从Cox PH模型预测概率 [英] Predict probability from Cox PH model

问题描述

我正在尝试使用cox模型来预测一段时间后发生故障的可能性（称为停止）。

I am trying to use cox model to predict the probability of failure after time (which is named stop) 3.

bladder1 <- bladder[bladder$enum < 5, ] 
coxmodel = coxph(Surv(stop, event) ~ (rx + size + number)  + 
        cluster(id), bladder1)
range(predict(coxmodel, bladder1, type = "lp"))
range(predict(coxmodel, bladder1, type = "risk"))
range(predict(coxmodel, bladder1, type = "terms"))
range(predict(coxmodel, bladder1, type = "expected"))

但是，预测函数的输出为都不在0-1范围内。是否有任何函数或如何使用lp预测和基线危害函数来计算概率？

However, the outputs of predict function are all not in 0-1 range. Is there any function or how can I use the lp prediction and baseline hazard function to calculate probability?

推荐答案

请阅读帮助页面为 predict.coxph 。这些都不应该是概率。一组特定的协变量的线性预测因子是相对于假设（且很可能不存在）情况的对数风险比，其中所有预测因子均值。 期望最接近概率，因为它是预测的事件数，但是它需要指定时间，然后用观察开始时的危险数除。

Please read the help page for predict.coxph. None of those are supposed to be probabilities. The linear predictor for a specific set of covariates is the log-hazard-ratio relative to a hypothetical (and very possibly non-existent) case with the mean of all the predictor values. The 'expected' comes the closest to a probability since it is a predicted number of events, but it would require specification of the time and then be divided by the number at risk at the beginning of observation.

在该帮助页面上提供的用于预测的示例中，您可以看到预测事件的总和与实际数字接近：

In the case of the example offered on that help page for predict, you can see that the sum of predicted events is close the the actual number:

> sum(predict(fit,type="expected"), na.rm=TRUE)
[1] 163

> sum(lung$status==2)
[1] 165

我怀疑您可能希望改为使用 survfit 函数，因为事件的概率是1个生存概率。

I suspect you may want to be working instead with the survfit function, since the probability of event is 1-probability of survival.

?survfit.coxph

一个类似问题的代码出现在这里：添加R

The code for a similar question appears here: Adding column of predicted Hazard Ratio to dataframe after Cox Regression in R

中的Cox回归后的预测危害与数据框的比率列由于您建议使用百事通1数据集，因此这将是指定时间的代码= 5

Since you suggested using the bladder1 dataset, then this would be the code for a specification of time=5

 summary(survfit(coxmodel), time=5)
#------------------
Call: survfit(formula = coxmodel)

 time n.risk n.event survival std.err lower 95% CI upper 95% CI
    5    302      26    0.928  0.0141        0.901        0.956

T帽子将以生存预测作为列表元素返回为列表，列表元素名为 $ surv ：

That would return as a list with the survival prediction as a list element named $surv:

> str(summary(survfit(coxmodel), time=5))
List of 14
 $ n       : int 340
 $ time    : num 5
 $ n.risk  : num 302
 $ n.event : num 26
 $ conf.int: num 0.95
 $ type    : chr "right"
 $ table   : Named num [1:7] 340 340 340 112 NA 51 NA
  ..- attr(*, "names")= chr [1:7] "records" "n.max" "n.start" "events" ...
 $ n.censor: num 19
 $ surv    : num 0.928
 $ std.err : num 0.0141
 $ lower   : num 0.901
 $ upper   : num 0.956
 $ cumhaz  : num 0.0744
 $ call    : language survfit(formula = coxmodel)
 - attr(*, "class")= chr "summary.survfit"
> summary(survfit(coxmodel), time=5)$surv
[1] 0.9282944

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