similar to: Coxph not converging with continuous variable

Displaying 20 results from an estimated 2000 matches similar to: "Coxph not converging with continuous variable"

2004 Apr 21
1
difference between coxph and cph
Hi. I am using Windows version of R 1.8.1. Being somewhat new to survival analysis, I am trying to compare cph (Design) with coxph (survival) for use with a survival data set. I was wondering why cph and coxph provide me with different confidence intervals for the hazard ratios for one of the variables. I was wondering if I am doing something wrong? Or if the two functions are calculating hazard
2008 May 09
1
predicting from coxph with pspline
Hello. I get a bit confused by the output from the predict function when used on an object from coxph in combination with p-spline, e.g. fit <- coxph(Surv(time1, time2, status)~pspline(x), Data) predict(fit, newdata=data.frame(x=1:2)) It seems like the output is somewhat independent of the x-values to predict at. For example x=1:2 gives the same result as x=21:22. Does the result span the
2017 Jul 28
0
problem with "unique" function
Most likely, previous computations have ended up giving slightly different values of say 0.13333. A pragmatic way out is to round to, say, 5 digits before applying unique. In this particular case, it seems like all numbers are multiples of 1/30, so another idea could be to multiply by 30, round, and divide by 30. -pd > On 28 Jul 2017, at 17:17 , li li <hannah.hlx at gmail.com> wrote:
2017 Jul 28
3
problem with "unique" function
I have the joint distribution of three discrete random variables z1, z2 and z3 which is captured by "z" and "prob" as described below. For example, the probability for z1=0.46667, z2=-1 and z3=-1 is 2.752e-13. Also, the probability adds up to 1. > head(z) z1 z2 z3 [1,] -0.46667 -1.0000 -1.0000 [2,] -0.33333 -0.9333 -0.9333 [3,] -0.20000 -0.8667 -0.8667
2010 Jun 11
3
Calculation of r squared from a linear regression
Hi, I'm trying to verify the calculation of coefficient of determination (r squared) for linear regression. I've done the calculation manually with a simple test case and using the definition of r squared outlined in summary(lm) help. There seems to be a discrepancy between the what R produced and the manual calculation. Does anyone know why this is so? What does the multiple r squared
2006 Jan 27
1
about lm restrictions...
Hello all R-users _question 1_ I need to make a statistical model and respective ANOVA table but I get distinct results for the T-test (in summary(lm.object) function) and the F-test (in anova(lm.object) ) shouldn't this two approach give me the same result, i.e to indicate the same significants terms in both tests??????? obs. The system has two restrictions: 1) sum( x_i ) = 1 2) sum(
2012 Jan 23
1
Jags problem
Hi, all: I met "Non-conforming parameters for function %*%" problem, when I run the Jags model in R. My model is like this: model{ for(i in 1:n){ for(j in 1:t[i]){ et[i,j]<-yt[i,j]-beta0+betax*xt[i,j]+betat*t[i,j] } for(a in 1:t[i]){ for(b in 1:t[i]){ sigma[i,a,b]<-pow(rho0,abs(t[a]-t[b])) } } phi[i]<-
2011 Mar 17
0
Gelman-Rubin convergence diagnostics via coda package
Dear, I'm trying to run diagnostics on MCMC analysis (fitting a log-linear model to rates data). I'm getting an error message when trying Gelman-Rubin shrink factor plot: >gelman.plot(out) Error in chol.default(W) : the leading minor of order 2 is not positive definite I take it that somewhere, somehow a matrix is singular, but how can that be remedied? My code: library(rjags)
2008 Nov 10
2
as.Data with minutes resolution
Hi, I have a vetor os dates with day and hour:minutes. > time1 <- c("03/08/08-11:00","03/08/08-11:10") > time1 <- as.Date(time1,"%d/%m/%y-%R") > summary(time1) Min. 1st Qu. Median Mean 3rd Qu. Max. "2008-08-03" "2008-08-03" "2008-08-03" "2008-08-03" "2008-08-03"
2012 Dec 17
3
simplifying code
Dear All,   I was wondering if you could help me with the following: I have the code:   tin <-0.5 tau <-24 output0 <-10 TIMELOW <-tin TIMEHIGH <-1*tau TIME1 <-c(seq(TIMELOW,TIMEHIGH, by = sign(TIMEHIGH-TIMELOW)*(tau-tin)/3))   then I would like to calculate:   cp1 <-output0*exp(-0.3*TIME1[1]) cp2 <-output0*exp(-0.3*TIME1[2]) cp3 <-output0*exp(-0.3*TIME1[3]) cp4
2010 Jan 12
0
Wishlist: Function 'difftime' to honor 'tzone' attribute (PR#14182)
Full_Name: Suharto Anggono Version: 2.8.1 OS: Windows Submission from: (NULL) (125.165.84.118) PR#14076 inspired me to write this. > t1 <- as.POSIXct("1970-01-01 00:00:00", tz="GMT") > t2 <- as.POSIXlt("1970-01-01 00:00:00", tz="GMT") > t1 - t2 Time difference of 7 hours Above, t1 and t2 represent the same time in the same specified
2008 Dec 28
2
zfs mount hangs
Hi, System: Netra 1405, 4x450Mhz, 4GB RAM and 2x146GB (root pool) and 2x146GB (space pool). snv_98. After a panic the system hangs on boot and manual attempts to mount (at least) one dataset in single user mode, hangs. The Panic: Dec 27 04:42:11 base ^Mpanic[cpu0]/thread=300021c1a20: Dec 27 04:42:11 base unix: [ID 521688 kern.notice] [AFT1] errID 0x00167f73.1c737868 UE Error(s) Dec 27
2004 Aug 26
2
text() with text, variables and math HOWTO?
Hello, One more question from the 'abusing R for blotting - particularly anally' department: How can I in the expression below make the '%~~%' show up as the aprrox-sign I want it to be? Thanks for any hint, Joh text( 500,1.5, cex=0.75, substitute( paste( OD[600][~nm], " of 1 at ", time1, " min ", "%~~%", time1h,
2008 Apr 09
2
fuzzy merge
Hi, I would like to merge two data frames. It is just that I want the merging to be done with some kind of a fuzzy criterion. Let me explain. My first data frame looks like this : ID1 time1 dt 1 2008-01-02 13:11 10 2 2008-01-02 14:20 20 3
2004 Aug 26
1
how to make lines() meet axis in autoscaled coordinate system?
Hello, I have an auto-scaled coordinate system and would like to add some clarifying lines to the graph - which ares supposed to meet the axis. >lines(c(0,time1,time1), c(1,1,0),lty=3) does what I want, BUT the second leg does not touch the x-axis since the auto-scaling of the y-axis does not start at'0' but slightly negative. I could now adjust the line length by 'trial and
2007 Jan 19
4
Newbie question: Statistical functions (e.g., mean, sd) in a "transform" statement?
Greetings listeRs - Given a data frame such as times time1 time2 time3 time4 1 70.408543 48.92378 7.399605 95.93050 2 17.231940 27.48530 82.962916 10.20619 3 20.279220 10.33575 66.209290 30.71846 4 NA 53.31993 12.398237 35.65782 5 9.295965 NA 48.929201 NA 6 63.966518 42.16304 1.777342 NA one can use "transform" to
2011 Jul 19
1
Measuring and comparing .C and .Call overhead
Further pursuing my curiosity to measure the efficiency of R/C++ interface, I conducted a simple matrix-vector multiplication test using .C and .Call functions in R. In each case, I measured the execution time in R, as well as inside the C++ function. Subtracting the two, I came up with a measure of overhead associated with each call. I assume that this overhead would be non-existent of the entire
2012 Jan 19
1
Legend problem in line charts
Hi all, Small problem in generating the line charts. Question: Legend for the first graph is coming wrong., for second graph correctly. Please fix the legend postion at the down of graph. Plesae give me the solution. Thank you Devarayalu Orange1 <- structure(list(REFID = c(7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9), ARM = c(1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2, 2,
2012 Dec 13
1
duplicated.data.frame() and POSIXct with DST shift
Hi, I encountered the behavior, that the duplicated method for data.frames gives "false positives" if there are columns of class POSIXct with a clock shift from DST to standard time. time <- as.POSIXct("2012-10-28 02:00", tz="Europe/Vienna") + c(0, 60*60) time [1] "2012-10-28 02:00:00 CEST" "2012-10-28 02:00:00 CET" df <-
2010 Nov 15
3
merge two dataset and replace missing by 0
Hi r users, I have two data sets (X1, X2). For example, time1<-c( 0, 8, 15, 22, 43, 64, 85, 106, 127, 148, 169, 190 ,211 ) outpue1<-c(171 ,164 ,150 ,141 ,109 , 73 , 47 ,26 ,15 ,12 ,6 ,2 ,1 ) X1<-cbind(time1,outpue1) time2<-c( 0 ,8 ,15 , 22 ,43 , 64 ,85 ,106 ,148) output2<-c( 5 ,5 ,4 ,5 ,5 ,4 ,1 ,2 , 1 ) X2<-cbind(time2,output2) I want to