CODEWITHSUNDEEP
rregex
lawyeR
lawyeR

Reputation: 7664

Convert long state names embedded with other text to two-letter state abbreviations

My objective is to identify US states written out in a character vector that has other text and convert the states to abbreviated form. For example, "North Carolina" to "NC". It is simple if the vector only has long-form state names. However, my vector has other text in random places, as in the example "states".

states <- c("Plano New Jersey", "NC", "xyz", "Alabama 02138", "Texas", "Town Iowa 99999")

From another post I found this:

state.abb[match(states, state.name)]

but it converts only the standalone Texas

> state.abb[match(states, state.name)]
[1] NA   NA   NA   NA   "TX"

and not the New Jersey, Alabama and Iowa strings.

From Fast grep with a vectored pattern or match, to return list of all matches I tried:

sapply(states, grep(pattern = state.name, x = states, value = TRUE))

but

Error in get(as.character(FUN), mode = "function", envir = envir) : 
  object 'Alabama 02138' of mode 'function' was not found
In addition: Warning message:
In grep(pattern = state.name, x = states, value = TRUE) :
  argument 'pattern' has length > 1 and only the first element will be used

Nor does this work:

sapply(states, function(x) state.abb[grep(state.name, states)])

This question did not help: regular expression to convert state names to abbreviations

How do I convert the embedded long names to the state abbreviation?

EDIT: I want to return the vector with the only change being that the long names of the states have been abbreviated, e.g., "Plano New Jersey" becomes "Plano NJ".

Thank you for correcting and/or educating me.

Upvotes: 4

Views: 1944

Answers (5)

rnso
rnso

Reputation: 24593

Try:

for(r in 1:nrow(states.list)) {
    states = gsub(states.list[r,1], states.list[r,2], states)
}

states
[1] "Plano NJ"      "NC"            "xyz"           "AL 02138"      "TX"            "Town IA 99999"

Data:

states <- c("Plano New Jersey", "NC", "xyz", "Alabama 02138", "Texas", "Town Iowa 99999")

states.list = structure(list(state.name = structure(c(4L, 1L, 5L, 2L, 3L), .Label = c("Alabama", 
"Iowa", "Minnesota", "New Jersey", "Texas"), class = "factor"), 
    state.abb = structure(c(4L, 1L, 5L, 2L, 3L), .Label = c("AL", 
    "IA", "MN", "NJ", "TX"), class = "factor")), .Names = c("state.name", 
"state.abb"), class = "data.frame", row.names = c(NA, -5L))

states.list
  state.name state.abb
1 New Jersey        NJ
2    Alabama        AL
3      Texas        TX
4       Iowa        IA
5  Minnesota        MN

Upvotes: 0

ebo
ebo

Reputation: 2747

If you do not want to use additional packages you can use the mapply function to apply gsub for all pairs of state.name and state.abb, e.g.:

mapply(gsub,state.name,state.abb,"ALABAMA 123",ignore.case=TRUE,USE.NAMES=FALSE)

The result of this is a list which could contain a replacement, e.g.:

 [1] "AL 123"      "ALABAMA 123" "ALABAMA 123" "ALABAMA 123" "ALABAMA 123" 
 [6] ...

by taking the shortest text from this list you can get the desired result. Thus we sort the list based on the length of the text and take the first element.

The complete code:

replaceState <- function(x) {  
     v = mapply(gsub,state.name,state.abb,x,ignore.case=TRUE, USE.NAMES=FALSE)
     v[order(nchar(v))][1] 
}

sapply(states, replaceState, USE.NAMES=FALSE)

Unfortunately, this approach only replaces the name of a single state (the longest). To replace multiple different states we need to iterate, e.g.:

replaceState <- function(x) {  
     v = mapply(gsub,state.name,state.abb,x,ignore.case=TRUE, USE.NAMES=FALSE)
     v[order(nchar(v))][1] 
}

replaceStates <- function(x) {
     newX = replaceState(x)

     # if they are different a state has been replaced, 
     # we try again to replace all states.
     if(newX != x){ 
          replaceStates(newX)
     } else {
          newX
     }
}

# Note the 'replaceStates'
sapply(states, replaceStates, USE.NAMES=FALSE)

Upvotes: 1

Tyler Rinker
Tyler Rinker

Reputation: 109984

Here's another approach:

library(qdap)
mgsub(state.name, state.abb, states)

## [1] "Plano NJ"      "NC"            "xyz"           "AL 02138"      
## "TX"            "Town IA 99999"

If you are uncertain that the states will be capitalized you may want to use:

mgsub(state.name, state.abb, states, ignore.case=TRUE, fixed=FALSE)

Upvotes: 3

akrun
akrun

Reputation: 887571

Try:

indx <- paste0(".*(", paste(state.name, collapse="|"), ").*")
v1 <- gsub(indx, "\\1", states)
ifelse( v1 %in% state.abb, v1, state.abb[match(v1, state.name)])
#[1] "NJ" "NC" NA   "AL" "TX" "IA"

If you want to just replace the states with the abbreviation and not the other text, you could also do:

indx1 <- paste(state.name, collapse="|")   
indx2 <- state.abb[match(v1, state.name)]

mapply(gsub, indx1, indx2, states, USE.NAMES=F)
#[1] "Plano NJ"      "NC"            "xyz"           "AL 02138"     
#[5] "TX"            "Town IA 99999"

Upvotes: 3

G. Grothendieck
G. Grothendieck

Reputation: 269885

It was not clear from the question what the expected result is to be but here we have assumed that you want to preserve the text in the input just replacing the fuil state names with the abbreviation.

Create a list, st, whose names are the full state names and whose values are the abbreviations. Then use paste(..., collapse = "|") to create a regular expression that matches any state and use gsubfn from the gsubfn package to perform the substitutions.

library(gsubfn)
st <- as.list(setNames(state.abb, state.name))
gsubfn(paste(state.name, collapse = "|"), st, states)

giving:

[1] "Plano NJ"      "NC"            "xyz"           "AL 02138"     
[5] "TX"            "Town IA 99999"

Upvotes: 1

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