r语言 - 拆分包含多个因子的输出列表

Question

假设我有这三个向量：

time <- c(306,455,1010,210,883,1022,310,361,218,166)
status <- c(0,1,0,1,0,0,1,0,1,1)
gender <- c("Male","Male","Female","Male","Male","Male","Female","Female","Female","Female")

我想做一个生存分析并获得摘要。

A <- survfit(Surv(time, status)~gender)
summary(A, censored = TRUE)

输出将如下所示：

> summary(A, censored = TRUE)
Call: survfit(formula = Surv(time, status) ~ gender)
gender=Female 
time n.risk n.event survival std.err lower 95% CI upper 95% CI
166      5       1      0.8   0.179        0.516            1
218      4       1      0.6   0.219        0.293            1
310      3       1      0.4   0.219        0.137            1
361      2       0      0.4   0.219        0.137            1
1010     1       0      0.4   0.219        0.137            1
gender=Male 
time n.risk n.event survival std.err lower 95% CI upper 95% CI
210      5       1    0.800   0.179        0.516            1
306      4       0    0.800   0.179        0.516            1
455      3       1    0.533   0.248        0.214            1
883      2       0    0.533   0.248        0.214            1
1022     1       0    0.533   0.248        0.214            1

我的问题是，有什么方法可以将输出分为男性和女性。例如：

output_Female <- ?????
output_Female
output_Female 
time n.risk n.event survival std.err lower 95% CI upper 95% CI
166      5       1      0.8   0.179        0.516            1
218      4       1      0.6   0.219        0.293            1
310      3       1      0.4   0.219        0.137            1
361      2       0      0.4   0.219        0.137            1
1010     1       0      0.4   0.219        0.137            1
output_Male <- ?????
output_Male
output_Male 
time n.risk n.event survival std.err lower 95% CI upper 95% CI
166      5       1      0.8   0.179        0.516            1
218      4       1      0.6   0.219        0.293            1
310      3       1      0.4   0.219        0.137            1
361      2       0      0.4   0.219        0.137            1
1010     1       0      0.4   0.219        0.137            1

Answer 1

这是一个使用tidy的选项

library(broom)
library(dplyr)
tidy(A, censored = TRUE) %>% 
split(.$strata)

---

或与base R

txt <- capture.output(summary(A, censored = TRUE))
ind <- cumsum(grepl("gender=", txt))
lst <- lapply(split(txt[ind >0], ind[ind >0]), function(x)
read.table(text = x[-(1:2)], header = FALSE))
nm1 <- scan(text= gsub("\s+[0-9]|%\s+", ".", txt[4]), quiet = TRUE, what = "")
lst <- lapply(lst, setNames, nm1)