search for: dataname

Displaying 20 results from an estimated 40 matches for "dataname".

2006 Mar 28
0
Help with the code
...s) ~ trt + strata(site), data=data1, iter.max=500) model1.surv=survfit(model1.cox) #### Cox proportional hazard model with dummy covariates for site model2.cox=coxph(Surv(time,status) ~ trt + factor(site), data=data1, iter.max=500) model2.surv=survfit(model2.cox) beta.fun1 = function(dataname) { cox = coef(coxph(Surv(dataname$time,dataname$status) ~ dataname$trt+strata(dataname$site))) } beta.fun2 = function(dataname) { cox =coef(coxph(Surv(dataname$time,dataname$status) ~ dataname$trt+factor(dataname$site))) #cox=cox[1] } dataname=data1 ## Calculate Cox's regressi...
2010 Jul 27
6
Eval() or parse() do not work inside function
I am writing a function where the arguments are names of objects or variable names in a data frame. To convert the strings to the objects I am using eval(parse(text=name)): f.graph.two.vbs<-function(dataname,v1){ val<-paste(dataname,v1,sep="$") val<-eval(parse(text=val)) val } However running this returns an error: >f.graph.two.vbs("data","RECORD") Error in data$RECORD : $ operator is invalid for atomic vectors Repeating the individual c...
2011 Nov 11
1
Formula variable help
I have an R script with the following applicable lines: xshort <- window(s, start=st, end=ed) . . . xshort <- ts(xshort, frequency=1, start=1) . . . m1 <- m2 <- m3 <- m4 <- m5 <- m6 <- NULL m1 <- tslm(xshort ~ trend) I get an error: Error in get(dataname) : object 'xshort' not found When I do traceback() I get: 3: get(dataname) 2: tslm(xshort ~ trend) at #19 1: model.cross.validation(l[["MEN"]]$series) Which points to the call to tslm above. Since I am not supply 'data' to the tslm call (in the forecast package),, I a...
2011 Sep 20
1
Data
Hey everybody, i am using the rugarch-package and its great! I have a pretty easy problem, but i just dont get it, so thanks if you can help me. Normally i use: / data(DATANAME) spec = ugarchspec() fit = ugarchfit(data = x[,1], spec = spec) fit slotNames(fit) names(fit at fit) coef(fit) infocriteria(fit) likelihood(fit) nyblom(fit) signbias(fit) head(as.data.frame(fit)) head(sigma(fit)) head(residuals(fit)) head(fitted(fit)) gof(fit,c(20,30,40,50)) uncmean(fit) uncvaria...
2000 May 02
2
Variable names in model formula
...n the fitted probability of >survival at a fixed time point t0 and covariate value Xi0. I tried to do >this in a for loop where the index is the variable name as follows > >covnames <- names(...) >for (covi in covnames) { > survout <- coxph(Surv(time,status) ~ covi, data = dataname) > print(summary(survfit(survout,newdata),t0)) >} You could try covnames <- names(...) form <- Surv(time, status) ~ dummy for(nam in covnames) { form[[3]] <- as.name(nam) survout <- coxph(form, data = dataname) print(summary(survfit(survout, newdata), t0)) } > >Unfort...
2011 Jan 21
3
confidence interval
Hi, I have a circular shaped set of point on the plane (X,Y) centered in zero. The distribution is more dense close to zero and less dense far from zero. I need to find the radius of a circle centered in zero that contains 65% of the points in the sample. Is there any R directive that can do this? I wanna start with 2D set of points, but the real case scenario is with a 5D set of points.
2017 Dec 04
3
Dynamic reference, right-hand side of function
...acro" (is that only a Stata-word?) in the right hand side of the functions. When I try to replace the static reference with a dynamic one (like in the left-hand side), the syntax just doesn't work. I would very much appreciate some help with this issue! All the best, Love year <- 2002 dataname <- paste0("aa_",year) assign(paste0(dataname), as.numeric(aa_2002[,1])) year <- 2003 assign(paste0("aa_",year), as.numeric(aa_2003)) year <- 2005 assign(paste0("aa_",year), aa_2005[,1]) list1 <- c(2000:2007) list1[c(7)] assign(paste0("aa_",list...
2017 Dec 04
0
Dynamic reference, right-hand side of function
The generic rule is that R is not a macro language, so looping of names of things gets awkward. It is usually easier to use compound objects like lists and iterate over them. E.g. datanames <- paste0("aa_", 2000:2007) datalist <- lapply(datanames, get) names(datalist) <- datanames col1 <- lapply(datalist, "[[", 1) colnum <- lapply(col1, as.numeric) (The 2nd line assumes that the damage has already been done so that you have aa_2000 ... aa_2007 in...
2017 Dec 04
3
Dynamic reference, right-hand side of function
...an at sociology.su.se> Kopia: r-help at r-project.org ?mne: Re: [R] Dynamic reference, right-hand side of function The generic rule is that R is not a macro language, so looping of names of things gets awkward. It is usually easier to use compound objects like lists and iterate over them. E.g. datanames <- paste0("aa_", 2000:2007) datalist <- lapply(datanames, get) names(datalist) <- datanames col1 <- lapply(datalist, "[[", 1) colnum <- lapply(col1, as.numeric) (The 2nd line assumes that the damage has already been done so that you have aa_2000 ... aa_2007 in y...
2009 Oct 08
3
error message - unexpected input
...program file folder. I have also tried changing the font the code is written in and turning off "smart quotes" in MS word. Additionally, I have right clicked on R and set myself as the only user on this computer to run this program. The syntax line that appears to be causing problems is: dataname="Igt2model.txt" and the error message is: Error: unexpected input in "dataname="" Does anyone have any suggestions? Thanks for all of your help, Danielle Danielle Dandreaux, Ph.D. Faculty Research Associate Arizona State University Department of Psychology...
2017 Dec 04
0
Dynamic reference, right-hand side of function
...Kopia: r-help at r-project.org > ?mne: Re: [R] Dynamic reference, right-hand side of function > > The generic rule is that R is not a macro language, so looping of names of things gets awkward. It is usually easier to use compound objects like lists and iterate over them. E.g. > > datanames <- paste0("aa_", 2000:2007) > datalist <- lapply(datanames, get) > names(datalist) <- datanames > col1 <- lapply(datalist, "[[", 1) > colnum <- lapply(col1, as.numeric) > > (The 2nd line assumes that the damage has already been done so that you...
2012 Feb 16
2
Problem building up ggplot graph in a loop.
Folks, I want to automate some graphing using ggplot. Here is my code graphChargeOffs2<-function(coffs) { ggplot(coffs, aes(levels)) dataNames<-names(coffs)[!names(coffs) == "levels"] for(i in dataNames) { thisData<-coffs[[i]] last_plot() + geom_line(aes(y = thisData, colour = i)) } last_plot() + ylab("Total Chargeoffs") } coffs is a data.frame. I get the following error: Error in eval(expr, envi...
2008 Jul 29
4
Graphics function question
Hello I have created a graph using the following commands: <<< startBReP3O1T <- diffs$BReP3O1T - diffs$diff_BReP3O1T endBReP3O1T <- diffs$BReP3O1T x <- seq(47,89, length = 10) ymin <- min(min(startBReP3O1T), min(endBReP3O1T)) ymax <- max(max(startBReP3O1T), max(endBReP3O1T)) y <- seq(ymin, ymax, length = 10) plot(x,y, type = 'n', xlab = 'Age', ylab =
2017 Dec 04
2
Dynamic reference, right-hand side of function
...Kopia: r-help at r-project.org > ?mne: Re: [R] Dynamic reference, right-hand side of function > > The generic rule is that R is not a macro language, so looping of names of things gets awkward. It is usually easier to use compound objects like lists and iterate over them. E.g. > > datanames <- paste0("aa_", 2000:2007) > datalist <- lapply(datanames, get) > names(datalist) <- datanames > col1 <- lapply(datalist, "[[", 1) > colnum <- lapply(col1, as.numeric) > > (The 2nd line assumes that the damage has already been done so that &g...
2010 Dec 29
1
Counting number of datasets and appending them
Hi there, I have a question on how to read a bunch of dataset, assign each of the dataset to a matrix in the memory, and append them. Suppose I have 20 dataset saved to different .rda files named gradeFileData1, gradeFileData2,...., gradeFileData20. And I would like to read them each into a dataset in the memory, then combine them. I wrote something like: e1<-new.env(parent=.GlobalEnv)
2017 Dec 05
3
Dynamic reference, right-hand side of function
....csv(text="1,0,1,1,0,0,0,0,0,0,1,0,0", header=FALSE)) bb_2001 <- as.matrix(read.csv( text="0,0,0,1,0,1,1,0,0,0,0,1,0,0", header=FALSE)) bb_2002 <- as.matrix(read.csv( text="1,0,0,0,0,0,0,0,1,0,0,0,0,1,0,0", header=FALSE)) is.vector(bb_2000) datanames <- paste0("bb_", 2000:2002) datalist <- lapply(datanames, get) is.vector(datalist[1]) I learned a lot of code today, and I really appreciate it! A million thanks! My R-superpowers are, well, not as minuscule as when I woke up this morning. All the best, Love (or maybe LoveR, my...
2017 Dec 04
0
Dynamic reference, right-hand side of function
...r-project.org >> ?mne: Re: [R] Dynamic reference, right-hand side of function >> >> The generic rule is that R is not a macro language, so looping of names of things gets awkward. It is usually easier to use compound objects like lists and iterate over them. E.g. >> >> datanames <- paste0("aa_", 2000:2007) >> datalist <- lapply(datanames, get) >> names(datalist) <- datanames >> col1 <- lapply(datalist, "[[", 1) >> colnum <- lapply(col1, as.numeric) >> >> (The 2nd line assumes that the damage has alread...
2017 Dec 05
0
Dynamic reference, right-hand side of function
...1,0,0,0,0,0,0,1,0,0", > header=FALSE)) > bb_2001 <- as.matrix(read.csv( text="0,0,0,1,0,1,1,0,0,0,0,1,0,0", > header=FALSE)) > bb_2002 <- as.matrix(read.csv( text="1,0,0,0,0,0,0,0,1,0,0,0,0,1,0,0", > header=FALSE)) > > is.vector(bb_2000) > > datanames <- paste0("bb_", 2000:2002) > datalist <- lapply(datanames, get) > is.vector(datalist[1]) > > > I learned a lot of code today, and I really appreciate it! A million > thanks! > My R-superpowers are, well, not as minuscule as when I woke up this > morning. &g...
2016 Dec 12
2
accessing data by packagename::dataname from within package code fails.
I have narrowed down the problem. The error Error : 'AminoAcids' is not an exported object from 'namespace:bibliospec' Error : unable to load R code in package 'bibliospec' occurs only if I try to access the data using bibliospec::AminoAcids within the initialize method of an R reference class. It does work, as far as I tested everywhere else. In other methods of a
2005 Feb 26
3
averaging within columns
...times between decisions in the second column. Since individuals make multiple decisions, I want the average for each individual. For example, the data might look like this name wtime jo 1 jo 2 jo 1 jo 3 tim 3 tim 2 tim 2 ro 1 ro 2 etc. I'm hoping there is something like mean(dataname$wtime[name]) which will just create a column with length equal to the number of different names (levels) and an average wtime for each. So far though, I haven't had much luck figuring that one out. Thanks. Thomas