similar to: survdiff

Folks, I'm running into something which looks like a bug in linearscan allocator. Of course I can't be 100% sure it's not some unobvious mistake on my part, so I'd like to hear your opinion. First, I attach two files -- LLVM asm and the asm for my target. The problem with assembler is: on line 171 it uses register gr2, which is copied from gr6 above, on line 161. The only

On Tue, 22 Jun 2004, Vladimir Prus wrote: > First, I attach two files -- LLVM asm and the asm for my target. The problem > with assembler is: on line 171 it uses register gr2, which is copied from gr6 > above, on line 161. The only predecessor of this basic block is jump on line > 90. The problem is that gr6 is not initialized in the interval from the > function entry till the

Dear R-bug-people I have encountered a problem with "unsplit", which I believe may be caused by a bug in the function. However, unexpericend with bug-reports I apologise if this is barely a user problem rather than a problem within R. The problem occurs if an object is split by several grouping factors with levels not occuring in the data, and using drop = TRUE. This may appear as

Chris Lattner wrote: > On Tue, 22 Jun 2004, Vladimir Prus wrote: > > First, I attach two files -- LLVM asm and the asm for my target. The > > problem with assembler is: on line 171 it uses register gr2, which is > > copied from gr6 above, on line 161. The only predecessor of this basic > > block is jump on line 90. The problem is that gr6 is not initialized in > >

Hi, looking at the fresh CVS state I see: class Register<string n> : RegisterBase<n> { list<RegisterBase> Aliases = []; } while previously the Register class did not require any parameters. The change log is just: * Target.td: Revamp the Register class, and allow the use of the RegisterGroup class to specify aliases directly in register definitions. and I

Dear R-users, I would like to maximize the function g above which depends on 4 parameters (2 vectors, 1 real number, and 1 matrix) using optim() and BFGS method. Here is my code: # fonction to maximize g=function(x) { x1 = x[1:ncol(X)] x2 = x[(ncol(X)+1)] x3 = matrix(x[(ncol(X)+2):(ncol(X)+1+ncol(X)*ncol(Y))],nrow=ncol(X),ncol=ncol(Y)) x4 = x[(ncol(X)+1+ncol(X)*ncol(Y)+1):length(x)]

On Tue, 16 Nov 2004, Vladimir Prus wrote: > and I could not find any discussions in the archives. > > Why the change was necessary? Writing: > > def gr0 : Register<"gr0">; > def gr1 : Register<"gr1">; > def gr2 : Register<"gr2">; > def gr3 : Register<"gr3">; > def gr4 :

Dear expeRts, Why does na.blank=TRUE not replace the NA's in the following LaTeX table? x <- matrix(1:72, ncol=4, nrow=8) colnames(x) <- c("gr1.sgr1", "gr1.sgr2", "gr2.sgr1", "gr2.sgr2") rn <- apply(expand.grid(beta=c(0.25, 0.75), n=c(100, 500), d=c(10, 100))[, 3:1], 2, rmNames) x <- cbind(rn, x) # append columns containing the row labels

When I create a lattice/Trellis type graph, I typically write a function that returns the graph, as in do.graph <- function(x, y, ...) { require(lattice) return(xyplot(y~x, ...)) } My question today is this: If I want two graphs on one page, one way of achieving it is to print the objects into defined areas, as in gr1 <- xyplot(rnorm(111) ~ runif(111)) gr2 <-

http://bugzilla.mindrot.org/show_bug.cgi?id=938 Summary: "AllowGroups" option and secondary user's groups limit Product: Portable OpenSSH Version: 3.9p1 Platform: ix86 OS/Version: Linux Status: NEW Severity: major Priority: P2 Component: PAM support AssignedTo: openssh-bugs at

Laurent: As I don't use dplyr, this won't help you, but I hope you and others may find it entertaining anyway. If I understand you correctly (and ignore this if I have not), there are a ton of ways to do this in base R, including using switch() along the lines you noted in your post. However, when the functions get sufficiently complicated or numerous, it may be useful to store them in a

Hi, The lab in which I send my samples return the results in a format that is difficult for me to run my analysis. The lab outputs the results where each parameter is its own row and it’s not consistently in the same order (and not each sample is tested for the same suite of variables). e.g. >dataset site parameter value a e2 3 a e1 1 a e3 5 b e3

Dear RHelp-list, ?? Using dplyr and the group_by approach on a dataframe, I want to be able to apply a specific action according to the group name. The code bellow works, but I am not able to write it in a more esthetic way using dplyr. Can somebody help me to find a better solution ? Thank you Best regards Laurent df_test <- data.frame( x1=1:9, x2=1:9,

Although there may well be many ways to do what is being asked for with the tidyverse, sometimes things are simple enough to do the old-fashioned way. The request seems to have been to do something to all rows in ONE specific group but was phrased in the sense of wanting to know which group your functionality is being called in. What grouping gains you is more worthwhile if you are interested in

Hi everyone!! I have dataset composed of a numbers of survival analyses. ( for batch survival analyses by using for-loop) . Here are code !! ####### dim(svsv) Num_t<-dim(svsv) Num<-Num_t[2] # These are predictors !! names=colnames(svsv) for (i in 1:Num ) { name_tt=names[i] survdiff(Surv(survival.m, survival) ~ names[i], data=svsv) fit.Group<-survfit(Surv(survival.m, survival) ~

Hi list, I want to use the p-value from the survdiff function (package survival) to reuse within a function in a Kaplan-Meier plot. The p-value is somehow not a component of the value list ?! Thanks in advance -- A. Goralczyk G?ttingen, Ger.

ExpeRts, I'm trying to compare three survival curves using the function "survdiff" in the survival package. Following is my code and corresponding error message. > survdiff(Surv(st_months, status) ~ strata(BOR), data=mydata) Error in survdiff(Surv(st_months, status) ~ strata(BOR), data = mydata) : No groups to test When I check the "strata" of the variable. I get .

Dear all, Does anyone knows how I could extract the p-value in: > survdiff(Surv(tempo,status) ~ grupo,data=dados1,rho=1) Call: survdiff(formula = Surv(tempo, status) ~ grupo, data = dados1, rho = 1) N Observed Expected (O-E)^2/E (O-E)^2/V grupo=1 21 5.12 12.00 3.94 14.5 grupo=2 21 14.55 7.68 6.16 14.5 Chisq= 14.5 on 1 degrees of freedom,

I read through Harrington and Fleming (1982) but it is beyond my statistical comprehension. I have survival data for insects that have a very finite expiration date. I'm trying to test for differences in survival distributions between different groups. I understand that the medical field is most often dealing with censored data and that survival analysis, at least in the package survival,

dear all, I would like to know whether survdiff and survReg function in the survival package work for left-truncated and right-censored data. If not, what other functions can i use to make comparison between two survival curves with LTRC data. thanks for any help given sing yee