Subset strings in R - r

Question

One of the strings in my vector (df$location1) is the following:
Potomac, MD 20854\n(39.038266, -77.203413)
Rest of the data in the vector follow same pattern. I want to separate each component of the string into a separate data element and put it in new columns like: df$city, df$state, etc.
So far I have been able to isolate the lat. long. data into a separate column by doing the following:
df$lat.long <- gsub('.*\\\n\\\((.*)\\\)','\\\1',df$location1)
I was able to make it work by looking at other codes online but I don't fully understand it. I understand the regex pattern but don't understand the "\\1" part. Since I don't understand it in full I have been unable to use it to subset other parts of this same string.
What's the best way to subset data like this?
Is using regex a good way to do this? What other ways should I be looking into?
I have looked into splitting the string after a comma, subset using regex, using scan() function and to many other variations. Now I am all confused. Thx

Answer 1

We can also use the separate function from the tidyr package (part of the tidyverse package).
library(tidyverse)
# Create example data frame
dat <- data.frame(Data = "Potomac, MD 20854\n(39.038266, -77.203413)",
stringsAsFactors = FALSE)
dat
# Data
# 1 Potomac, MD 20854\n(39.038266, -77.203413)
# Separate the Data column
dat2 <- dat %>%
separate(Data, into = c("City", "State", "Zip", "Latitude", "Longitude"),
sep = ", |\\\n\\(|\\)|[[:space:]]")
dat2
# City State Zip Latitude Longitude
# 1 Potomac MD 20854 39.038266 -77.203413

Answer 2

You can try strsplit or data.table::tstrsplit(strsplit + transpose):
> x <- 'Potomac, MD 20854\n(39.038266, -77.203413)'
> data.table::tstrsplit(x, ', |\\n\\(|\\)')
[[1]]
[1] "Potomac"
[[2]]
[1] "MD 20854"
[[3]]
[1] "39.038266"
[[4]]
[1] "-77.203413"
More generally, you can do this:
library(data.table)
df[c('city', 'state', 'lat', 'long')] <- tstrsplit(df$location1, ', |\\n\\(|\\)')
The pattern ', |\\n\\(|\\)' tells tstrsplit to split by ", ", "\n(" or ")".
In case you want to sperate state and zip and cite names may contain spaces, You can try a two-step way:
# original split (keep city names with space intact)
df[c('city', 'state', 'lat', 'long')] <- tstrsplit(df$location1, ', |\\n\\(|\\)')
# split state and zip
df[c('state', 'zip')] <- tstrsplit(df$state, ' ')

Answer 3

Here is an option using base R
read.table(text= trimws(gsub(",+", " ", gsub("[, \n()]", ",", dat$Data))),
header = FALSE, col.names = c("City", "State", "Zip", "Latitude", "Longitude"),
stringsAsFactors = FALSE)
# City State Zip Latitude Longitude
#1 Potomac MD 20854 39.03827 -77.20341

Answer 4

So this process might be a little longer, but for me it makes things clear. As opposed to using breaks, below I identify values by using a specific regex for each value I want. I make a vector of regex to extract each value, a vector for the variable names, then use a loop to extract and create the dataframe from those vectors.
library(stringi)
library(dplyr)
library(purrr)
rgexVec <- c("[\\w\\s-]+(?=,)",
"[A-Z]{2}",
"\\d+(?=\\n)",
"[\\d-\\.]+(?=,)",
"[\\d-\\.]+(?=\\))")
varNames <- c("city",
"state",
"zip",
"lat",
"long")
map2_dfc(varNames, rgexVec, function(vn, rg) {
extractedVal <- stri_extract_first_regex(value, rg) %>% as.list()
names(extractedVal) <- vn
extractedVal %>% as_tibble()
})

Answer 5

\\1 is a back reference in regex. It is similar to a wildcard (*) that will grab all instances of your search term, not just the first one it finds.