如果我使用mlr中5倍交叉验证的重采样训练cox模型，则通过打印每个折叠的cox模型摘要输出的Concordance值与由mlr计算的cindex值不同。我的解释有错吗？还是我使用了太多的预测因素？如果是这样，为什么会造成这种差异？
在下面的示例中，mlr为第一次折叠返回0.5093809的cindex值，但cox摘要输出报告的Concordance为0.76。我的数据可以在这里下载：https://www.dropbox.com/s/nt9s3p1rdafq465/test_data.csv?dl=0
重新采样：
library(survival)
library(mlr)
mydata <- read.csv(file="test_data.csv", header=TRUE, sep=",",row.names=NULL)
surv.task <- makeSurvTask(data = mydata, target = c("timeToEvent", "status"))
rdesc <- makeResampleDesc(method="CV", iters=5, stratify=TRUE)
r = resample("surv.coxph", surv.task, rdesc, models=TRUE)
r
Resample Result
Task: mydata
Learner: surv.coxph
Aggr perf: cindex.test.mean=0.5999838
Runtime: 0.151174
r$measures.test
iter cindex
1 1 0.5093809
2 2 0.7324649
3 3 0.4984653
4 4 0.6461876
5 5 0.6134201
检查第一次折叠的Cox模型摘要：
summary(getLearnerModel(r$models[[1]]))
Call:
survival::coxph(formula = f, data = data)
n= 698, number of events= 65
coef exp(coef) se(coef) z Pr(>|z|)
V1 -0.1225832 0.8846323 0.1833418 -0.669 0.503748
V2 -1.9815012 0.1378621 2.9565667 -0.670 0.502728
V3 -0.5894775 0.5546170 1.9276623 -0.306 0.759758
V4 0.5005582 1.6496418 0.9433060 0.531 0.595667
V5 0.0179647 1.0181271 1.9273040 0.009 0.992563
V6 0.7309210 2.0769926 1.9361340 0.378 0.705790
V7 -0.0012070 0.9987937 0.0890533 -0.014 0.989186
V8 0.1029020 1.1083828 0.0356533 2.886 0.003899 **
V9 -0.2728561 0.7612023 0.2311420 -1.180 0.237813
V10 -0.0213663 0.9788604 0.0133210 -1.604 0.108725
V11 0.2416705 1.2733746 0.2113099 1.144 0.252757
V12 -0.0021392 0.9978631 0.0550684 -0.039 0.969014
V13 -0.0047373 0.9952739 0.0073776 -0.642 0.520794
V14 0.0119084 1.0119796 0.0036098 3.299 0.000971 ***
V15 -6.6529859 0.0012902 2.8566451 -2.329 0.019862 *
V16 -0.0005712 0.9994290 0.0015808 -0.361 0.717842
V17 -0.0058360 0.9941810 0.0970749 -0.060 0.952062
V18 -0.0095129 0.9905322 0.0072980 -1.304 0.192402
V19 0.0004149 1.0004150 0.0002001 2.074 0.038107 *
V20 0.0001584 1.0001584 0.0002319 0.683 0.494487
V21 -0.0010930 0.9989076 0.0045039 -0.243 0.808255
V22 -0.0015312 0.9984700 0.0023389 -0.655 0.512699
V23 -0.0441918 0.9567705 0.0936314 -0.472 0.636944
V24 0.0475120 1.0486588 0.0681332 0.697 0.485590
V25 0.1637753 1.1779496 0.1177553 1.391 0.164283
V26 -0.0296841 0.9707521 0.0460953 -0.644 0.519593
V27 -0.1181631 0.8885511 0.0824113 -1.434 0.151623
V28 0.0081237 1.0081568 0.0106226 0.765 0.444419
V29 -0.0409860 0.9598425 0.0282858 -1.449 0.147339
V30 0.0006100 1.0006102 0.0002408 2.533 0.011293 *
V31 -0.0016426 0.9983587 0.0054629 -0.301 0.763655
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
exp(coef) exp(-coef) lower .95 upper .95
V1 0.88463 1.1304 6.176e-01 1.2671
V2 0.13786 7.2536 4.196e-04 45.2980
V3 0.55462 1.8030 1.268e-02 24.2562
V4 1.64964 0.6062 2.597e-01 10.4793
V5 1.01813 0.9822 2.330e-02 44.4965
V6 2.07699 0.4815 4.671e-02 92.3581
V7 0.99879 1.0012 8.388e-01 1.1893
V8 1.10838 0.9022 1.034e+00 1.1886
V9 0.76120 1.3137 4.839e-01 1.1974
V10 0.97886 1.0216 9.536e-01 1.0048
V11 1.27337 0.7853 8.416e-01 1.9267
V12 0.99786 1.0021 8.958e-01 1.1116
V13 0.99527 1.0047 9.810e-01 1.0098
V14 1.01198 0.9882 1.005e+00 1.0192
V15 0.00129 775.0952 4.776e-06 0.3485
V16 0.99943 1.0006 9.963e-01 1.0025
V17 0.99418 1.0059 8.219e-01 1.2025
V18 0.99053 1.0096 9.765e-01 1.0048
V19 1.00041 0.9996 1.000e+00 1.0008
V20 1.00016 0.9998 9.997e-01 1.0006
V21 0.99891 1.0011 9.901e-01 1.0078
V22 0.99847 1.0015 9.939e-01 1.0031
V23 0.95677 1.0452 7.964e-01 1.1495
V24 1.04866 0.9536 9.176e-01 1.1985
V25 1.17795 0.8489 9.352e-01 1.4837
V26 0.97075 1.0301 8.869e-01 1.0625
V27 0.88855 1.1254 7.560e-01 1.0443
V28 1.00816 0.9919 9.874e-01 1.0294
V29 0.95984 1.0418 9.081e-01 1.0146
V30 1.00061 0.9994 1.000e+00 1.0011
V31 0.99836 1.0016 9.877e-01 1.0091
Concordance= 0.76 (se = 0.037 )
Rsquare= 0.087 (max possible= 0.68 )
Likelihood ratio test= 63.69 on 31 df, p=5e-04
Wald test = 67.74 on 31 df, p=2e-04
Score (logrank) test = 70.07 on 31 df, p=7e-05
cox模型的一致性指数是用训练数据计算的，mlr用每个折叠的样本外数据计算。这就是区别，毫无疑问，在样本之外，情况要糟糕得多。(