similar to: cnfidence intervals for survfit()

The first thing you are missing is the documentation -- try ?survfit.object. fit <- survfit(Surv(time,status)~1,data) fit$std.err will contain the standard error of the cumulative hazard or -log(survival) The standard error of the survival curve is approximately S(t) * std(hazard), by the delta method. This is what is printed by the summary function, because it is what user's

http://bugzilla.mindrot.org/show_bug.cgi?id=341 ------- Additional Comments From norbert.weuster at mgi.de 2002-07-09 15:44 ------- the optinal Flags -n and/or -T doesn't change the situation ------- You are receiving this mail because: ------- You are the assignee for the bug, or are watching the assignee.

Dear R-Helpers, I've looked high and low for a function that provides frequencies, proportions and cumulative proportions side-by-side. Below is the table I need. Is there a function that already does it? Thanks, Bob > # Generate some test scores > myValues <- c(70:95) > Score <- ( sample( myValues, size=1000, replace=TRUE) ) > head(Score) [1] 77 71 81 88 83 93 > >

Hi, When I run the survfit function, I want to get the restricted mean value and the standard error also. I found out using the "print" function to do so, as shown below, print(km.fit,print.rmean=TRUE) Call: survfit(formula = Surv(diff, status) ~ 1, type = "kaplan-meier") records n.max n.start events *rmean *se(rmean) median 200.000

Hi R-experts, suppose I have a list with containing data frame elements: [[1]] (Intercept) y1 y2 y3 y4 -6.64 0.761 0.383 0.775 0.163 [[2]] (Intercept) y2 y3 -3.858 0.854 0.834 Now I want to put them into ONE dataframe like this: (Intercept) y1

http://bugzilla.mindrot.org/show_bug.cgi?id=341 Summary: Return Code unpredictable Product: Portable OpenSSH Version: -current Platform: Other OS/Version: AIX Status: NEW Severity: normal Priority: P3 Component: ssh AssignedTo: openssh-unix-dev at mindrot.org ReportedBy: norbert.weuster at

http://bugzilla.mindrot.org/show_bug.cgi?id=400 Summary: ssh-keygen hangs Product: Portable OpenSSH Version: -current Platform: All URL: http://www.mgi-networks.com/ OS/Version: AIX Status: NEW Severity: normal Priority: P2 Component: ssh-keygen AssignedTo: openssh-unix-dev at

Hi all My S4 is configured as a file server. The AD DC is a 2003 win server. My users are talking about slow speed on open pst file or other largest file. I have this in my smb.conf socket options = TCP_NODELAY SO_RCVBUF=32768 SO_SNDBUF=32768 I tranfer a 3GB file on the server and I see variations in speed transfer. Generally it cost 28% of my bandwith (1 Gb/s) but, after a while, it

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)*

Dear r-helpers, xx <- c(0.000, 0.210, 0.714, 0.514, 1.000, 0.190, 0.590, 0.152) yy <- c(0.000, 0.265, 0.256, 0.521, 0.538, 0.761, 0.821, 1.000) aa <- c(19, 19, 19, 21, 19, 21, 21, 21) x0 <- xx[c(1, 1, 2, 2, 2, 3, 3, 4, 4, 4, 5, 6, 6, 7)] y0 <- yy[c(1, 1, 2, 2, 2, 3, 3, 4, 4, 4, 5, 6, 6, 7)] x1 <- xx[c(2, 3, 3, 4, 6, 4, 5, 5, 6, 7, 7, 7, 8, 8)] y1 <- yy[c(2, 3, 3, 4, 6, 4, 5,

Hi, I am trying to make a multi-class classification tree by using rpart. I used MASS package'd data: fgl to test and it works well. However, when I used my small-sampled data as below, the program seems to take forever. I am not sure if it is due to slowness or there is something wrong with my codes or data manipulation. Please be advised ! The data is described as the output from str()

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 R People: Here is a small data frame and two particular formulas: > test.df y x 1 -0.9261650 1 2 1.5702700 2 3 0.1673920 3 4 0.7893085 4 5 0.3576875 5 6 -1.4620915 6 7 -0.5506215 7 8 -0.3480292 8 9 -1.2344036 9 10 0.8502660 10 > summary(lm(exp(y)~x)) Call: lm(formula = exp(y) ~ x) Residuals: Min 1Q Median 3Q Max -1.6360 -0.6435

Dear All, I have just estimated this model: ----------------------------------------------------------- Logistic Regression Model lrm(formula = Y ~ X16, x = T, y = T) Model Likelihood Discrimination Rank Discrim. Ratio Test Indexes Indexes Obs 82 LR chi2 5.58 R2 0.088 C 0.607 0

Hi! I came across R just a few days ago since I was looking for a toolbox for cox-regression. I?ve read "Cox Proportional-Hazards Regression for Survival Data Appendix to An R and S-PLUS Companion to Applied Regression" from John Fox. As described therein plotting survival-functions works well (plot(survfit(model))). But I?d like to do some manipulation with the survival-functions

Consider the following example (based on an example in Pat Altham's GLM notes) pyears <- scan() 18793 52407 10673 43248 5710 28612 2585 12663 1462 5317 deaths <- scan() 2 32 12 104 28 206 28 186 31 102 Smoke <- gl(2,1,10,labels=c("No","Yes")) Age <- gl(5,2,10,labels=c("35-44","45-54","55-64","65-74","75-84"),

http://bugzilla.mindrot.org/show_bug.cgi?id=341 ------- Additional Comments From djm at mindrot.org 2003-01-07 18:26 ------- Please verify that this is still the case with 3.5p1 (2.9 is ooold) ------- You are receiving this mail because: ------- You are the assignee for the bug, or are watching the assignee.

Hi, I have searched the archives and can't quite confirm the answer to this. I appreciate your time... I have 4 treatments (fixed) and I would like to know if there is a significant difference in metal volume (metal) between the treatments. The experiment has 5 blocks (random) in each treatment and no block is repeated across treatments. Within each plot there are varying numbers of

Using the small reproducible example below, I'd like to know if one can somehow use the matrix "sig" (defined below) to add a black outline (with lwd=2) to all pixels with a corresponding value of 1 in the matrix 'sig'? So for example, in the ggplot2 plot below, the pixel located at [1,3] would be outlined by a black square since the value at sig[1,3] == 1. This is my first

Hi I am working on corpora of automatically recognized utterances, looking for features that predict error in the hypothesis the recognizer is proposing. I am using the glm functions to do logistic regression. I do this type of thing: * logistic.model = glm(formula = similarity ~., family = binomial, data = data) and end up with a model: > summary(logistic.model) Call: