State name to abbreviation - r

Question

I have a large file with a variable state that has full state names. I would like to replace it with the state abbreviations (that is "NY" for "New York"). Is there an easy way to do this (apart from using several if-else commands)? May be using replace() statement?

Answer 1

R has two built-in constants that might help: state.abb with the abbreviations, and state.name with the full names. Here is a simple usage example:
> x <- c("New York", "Virginia")
> state.abb[match(x,state.name)]
[1] "NY" "VA"

Answer 2

1) grep the full name from state.name and use that to index into state.abb:
state.abb[grep("New York", state.name)]
## [1] "NY"
1a) or using which:
state.abb[which(state.name == "New York")]
## [1] "NY"
2) or create a vector of state abbreviations whose names are the full names and index into it using the full name:
setNames(state.abb, state.name)["New York"]
## New York
## "NY"
Unlike (1), this one works even if "New York" is replaced by a vector of full state names, e.g. setNames(state.abb, state.name)[c("New York", "Idaho")]

Answer 3

Old post I know, but wanted to throw mine in there. I learned on tidyverse, so for better or worse I avoid base R when possible. I wanted one with DC too, so first I built the crosswalk:
library(tidyverse)
st_crosswalk <- tibble(state = state.name) %>%
bind_cols(tibble(abb = state.abb)) %>%
bind_rows(tibble(state = "District of Columbia", abb = "DC"))
Then I joined it to my data:
left_join(data, st_crosswalk, by = "state")

Answer 4

I found the built-in state.name and state.abb have only 50 states. I got a bigger table (including DC and so on) from online (e.g., this link: http://www.infoplease.com/ipa/A0110468.html) and pasted it to a .csv file named States.csv. I then load states and abbr. from this file instead of using the built-in. The rest is quite similar to #Aniko 's
library(dplyr)
library(stringr)
library(stringdist)
setwd()
# load data
data = c("NY", "New York", "NewYork")
data = toupper(data)
# load state name and abbr.
State.data = read.csv('States.csv')
State = toupper(State.data$State)
Stateabb = as.vector(State.data$Abb)
# match data with state names, misspell of 1 letter is allowed
match = amatch(data, State, maxDist=1)
data[ !is.na(match) ] = Stateabb[ na.omit( match ) ]
There's a small difference between match and amatch in how they calculate the distance from one word to another. See P25-26 here http://cran.r-project.org/doc/contrib/de_Jonge+van_der_Loo-Introduction_to_data_cleaning_with_R.pdf

Answer 5

You can also use base::abbreviate if you don't have US state names. This won't give you equally sized abbreviations unless you increase minlength.
state.name %>% base::abbreviate(minlength = 1)

Answer 6

Here is another way of doing it in case you have more than one state in your data and you want to replace the names with the corresponding abbreviations.
#creating a list of names
states_df <- c("Alabama","California","Nevada","New York",
"Oregon","Texas", "Utah","Washington")
states_df <- as.data.frame(states_df)
The output is
> print(states_df)
states_df
1 Alabama
2 California
3 Nevada
4 New York
5 Oregon
6 Texas
7 Utah
8 Washington
Now using the state.abb function you can easily convert the names into abbreviations, and vice-versa.
states_df$state_code <- state.abb[match(states_df$states_df, state.name)]
> print(states_df)
states_df state_code
1 Alabama AL
2 California CA
3 Nevada NV
4 New York NY
5 Oregon OR
6 Texas TX
7 Utah UT
8 Washington WA

Answer 7

If matching state names to abbreviations or the other way around is something you have to frequently, you could put Aniko's solution in a function in a .Rprofile or a package:
state_to_st <- function(x){
c(state.abb, 'DC')[match(x, c(state.name, 'District of Columbia'))]
}
st_to_state <- function(x){
c(state.name, 'District of Columbia')[match(x, c(state.abb, 'DC'))]
}
Using that function as a part of a dplyr chain:
enframe(state.name, value = 'state_name') %>%
mutate(state_abbr = state_to_st(state_name))