Get the species list from a dataframe based on rFishStatus template

This function uses a dataframe of scientific names from organized a list of species names with their status.

Usage

rfs_get_species_list(df, summary = FALSE)

Arguments

df

A data frame based on the rFishStatus template (rFishStatus::data_template_ref).

Value

A dataframe with the species list.

Examples

df <- rFishStatus::data_template_ref
df_list <- rfs_get_species_list(df)
#> ℹ Creating species list from the input dataset.
#> Looking for uncertain species.
#> Looking for valid species.
#> Looking for all previous valid species names.
#> Filtering valid species.
#> Error in dplyr::filter(dplyr::mutate(dplyr::mutate(species_df, genus = stringr::str_extract(scientific_name,     "\\w+"), epithet = stringr::str_extract(scientific_name,     stringr::regex("(?<=\\s)[a-z]+(\\s(?![a-z]*\\.)[a-z]+)*"))),     species = paste(genus, epithet, sep = " "), year = as.numeric(stringr::str_extract(scientific_name_author,         "\\d{4}")), authors = stringr::str_replace(scientific_name_author,         paste0(" ", year), ""), scientific_name = stringr::str_replace_all(scientific_name,         "\\[.*?\\]", ""), scientific_name = stringr::str_trim(scientific_name,         side = "both"), status = dplyr::if_else(scientific_name %in%         valid_spp_list, "Valid", status), status = dplyr::if_else(scientific_name ==         valid_scientific_name, "Valid", status)), !(status ==     "Valid" & scientific_name != valid_scientific_name), year !=     "Invalid Number", year <= lubridate::year(Sys.Date()), !grepl("^\\w+ & \\w+ \\d{4}$",     scientific_name), !grepl("^[a-z]", scientific_name), !stringr::str_detect(scientific_name,     " \\[ref. "), !stringr::str_detect(scientific_name, "^\\["),     !stringr::str_detect(scientific_name, ";")): ℹ In argument: `year <= lubridate::year(Sys.Date())`.
#> Caused by error in `loadNamespace()`:
#> ! there is no package called ‘lubridate’