similar to: survfit function?

Hi All I found survfit function was very slow for a large dataset and I am looking for an alternative way to quickly get the predicted survival probabilities. My historical data set is a pool of loans with monthly observed default status for 24 months. I would like to fit the proportional hazard model with time varying covariate such as unemployment rates and time constant variables at loan

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

Question: When survfit() function is used upon a coxph object, the 'n' returned is vastly smaller (n=6) than the number of distinct loans in the dataset used. I am trying to estimate a Cox proportional hazards model for a set of loans (over 6000) using using time varying covariates. For this 6000+ loans, I have some 62,000 different vectors representing the loans at different periods of

Hey all, first time poster here. I'm new to R and working on my first real programming and forecasting asignment. I'm using unemployment data from 1948-2012. I successfully completed part a and the linear fit for part b, but i am really struggling fitting a polynomial with a power greater than 2 to my forecast. I'll upload my R code at the bottom. Any help is very much appreciated!

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Hello, I'm interested in correcting for and measuring unobserved heterogeneity ("missing variables") using R. In particular, I'm searching for a simple way to measure the amount of unobserved heterogeneity remaining in a series of increasingly complex models (adding additional variables to each new model) on the same data. I have a static database of 400,000 or

my models borrower ----- has_many :loans loan ----- belongs_to :borrower my loans _controller def new @borrower = Borrower.find(params[:borrower_id]) logger.debug '' @borrower.id'' logger.debug @borrower.id @loan = Loan.new respond_to do |format| format.html # new.html.erb format.xml { render :xml => @loan } end

In Germany the Unemployment Agency uses a sectioning of the german map that is different from the usual Administrative Boundaries. Some demographic data are available in Administrative Boundaries only, some in Unemployment Boundaries only. I would like to generate estimates in one boundary system of data availabe in the other boundary system, and would appreciate advice concerning the following

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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 trying to remove some text after the period (a decimal point) in the data frame 'hi', below. This is one step in formatting a table. So I would like e.g. "2.0" to become "2" and "5.3" to be "5.3", where the variable digordered contains the number of digits after the decimal that I would like to display, in the same order in which

Terry, I recently noticed the censor argument of survfit. For some analyses it greatly reduces the size of the resulting object, which is a nice feature. However, when combined with the id argument, only 1 prediction is made. Predictions can be made individually but I'd prefer to do them all at once if that change can be made. Chris ##################################### # CODE # create

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)

Dear all, I'm having difficulty getting access to data generated by survfit and print.survfit when they are using with a Cox model (survfit.coxph). I would like to programmatically access the median survival time for each strata together with the 95% confidence interval. I can get it on screen, but can't get to it algorithmically. I found myself examining the source of print.survfit to

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

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.

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

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 =

For a fitted Cox model, one can either produce the predicted survival curve for a particular "hypothetical" subject (survfit), or the predicted curve for a particular cohort of subjects (survexp). See chapter 10 of Therneau and Grambsch for a long discussion of the differences between these, and the various pitfalls. By default, survfit produces the curve for a hypothetical