Extracting cities from string vector in R - r

Question

I have a column in my dataset db, say db$affiliation, which looks like:
**db$affiliation**
[1] "[SCOTT, ALLEN J.] UNIV CALIF LOS ANGELES, DEPT GEOG, LOS ANGELES, CA 90095 USA"
[2] "[VAN DUINEN, RIANNE; VAN DER VEEN, ANNE] UNIV TWENTE, DEPT WATER ENGN & MANAGEMENT, DRIENERLOLAAN 5,POB 217, NL-7500 AE ENSCHEDE, NETHERLANDS."
[3] "[ANANTSUKSOMSRI, SUTEE] CHULALONGKORN UNIV, FAC ARCHITECTURE, BANGKOK, THAILAND."
[4] ...
I would like to create a column within the same dataset containing only the name of the city in db$affiliation, such as
**db$cities**
[1] LOS ANGELES
[2] TWENTE
[3] BANGKOK
[4] ...
If multiple city names are available, I'd like the command to return only the last one, if no city names are available I'd like to have NA. How can I do that?
I thought that I could use world.cities$name in data(world.cities) in the maps package but I can not figure out how.
I even tried to split the db$affiliation column such as:
db$affiliation <- gsub("\\[[^\\]]*\\]", "", db$affiliation, perl=TRUE) # remove content within brackets
db$affiliation[2] # check the separator
db <- cSplit(db, 'affiliation', sep=c(", "), type.convert=FALSE) # split after comma
Which results (I've truncated it after affiliation_3) in:
affiliation_1 affiliation_2 affiliation_3
[1] UNIV CALIF LOS ANGELES DEPT GEOG LOS ANGELES
[2] UNIV TWENTE DEPT WATER ENGN & MANAGEMENT DRIENERLOLAAN
[3] CHULALONGKORN UNIV FAC ARCHITECTURE BANGKOK
And then pass:
db$cities <- lapply(db$affiliation_1, function(x)x[which(x %in% world.cities$name)])
But I get an empty column.
Thanks for the help!

Answer 1

There are many cities in your sample string so you may need to think again if you still want to fetch the 'last city' in case multiple cities are found in affiliation column.
library(maps)
data(world.cities)
#sample data
df <- data.frame(affiliation = c("[SCOTT, ALLEN J.] UNIV CALIF LOS ANGELES, DEPT GEOG, LOS ANGELES, CA 90095 USA",
"[VAN DUINEN, RIANNE; VAN DER VEEN, ANNE] UNIV TWENTE, DEPT WATER ENGN & MANAGEMENT, DRIENERLOLAAN 5,POB 217, NL-7500 AE ENSCHEDE, NETHERLANDS.",
"[ANANTSUKSOMSRI, SUTEE] CHULALONGKORN UNIV, FAC ARCHITECTURE, BANGKOK, THAILAND.",
"Prem"), stringsAsFactors = F)
#fetch city and it's respective country from 'affiliation' column
cities_country <- lapply(gsub("\\[|\\]|[,;]|\\.","",df$affiliation), function(x)
paste(as.character(world.cities$name[sapply(world.cities$name, grepl, x, ignore.case=T)]),
as.character(world.cities$country.etc[sapply(world.cities$name, grepl, x, ignore.case=T)]),
sep="_"))
df$cities_country <- lapply(cities_country, function(x) if(identical(x, character(0))) NA_character_ else x)
df
Output is:
affiliation
1 [SCOTT, ALLEN J.] UNIV CALIF LOS ANGELES, DEPT GEOG, LOS ANGELES, CA 90095 USA
2 [VAN DUINEN, RIANNE; VAN DER VEEN, ANNE] UNIV TWENTE, DEPT WATER ENGN & MANAGEMENT, DRIENERLOLAAN 5,POB 217, NL-7500 AE ENSCHEDE, NETHERLANDS.
3 [ANANTSUKSOMSRI, SUTEE] CHULALONGKORN UNIV, FAC ARCHITECTURE, BANGKOK, THAILAND.
4 Prem
cities_country
1 Al_Norway, Alle_Switzerland, Allen_Philippines, Allen_USA, Angeles_Costa Rica, Angeles_Philippines, Cali_Colombia, Cot_Costa Rica, Li_Norway, Los Angeles_Chile, Los Angeles_USA, Os_Kyrgyzstan, Os_Norway, U_Micronesia, Usa_Japan
2 Ae_Marshall Islands, Ede_Netherlands, Ede_Nigeria, Enschede_Netherlands, Hede_China, Ine_Marshall Islands, Laa_Austria, Lola_Guinea, Man_Ivory Coast, Mana_French Guiana, Manage_Belgium, Nagem_Luxembourg, Ob_Russia, Ola_Panama, Po_Burkina Faso, U_Micronesia, Van_Turkey, Wa_Ghana, We_New Caledonia
3 Aila_Estonia, Al_Norway, Anan_Japan, Ba_Fiji, Bangkok_Thailand, Hit_Iraq, Ila_Nigeria, Ilan_Taiwan, Long_Thailand, Nan_Thailand, Tsu_Japan, U_Micronesia, Ula_Turkey
4 NA
(Note that in above output I have kept all occurrences of cities and for convenience also suffixed it with their respective countries)

Answer 2

From the few lines you have shown it looks like you might be able to do the following (note you missed aligning the casing):
tmpVec <- sapply(strsplit(db$affiliation, split = ","), function(x) {
cleanVec <- toupper(trimws(x))
cleanVec[max(which(cleanVec %in% toupper(maps::world.cities$name)))]
})
Or put a bit more code into the function to avoid the ugly warnings.

Answer 3

Let me leave a part of a solution. As far as I can tell from my own research, letters in the square brackets seem to indicate personal names. For example, I found that Sutee Anantsuksomsri is an actual name. This observation suggests that we probably want to remove texts in the brackets.
Once I removed the texts in the square brackets, I split the words using unnest_tokens() in the tidytext package. Note that the function converts all letters to small letters. If you do not like it, you can change that by specifying to_lower = FALSE. First, I split each city name into word. I also assigned an ID number for each city. Second, I cleaned up your data. As I said earlier, I removed texts in square brackets using gsub(). Then, I applied unnest_tokens() to the data. I subset words using the words from cities in filter(). The result we get up to this point is the following. Obviously, you have more work to do. I leave the sampling data, mydf below. I hope you can move on from here.
data(world.cities)
cities <- world.cities %>%
mutate(id = 1:n()) %>%
unnest_tokens(input = name, output = word, token = "words")
temp <- mydf %>%
mutate(affiliation = gsub(x = affiliation, pattern = "\\[.*\\]", replacement = "")) %>%
unnest_tokens(input = affiliation, output = word, token = "words") %>%
filter(word %in% cities$word)
id word
1 1 los
2 1 angeles
3 1 los
4 1 angeles
5 1 ca
6 1 usa
7 2 water
8 2 ae
9 2 enschede
10 3 bangkok
DATA
mydf <- structure(list(id = 1:3, affiliation = c("[SCOTT, ALLEN J.] UNIV CALIF LOS ANGELES, DEPT GEOG, LOS ANGELES, CA 90095 USA",
"[VAN DUINEN, RIANNE; VAN DER VEEN, ANNE] UNIV TWENTE, DEPT WATER ENGN & MANAGEMENT, DRIENERLOLAAN 5,POB 217, NL-7500 AE ENSCHEDE, NETHERLANDS.",
"[ANANTSUKSOMSRI, SUTEE] CHULALONGKORN UNIV, FAC ARCHITECTURE, BANGKOK, THAILAND."
)), .Names = c("id", "affiliation"), row.names = c(NA, -3L), class = "data.frame")