I have a data frame
df = data.frame("A" = c("a","b","c","d"), "B" = c(1,2,3,4), "link" = c("www.a.com", "www.b.com", "www.c.com", "www.d.com"))
A B link
a 1 www.a.com
b 2 www.b.com
c 3 www.c.com
d 4 www.d.com
I make the format table
dt.ft <- regulartable(data = dt[, list(A, B, link)])
I want to have the values in column "A" hyperlinked with corresponding values in "link" column.
I tried
compose(x = dt.ft, j = "A", value = as_paragraph( hyperlink_text(x = A, url = link)))
and I got following error:
Error in
$<-.data.frame(*tmp*, "url", value = c(1L, 2L, 3L, 4L, 1L, : replacement has 16 rows, data has 4
The above R statement works if there is only 1 row in the table, but fails to work on multiple rows. Can you please help me. Also, is there any way to hide/remove the column "link", after linking.
First, there was an issue, the version 0.5.2 should fix it.
library(flextable)
df = data.frame("A" = c("a","b","c","d"), "B" = c(1,2,3,4), "link" = c("www.a.com", "www.b.com", "www.c.com", "www.d.com"))
dt.ft <- flextable(data = df, col_keys = c("A", "B"))
dt.ft <- compose(x = dt.ft, j = 1, value = as_paragraph( hyperlink_text(x = A, url = link)))
dt.ft
The example show also how to select columns you want to be displayed by using argument col_keys.