obscure error with indexing st_geometry on tibble

[timelyportfolio]

I hesitated to even file this issue as it seems to be an obscure edge case, but in the interest of robustness of sf I elected to file. Feel free to close if too obscure. I discovered researching r-spatial/mapedit#32.

generalized issue

Probably easier to understand below as a reproducible example. The error only happens with assignment using st_geometry(my_sf[####,]) on a sf tibble object with a column that is after the geometry column.

reproducible example

Everything works exactly as I would expect except the one commented case below.

library(sf)
library(tibble)

geom <- st_sfc(
  st_polygon(
    list(
      cbind(x = c(0.69, 0.8, 1.1, 1.23, 0.69), 
            y = c(0, 0.6, 0.63, 0.3, 0))
    )
  )
)

feat <- st_sf(ltr="a",geom)

# add column when as data.frame
feat$id <- 1
# demonstrate st_geometry assignment works with data.frame
# works
st_geometry(feat) <- geom
feat
# works
st_geometry(feat[1,]) <- geom
feat
# works
st_geometry(feat)[1] <- geom
feat

# do same as above but this time as tibble
feat_tbl <- st_sf(tibble(ltr="a",geom=geom))
# add a column when as tibble
feat_tbl$id <- 1
# works
st_geometry(feat_tbl) <- geom
feat_tbl
################# does not work ##########################
feat_tbl <- st_sf(tibble(ltr="a",geom=geom))
feat_tbl$id <- 1
st_geometry(feat_tbl[1,]) <- geom
feat_tbl
##########################################################
# works
feat_tbl <- st_sf(tibble(ltr="a",geom=geom))
feat_tbl$id <- 1
st_geometry(feat_tbl)[1] <- geom
feat_tbl
# works
feat_tbl <- st_sf(tibble(ltr="a",geom=geom))
feat_tbl$id <- 1
# move id column before geometry column
feat_tbl <- feat_tbl[,c("ltr","id","geom")]
st_geometry(feat_tbl[1,]) <- geom
feat_tbl

[edzer]

p1 is not defined, afaics.

[timelyportfolio]

@edzer sorry about that. Fixed example above.

[edzer]

This is a nested replacement function; I'm afraid this is never going to work. A similar example from base R, which also doesn't work, is this:

df = data.frame(a = 3, id = 1)
row.names(df[1,]) = "a"
row.names(df)
# [1] "1"
row.names(df[1,])
# [1] "1"

I'll try to find out why the object returned is invalid - that shouldn't happen.

[timelyportfolio]

No problem and thanks. I just thought it strange that all cases worked except the one. Thanks again!

[edzer]

Sure; I'll keep it open until I'm convinced it doesn't reveal a problem in sf.