search for: meiers

Is there anything I can do about it? BTW: gcc is loosing the AVR backend, so I would assume, there will be a greater interest to this in llvm compared to the past. Thanks, Wilhelm Am 03.04.20 um 15:09 schrieb Wilhelm Meier via llvm-dev: > Should I create an issue in bugzilla for this? Just to be reminded ... > > Am 31.03.20 um 09:34 schrieb Wilhelm Meier via llvm-dev: >> Hi

Hi Dylan, looks ok now. One thing: the ISR is now: __vector_21: ; @__vector_21 __vector_21$local: sei push r0 push r1 in r0, 63 push r0 clr r0 push r24 lds r24, v1 sts v2, r24 pop r24 pop r0 out 63, r0 pop r1 pop r0 reti There are unneccessary push/pops of r1 and r0 too, since the clr is useless ... GCC had the same

Hey Wilhelm, That's a bug, the "interrupt" attribute is not being recognized by the backend. I have fixed it in https://github.com/llvm/llvm-project/commit/339b34266c1b54a9b5ff2f83cfb1da9cd8c9d90a Pull the latest LLVM and it should be fixed. On Tue, Mar 31, 2020 at 8:00 AM Wilhelm Meier <wilhelm.meier at hs-kl.de> wrote: > Hi Dylan, > > I used the following

Hi all, I am trying to draw a Kaplan-Meier curve and I found online that Kaplan - Meier estimates are computed with a function called km in the event package. Is there an update for that because when I choose to download packages in R,. there is no package called event, even though I have selected all the repositories. Thanks in advance, Eleni [[alternative HTML version deleted]]

Hey Wilhelm, Could you post the LLVM IR generated from your C++ file? This can be achieved with 'clang -S -emit-llvm' Cheers On Sat, Mar 28, 2020 at 6:36 PM Wilhelm Meier <wilhelm.meier at hs-kl.de> wrote: > Answering partly to myself there was a extern "C" missing. > > But the register pushes ans reti are still missing. > > Whats wrong? > > Am

Dear all, I have data from 1970 to 1990 for people above age 50. Now I want to calculate survival curves by age starting at age 50 using the Kaplan Meier Estimator. The problem I have is that there are already people in 1970 who are older than 50 years. I guess this is called delayed entry or left truncation (?). I thought the code would be: roland <- survfit(Surv(time=age.enter,

Is it possible to calculate Kaplan-Meier for left-truncated, right-censored data using survival5? -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request at

Hello, I have a question about the function lifetab in package KMsurv. The description of the output value surv says "the estimated survival function at the start of the intervals". Are these estimates the ones calculated via Kaplan-Meier probability of survival ? Thanks in advance! -- View this message in context:

When I try to the code from library(survival) of library(ISwR), the following code survfit(Surv(days,status==1)) that could produce Kaplan-Meier estimates shows the following error "Error in survfit(Surv(days, status == 1)) : Survfit requires a formula or a coxph fit as the first argument" How it can be done in R.2.10 -- View this message in context:

Dear help news reader, I'm trying to draw a Kaplan-Meier curve and would like to ask the news group for some help Supposing I have study comapring two drugs, "A", and "B" and I recorde the time to get to the clinical endpoint (Time), in my case becommming virus free. I have setup the following frame: Time c Drug 1 5 1 A 2 7 1 B 3 2 1 A 4 10 1

Hi Fello: I am asked to compute the Kaplan-Meier estimator of data with right censoring without using surfit(). Does anyone know how to use R to compute the estimators? The data should input X: vector of right-censored observed time for n individuals, and d: vector of failure time indicators (0=censored individual;1=unconsored individual) and the function should return with t: vector of sorted

Dear all, I have a stat question that may not be related to R, but I would like to have your advice. I have just read a medical paper in which the authors report the 1-p (where p is the cumulative survival probability from the Kaplan Meier curve) as incidence of disease. Specifically, the study followed ~12000 women on drug A and ~20000 women on drug B for 12 months. During that period

Dear R-users I am trying to make an adjusted Kaplan-Meier curve (using the Survival package) but I am having difficulty with plotting it so that the plot only shows the curves for the adjusted results. My data come from a randomised controlled trial, and I would like the adjusted Kaplan-Meier curve to only show two curves for the adjusted survival: one for those on treatment (Treatment==1)

Hi Dylan, the following code volatile uint8_t v1; volatile uint8_t v2; __attribute__((interrupt)) void __vector_21(void) { v2 = v1; } produces in C mode: 00000092 <__vector_21>: 92: 80 91 61 00 lds r24, 0x0061 ; 0x800061 <v1> 96: 80 93 60 00 sts 0x0060, r24 ; 0x800060 <__data_end> 9a: 08 95 ret and in C++ mode: 00000074

This is more of a general question without data. After doing 'survdiff', from the 'survival' package, on strata including four groups (so 4 curves on a Kaplan Meier curve) you get a chi squared p-value whether to reject the null hypothesis or not. Is there a method to followup with pairwise testing on the respective groups? I have searched the library but have come up with

Hello R users I have a question with Kaplan-Meier Curve with respect to my research. We have done a retrospective study on fillings in the tooth and their survival in relation to the many influencing factors. We had a long follow-up time (upto 8yrs for some variables). However, we decided to stop the analysis at the 6year follow up time, so that we can have uniform follow-up time for all the

Helo: After changing "involuntarily" some of the graphics parameters with the command par() (I did not know that changes with this command are permanent), now when I made a plot of the survival Kaplan-Meier function, the Y axis does not start at 1, and the X axis does starts at 0. The commands that I use are: library(survival) BROWN.SPV = Surv(BROWN$TEMPS, BROWN$DEF)

What is the best way to report the standard error when publishing Kaplan-Meier plots? In my field (Vascular Surgery), practitioners loosely refer to the "10% error" cutoff as the point at which to stop drawing the KM curve. I am interpreting this as the *standard error of the cumulative hazard*, although I'm having a difficult time finding some guidelines about this (perhaps I am

Thanks! The new are of xmega3 architecture, which is already included. So this should be simple. Where is the information about ISR-vector table, SRAM addresses and so on stored? -- Wilhelm Am 04.03.20 um 11:03 schrieb Dylan McKay: > Hey Wilhelm, > > This should be possible by editing the 'AVRDevices.td' [1]TableGen > definitions to add an entry for the newer chip types.

Hi, all dear R experts, I am really stuck by how to get the 5th percentile( with a 95% CI )of the Kaplan-meier estimator of survival function P(T>t). I have a simulated sample of lifetime Ts, then I set the KM estimator < km.fit<-survfit(Surv(T),type="kaplan-meier",data=Surv(T)) <quantile(km.fit,.05) However, it gave "Error in order...". Does someone have some