similar to: survfit & number of variables != number of variable names

Hello, Sorry if this has been still answered, I haven''t found nothing on it. I would love to know why ActiveRecord::Base#last doesn''t return an ActiveRecord::Relation just like all or where since an ActiveRecord::Relation can act more or less like an array (as specified here<https://github.com/rails/rails/commit/0a6833b6f701c8c8febadfe2f45e25df29493602> )? Thanks, have

Hi! First I would like to thank people working on the xapian project. I have been trying hard to find out how to access the indexed documents weight vectors without success. I found the query weights and the total weights of documents, but not the individual weights. Could somebody give me a hint? Georges Dupret

Hi! First I would like to thank people working on the xapian project. I have been trying hard to find out how to access the indexed documents weight vectors without success. I found the query weights and the total weights of documents, but not the individual weights. Could somebody give me a hint? Georges Dupret

Hi, In the following results I interpret exp(coef) as the factor that multiplies the base hazard rate if the corresponding variable is TRUE. For example, when the bucket is ks008 and fidelity <= 3, then the rate, compared to the base rate h_0(t), is h(t) = 0.200 h_0(t). My question is then, to what case does the base hazard rate correspond to? I would expect the reference to be the first

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

I am confused when trying the function survfit. my question is: what does the survival curve given by plot.survfit mean? is it the survival curve with different covariates at different points? or just the baseline survival curve? for example, I run the following code and get the survival curve #### library(survival) fit<-coxph(Surv(futime,fustat)~resid.ds+rx+ecog.ps,data=ovarian)

I've a data set with 60000 rows of data representing 6000+ distinct loans. I did a coxph() regression on it (see call below), but a subsequent survfit() call on the coxph object is almost certainly wrong. It gives n=6 when it should be more like 6000+ (I think) > survfit(resultag) Call: survfit.coxph(object = resultag) n events median 0.95LCL 0.95UCL 6 489 Inf

Hello, I create a plot from a coxph object called fit.ads4: plot(survfit(fit.ads4)) plot is located at: https://www.dropbox.com/s/9jswrzid7mp1u62/survfit%20plot.png I also create the following survfit statistics: > print(survfit(fit.ads4),print.rmean=T) Call: survfit(formula = fit.ads4) records n.max n.start events *rmean *se(rmean) median 0.95LCL 0.95UCL 203.0

Hi everyone, I am not new to R, but new to running survival models in R. I am trying to create some basic KM curves, using the following code: library(survival) library(KMsurv) (import data etc - basic right censored, with continuously observed time of death) sleepfit <- survfit(Surv(timeb, death), data = sleep) Here timeb is measured is survival in years, death is a 1/0 indicator (1 =

Hello, I am trying to compute the mdeian of the survival time from the function survfit: > fit <- survfit(Surv(time, status) ~ 1) > fit Call: survfit(formula = Surv(time, status) ~ 1) records n.max n.start events median 0.95LCL 0.95UCL 111 111 111 20 NA NA NA The results is NA? the fit$surv gives values between 1 and 0.749! Am I doing this correct?

Hi! I would like to use the lists of stopwords provided with Xapian. Are there some standard way to remove stopwords automatically, or should I implement it mysel in the indexer? Regards, Georges Dupret

Hi, I am attempting to graph a Kaplan Meier estimate for some claims using the survfit function. However, I was wondering if it is possible to plot a cdf of the kaplan meier rather than the survival function. Here is some of my code: library(survival) Surv(claimj,censorj==0) survfit(Surv(claimj,censorj==0)~1) surv.all<-survfit(Surv(claimj,censorj==0)~1) summary(surv.all) plot(surv.all)

Hi, I am wondering how the confidence interval for Kaplan-Meier estimator is calculated by survfit(). For example,  > summary(survfit(Surv(time,status)~1,data),times=10) Call: survfit(formula = Surv(rtime10, rstat10) ~ 1, data = mgi)  time n.risk n.event survival std.err lower 95% CI upper 95% CI    10    168      55    0.761  0.0282        0.707        0.818 I am trying to reproduce the

The answer to this may be obvious, but I was wondering in the rms package and the survfit(), how you can plot the censored time points as ticks. Take for example, library(survival) library(rms) foo <- data.frame(Time=c(1,2,3,4,5,6,10), Status=c(1,1,0,0,1,1,1)) answer <- survfit(Surv(foo$Time, foo$Status==1) ~1) # this shows the censored time points as ticks at Time = 3 and 4 plot(answer)

Hi All, Please pardon me if I am missing something obvious here. How do I get the Kaplan-Meier estimate function that is created by survfit and plotted by the code. fit <- survfit(Surv(time, status) , data=aml) plot(fit) That is, I need a function that will give me the survival estimate at a given time: \hat{S}(t). Thanks in advance. Ritwik Sinha ritwik.sinha at gmail.com | +12033042111 |

> Further I appreciate your new function survmean(). At the moment it > seems to be intended as internal, and not documented in the help. The computations done by print.survfit are now a part of the results returned by summary.survfit. See 'table' in the output list of ?summary.survfit. Both call an internal survmean() function to ensure that any future updates stay in

Dear All, Maybe it is a silly question. But I wasn't able to find it from manual or R site search. I was wondering what is the corresponding time axis for survival function outputs in "survfit". I think it is "survfit(.......)$time", but not 100% sure. If it is, is it possible we could make survival function outputs on the pre-specified time grid with fixed increment and

Hello, I'm trying to estimate a Cox proportional hazard model with time-varying covariates using coxph. The parameter estimates are fine but there is something wrong with the survival curves I get with survfit (results are not plausible). Let me explain why I think something's wrong. To make sure I'm setting up my data correctly to estimate a model with time-varying covariates, I

I have a function that generates data and tries to apply the survfit within it. I have tried: function(..){ x ..... survfit(Surv(tim,sts) ~ grp, data = x, envir = sys.frame(sys.parent())) } but I still get "Error: Object "x" not found". Is there a fix? (I use R-0.63.2 with the current survival4 package) Thanks E. S. Venkatraman

Full_Name: Jerome Asselin Version: 1.6.0 OS: RedHat 7.2 Submission from: (NULL) (24.83.203.63) Hello, I am trying to draw a legend on top of survival curves using plot.survfit(). As in the example below, I would like to specify a large interval for the x-axis. I can achieve such result using "xlim". However, an error occurs if I use the legend.pos and legend.text parameters as well.