在R中按层将具有不等矢量长度的列表转换为数据帧

Question

I have the output from a coxph function, which is estimated by strata. I would like to transform this output from a list into a data frame. The code I ran for coxph is below:

k <- coxph(Surv(cum.goodp, dlq.next) ~ rpc.length + cluster(itemcode) + strata(sector), data = nr.sample)
m <- summary(survfit(k))

There are twenty different strata used to estimate the coxph. Here is the structure of the list

List of 16
$ n        : int [1:20] 870 843 2278 603 6687 8618 15155 920 2598 654 ...
$ time     : num [1:870] 1 2 3 4 5 6 7 8 9 10 ...
$ n.risk   : num [1:870] 870 592 448 361 320 286 232 214 196 186 ...
$ n.event  : num [1:870] 246 126 77 34 33 25 18 18 8 6 ...
$ n.censor : num [1:870] 32 18 10 7 1 29 0 0 2 0 ...
$ strata   : Factor w/ 20 levels "sector=11","sector=21",..: 1 1 1 1 1 1 1 1 1 1 ...
$ surv     : num [1:870] 0.725 0.571 0.471 0.425 0.379 ...
$ type     : chr "right"
$ cumhaz   : num [1:870] 0.322 0.561 0.754 0.856 0.971 ...
$ std.err  : num [1:870] 0.015 0.017 0.0174 0.0174 0.0173 ...
$ upper    : num [1:870] 0.755 0.605 0.506 0.46 0.414 ...
$ lower    : num [1:870] 0.696 0.538 0.438 0.392 0.347 ...
$ conf.type: chr "log"
$ conf.int : num 0.95
$ call     : language survfit(formula = k)
$ table    : num [1:20, 1:7] 870 843 2278 603 6687 ...
..- attr(*, "dimnames")=List of 2
.. ..$ : chr [1:20] "sector=11" "sector=21" "sector=22" "sector=23" ...
.. ..$ : chr [1:7] "records" "n.max" "n.start" "events" ...
- attr(*, "class")= chr "summary.survfit"

I have done this before, but without strata. When I did not have strata I used the following approach:

col <- lapply(c(1 : 7), function(x) m[x])
tbl <- do.call(data.frame, col)

However, when I try that approach here, I get the familiar error:

cannot coerce class "c("survfit.cox", "survfit")" to a data.frame

All columns have the same name, but they are of different length. If possible, I would like to add a column to the final data frame that contains the particular strata that the results are for. Is there a way to do this? It doesn't have to be in base R. Any help would be much appreciated. Thanks so much.

Answer 1

This problem can be solved via the tidy function in the broom package. For the example above, the code is:

n <- survfit(k)
df <- tidy(n)

The tidy function produces a data frame with a variable "strata". It does not, however, provide the median and mean, but they can be estimated from the data frame df if one were so inclined. If the survfit object has multiple strata, the glance(list) cannot provide the median or mean.