436 lines
17 KiB
Plaintext
436 lines
17 KiB
Plaintext
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R Under development (unstable) (2024-04-17 r86441) -- "Unsuffered Consequences"
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Copyright (C) 2024 The R Foundation for Statistical Computing
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Platform: aarch64-unknown-linux-gnu
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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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> options(na.action=na.exclude) # preserve missings
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> options(contrasts=c('contr.treatment', 'contr.poly')) #ensure constrast type
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> library(survival)
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>
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> #
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> # Reproduce example 1 in the SAS lifereg documentation
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> #
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>
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> # this fit doesn't give the same log-lik that they claim
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> fit1 <- survreg(Surv(time, status) ~ I(1000/(273.2+temp)), imotor,
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+ subset=(temp>150), dist='lognormal')
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> summary(fit1)
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Call:
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survreg(formula = Surv(time, status) ~ I(1000/(273.2 + temp)),
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data = imotor, subset = (temp > 150), dist = "lognormal")
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Value Std. Error z p
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(Intercept) -10.471 2.772 -3.78 0.00016
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I(1000/(273.2 + temp)) 8.322 1.284 6.48 9.1e-11
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Log(scale) -0.504 0.183 -2.75 0.00596
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Scale= 0.604
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Log Normal distribution
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Loglik(model)= -145.9 Loglik(intercept only)= -155
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Chisq= 18.3 on 1 degrees of freedom, p= 1.9e-05
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Number of Newton-Raphson Iterations: 6
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n= 30
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>
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> # This one, with the loglik on the transformed scale (the inappropriate
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> # scale, Ripley & Venables would argue) does agree.
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> # All coefs are of course identical.
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> fit2 <- survreg(Surv(log(time), status) ~ I(1000/(273.2+temp)), imotor,
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+ subset=(temp>150), dist='gaussian')
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>
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>
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> # Give the quantile estimates, which is the lower half of "output 48.1.5"
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> # in the SAS 9.2 manual
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>
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> pp1 <- predict(fit1, newdata=list(temp=c(130,150)), p=c(.1, .5, .9),
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+ type='quantile', se=T)
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> pp2 <- predict(fit1, newdata=list(temp=c(130,150)), p=c(.1, .5, .9),
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+ type='uquantile', se=T)
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> pp1
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$fit
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[,1] [,2] [,3]
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[1,] 12033.185 26095.677 56592.20
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[2,] 4536.877 9838.864 21336.98
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$se.fit
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[,1] [,2] [,3]
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[1,] 5482.338 11359.450 26036.917
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[2,] 1443.072 2901.155 7172.343
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>
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> temp130 <- matrix(0, nrow=3, ncol=6)
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> temp130[,1] <- pp1$fit[1,]
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> temp130[,2] <- pp1$se.fit[1,]
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> temp130[,3] <- pp2$fit[1,]
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> temp130[,4] <- pp2$se.fit[1,]
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> temp130[,5] <- exp(pp2$fit[1,] - 1.64*pp2$se.fit[1,])
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> temp130[,6] <- exp(pp2$fit[1,] + 1.64*pp2$se.fit[1,])
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> dimnames(temp130) <- list(c("p=.1", "p=.2", "p=.3"),
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+ c("Time", "se(time)", "log(time)", "se[log(time)]",
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+ "lower 90", "upper 90"))
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> print(temp130)
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Time se(time) log(time) se[log(time)] lower 90 upper 90
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p=.1 12033.18 5482.338 9.395424 0.4556015 5700.089 25402.68
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p=.2 26095.68 11359.450 10.169525 0.4353001 12779.950 53285.37
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p=.3 56592.20 26036.917 10.943626 0.4600796 26611.422 120349.71
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>
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> # A set of examples, copied from the manual pages of SAS procedure
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> # "reliability", which is part of their QC product.
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> #
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>
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> color <- c("black", "red", "green", "blue", "magenta", "red4",
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+ "orange", "DarkGreen", "cyan2", "DarkViolet")
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> palette(color)
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> pdf(file='reliability.pdf')
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>
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> #
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> # Insulating fluids example
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> #
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>
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> # Adding a -1 to the fit just causes the each group to have it's own
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> # intercept, rather than a global intercept + constrasts. The strata
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> # statement allows each to have a separate scale
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> ffit <- survreg(Surv(time) ~ factor(voltage) + strata(voltage) -1, ifluid)
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>
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> # Get predicted quantiles at each of the voltages
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> # By default predict() would give a line of results for each observation,
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> # I only want the unique set of x's, i.e., only 4 cases
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> uvolt <- sort(unique(ifluid$voltage)) #the unique levels
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> plist <- c(1, 2, 5, 1:9 *10, 95, 99)/100
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> pred <- predict(ffit, type='quantile', p=plist,
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+ newdata=data.frame(voltage=uvolt))
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> tfun <- function(x) log(-log(1-x))
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>
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> matplot(t(pred), tfun(plist), type='l', log='x', lty=1,
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+ col=1:4, yaxt='n',
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+ xlab="Predicted", ylab="Quantile")
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> axis(2, tfun(plist), format(100*plist), adj=1)
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>
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> kfit <- survfit(Surv(time) ~ voltage, ifluid, type='fleming') #KM fit
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> for (i in 1:4) {
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+ temp <- kfit[i]
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+ points(temp$time, tfun(1-temp$surv), col=i, pch=i)
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+ }
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>
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> # Now a table
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> temp <- array(0, dim=c(4,4,4)) #4 groups by 4 parameters by 4 stats
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> temp[,1,1] <- ffit$coef # "EV Location" in SAS manual
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> temp[,2,1] <- ffit$scale # "EV scale"
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> temp[,3,1] <- exp(ffit$coef) # "Weibull Scale"
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> temp[,4,1] <- 1/ffit$scale # "Weibull Shape"
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>
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> temp[,1,2] <- sqrt(diag(ffit$var))[1:4] #standard error
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> temp[,2,2] <- sqrt(diag(ffit$var))[5:8] * ffit$scale
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> temp[,3,2] <- temp[,1,2] * temp[,3,1]
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> temp[,4,2] <- temp[,2,2] / (temp[,2,1])^2
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>
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> temp[,1,3] <- temp[,1,1] - 1.96*temp[,1,2] #lower conf limits
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> temp[,1,4] <- temp[,1,1] + 1.96*temp[,1,2] # upper
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> # log(scale) is the natural parameter, in which the routine did its fitting
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> # and on which the std errors were computed
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> temp[,2, 3] <- exp(log(ffit$scale) - 1.96*sqrt(diag(ffit$var))[5:8])
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> temp[,2, 4] <- exp(log(ffit$scale) + 1.96*sqrt(diag(ffit$var))[5:8])
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>
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> temp[,3, 3:4] <- exp(temp[,1,3:4])
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> temp[,4, 3:4] <- 1/temp[,2,4:3]
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>
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> dimnames(temp) <- list(uvolt, c("EV Location", "EV Scale", "Weibull scale",
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+ "Weibull shape"),
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+ c("Estimate", "SE", "lower 95% CI", "uppper 95% CI"))
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> print(aperm(temp, c(2,3,1)), digits=5)
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, , 26
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Estimate SE lower 95% CI uppper 95% CI
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EV Location 6.86249 1.10404 4.69857 9.0264
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EV Scale 1.83423 0.96114 0.65677 5.1227
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Weibull scale 955.74665 1055.18620 109.78973 8320.0103
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Weibull shape 0.54519 0.28568 0.19521 1.5226
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, , 30
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Estimate SE lower 95% CI uppper 95% CI
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EV Location 4.35133 0.30151 3.76037 4.9423
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EV Scale 0.94446 0.22544 0.59156 1.5079
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Weibull scale 77.58159 23.39176 42.96420 140.0911
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Weibull shape 1.05881 0.25274 0.66318 1.6904
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, , 34
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Estimate SE lower 95% CI uppper 95% CI
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EV Location 2.50326 0.31476 1.88632 3.1202
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EV Scale 1.29732 0.22895 0.91796 1.8334
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Weibull scale 12.22222 3.84707 6.59509 22.6506
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Weibull shape 0.77082 0.13603 0.54542 1.0894
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, , 38
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Estimate SE lower 95% CI uppper 95% CI
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EV Location 0.00092629 0.27318 -0.53450 0.53635
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EV Scale 0.73367610 0.20380 0.42565 1.26460
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Weibull scale 1.00092672 0.27343 0.58596 1.70976
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Weibull shape 1.36299929 0.37861 0.79077 2.34933
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>
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> rm(temp, uvolt, plist, pred, ffit, kfit)
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>
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> #####################################################################
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> # Turbine cracks data
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> crack2 <- with(cracks, data.frame(day1=c(NA, days), day2=c(days, NA),
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+ n=c(fail, 167-sum(fail))))
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> cfit <- survreg(Surv(day1, day2, type='interval2') ~1,
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+ dist='weibull', data=crack2, weights=n)
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>
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> summary(cfit)
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Call:
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survreg(formula = Surv(day1, day2, type = "interval2") ~ 1, data = crack2,
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weights = n, dist = "weibull")
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Value Std. Error z p
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(Intercept) 7.6880 0.0744 103.30 < 2e-16
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Log(scale) -0.3953 0.0987 -4.01 6.2e-05
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Scale= 0.674
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Weibull distribution
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Loglik(model)= -309.6 Loglik(intercept only)= -309.6
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Number of Newton-Raphson Iterations: 5
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n= 9
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> #Their output also has Wiebull scale = exp(cfit$coef), shape = 1/(cfit$scale)
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>
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> # Draw the SAS plot
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> # The "type=fleming" argument reflects that they estimate hazards rather than
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> # survival, and forces a Nelson-Aalen hazard estimate
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> #
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> plist <- c(1, 2, 5, 1:8 *10)/100
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> plot(qsurvreg(plist, cfit$coef, cfit$scale), tfun(plist), log='x',
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+ yaxt='n', type='l',
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+ xlab="Weibull Plot for Time", ylab="Percent")
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> axis(2, tfun(plist), format(100*plist), adj=1)
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>
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> kfit <- survfit(Surv(day1, day2, type='interval2') ~1, data=crack2,
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+ weight=n, type='fleming')
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> # Only plot point where n.event > 0
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> # Why? I'm trying to match them. Personally, all should be plotted.
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> who <- (kfit$n.event > 0)
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> points(kfit$time[who], tfun(1-kfit$surv[who]), pch='+')
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> points(kfit$time[who], tfun(1-kfit$upper[who]), pch='-')
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> points(kfit$time[who], tfun(1-kfit$lower[who]), pch='-')
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>
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> text(rep(3,6), seq(.5, -1.0, length=6),
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+ c("Scale", "Shape", "Right Censored", "Left Censored",
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+ "Interval Censored", "Fit"), adj=0)
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> text(rep(9,6), seq(.5, -1.0, length=6),
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+ c(format(round(exp(cfit$coef), 2)),
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+ format(round(1/cfit$scale, 2)),
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+ format(tapply(crack2$n, cfit$y[,3], sum)), "ML"), adj=1)
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>
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> # Now a portion of his percentiles table
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> # I don't get the same SE as SAS, I haven't checked out why. The
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> # estimates and se for the underlying Weibull model are the same.
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> temp <- predict(cfit, type='quantile', p=plist, se=T)
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> tempse <- sqrt(temp$se[1,])
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> mat <- cbind(temp$fit[1,], tempse,
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+ temp$fit[1,] -1.96*tempse, temp$fit[1,] + 1.96*tempse)
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> dimnames(mat) <- list(plist*100, c("Estimate", "SE", "Lower .95", "Upper .95"))
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> print(mat)
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Estimate SE Lower .95 Upper .95
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1 98.47183 5.321676 88.04134 108.9023
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2 157.59360 6.186246 145.46856 169.7186
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5 295.17088 7.377034 280.71189 309.6299
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10 479.31737 8.227489 463.19149 495.4432
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20 794.55351 8.951640 777.00829 812.0987
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30 1089.70373 9.404613 1071.27068 1108.1368
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40 1387.95516 9.983696 1368.38712 1407.5232
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50 1704.71015 10.888487 1683.36872 1726.0516
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60 2057.23906 12.217230 2033.29329 2081.1848
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70 2472.58593 14.035515 2445.07632 2500.0955
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80 3006.43572 16.510569 2974.07501 3038.7964
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>
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> #
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> # The cracks data has a particularly easy estimate, so use
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> # it to double check code
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> time <- c(crack2$day2[1], (crack2$day1 + crack2$day2)[2:8]/2,
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+ crack2$day1[9])
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> cdf <- cumsum(crack2$n)/sum(crack2$n)
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> all.equal(kfit$time, time)
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[1] TRUE
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> all.equal(kfit$surv, 1-cdf[c(1:8,8)])
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[1] TRUE
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> rm(time, cdf, kfit)
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>
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>
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> #######################################################
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> #
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> # Replacement of valve seats in diesel engines
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> # The input data has id, time, and an indicator of whether there was an
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> # event at that time: 0=no, 1=yes. No one has an event at their last time.
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> # The input data has two engines with dual failures: 328 loses 2 valves at
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> # time 653, and number 402 loses 2 at time 139. For each, fudge the first
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> # time to be .1 days earlier.
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> #
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> ties <- which(diff(valveSeat$time)==0 & diff(valveSeat$id)==0)
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> temp <- valveSeat$time
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> temp[ties] <- temp[ties] - .1
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> n <- length(temp)
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> first <- !duplicated(valveSeat$id)
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> vtemp <- with(valveSeat, data.frame(id =id,
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+ time1= ifelse(first, 0, c(0, temp[-n])),
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+ time2= temp, status=status))
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>
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> kfit <- survfit(Surv(time1, time2, status) ~1, vtemp, id=id)
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>
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> plot(kfit, fun='cumhaz', ylab="Sample Mean Cumulative Failures", xlab='Time')
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> title("Valve replacement data")
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>
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> # The summary.survfit function doesn't have an option for printing out
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> # cumulative hazards instead of survival --- need to add that
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> # so I just reprise the central code of print.summary.survfit
|
||
|
|
> xx <- summary(kfit)
|
||
|
|
> temp <- cbind(xx$time, xx$n.risk, xx$n.event, xx$cumhaz,
|
||
|
|
+ xx$std.chaz, -log(xx$upper), -log(xx$lower))
|
||
|
|
> dimnames(temp) <- list(rep("", nrow(temp)),
|
||
|
|
+ c("time", "n.risk", "n.event", "Cum haz", "std.err",
|
||
|
|
+ "lower 95%", "upper 95%"))
|
||
|
|
> print(temp, digits=2)
|
||
|
|
time n.risk n.event Cum haz std.err lower 95% upper 95%
|
||
|
|
61 41 1 0.024 0.024 0.0000 0.073
|
||
|
|
76 41 1 0.049 0.034 0.0000 0.117
|
||
|
|
84 41 1 0.073 0.041 0.0000 0.156
|
||
|
|
87 41 1 0.098 0.046 0.0057 0.192
|
||
|
|
92 41 1 0.122 0.051 0.0208 0.226
|
||
|
|
98 41 1 0.146 0.055 0.0373 0.259
|
||
|
|
120 41 1 0.171 0.059 0.0548 0.291
|
||
|
|
139 41 1 0.195 0.062 0.0732 0.322
|
||
|
|
139 41 1 0.220 0.073 0.0750 0.369
|
||
|
|
165 41 1 0.244 0.075 0.0954 0.398
|
||
|
|
166 41 1 0.268 0.077 0.1163 0.427
|
||
|
|
202 41 1 0.293 0.079 0.1376 0.455
|
||
|
|
206 41 1 0.317 0.088 0.1452 0.497
|
||
|
|
249 41 1 0.341 0.089 0.1675 0.524
|
||
|
|
254 41 1 0.366 0.090 0.1903 0.551
|
||
|
|
258 41 1 0.390 0.090 0.2133 0.577
|
||
|
|
265 41 1 0.415 0.091 0.2368 0.603
|
||
|
|
276 41 1 0.439 0.098 0.2479 0.641
|
||
|
|
298 41 1 0.463 0.110 0.2490 0.689
|
||
|
|
323 41 1 0.488 0.110 0.2734 0.714
|
||
|
|
326 41 1 0.512 0.110 0.2981 0.739
|
||
|
|
328 41 1 0.537 0.115 0.3124 0.774
|
||
|
|
344 41 1 0.561 0.115 0.3376 0.798
|
||
|
|
348 41 1 0.585 0.124 0.3430 0.842
|
||
|
|
349 41 1 0.610 0.124 0.3686 0.866
|
||
|
|
367 41 1 0.634 0.123 0.3945 0.890
|
||
|
|
377 41 1 0.659 0.132 0.4018 0.932
|
||
|
|
404 40 1 0.684 0.136 0.4189 0.965
|
||
|
|
408 40 1 0.709 0.140 0.4365 0.998
|
||
|
|
410 40 1 0.734 0.139 0.4631 1.022
|
||
|
|
449 40 1 0.759 0.143 0.4813 1.055
|
||
|
|
479 40 1 0.784 0.146 0.4998 1.087
|
||
|
|
497 40 1 0.809 0.149 0.5187 1.118
|
||
|
|
538 40 1 0.834 0.152 0.5380 1.150
|
||
|
|
539 40 1 0.859 0.155 0.5576 1.181
|
||
|
|
561 40 1 0.884 0.162 0.5696 1.220
|
||
|
|
563 40 1 0.909 0.164 0.5898 1.250
|
||
|
|
570 40 1 0.934 0.170 0.6029 1.287
|
||
|
|
573 40 1 0.959 0.169 0.6310 1.310
|
||
|
|
581 38 1 0.985 0.171 0.6532 1.341
|
||
|
|
586 34 1 1.014 0.174 0.6777 1.376
|
||
|
|
604 22 1 1.060 0.185 0.7009 1.446
|
||
|
|
621 17 1 1.119 0.208 0.7144 1.554
|
||
|
|
635 16 1 1.181 0.207 0.7795 1.618
|
||
|
|
640 16 1 1.244 0.226 0.8033 1.723
|
||
|
|
646 13 1 1.320 0.229 0.8775 1.809
|
||
|
|
653 9 1 1.432 0.252 0.9394 1.983
|
||
|
|
653 9 1 1.543 0.312 0.9201 2.238
|
||
|
|
>
|
||
|
|
> # Note that I have the same estimates but different SE's than SAS. We are using
|
||
|
|
> # a different estimator. It's a statistical argument as to which is
|
||
|
|
> # better (one could defend both sides): SAS the more standard estimate found
|
||
|
|
> # in the reliability literature, mine the estimate from statistics literature
|
||
|
|
> rm(temp, kfit, xx)
|
||
|
|
>
|
||
|
|
> ######################################################
|
||
|
|
> # Turbine data, lognormal fit
|
||
|
|
> turbine <- read.table('data.turbine',
|
||
|
|
+ col.names=c("time1", "time2", "n"))
|
||
|
|
>
|
||
|
|
> tfit <- survreg(Surv(time1, time2, type='interval2') ~1, turbine,
|
||
|
|
+ dist='lognormal', weights=n, subset=(n>0))
|
||
|
|
>
|
||
|
|
> summary(tfit)
|
||
|
|
|
||
|
|
Call:
|
||
|
|
survreg(formula = Surv(time1, time2, type = "interval2") ~ 1,
|
||
|
|
data = turbine, weights = n, subset = (n > 0), dist = "lognormal")
|
||
|
|
Value Std. Error z p
|
||
|
|
(Intercept) 3.6999 0.0708 52.23 <2e-16
|
||
|
|
Log(scale) -0.3287 0.1232 -2.67 0.0076
|
||
|
|
|
||
|
|
Scale= 0.72
|
||
|
|
|
||
|
|
Log Normal distribution
|
||
|
|
Loglik(model)= -190.7 Loglik(intercept only)= -190.7
|
||
|
|
Number of Newton-Raphson Iterations: 6
|
||
|
|
n= 21
|
||
|
|
|
||
|
|
>
|
||
|
|
> # Now, do his plot, but put bootstrap confidence bands on it!
|
||
|
|
> # First, make a simple data set without weights
|
||
|
|
> tdata <- turbine[rep(1:nrow(turbine), turbine$n),]
|
||
|
|
>
|
||
|
|
> qstat <- function(data) {
|
||
|
|
+ temp <- survreg(Surv(time1, time2, type='interval2') ~1, data=data,
|
||
|
|
+ dist='lognormal')
|
||
|
|
+ qsurvreg(plist, temp$coef, temp$scale, dist='lognormal')
|
||
|
|
+ }
|
||
|
|
>
|
||
|
|
> {if (exists('bootstrap')) {
|
||
|
|
+ set.seed(1953) # a good year :-)
|
||
|
|
+ bfit <- bootstrap(tdata, qstat, B=1000)
|
||
|
|
+ bci <- limits.bca(bfit, probs=c(.025, .975))
|
||
|
|
+ }
|
||
|
|
+ else {
|
||
|
|
+ values <- matrix(0, nrow=1000, ncol=length(plist))
|
||
|
|
+ n <- nrow(tdata)
|
||
|
|
+ for (i in 1:1000) {
|
||
|
|
+ subset <- sample(1:n, n, replace=T)
|
||
|
|
+ values[i,] <- qstat(tdata[subset,])
|
||
|
|
+ }
|
||
|
|
+ bci <- t(apply(values,2, quantile, c(.05, .95)))
|
||
|
|
+ }
|
||
|
|
+ }
|
||
|
|
> xmat <- cbind(qsurvreg(plist, tfit$coef, tfit$scale, dist='lognormal'),
|
||
|
|
+ bci)
|
||
|
|
>
|
||
|
|
>
|
||
|
|
> matplot(xmat, qnorm(plist),
|
||
|
|
+ type='l', lty=c(1,2,2), col=c(1,1,1),
|
||
|
|
+ log='x', yaxt='n', ylab='Percent',
|
||
|
|
+ xlab='Time of Cracking (Hours x 100)')
|
||
|
|
> axis(2, qnorm(plist), format(100*plist), adj=1)
|
||
|
|
> title("Turbine Data")
|
||
|
|
> kfit <- survfit(Surv(time1, time2, type='interval2') ~1, data=tdata)
|
||
|
|
> points(kfit$time, qnorm(1-kfit$surv), pch='+')
|
||
|
|
>
|
||
|
|
> dev.off() #close the plot file
|
||
|
|
null device
|
||
|
|
1
|
||
|
|
>
|
||
|
|
>
|
||
|
|
> proc.time()
|
||
|
|
user system elapsed
|
||
|
|
1.224 0.016 1.238
|