Cannot add 2nd Surv object to data frame
Closed this issue · 4 comments
Surv objects are not correctly adding to a data frame while using R version 4.2.0 and survival 3.3-1.
Reproducible example:
df <- data.frame(id = c(1:10)
, cen1_lo = c(0, 0, 2, 0, 2, 6, 2, 0, 2, 2)
, cen1_hi = c(6, 0, 6, 0, 2, 6, 2, 0, 6, 3)
, cen2_lo = c(1, 1, 3, 2, 3, 6, 1, 3, 1, 0)
, cen2_hi = c(6, 3, 6, 2, 4, 6, 2, 5, 6, 3))
df$cen1 <- Surv(df$cen1_lo, df$cen1_hi, type = "interval2")
This yields an expected outcome
id cen1_lo cen1_hi cen2_lo cen2_hi cen1
1 1 0 6 1 6 [0, 6]
2 2 0 0 1 3 0
3 3 2 6 3 6 [2, 6]
4 4 0 0 2 2 0
5 5 2 2 3 4 2
6 6 6 6 6 6 6
7 7 2 2 1 2 2
8 8 0 0 3 5 0
9 9 2 6 1 6 [2, 6]
10 10 2 3 0 3 [2, 3]
However, viewing in RStudio seems off--notice last column is labeled cols instead of cen1
Proceeding to add 2nd censored variable:
df$cen2 <- Surv(df$cen2_lo, df$cen2_hi, type = "interval2")
yields the error message:
Error in .subset2(x, i, exact = exact) : subscript out of bounds
But the listing, i.e., df appears ok:
id cen1_lo cen1_hi cen2_lo cen2_hi cen1 cen2
1 1 0 6 1 6 [0, 6] [1, 6]
2 2 0 0 1 3 0 [1, 3]
3 3 2 6 3 6 [2, 6] [3, 6]
4 4 0 0 2 2 0 2
5 5 2 2 3 4 2 [3, 4]
6 6 6 6 6 6 6 6
7 7 2 2 1 2 2 [1, 2]
8 8 0 0 3 5 0 [3, 5]
9 9 2 6 1 6 [2, 6] [1, 6]
10 10 2 3 0 3 [2, 3] [0, 3]
However, the viewing in RStudio, i.e., View(df) gives the below error
Error in View : subscript out of bounds
Everything works fine for me. The text below is a simple cut/paste from my terminal. This is of course a clean session, with no extra libraries loaded except survival. (An important pre-condition to any bug report). What else to you have loaded?
> df <- data.frame(id = c(1:10)
+ , cen1_lo = c(0, 0, 2, 0, 2, 6, 2, 0, 2, 2)
+ , cen1_hi = c(6, 0, 6, 0, 2, 6, 2, 0, 6, 3)
+ , cen2_lo = c(1, 1, 3, 2, 3, 6, 1, 3, 1, 0)
+ , cen2_hi = c(6, 3, 6, 2, 4, 6, 2, 5, 6, 3))
>
> df$cen1 <- Surv(df$cen1_lo, df$cen1_hi, type = "interval2")
>
> df$cen2 <- Surv(df$cen2_lo, df$cen2_hi, type="interval2")
> df
id cen1_lo cen1_hi cen2_lo cen2_hi cen1 cen2
1 1 0 6 1 6 [0, 6] [1, 6]
2 2 0 0 1 3 0 [1, 3]
3 3 2 6 3 6 [2, 6] [3, 6]
4 4 0 0 2 2 0 2
5 5 2 2 3 4 2 [3, 4]
6 6 6 6 6 6 6 6
7 7 2 2 1 2 2 [1, 2]
8 8 0 0 3 5 0 [3, 5]
9 9 2 6 1 6 [2, 6] [1, 6]
10 10 2 3 0 3 [2, 3] [0, 3]
>
Thanks for your quick response and confirmation the issues are elsewhere.