Removing row in R based on a “typeof” list column

Question

I have looked through a few other solutions on how to remove a row based on a column value, but have not been able to do it for a "list" typeof . The following tibble is a shapefile built for classification training. To rasterize the shapefile, I need to remove rows that contain no geometry data. ie row 8. I can simply perform df <- df$geometry[-8, ] , however that would be inefficient for large data sets.

The geometry column also reads a bit different in the R dataframe (for some reason, this is what the tibble function outputted). The geometry column should read with a c (as in vector). ie c(-123.1166, 44.67333) . So the EMPTY actually reads c(NaN, NaN) .

I was thinking about turning the typeof "list" into a "string" and removing it like that??? Any suggestions?

# A tibble: 102 x 3
   Class Names2             geometry
   <dbl> <fct>           <POINT [°]>
 1     1 Hole   (-123.1166 44.67333)
 2     1 Hole   (-123.1166 44.67333)
 3     1 Hole   (-123.1166 44.67332)
 4     1 Hole    (-123.1167 44.6734)
 5     1 Hole    (-123.1167 44.6734)
 6     1 Hole   (-123.1166 44.67344)
 7     1 Hole   (-123.1165 44.67358)
 8     1 Hole                  EMPTY
 9     1 Hole   (-123.1167 44.67367)
10     1 Hole   (-123.1167 44.67367)
# ... with 92 more rows

Answer 1

Here is one option with map/filter , where we loop over the list column 'geometry with map , check if not all the values are NaN , to return a logical vector in filter to remove those with all NaN values

library(dplyr)
library(purrr)
df1 %>%
      filter(map_lgl(geometry, ~ !all(is.nan(.x)))

---

If it is an sf object, an option is st_is_empty

library(sf)
df1 %>% 
    filter(!st_is_empty(geometry)]

Or it could be

df1 %>%
     st_is_empty(.) %>%
     `!` %>%
     magrittr::extract(df1, ., )