Filling rows of multiple columns based on multiple conditions - r

Question

I have example data as follows:
dat <- structure(list(
zipcode = c(1001, 1002, 1003, 1004, 1101, 1102, 1103, 1104, 1201, 1202, 1203, 1302),
areacode = c(4, 4, NA, 4, 4, 4, NA, 1, 4, 4, NA, 4),
type = structure(c(1L, 1L, NA, 1L, 2L, 2L, NA, 1L, 1L, 1L, NA, 1L),
.Label = c("clay", "sand"), class = "factor"),
region = c(3, 3, NA, 3, 3, 3, NA, 3, 3, 3, NA, 3),
do_not_fill = c(1, NA, NA, 1, 1, NA, NA, 1, NA, NA, NA, 1)),
class = c("data.table", "data.frame"), row.names = c(NA, -4L))
zipcode areacode type region do_not_fill
1: 1001 4 clay 3 1
2: 1002 4 clay 3 NA
3: 1003 NA <NA> NA NA
4: 1004 4 clay 3 1
5: 1101 4 sand 3 1
6: 1102 4 sand 3 NA
7: 1103 NA <NA> NA NA
8: 1104 1 clay 3 1
9: 1201 4 clay 3 NA
10: 1202 4 clay 3 NA
11: 1203 NA <NA> NA NA
12: 1302 4 clay 3 1
I want to fill ONLY the columns areacode, type and region based on two conditions.
The areacode has to be the same before and after the NA.
The first two digits of the zipcode have to be the same before and after the NA.
Based on this solution, and this solution, I attempted following (however data.table solutions are welcomed and even preferred):
library(dplyr)
dat |>
mutate(type = as.character(type)) |>
mutate(across(1:4,
~ ifelse(is.na(.) & lag(areacode) == lead(areacode) &
lag(as.numeric(substr(zipcode, 1, 2))) == lead(as.numeric(substr(zipcode, 1, 2))),
lag(.),
.)))
But somewhere I am doing something wrong, because I get:
Error:
! `n` and `row.names` must be consistent.
Run `rlang::last_error()` to see where the error occurred.
Desired output:
zipcode areacode type region do_not_fill
1: 1001 4 clay 3 1
2: 1002 4 clay 3 NA
3: 1003 4 clay 3 NA
4: 1004 4 clay 3 1
5: 1101 4 sand 3 1
6: 1102 4 sand 3 NA
7: 1103 NA <NA> NA NA
8: 1104 1 clay 3 1
9: 1201 4 clay 3 NA
10: 1202 4 clay 3 NA
11: 1203 NA <NA> NA NA
12: 1302 4 clay 3 1
EDIT
as_tibble(dat) |>
mutate(type = as.character(areacode)) |>
mutate(across(1:4,
~ ifelse(is.na(.) & lag(areacode) == lead(areacode) &
lag(as.numeric(substr(zipcode, 1, 2))) == lead(as.numeric(substr(zipcode, 1, 2))),
lag(.),
.)))
# A tibble: 12 x 5
zipcode areacode type region do_not_fill
<dbl> <dbl> <chr> <dbl> <dbl>
1 1001 4 4 3 1
2 1002 4 4 3 NA
3 1003 4 4 3 NA
4 1004 4 4 3 1
5 1101 4 4 3 1
6 1102 4 4 3 NA
7 1103 NA NA NA NA
8 1104 1 1 3 1
9 1201 4 4 3 NA
10 1202 4 4 3 NA
11 1203 NA NA NA NA
12 1302 4 4 3 1

Answer 1

You need to convert it to a tibble first. I think this is because data.table has extra attributes
Have a look at the rownames,
rownames(as_tibble(dat))
[1] "1" "2" "3" "4" "5" "6" "7" "8" "9" "10" "11" "12"
rownames(dat)
[1] "1" "2" "3" "4"
as_tibble(dat) |>
mutate(type = as.character(type)) |>
mutate(across(1:4,
~ ifelse(is.na(.) & lag(areacode) == lead(areacode) &
lag(as.numeric(substr(zipcode, 1, 2))) == lead(as.numeric(substr(zipcode, 1, 2))),
lag(.),
.)))
# A tibble: 12 x 5
zipcode areacode type region do_not_fill
<dbl> <dbl> <chr> <dbl> <dbl>
1 1001 4 clay 3 1
2 1002 4 clay 3 NA
3 1003 4 clay 3 NA
4 1004 4 clay 3 1
5 1101 4 sand 3 1
6 1102 4 sand 3 NA
7 1103 NA NA NA NA
8 1104 1 clay 3 1
9 1201 4 clay 3 NA
10 1202 4 clay 3 NA
11 1203 NA NA NA NA
12 1302 4 clay 3 1

Answer 2

This can be done in data.table using the same code:
dat[, c(lapply(.SD, \(v) {fifelse(
is.na(areacode) & lag(areacode) == lead(areacode) &
lag(as.numeric(substr(zipcode, 1, 2))) == lead(as.numeric(substr(zipcode, 1, 2))), lag(v), v)}),
.SD[, .(do_not_fill)]), .SDcols = !patterns("do_not_fill")]
zipcode areacode type region do_not_fill
<num> <num> <fctr> <num> <num>
1: 1001 4 clay 3 1
2: 1002 4 clay 3 NA
3: 1004 4 clay 3 NA
4: 1004 4 clay 3 1
5: 1101 4 sand 3 1
6: 1102 4 sand 3 NA
7: 1103 NA <NA> NA NA
8: 1104 1 clay 3 1
9: 1201 4 clay 3 NA
10: 1202 4 clay 3 NA
11: 1203 NA <NA> NA NA
12: 1302 4 clay 3 1