Select segments and columns of a triangle by name

Keeps the segments whose key values match and the measure columns named: subset(tri, lob = "auto", state = c("CA", "NY"), columns = "paid"). Each key argument gives the values to keep (compared as character, as keys are stored); segments must match every key argument and keep their order. Columns are kept in the order given. This is Python's Triangle.select(columns=None, **keys).

Arguments

x

A triangle.

...

Key conditions as key = values.

columns

Column names to keep, in order, or NULL for all.

Value

A triangle.

Details

The method is on base R's subset() generic rather than a new verb, so it does not mask dplyr::filter() or dplyr::select(). A key named x or columns cannot be selected this way.

Errors: a key, value or column that does not exist or is given twice, an empty set of values, an unnamed argument, or a selection that matches no segment.

See also

aggregate() to sum segments over keys.

Examples

long <- data.frame(lob = rep(c("auto", "home"), each = 3), year = c(2020, 2020, 2021),
                   age = c(12, 24, 12), paid = 1:6, incurred = 2 * (1:6))
tri <- triangle(long, "year", "age", c("paid", "incurred"), keys = "lob")
subset(tri, lob = "home", columns = "incurred")@values
#> , , origin = 2020, development = 12
#> 
#>       column
#> index  incurred
#>   home        8
#> 
#> , , origin = 2021, development = 12
#> 
#>       column
#> index  incurred
#>   home       12
#> 
#> , , origin = 2020, development = 24
#> 
#>       column
#> index  incurred
#>   home       10
#> 
#> , , origin = 2021, development = 24
#> 
#>       column
#> index  incurred
#>   home       NA
#> 
subset(tri, columns = c("incurred", "paid"))@columns
#> [1] "incurred" "paid"