60 lines
2.2 KiB
Plaintext
60 lines
2.2 KiB
Plaintext
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R Under development (unstable) (2019-08-23 r77061) -- "Unsuffered Consequences"
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Copyright (C) 2019 The R Foundation for Statistical Computing
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Platform: x86_64-pc-linux-gnu (64-bit)
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R is free software and comes with ABSOLUTELY NO WARRANTY.
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You are welcome to redistribute it under certain conditions.
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Type 'license()' or 'licence()' for distribution details.
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R is a collaborative project with many contributors.
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Type 'contributors()' for more information and
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'citation()' on how to cite R or R packages in publications.
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Type 'demo()' for some demos, 'help()' for on-line help, or
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'help.start()' for an HTML browser interface to help.
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Type 'q()' to quit R.
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> #
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> # Make sure that useless intervals do not cause issues, i.e., any that do
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> # not overlap at least one event time
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> #
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> library(survival)
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> test2 <- data.frame(time1 =c(1, 2, 5, 2, 1, 7, 3, 4, 8, 8, 3),
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+ time2 =c(2, 3, 6, 7, 8, 9, 9, 9,14,17, 5),
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+ event =c(1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0),
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+ x =c(1, 0, 0, 1, 0, 1, 1, 1, 0, 0, 500) )
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>
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> # The data set is the same as book3.R, except for the wild observation
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> # with x=500 whose time interval of (4,5) overlaps no events.
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>
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> fit1 <- coxph(Surv(time1, time2, event) ~ x, test2, subset=(x<100))
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> fit2 <- coxph(Surv(time1, time2, event) ~ x, test2)
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>
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> ii <- match(c("coefficients", "var", "loglik", "score", "iter",
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+ "wald.test", "concordance"), names(fit1))
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> all.equal(fit1[ii], fit2[ii])
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[1] TRUE
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> all.equal(c(fit1$residuals,0), fit2$residuals, check.attributes=FALSE)
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[1] TRUE
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>
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> # The mean differs condiderably, and so to the linear predictors
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>
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> # Now the same with a penalized model
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> fit3 <- coxph(Surv(time1, time2, event) ~ ridge(x, theta=.1), test2,
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+ subset= (x< 100))
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> fit4 <- coxph(Surv(time1, time2, event) ~ ridge(x, theta=.1), test2)
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> fit5 <- coxph(Surv(time1,time2, event) ~ x, test2,
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+ iter=0, init=fit4$coef)
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>
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> all.equal(fit3[ii], fit4[ii])
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[1] TRUE
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> all.equal(c(fit3$residuals,0), fit4$residuals, check.attributes=FALSE)
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[1] TRUE
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> all.equal(fit4$residuals, fit5$residuals)
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[1] TRUE
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>
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> proc.time()
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user system elapsed
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0.997 0.056 1.336
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