我正在从survminer
包中运行一个非常简单的cox模型。
surv_object <- Surv(time, event)
model <- coxph(surv_object ~ female + age + ethnicity + imd, data = df)
我需要运行多个Cox模型,并且对于每个模型,我的预测变量都会发生变化。我将所有预测变量存储在这样的单独数据框中(我们将其称为pred_df
):
> pred_df
# A tibble: 4 x 2
predictor endpoint
<chr> <chr>
1 female Mortality
2 age Mortality
3 ethnicity Mortality
4 imd Mortality
是否有一种简单的方法将项目从predictor
列传递到coxph()
?像这样:
coxph(surv_object ~ predictors, data = df)
我已经尝试过的内容:
我已经尝试过以下方法:
pred_vars <- pred_df %>%
pull(predictor) %>% # extract column values as a vector
paste(collapse = " + ") %>% # combine values in a string
parse(text = . ) # parse the string as an expression
model <- coxph(surv_object ~ eval(pred_vars), data = df)
R实际上了解这一点并运行模型。但是输出是无法解释的。该模型似乎正在运行,但是没有输出单独的预测变量,即female
,age
,ethnicity
和imd
。相反,它仅输出eval(pred_vars)
Call:
coxph(formula = Surv(time, event) ~ eval(pred_vars), data = df)
n= 62976, number of events= 12882
(3287 observations deleted due to missingness)
coef exp(coef) se(coef) z Pr(>|z|)
eval(pred_vars) 3.336e-05 1.000e+00 5.339e-06 6.249 4.14e-10 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
exp(coef) exp(-coef) lower .95 upper .95
eval(pred_vars) 1 1 1 1
Concordance= 0.515 (se = 0.003 )
Rsquare= 0.001 (max possible= 0.989 )
Likelihood ratio test= 38.28 on 1 df, p=6e-10
Wald test = 39.04 on 1 df, p=4e-10
Score (logrank) test = 39.07 on 1 df, p=4e-10
必须有一种更简单的方法吗?
答案 0 :(得分:1)
尝试重新配制。
public class Toto {
[Display(Name = "Identifier", Order = 2)
public int Id { get; set; }
[Display(Name = "Description", Order = 1)
public string Label {get; set; }
}
答案 1 :(得分:0)
您可以在as.formula
和paste(..., collapse = " + ")
中以R为基数,例如...
foo <- as.formula(paste0("Surv(time, event) ~ ", paste(pred_df$predictors, collapse = " + ")))
该行的结果:
> foo
Surv(time, event) ~ female + age + ethnicity + imd
然后您只需将foo
传递给对coxph
的呼叫即可。