search for: 0.0556

Greetings, I am having a problem with DescTools::Quantile (a function computing quantiles from weighted samples): # these sum to one probWeights = c( 0.0043, 0.0062, 0.0087, 0.0119, 0.0157, 0.0204, 0.0257, 0.0315, 0.0378, 0.0441, 0.0501, 0.0556, 0.06, 0.0632, 0.0648, 0.0648, 0.0632, 0.06, 0.0556, 0.0501, 0.0441, 0.0378, 0.0315, 0.0257, 0.0204, 0.0157, 0.0119, 0.0087,

OS X R 4.3.3 Colleagues I have created objects using the Surv function in the survival package: > FIT.1 Call: survfit(formula = FORMULA1) n events median 0.95LCL 0.95UCL SUBDATA$ARM=1, SUBDATA[, EXP.STRAT]=0 18 13 345 156 NA SUBDATA$ARM=2, SUBDATA[, EXP.STRAT]=1 13 5 NA 186 NA SUBDATA$ARM=2, SUBDATA[, EXP.STRAT]=2 5

Hi Dennis, look at the help page for summary.survfit, the Value n.event. G?ran On 2024-05-15 22:41, Dennis Fisher wrote: > OS X > R 4.3.3 > > Colleagues > > I have created objects using the Surv function in the survival package: >> FIT.1 > Call: survfit(formula = FORMULA1) > > n events median 0.95LCL 0.95UCL >

It looks like a homework assignment. It also looks like you didn't read the documentation carefully enough. The 'len.out' argument in seq is solely for specifying the length of a sequence. The 'quantile' function omputes the empirical quantile of raw data in the vector 'x' at cumulative probabilit(y)(ies) given in the weights' argument, with interpolation I'm

Hello: In the example below (or for a censored data) using survfit.coxph, can anyone point me to a link or a pdf as to how the probabilities appearing in bold under "summary(pred$surv)" are calculated? Do these represent acumulative probability distribution in time (not including censored time)? Thanks very much, parmee *fit <- coxph(Surv(futime, fustat) ~ age, data = ovarian)*

I have a fairly large set of data with the following attributes: >str(raw.data) `data.frame': 1429 obs. of 16 variables: $ TStamp :`POSIXlt', format: chr "2001-11-25 02:00:00" "2001-11-25 01:55:00" "2001-11-25 01:50:00" "2001-11-25 01:45:00" ... $ iPDT.AHU14.14: num 0.0122 0.0125 0.0120 0.0120 0.0122 ... $ iPDT.AHU14.15: num 0.0121

The function 'wilcox.test' in R and S gives (almost) identical results (see below). 'qwilcox' however, does not: > qwilcox(p,5,5) p: 0.025 0.975 -------------------- R> 3 22 S> 18 37 I originally wanted to ask a questions, but then I found the answer. Given the confusion I run into, I wonder if this experience is worth reporting. The

Having spent the last few weeks trying to decipher R, I feel I may finally be getting somewhere, but i'M still in need of some advice and all my tutors seem to be on holiday! Basically a bit of background, I have data collected on a population of Lizards which includes age,sex, and body condition. I collected data myself this year and I have data previously collected from 1999, 2002 and

Dear UseRs, I'm running a chi-squared test where the expected matrix is the same as the observed, after rounding. R reports a X-squared of zero with a p value of one. I can justify this because any other result will deviate at least as much from the expected because what we observe is the expected, after rounding. But the formula for X-squared, sum (O-E)^2/E gives a positive value. What

require(randomForest) rf.pred<-predict(fit, valid, type="prob") > rf.pred[1:20, ] 0 1 16 0.0000 1.0000 23 0.3158 0.6842 43 0.3030 0.6970 52 0.0886 0.9114 55 0.1216 0.8784 75 0.0920 0.9080 82 0.4332 0.5668 120 0.2302 0.7698 128 0.1336 0.8664 147 0.4272 0.5728 148 0.0490 0.9510 153 0.0556 0.9444 161 0.0760 0.9240 162 0.4564 0.5436 172 0.5148 0.4852 176 0.1730

Dear list, I am facing an unusual situation where I need to create two sets of legends based on the color mapping. Can't get exactly what I want and really appreciate any advice from ggplot experts. Let's say I have the first dataset "df1" that draws some points and based on which a "loess" line with confidence interval is added. Then the second dataset

I am using the coxph, survfit and summary.survfit functions to calculate an estimate of predicted survival with confidence interval for future patients based on the survival distribution of an existing cohort of subjects. I am trying to understand the calculation and interpretation of the std.err and confidence intervals printed by the summary.survfit function. Using the default confidence

Dear all, I'm using R 2.8.1 under Windows Vista on a dual core 2,4 GhZ with 4 GB of RAM. I'm trying to reproduce a result out of "Analysis of Financial Time Series" by Ruey Tsay. In R I'm using the fGarch library. After fitting a ar(3)-garch(1,1)-model > model<-garchFit(~arma(3,0)+garch(1,1), analyse) I'm saving the results via > result<-model

Hi, Sean: I'm sorry I lie. I didn't mean to lie. I did try to avoid making a *BIG* change to the IPO pass-ordering for now. However, when I make a minor change to populateLTOPassManager() by separating module-pass and non-module-passes, I saw quite a few performance difference, most of them are degradations. Attacking these degradations one by one in a piecemeal manner is wasting

Andy and I briefly discussed this the other day, we have not yet got chance to list a detailed pass order for the pre- and post- IPO scalar optimizations. This is wish-list in our mind: pre-IPO: based on the ordering he propose, get rid of the inlining (or just inline tiny func), get rid of all loop xforms... post-IPO: get rid of inlining, or maybe we still need it, only