Splitting full address column in multiple columns - r

Question

I have a dataframe with the following column structure (over 1000+ rows total):
addressfull
POINT(3.124537653 32.179354012)||DEF_32||molengraaf 20, 1689 GL Utrecht, Netherlands||15||map
POINT(3.124537680 32.179354014)||DEF_32||winkellaan 67, 5788 BG Amsterdam, Netherlands||13||map
POINT(3.124537653 32.179354012)||DEF_32||vermeerstraat 18, 0932 DC Rotterdam, Netherlands||11||map
POINT(2.915206183 24.315583523)||DEF_32||--||13||map
POINT (2.900824999999923 34.3175721)||DEF_84||Zandhorstlaan 122, 0823 GT Ochtrup, Germany||17||map
structure(list(addressfull = structure(c(3L, 5L, 4L, 2L, 1L), .Label = c("POINT (2.900824999999923 34.3175721)||DEF_84||Zandhorstlaan 122, 0823 GT Ochtrup, Germany||17||map",
"POINT(2.915206183 24.315583523)||DEF_32||--||13||map", "POINT(3.124537653 32.179354012)||DEF_32||molengraaf 20, 1689 GL Utrecht, Netherlands||15||map",
"POINT(3.124537653 32.179354012)||DEF_32||vermeerstraat 18, 0932 DC Rotterdam, Netherlands||11||map",
"POINT(3.124537680 32.179354014)||DEF_32||winkellaan 67, 5788 BG Amsterdam, Netherlands||13||map"
), class = "factor")), class = "data.frame", row.names = c(NA,
-5L))
The column contains an location, street, housenumber, zip code, city, and country. I want to split the column addressfull with R in multiple columns, as example:
street house number zip city country
molengraaf 20 1689 GL Utrecht Netherlands
winkellaan 67 5788 BG Amsterdam Netherlands
vermeerstraat 18 0932 DC Rotterdam Netherlands
na na na na na
Zandhorstlaan 122 0823 GT Ochtrup Germany
I have readed the tidyr and stringr documentation. I can see an pattern for splitting (by ")", "| from position x", and ",". but i can't figure out the correct code to split the column into multiple columns.
Can someone help me?

Answer 1

You could brute force it using sub for a base R approach:
df$steet <- sub("^(\\S+)\\s+.*$", "\\1", df$adressfull)
df$`house number` <- sub("^\\S+\\s+(\\d+).*$", "\\1", df$adressfull)
df$zip <- sub("^\\S+\\s+\\d+,\\s*(\\d+\\s+[A-Z]+).*$", "\\1", df$adressfull)
df$city <- sub("^.*?(\\S+),\\s*\\S+$", "\\1", df$adressfull)
df$country <- sub("^.*,\\s*(\\S+)$", "\\1", df$adressfull)
df
adressfull steet house number zip
1 molengraaf 20, 1689 GL Utrecht, Netherlands molengraaf 20 1689 GL
city country
1 Utrecht Netherlands
Data:
df <- data.frame(adressfull=c("molengraaf 20, 1689 GL Utrecht, Netherlands"),
stringsAsFactors=FALSE)
This assumes that we have already isolated just the address text. To do that, conisder:
text <- "POINT(3.124537653 32.179354012)||DEF_32||molengraaf 20, 1689 GL Utrecht, Netherlands||15||map"
addresfull <- unlist(strsplit(text, "\\|\\|"))[3]
addresfull
[1] "molengraaf 20, 1689 GL Utrecht, Netherlands"

Answer 2

A stringrsolution is this:
addresssplit <- data.frame(
street = str_extract(addressfull$addressfull, "(?<=DEF_\\d{2}\\|\\|)\\w+\\b"),
number = str_extract(addressfull$addressfull, "\\d{1,}(?=,)"),
zip = str_extract(addressfull$addressfull, "(?<=\\s)\\d{4}\\s[A-Z]{2}"),
city = str_extract(addressfull$addressfull, "(?<=\\d{4}\\s[A-Z]{2}\\s)\\w+"),
country = str_extract(addressfull$addressfull, "(?<=[a-z]\\b,\\s)\\w+\\b")
)
RESULT:
addresssplit
street number zip city country
1 molengraaf 20 1689 GL Utrecht Netherlands
2 winkellaan 67 5788 BG Amsterdam Netherlands
3 vermeerstraat 18 0932 DC Rotterdam Netherlands
4 <NA> <NA> <NA> <NA> <NA>
5 Zandhorstlaan 122 0823 GT Ochtrup Germany
DATA:
addressfull <- structure(list(addressfull = structure(c(3L, 5L, 4L, 2L, 1L), .Label = c("POINT (2.900824999999923 34.3175721)||DEF_84||Zandhorstlaan 122, 0823 GT Ochtrup, Germany||17||map",
"POINT(2.915206183 24.315583523)||DEF_32||--||13||map", "POINT(3.124537653 32.179354012)||DEF_32||molengraaf 20, 1689 GL Utrecht, Netherlands||15||map",
"POINT(3.124537653 32.179354012)||DEF_32||vermeerstraat 18, 0932 DC Rotterdam, Netherlands||11||map",
"POINT(3.124537680 32.179354014)||DEF_32||winkellaan 67, 5788 BG Amsterdam, Netherlands||13||map"
), class = "factor")), class = "data.frame", row.names = c(NA,
-5L))

Answer 3

This would be a tidyverse approach to the problem:
library(tidyverse)
df <- structure(list(addressfull = structure(c(3L, 5L, 4L, 2L, 1L), .Label = c("POINT (2.900824999999923 34.3175721)||DEF_84||Zandhorstlaan 122, 0823 GT Ochtrup, Germany||17||map",
"POINT(2.915206183 24.315583523)||DEF_32||--||13||map", "POINT(3.124537653 32.179354012)||DEF_32||molengraaf 20, 1689 GL Utrecht, Netherlands||15||map",
"POINT(3.124537653 32.179354012)||DEF_32||vermeerstraat 18, 0932 DC Rotterdam, Netherlands||11||map",
"POINT(3.124537680 32.179354014)||DEF_32||winkellaan 67, 5788 BG Amsterdam, Netherlands||13||map"
), class = "factor")), class = "data.frame", row.names = c(NA,
-5L))
df %>% separate(addressfull, sep = "\\|\\|", into = c("Coords", "DEF", "ADDRESS"),extra = "drop") %>%
select(ADDRESS) %>%
separate(ADDRESS, sep = ",", into = c("street", "city", "country")) %>%
separate(street, sep = "(?= \\d)", into = c("street", "house_number")) %>%
separate(city, sep = "(?<=[A-Z][A-Z])", into = c("zip", "city"))
#> Warning: Expected 3 pieces. Missing pieces filled with `NA` in 1 rows [4].
#> Warning: Expected 2 pieces. Missing pieces filled with `NA` in 1 rows [4].
#> street house_number zip city country
#> 1 molengraaf 20 1689 GL Utrecht Netherlands
#> 2 winkellaan 67 5788 BG Amsterdam Netherlands
#> 3 vermeerstraat 18 0932 DC Rotterdam Netherlands
#> 4 -- <NA> <NA> <NA> <NA>
#> 5 Zandhorstlaan 122 0823 GT Ochtrup Germany
Created on 2020-02-27 by the reprex package (v0.3.0)