R语言如何在生存分析与Cox回归中计算IDI，NRI指标

原文链接：http://tecdat.cn/?p=6095

读取样本数据

## [1] 416 7

查询部分数据（结果和预测因子）

head(D)

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## time status age albumin edema protime bili ## 1 400 1 58.76523 2.60 1.0 12.2 14.5 ## 2 4500 0 56.44627 4.14 0.0 10.6 1.1 ## 3 1012 1 70.07255 3.48 0.5 12.0 1.4 ## 4 1925 1 54.74059 2.54 0.5 10.3 1.8 ## 5 1504 0 38.10541 3.53 0.0 10.9 3.4 ## 6 2503 1 66.25873 3.98 0.0 11.0 0.8

模型0和模型1的结果数据和预测变量集

<-as.matrix(D[,c(-1,-2)])\ncovs0<-as.matrix(D[,c(-1,-2, -7)])\n\nhead(outcome)","classes":{"has":1},"lang":""}" data-cke-widget-upcasted="1" data-cke-widget-keep-attr="0" data-widget="codeSnippet">outcome=D[,c(1,2)] covs1<-as.matrix(D[,c(-1,-2)]) covs0<-as.matrix(D[,c(-1,-2, -7)]) head(outcome)

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## time status ## 1 400 1 ## 2 4500 0 ## 3 1012 1 ## 4 1925 1 ## 5 1504 0 ## 6 2503 1

head(covs0)

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## age albumin edema protime ## 1 58.76523 2.60 1.0 12.2 ## 2 56.44627 4.14 0.0 10.6 ## 3 70.07255 3.48 0.5 12.0 ## 4 54.74059 2.54 0.5 10.3 ## 5 38.10541 3.53 0.0 10.9 ## 6 66.25873 3.98 0.0 11.0

head(covs1)

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## age albumin edema protime bili ## 1 58.76523 2.60 1.0 12.2 14.5 ## 2 56.44627 4.14 0.0 10.6 1.1 ## 3 70.07255 3.48 0.5 12.0 1.4 ## 4 54.74059 2.54 0.5 10.3 1.8 ## 5 38.10541 3.53 0.0 10.9 3.4 ## 6 66.25873 3.98 0.0 11.0 0.8

推理

t0=365*5\nx<-IDI (outcome, covs0, covs1, t0, npert=200) ;","classes":{"has":1},"lang":""}" data-cke-widget-upcasted="1" data-cke-widget-keep-attr="0" data-widget="codeSnippet">t0=365*5 x<-IDI (outcome, covs0, covs1, t0, npert=200) ;

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输出

## Est. Lower Upper p-value ## M1 0.090 0.052 0.119 0 ## M2 0.457 0.340 0.566 0 ## M3 0.041 0.025 0.062 0

M1表示IDI

M2表示NRI

M3表示中位数差异

图形演示

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