# tests/fr_resid.R In survival: Survival Analysis

```options(na.action=na.exclude) # preserve missings
options(contrasts=c('contr.treatment', 'contr.poly')) #ensure constrast type
library(survival)

#
# The residual methods treat a sparse frailty as a fixed offset with
#   no variance
#
aeq <- function(x,y, ...) all.equal(as.vector(x), as.vector(y), ...)

kfit1 <- coxph(Surv(time, status) ~ age + sex +
frailty(id, dist='gauss'), kidney)
tempf <- predict(kfit1, type='terms')[,3]
temp  <- kfit1\$frail[match(kidney\$id, sort(unique(kidney\$id)))]
#all.equal(unclass(tempf), unclass(temp))
all.equal(as.vector(tempf), as.vector(temp))

# Now fit a model with explicit offset
kfitx <- coxph(Surv(time, status) ~ age + sex + offset(tempf),kidney,
eps=1e-7)

# These are not always precisely the same, due to different iteration paths
aeq(kfitx\$coef, kfit1\$coef)

# This will make them identical
kfitx <- coxph(Surv(time, status) ~ age + sex  + offset(temp),kidney,
iter=0, init=kfit1\$coef)
aeq(resid(kfit1), resid(kfitx))
aeq(resid(kfit1, type='score'), resid(kfitx, type='score'))
aeq(resid(kfit1, type='schoe'), resid(kfitx, type='schoe'))

# These are not the same, due to a different variance matrix
#  The frailty model's variance is about 2x the naive "assume an offset" var
# Expect a value of about 0.5
aeq(resid(kfit1, type='dfbeta'), resid(kfitx, type='dfbeta'))

# Force equality
zed <- kfitx
zed\$var <- kfit1\$var
aeq(resid(kfit1, type='dfbeta'), resid(zed, type='dfbeta'))

# The score residuals are equal, however.

temp1 <- resid(kfit1, type='score')
temp2 <- resid(kfitx, type='score')
aeq(temp1, temp2)

#
# Now for some tests of predicted values
#
aeq(predict(kfit1, type='expected'), predict(kfitx, type='expected'))
aeq(predict(kfit1, type='lp'), predict(kfitx, type='lp'))

temp1 <- predict(kfit1, type='terms', se.fit=T)
temp2 <- predict(kfitx, type='terms', se.fit=T)
aeq(temp1\$fit[,1:2], temp2\$fit)
# the next is not equal, all.equal returns a character string in that case
is.character(aeq(temp1\$se.fit[,1:2], temp2\$se.fit))
mean(temp1\$se.fit[,1:2]/ temp2\$se.fit)
aeq(as.vector(temp1\$se.fit[,3])^2,
as.vector(kfit1\$fvar[match(kidney\$id, sort(unique(kidney\$id)))]))

print(temp1)
kfit1
kfitx

rm(temp1, temp2, kfitx, zed, tempf)
#
# The special case of a single sparse frailty
#

kfit1 <- coxph(Surv(time, status) ~ frailty(id, dist='gauss'), kidney)
tempf <- predict(kfit1, type='terms')
temp  <- kfit1\$frail[match(kidney\$id, sort(unique(kidney\$id)))]
all.equal(as.vector(tempf), as.vector(temp))

# Now fit a model with explicit offset
kfitx <- coxph(Surv(time, status) ~ offset(tempf),kidney, eps=1e-7)

aeq(resid(kfit1), resid(kfitx))
aeq(resid(kfit1, type='deviance'), resid(kfitx, type='deviance'))

#
# Some tests of predicted values
#
aeq <- function(x,y) all.equal(as.vector(x), as.vector(y))
aeq(predict(kfit1, type='expected'), predict(kfitx, type='expected'))
aeq(predict(kfit1, type='lp'), predict(kfitx, type='lp'))

temp1 <- predict(kfit1, type='terms', se.fit=T)
aeq(temp1\$fit, kfitx\$linear)
aeq(temp1\$se.fit^2,
kfit1\$fvar[match(kidney\$id, sort(unique(kidney\$id)))])

temp1
kfit1
```

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survival documentation built on Aug. 24, 2021, 5:06 p.m.