2f", a*0.2)
(or put noquote() around it if you don't like the quoted result).
Duncan Murdoch
I need to round the number also with the zero.
Thanks in advance,
Alfredo
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Re: [R] round number
Hi Alfredo,
Try
noquote(sprintf("%.2f", a*.2))
HTH,
Jorge
On Sat, Mar 26, 2011 at 2:05 PM, Alfredo Alessandrini <> wrote:
> Hi,
>
> > a <- 4
>
> > a*0.2
> [1] 0.8
>
> ok!!
>
> Is there a method to obtain this:
>
> > a*0.2
> [1] 0.80
>
> I need to round the number also with the zero.
>
>
>
> Thanks in advance,
>
> Alfredo
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
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Re: [R] round number
On Sat, Mar 26, 2011 at 07:05:40PM +0100, Alfredo Alessandrini wrote: > Hi, > > > a <- 4 > > > a*0.2 > [1] 0.8 > > ok!! > > Is there a method to obtain this: > > > a*0.2 > [1] 0.80 > > I need to round the number also with the zero. Hi. Try the following formatC(a*0.2, digits=2, format="f") [1] "0.80" noquote(formatC(a*0.2, digits=2, format="f")) [1] 0.80 The two-digit form 0.80 is used also if a vector is printed and some other component requires 2 digits. For example c(a*0.2, 0.01) [1] 0.80 0.01 Hope this helps. Petr Savicky. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] line graph question
Hi, I am working on some line charts and although I have a lot of resources, I cannot seem to find an answer to this question: how can I set the incrementation of values on the x-axis values so that I can see all the groups on this axis. Right now, R puts them at increments of 2 (i.e., 2, 4, 6, 8, 10, 12). I need to see all, from 1 to 12. Thank you! -- BÜLENT [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] line graph question
On 03/27/2011 09:21 AM, Bulent Arikan wrote: Hi, I am working on some line charts and although I have a lot of resources, I cannot seem to find an answer to this question: how can I set the incrementation of values on the x-axis values so that I can see all the groups on this axis. Right now, R puts them at increments of 2 (i.e., 2, 4, 6, 8, 10, 12). I need to see all, from 1 to 12. Hi Bulent, First try: plot(...,xaxt="n",...) axis(1,at=1:12,labels=1:12) and if they are still not there, look at staxlab in the plotrix package. Jim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] line graph question
Bulent - Many characteristics of plots are determined by the graphical parameters documented in the help file for the par function. In this case, the xaxp parameter should be useful. Compare plot(1:12,1:12) with plot(1:12,1:12,xaxp=c(0,12,12)) - Phil Spector Statistical Computing Facility Department of Statistics UC Berkeley spec...@stat.berkeley.edu On Sat, 26 Mar 2011, Bulent Arikan wrote: Hi, I am working on some line charts and although I have a lot of resources, I cannot seem to find an answer to this question: how can I set the incrementation of values on the x-axis values so that I can see all the groups on this axis. Right now, R puts them at increments of 2 (i.e., 2, 4, 6, 8, 10, 12). I need to see all, from 1 to 12. Thank you! -- B?LENT [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] simple if question
Hi everyone,
I have just got different samples from a dataframe (independent and
exclusive, there aren't common elements among them). I want to create a
variable that indicate the sampling selection of the elements in the
original dataframe (for example, 0 = no selected, 1= sample 1, 2=sample
2, etc.).
I have tried to do it with ifelse command, but the problem is that the
second line replaces the values of the first line, and I haven't been
able to do it with the if command (I got this error: In if (data$school
%in% sample1) { :
the condition has length > 1 and only the first element will be used)
data$selection <- ifelse(data$school %in% sample1, 1, 0)
data$selection <- ifelse(data$school %in% sample2, 2, 0)
Any ideas?
Thank you in advance.
--
Sebastián Daza
sebastian.d...@gmail.com
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Re: [R] Two matrix loop
This way uses a three-dimensional array instead of the nested apply. It seems to take the same amount of time, even on larger datasets, but it may give you ideas. distance=function(x) daisy(x, metric = 'gower') persons=array(dim=c(2,nrow(donor)*nrow(receiver),ncol(receiver))) persons[1,,]=donor[rep(1:nrow(donor),each=nrow(receiver)),] persons[2,,]=receiver[rep(1:nrow(receiver),nrow(donor)),] matrix(apply(persons,2,distance),,nrow(donor)) Tom On Thu, Mar 24, 2011 at 8:23 AM, Stefan Petersson wrote: > > Hi, > > I'm trying to create a distance matrix. And it works out somewhat ok. > However, I suspect that there are > some efficiency issues with my efforts. Plz have a look at this: > > donor <- matrix(c(3,1,2,3,3,1,4,3,5,1,3,2), ncol=4) > receiver <- > matrix(c(1,4,3,2,4,3,1,5,1,3,2,1,4,5,3,5,1,3,2,4,5,1,2,3,1,4,5,5,1,2,1,3,4,3,2,5,5,1,4,2,5,4,3,2), > ncol=4) > > The above creates my two matrices. I have three donors, and eleven receivers > (rows), with four > measures (columns) in each matrix. > > And now, I want to apply the daisy() function from the cluster library, to > calculate distances between my > three donors, and eleven receivers. The end result should be a 11x3 matrix > with distances between the > units from the two matrices. I can calculate one distance measure (ie donor 1 > and receiver 1). Like this: > > library(cluster) > daisy(rbind(donor[1,], receiver[1,]), metric = 'gower') > > My first attempt was a simple nested for-loop. But that one was discarded > after reading up on efficiency > issues with for-looping. So I turned to 'apply' with this result: > > apply(donor, 1, function(b) apply(receiver, 1, function(a) daisy(rbind(b, a), > metric = 'gower'))) > > [,1] [,2] [,3] > [1,] 1.00 0.50 0.75 > [2,] 1.00 0.75 0.75 > [3,] 0.75 1.00 1.00 > [4,] 0.50 0.75 0.75 > [5,] 0.75 1.00 0.75 > [6,] 0.75 1.00 0.50 > [7,] 0.75 0.50 0.75 > [8,] 1.00 1.00 1.00 > [9,] 1.00 0.75 1.00 > [10,] 0.75 0.50 1.00 > [11,] 0.75 1.00 0.25 > > However, something tells me that there is a simpler (more efficient) way of > doing this. I've been reading > up on the Matrix library, but I'm having trouble understanding the > functions... > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] simple if question
On Mar 26, 2011, at 7:05 PM, Sebastián Daza wrote:
Hi everyone,
I have just got different samples from a dataframe (independent and
exclusive, there aren't common elements among them). I want to
create a variable that indicate the sampling selection of the
elements in the original dataframe (for example, 0 = no selected, 1=
sample 1, 2=sample 2, etc.).
I have tried to do it with ifelse command, but the problem is that
the second line replaces the values of the first line, and I haven't
been able to do it with the if command (I got this error: In if (data
$school %in% sample1) { :
the condition has length > 1 and only the first element will be used)
Nested ifelse's should work. There may be a limit to nested depth.
data$selection <- ifelse(data$school %in% sample1, 1, # else
ifelse(data$school %in% sample2, 2, #else
0) )
Any ideas?
Thank you in advance.
--
Sebastián Daza
sebastian.d...@gmail.com
__
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and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
West Hartford, CT
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and provide commented, minimal, self-contained, reproducible code.
[R] Multiple plots with one legend
Hi, I've created 3 plots one under the other, and want to include a legend on the right that spans the height of all 3 plots. How can I do this? Thanks! -- View this message in context: http://r.789695.n4.nabble.com/Multiple-plots-with-one-legend-tp3408537p3408537.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] simple if question
Posting some sample data would help, but I think something like this
is what you want
data[data$school=='Cornell University',]
For example
CO2[CO2$Type=='Quebec',]
Tom
2011/3/26 Sebastián Daza :
> Hi everyone,
> I have just got different samples from a dataframe (independent and
> exclusive, there aren't common elements among them). I want to create a
> variable that indicate the sampling selection of the elements in the
> original dataframe (for example, 0 = no selected, 1= sample 1, 2=sample 2,
> etc.).
>
> I have tried to do it with ifelse command, but the problem is that the
> second line replaces the values of the first line, and I haven't been able
> to do it with the if command (I got this error: In if (data$school %in%
> sample1) { :
> the condition has length > 1 and only the first element will be used)
>
> data$selection <- ifelse(data$school %in% sample1, 1, 0)
> data$selection <- ifelse(data$school %in% sample2, 2, 0)
>
> Any ideas?
> Thank you in advance.
>
> --
> Sebastián Daza
> sebastian.d...@gmail.com
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
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Re: [R] Multiple plots with one legend
On Mar 26, 2011, at 7:06 PM, mavkoup wrote: Hi, I've created 3 plots one under the other, and want to include a legend on the right that spans the height of all 3 plots. ?mtext # with the `las` parameter for rotation Or you can use: text(x,y, "some text", srt=-90,xpd=NA ) -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Multiple plots with one legend
Ok I think I can figure that out. Which leads me to a further question. My 3 plots contain 10 time series, each with a different name and color. Can I create the legend such that it has a line of the correct color follow by the name of the series? I.e. line(with color1) NAME 1 line(with color2) NAME 2 ... I have a vector of colors I'm using and a vector of names. Thanks -- View this message in context: http://r.789695.n4.nabble.com/Multiple-plots-with-one-legend-tp3408537p3408646.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Coordinates of the text region
Hello R people Is there a way to get the coordinates of the text region (coordinates of the four corners for example) when using the text function? I'm looking for a way that does not make use of interactive function like locator. My goal is to determine the position of other structures in a graphic so that they don't overlap with the text. Thanks Francois Rousseu [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Multiple plots with one legend
On Mar 26, 2011, at 8:27 PM, mavkoup wrote: Ok I think I can figure that out. Which leads me to a further question. My 3 plots contain 10 time series, each with a different name and color. Can I create the legend such that it has a line of the correct color follow by the name of the series? I.e. "plots" and therefore "legends" in R can take many forms. Have your read the Posting Guide's recommendations on how to submit a reproducible example yet? That would be particularly important if you have three different plots on the same page. line(with color1) NAME 1 line(with color2) NAME 2 ... I have a vector of colors I'm using and a vector of names. -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Coordinates of the text region
On Mar 26, 2011, at 8:39 PM, Francois Rousseu wrote: Hello R people Is there a way to get the coordinates of the text region (coordinates of the four corners for example) What "text region" are you talking about? when using the text function? I'm looking for a way that does not make use of interactive function like locator. My goal is to determine the position of other structures in a graphic so that they don't overlap with the text. The plotrix package and the rms package have functions for avoiding plotted points but you seem to be looking for something else. -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Coordinates of the text region
Let's say I do the following: plot(0,0,type="n",xlim=c(0,10),ylim=c(0,10)) rect(4,5,5,6,border="black",col="white") text(4.5,5.5,"species",cex=2) I would like to be able to determine a rectangle size that will be able to contain the text species. I'm working on a function using multiple lines and boxes where text of varying length has to be written. Francois > > On Mar 26, 2011, at 8:39 PM, Francois Rousseu wrote: > > > > > Hello R people > > > > Is there a way to get the coordinates of the text region > > (coordinates of the four corners for example) > > What "text region" are you talking about? > > > when using the text function? I'm looking for a way that does not > > make use of interactive function like locator. My goal is to > > determine the position of other structures in a graphic so that they > > don't overlap with the text. > > The plotrix package and the rms package have functions for avoiding > plotted points but you seem to be looking for something else. > > -- > > David Winsemius, MD > West Hartford, CT > [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Coordinates of the text region
take a look at 'textbox' in plotrix. On Sat, Mar 26, 2011 at 9:05 PM, Francois Rousseu wrote: > > Let's say I do the following: > > plot(0,0,type="n",xlim=c(0,10),ylim=c(0,10)) > rect(4,5,5,6,border="black",col="white") > text(4.5,5.5,"species",cex=2) > > I would like to be able to determine a rectangle size that will be able to > contain the text species. I'm working on a function using multiple lines and > boxes where text of varying length has to be written. > > Francois > > > >> >> On Mar 26, 2011, at 8:39 PM, Francois Rousseu wrote: >> >> > >> > Hello R people >> > >> > Is there a way to get the coordinates of the text region >> > (coordinates of the four corners for example) >> >> What "text region" are you talking about? >> >> > when using the text function? I'm looking for a way that does not >> > make use of interactive function like locator. My goal is to >> > determine the position of other structures in a graphic so that they >> > don't overlap with the text. >> >> The plotrix package and the rms package have functions for avoiding >> plotted points but you seem to be looking for something else. >> >> -- >> >> David Winsemius, MD >> West Hartford, CT >> > > [[alternative HTML version deleted]] > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > -- Jim Holtman Data Munger Guru What is the problem that you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Multiple plots with one legend
You can use 'layout' to create 4 plot areas: the 3 plot you currently have, and one for the legend. Try this to see what happens: > x <- cbind(rbind(1,2,3), 4) > layout(x, width = c(5,1)) > layout.show(4) On Sat, Mar 26, 2011 at 7:06 PM, mavkoup wrote: > Hi, > > I've created 3 plots one under the other, and want to include a legend on > the right that spans the height of all 3 plots. > > How can I do this? > > Thanks! > > -- > View this message in context: > http://r.789695.n4.nabble.com/Multiple-plots-with-one-legend-tp3408537p3408537.html > Sent from the R help mailing list archive at Nabble.com. > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > -- Jim Holtman Data Munger Guru What is the problem that you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] library(foreign) read.spss warning
There is some information about this subtype in the PSPP source code,
and for other subtypes not yet implemented by read.spss. The PSPP source
code indicates that this subtype consists of "Value labels for long
strings", which isn't very illuminating to me (probably because I don't
use PSPP, or SPSS, though I increasingly have need to import SPSS data
files). Copied below are the relevant bits.
-Matt
>From (the PSPP source file) src/data/sys-file-reader.c:
enum
{
/* subtypes 0-2 unknown */
EXT_INTEGER = 3, /* Machine integer info. */
EXT_FLOAT = 4, /* Machine floating-point info. */
EXT_VAR_SETS = 5, /* Variable sets. */
EXT_DATE = 6, /* DATE. */
EXT_MRSETS= 7, /* Multiple response sets. */
EXT_DATA_ENTRY= 8, /* SPSS Data Entry. */
/* subtypes 9-10 unknown */
EXT_DISPLAY = 11, /* Variable display parameters. */
/* subtype 12 unknown */
EXT_LONG_NAMES= 13, /* Long variable names. */
EXT_LONG_STRINGS = 14, /* Long strings. */
/* subtype 15 unknown */
EXT_NCASES= 16, /* Extended number of cases. */
EXT_FILE_ATTRS= 17, /* Data file attributes. */
EXT_VAR_ATTRS = 18, /* Variable attributes. */
EXT_MRSETS2 = 19, /* Multiple response sets (extended). */
EXT_ENCODING = 20, /* Character encoding. */
EXT_LONG_LABELS = 21 /* Value labels for long strings. */
};
and
static const struct extension_record_type types[] =
{
/* Implemented record types. */
{ EXT_INTEGER, 4, 8 },
{ EXT_FLOAT,8, 3 },
{ EXT_MRSETS, 1, 0 },
{ EXT_DISPLAY, 4, 0 },
{ EXT_LONG_NAMES, 1, 0 },
{ EXT_LONG_STRINGS, 1, 0 },
{ EXT_NCASES, 8, 2 },
{ EXT_FILE_ATTRS, 1, 0 },
{ EXT_VAR_ATTRS,1, 0 },
{ EXT_MRSETS2, 1, 0 },
{ EXT_ENCODING, 1, 0 },
{ EXT_LONG_LABELS, 1, 0 },
/* Ignored record types. */
{ EXT_VAR_SETS, 0, 0 },
{ EXT_DATE, 0, 0 },
{ EXT_DATA_ENTRY, 0, 0 },
};
On Fri, 2011-03-25 at 18:39 -0500, Robert Baer wrote:
> I got the following:
> > library(foreign)
> > swal = read.spss("swallowing.sav", to.data.frame =TRUE)
> Warning message:
> In read.spss("swallowing.sav", to.data.frame = TRUE) :
> swallowing.sav: Unrecognized record type 7, subtype 21 encountered in
> system file
> >
>
> The bulk of the data seems to read in a usable form, but I'm curious about
> what might be getting lost because I don't know how to translate type 7,
> subtype 21. I did not generate the SPSS data so I'm not certain of the
> version, but I'm assuming version 18 or 19. I did a quick Find on the PSPP
> manual for Type 7 and subtype 21 and came up dry.
>
> Any insights or clues how I might learn more?
>
> Thanks,
> Rob
>
>
> > R.Version()
> $platform
> [1] "i386-pc-mingw32"
>
> $arch
> [1] "i386"
>
> $os
> [1] "mingw32"
>
> $system
> [1] "i386, mingw32"
>
> $status
> [1] ""
>
> $major
> [1] "2"
>
> $minor
> [1] "12.2"
>
> $year
> [1] "2011"
>
> $month
> [1] "02"
>
> $day
> [1] "25"
>
> $`svn rev`
> [1] "54585"
>
> $language
> [1] "R"
>
> $version.string
> [1] "R version 2.12.2 (2011-02-25)"
>
>
>
> --
> Robert W. Baer, Ph.D.
> Professor of Physiology
> Kirksville College of Osteopathic Medicine
> A. T. Still University of Health Sciences
> Kirksville, MO 63501
> 660-626-232
> FAX 660-626-2965
>
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Building a matrix so that matrix(r, c)<-matrix(c, r) with No For Loops
Thanks, David. This works splendidly. Thanks for your help.
Sincerely,
Brian
>>> David Winsemius 03/26/11 3:06 PM >>>
On Mar 26, 2011, at 9:44 AM, Brian Pellerin wrote:
> Hello,
>
> I would like to take advantage of the upper.tri() function here but
> I don't
> know exactly. Here is some working code...
> i<-5
> fi<-matrix(0,nrow=i,ncol=i)
> for(r in 1:i){
> for(c in 1:i){
> if(r==c){
> fi[r,c]<-1
> }else if(r fi[r,c]<-1-runif(1)^.5
> }else{
> fi[r,c]<-fi[c,r]
> }
> }
> }
>
> So far I know I can simplify this code to 5 lines (no for loops):
> i<-5
> fi<-matrix(nrow=i,ncol=i)
> fi[upper.tri(fi)]<-1-runif(length(fi[upper.tri(fi)]))^.5
> diag(fi)<-1
> fi[lower.tri(fi)]<-fi[upper.tri(fi)]#This entry is not correct.
> fi[r,c] ! ==
> fi[c,r]
I've always found using the upper.tri and lower.tri functions error
prone in my hands, because they are really logical matrices for
selection rather than returning values as I naively expect. Try this:
i<-5
fi<-diag(1,i,i)
fi[upper.tri(fi)]<-1-runif(length(fi[upper.tri(fi)]))^.5
fi[lower.tri(fi)]<-t(fi)[lower.tri(fi)]
fi
It may seem odd to use lower.tri(fi) inside `[ ]` since the values of
`fi` in the lower triangle are all zero, but you are really just using
it to extract from `t(fi)`.
--
David.
>
> Any suggestions?
>
> Sincerely,
> Brian
>
> [[alternative HTML version deleted]]
>
> __
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> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
West Hartford, CT
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Re: [R] Coordinates of the text region
Look at the strwidth and strheight functions. Also note the adj parameter (see ?par for details). -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 > -Original Message- > From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- > project.org] On Behalf Of Francois Rousseu > Sent: Saturday, March 26, 2011 6:40 PM > To: r-help@r-project.org > Subject: [R] Coordinates of the text region > > > Hello R people > > Is there a way to get the coordinates of the text region (coordinates > of the four corners for example) when using the text function? I'm > looking for a way that does not make use of interactive function like > locator. My goal is to determine the position of other structures in a > graphic so that they don't overlap with the text. > > Thanks > Francois Rousseu > > [[alternative HTML version deleted]] > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting- > guide.html > and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Multiple plots with one legend
Yes that's what I had managed to generate too. I can produce my 3 plots. Each plot has 10 colored lines say. I want to place the legend in the 4th spot listing the name of the 10 colored lines, and their color. -- View this message in context: http://r.789695.n4.nabble.com/Multiple-plots-with-one-legend-tp3408537p3408850.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Bootstrap 95% confidence intervals for splines
There appear to be reports in the literature that transform continuous independent variablea by the use of splines, e.g., assume the dependent variable is hot dogs eaten per week (HD) and the independent variable is waistline (WL), a normal linear regression model would be: nonconfusing_regression <- lm(HD ~ WL) One might use a spline, confusion_inducing_regression_with_spline <- lm(HD ~ ns(WL, df = 4) ) Now is where the problem starts. >From nonconfusing_regression , I get, say 2 added hot dogs per week for each centimeter of waistline along with a s.e. of 0.5 hot dogs per week, which I multiply by 1.96 to garner each side of the 95% c.i. If I want to show what the difference between the 75th percentile (say 100 cm) and 25th percentile (say 80 cm) waistlines are, I multiply 2 by 100-80=20 and get 40 hot dogs per week as the point estimate with a similar bumping of the s.e. to 10 hot dogs per week. What do I do to get the point estimate and 95% confidence interval for the difference between 100 cm persons and 80 cm persons with confusion_inducing_regression_with_spline ? Best regards. Mitchell S. Wachtel, MD -- View this message in context: http://r.789695.n4.nabble.com/Bootstrap-95-confidence-intervals-for-splines-tp3408813p3408813.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Garchoxfit package
Dear List, I'm now using Ubuntu 10.10 and I want to use the garchoxfit function.It seems that I need to download the package. While after installing the package,I still can't use the garchoxfit function.What's the reason and how to fix that? Thanks for your time! Best, Ning __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] run function on subsets of matrix
I was wondering if it is possible to do the following in a smarter way. I want get the mean value across the columns of a matrix, but I want to do this on subrows of the matrix, given by some vector(same length as the the number of rows). Something like nObs<- 6 nDim <- 4 m <- matrix(rnorm(nObs*nDim),ncol=nDim) fac<-sample(1:(nObs/2),nObs,rep=T) ##loop trough different 'factor' levels for (i in unique(fac)) print(apply(m[fac==i,],2,mean)) Now, the problem is that if a value in 'fac' only occurs once, the 'apply' function will complain. But I think, this can also be written more cleverly using some R style construct using factorlevels. Thanks __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Garchoxfit package
On Mar 26, 2011, at 11:16 PM, Ning Cheng wrote: Dear List, I'm now using Ubuntu 10.10 and I want to use the garchoxfit function.It seems that I need to download the package. While after installing the package,I still can't use the garchoxfit function.What's the reason and how to fix that? A typical beginner error is to forget to also use either library() or require() to load an installed package. Installation only puts the files in the library. It does not load it into the workspace. You should in the future offer your session which will make comments more specific than this guessing game you have requested. -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] run function on subsets of matrix
On Mar 26, 2011, at 10:26 PM, fisken wrote: I was wondering if it is possible to do the following in a smarter way. I want get the mean value across the columns of a matrix, but I want to do this on subrows of the matrix, given by some vector(same length as the the number of rows). Something like nObs<- 6 nDim <- 4 m <- matrix(rnorm(nObs*nDim),ncol=nDim) fac<-sample(1:(nObs/2),nObs,rep=T) ##loop trough different 'factor' levels for (i in unique(fac)) print(apply(m[fac==i,],2,mean)) This would be a lot simpler and faster: colMeans(m[unique(fac),]) #[1] 1.3595197 -0.1374411 0.1062527 -0.3897732 Now, the problem is that if a value in 'fac' only occurs once, the 'apply' function will complain. Because "[" will drop single dimensions and so the matrix becomes a vector and looses the number-2 margin. Use drop=FALSE to prevent this, and note the extra comma: print(apply(m[1, , drop=FALSE],2,mean)) -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
