Bug Report: tbl_summary throws error when grouping and data contains a factor filled with NAs

[erikvona]

I'm using tbl_summary programmatically.
If I have a table with a factor variable and all values happen to be NA, an error occurs:

Reprex:

library(gtsummary)
trial2 <- trial %>% select(trt, age, grade)
trial2$has_banana <- factor(NA) # We don't know which patients have a banana
trial2 %>% tbl_summary(by = trt) # Error :(

Error: Problem with mutate() column tbl_stats.
i tbl_stats = pmap(...).
x Can't subset columns that don't exist.
x Columns stat_1 and stat_2 don't exist.

If the factor has levels, tbl_summary works, but add_p fails:

library(gtsummary)
trial3 <- trial %>% select(trt, age, grade)
trial3$has_banana <- factor(NA, levels = c("Yes", "No")) # We don't know which patients have a banana
trial3 %>% tbl_summary(by = trt) %>% add_p() # Error :(

Error: Problem with mutate() column test.
i test = map2(...).
x missing value where TRUE/FALSE needed

(I tend to run a fct_drop across all factors so mine don't have levels, but it would be nice if both could get fixed so tbl_summary and add_p could produce valid output for any factor).

[ddsjoberg]

Thank you @erikvona for reporting the bug! I am surprised this hasn't been addressed previously. We'll ensure this fix makes it into the next release

[ddsjoberg]

@erikvona

The issue here is that the factor variable has no levels, FYI. If there is a level, the function doesn't return an error.

trial %>% 
  dplyr::mutate(
    has_banana = factor(NA, levels = "level") # We don't know which patients have a banana
  ) %>% 
  tbl_summary(by = trt, include = has_banana)

[erikvona]

@ddsjoberg That's true, but if the factor has levels, add_p() still fails, see the second example. If desired I could split this in two reports, since they could be considered separate issues.

[ddsjoberg]

Thanks for adding the second example. I am surprised to see that the fisher.test() returns a p-value when all observations are missing! 😨 😨 😨

> with(df, table(has_banana, trt))
          trt
has_banana Drug A Drug B
       Yes      0      0
       No       0      0

> with(df, table(has_banana, trt)) %>%
+   fisher.test()

	Fisher's Exact Test for Count Data

data:  .
p-value = 1
alternative hypothesis: true odds ratio is not equal to 1
95 percent confidence interval:
   0 Inf
sample estimates:
odds ratio 
         0