问题
I am trying to plot Kaplan-Meyer curve using ggsurvplot from survminer package. I'm unable to plot it when I pass a survfit object saved in a list.
Let me use lung dataset as a example. Everything works below:
library("survival")
library("survminer")
fit <- survfit(Surv(time, status) ~ sex, data = lung)
ggsurvplot(fit,
conf.int = TRUE,
risk.table.col = "strata",
palette = c("#E7B800", "#2E9FDF"),
xlim = c(0, 600))
Now I do survfit on two variables and save the model result in a list. Then tried to make KM plot with ggsurvplot.
vars <- c('sex', 'ph.ecog')
l<- map (vars, ~survfit(Surv(time, status)~ get(.x),data = lung ))
l<- set_names(l, vars)
ggsurvplot(l$sex,
conf.int = TRUE,
risk.table.col = "strata",
palette = c("#E7B800", "#2E9FDF"),
xlim = c(0, 600))
I got error message like this:
Error in eval(inp, data, env) : object '.x' not found
Does someone know why? How can I fix this problem? Thanks a lot!
回答1:
First one would need to load the package or packages needed. I suppose these days many users think that running R means that everyone is presumed to have the tidyverse in place, but that is NOT true.
library(tidyverse)
# run both your code segments, since you will need a small piece of first one
str(l$sex)
List of 14
$ n : int [1:2] 138 90
$ time : num [1:206] 11 12 13 15 26 30 31 53 54 59 ...
$ n.risk : num [1:206] 138 135 134 132 131 130 129 128 126 125 ...
$ n.event : num [1:206] 3 1 2 1 1 1 1 2 1 1 ...
$ n.censor : num [1:206] 0 0 0 0 0 0 0 0 0 0 ...
$ surv : num [1:206] 0.978 0.971 0.957 0.949 0.942 ...
$ type : chr "right"
$ strata : Named int [1:2] 119 87
..- attr(*, "names")= chr [1:2] "get(.x)=1" "get(.x)=2"
$ std.err : num [1:206] 0.0127 0.0147 0.0181 0.0197 0.0211 ...
$ upper : num [1:206] 1 0.999 0.991 0.987 0.982 ...
$ lower : num [1:206] 0.954 0.943 0.923 0.913 0.904 ...
$ conf.type: chr "log"
$ conf.int : num 0.95
$ call : language survfit(formula = Surv(time, status) ~ get(.x), data = lung)
- attr(*, "class")= chr "survfit"
So when you see that strata
"names"-attribute, it's got a get(
-call in it and that appears to choke the logic of ggsurvplot
. Use attr<-
to replace it with something more informative (and less-"language-y").
attr(l[['sex']][['strata']], "names") <- c("sex=1", "sex=2")
That expression is in the "call"-leaf as well, so you will need to replace it with something more tractable. I think that's easy to do by replacing it with the "call" leaf from the first
fit`-object yoiu made:
l$sex$call <- fit$call
ggsurvplot(l$sex,
conf.int = TRUE,
risk.table.col = "strata",
palette = c("#E7B800", "#2E9FDF"),
xlim = c(0, 600))
来源:https://stackoverflow.com/questions/45872703/unable-to-plot-kaplan-meier-curve-with-survfit-object-from-a-list-using-ggsurvpl