knitr::opts_chunk$set( collapse = TRUE, comment = "#>", fig.path = "man/figures/README-" ) version <- as.vector(read.dcf('DESCRIPTION')[, 'Version']) version <- gsub('-', '.', version)
This package has merged with the abscorr
package by the R Users' Network for
Australian Public Policy. The merged package is called strayr
. Find the merged package at runapp's GitHub.
strayr is a simple tool to wrangle messy Australian state names and/or abbreviations into a consistent format.
Install from GitHub with:
# if you don't have devtools installed, first run: # install.packages("devtools") devtools::install_github("mattcowgill/strayr")
Let's start with a character vector that includes some misspelled State names, some correctly spelled state names, as well as some abbreviations both malformed and correctly formed.
x <- c("western Straya", "w. A ", "new soth wailes", "SA", "tazz", "Victoria", "northn territy")
To convert this character vector to a vector of abbreviations for State names,
simply use the strayr()
function:
library(strayr) strayr(x)
If you want full names for the states rather than abbreviations:
strayr(x, to = "state_name")
By default, strayr()
uses fuzzy or approximate string matching to match the
elements in your character vector to state names/abbreviations. If you only want
to permit exact matching, you can disable fuzzy matching. This means you will
never get false matches, but you will also fail to match misspelled state names
or malformed abbreviations; you'll get an NA
if no match can be found.
strayr(x, fuzzy_match = FALSE)
If your data is in a data frame, strayr()
works well within a dplyr::mutate()
call:
x_df <- data.frame(state = x, stringsAsFactors = FALSE) library(dplyr) x_df %>% mutate(state_abbr = strayr(state))
This package includes the auholidays
dataset from the Australian Public Holidays Dates Machine Readable Dataset as well as a helper function is_holiday
:
str(auholidays) is_holiday('2020-01-01') is_holiday('2019-05-27', jurisdictions=c('ACT', 'TAS')) h_df <- data.frame(dates = c('2020-01-01', '2020-01-10')) h_df %>% mutate(IsHoliday = is_holiday(dates))
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