Best way to extract longitude/latitude from a character vector in R

If you have multiple strings, you can use strcapture to return a data.frame:

vec <- c("POINT (-90.10051372 29.97596117)", "POINT (-91.10051372 28.97596117)")
strcapture("\\(([-0-9.]+)\\s+([-0-9.]+)", vec, proto = list(lon = 1,lat = 1))
#         lon      lat
# 1 -90.10051 29.97596
# 2 -91.10051 28.97596

Walk-through:

• Pattern:

  \\(([-0-9.]+)\\s+([-0-9.]+)
  ^^^                         literal left-paren
     ^^^^^^^^^^    ^^^^^^^^^^ groups of pos/neg numbers
               ^^^^           blank-space
  

  We can add a right-paren \\) to the end of the pattern for good measure, not sure if it adds much.

• proto= is purely to match each (..) pattern-group with a column name, and the values in the proto define the class/type, the actual values here don't matter (proto=c(lon=999,lat=198282) produce the same results; proto=c(lon="",lat=9) would produce a string class for lon, not what we want/need). If the class used within proto= does not work, for instance the pattern group captures something non-number-like, then it will be NA.

Since you are probably dealing with an sf object, here is an sf solution:

library(dplyr)
library(sf)
data.frame(a = "POINT (-90.10051372 29.97596117)") %>% 
  st_as_sf(wkt = "a") %>% 
  st_coordinates()

#          X        Y
#1 -90.10051 29.97596

Or, more lengthy but maybe more flexible:

st_as_sf(data.frame(a = "POINT (-90.10051372 29.97596117)"),
         wkt = "a") %>% 
  mutate(lon = st_coordinates(.)[,1],
        lat = st_coordinates(.)[,2]) %>% 
  st_drop_geometry()

The trick is in the regex.

x <- "POINT (-90.10051372 29.97596117)"
y <- sub("^[^\\(]+\\(([^\\)]+)\\)$", "\\1", x)
p <- as.numeric(strsplit(y, " ")[[1]])
p
#> [1] -90.10051  29.97596

^{Created on 2022-09-23 with reprex v2.0.2}

Explanation:

Another approach:

gsub('([A-Z]+ \\()(-?[0-9.]+)\\s([0-9.]+)\\)', '\\2 \\3', "POINT (-90.10051372 29.97596117)")
[1] "-90.10051372 29.97596117"