我正在对竞争风险比例风险模型进行交叉验证。在mstatepacakge 的帮助下，我已经准备好我的数据并正在将其与survival::coxph. 我为我的训练数据获得了一个合适的 Cox 模型对象，但我想用我的测试数据评估我的训练系数的部分可能性。
如果需要，我会自己编写部分似然函数，但我宁愿不写（尽管它可能对我有好处）。生存包在此 C 代码中计算，但似然计算嵌入在拟合函数中。也许有一种方法可以修复参数，或者其他一些工具可以轻松获得部分可能性？
最小工作示例
# Adapted from examples in the mstate vignette
# http://cran.r-project.org/web/packages/mstate/vignettes/Tutorial.pdf
# beginning at the bottom of page 28
library(mstate)
library(survival)
# Get data. I add a second explanatory variable (badx) for illustration
# Also divide the data by subject into training and test sets.
data(aidssi)
si <- aidssi # Just a shorter name
si$badx <- sample(c("A", "B"), size = nrow(si), replace = TRUE)
si$fold <- sample(c("train", "test"), size = nrow(si), replace = TRUE, prob = c(0.7, 0.3))
tmat <- trans.comprisk(2, names = c("event-free", "AIDS", "SI"))
si$stat1 <- as.numeric(si$status == 1)
si$stat2 <- as.numeric(si$status == 2)
# Convert the data to a long competing risks format
silong <- msprep(time = c(NA, "time", "time"),
status = c(NA,"stat1", "stat2"),
data = si, keep = c("ccr5", "badx", "fold"), trans = tmat)
silong <- na.omit(silong)
silong <- expand.covs(silong, c("ccr5", "badx"))
train.dat <- subset(silong, fold == "train")
test.dat <- subset(silong, fold == "test")
数据如下所示：
> head(silong)
An object of class 'msdata'
Data:
id from to trans Tstart Tstop time status ccr5 badx fold ccr5WM.1 ccr5WM.2 badxB.1 badxB.2
1 1 1 2 1 0 9.106 9.106 1 WW A train 0 0 0 0
2 1 1 3 2 0 9.106 9.106 0 WW A train 0 0 0 0
3 2 1 2 1 0 11.039 11.039 0 WM B train 1 0 1 0
4 2 1 3 2 0 11.039 11.039 0 WM B train 0 1 0 1
5 3 1 2 1 0 2.234 2.234 1 WW B train 0 0 1 0
6 3 1 3 2 0 2.234 2.234 0 WW B train 0 0 0 1
现在，ccr5可以将变量建模为特定于转换的变量，或者建模为对所有转换具有相等的比例效应。这些型号是：
train.mod.equal <- coxph(Surv(time, status) ~ ccr5 + badx + strata(trans),
data = train.dat)
train.mod.specific <- coxph(Surv(time, status) ~ ccr5WM.1 + ccr5WM.2 + badx + strata(trans),
data = train.dat)
现在我想使用测试数据来评估变量选择是否ccr5应该是特定于转换的。我有一个庞大的数据集和许多变量——大部分但不是所有的分类变量——这两种方式都可以。评估是我卡住的地方。
# We can fit the same models to the test data,
# this yields new parameter estimates of course,
# but the model matrices might be useful
test.mod.equal <- coxph(Surv(time, status) ~ ccr5 + badx + strata(trans),
data = test.dat)
test.mod.specific <- coxph(Surv(time, status) ~ ccr5WM.1 + ccr5WM.2 + badx + strata(trans),
data = test.dat)
test.eq.mm <- model.matrix(test.mod.equal)
test.sp.mm <- model.matrix(test.mod.specific)
# We can use these to get the first part of the sum of the partial likelihood:
xbeta.eq <- test.eq.mm[test.dat$status == 1, ] %*% coef(train.mod.equal)
xbeta.sp <- test.sp.mm[test.dat$status == 1, ] %*% coef(train.mod.specific)
# We can also get linear predictors
lp.eq <- predict(train.mod.equal, newdata = test.dat, type = "lp")
lp.sp <- predict(train.mod.specific, newdata = test.dat, type = "lp")
我希望用训练系数估计来计算测试数据上每个模型的部分可能性。也许我应该将问题移至交叉验证并询问线性预测变量的总和（或不包括审查案例的线性预测变量的总和）是否足够接近等效度量。