Extract House Number from (address) string using r - r

Question

I want to parse apart (extract) addresses into HouseNumber and Streetname.
I should later be able to write the extracted "values" into new columns (shops$HouseNumber and shops$Streetname).
So lets say I have a data frame called "shops":
> shops
Name city street
1 Something Fakecity New Street 3
2 SomethingOther Fakecity Some-Complicated-Casestreet 1-3
3 SomethingDifferent Fakecity Fake Street 14a
So is there a way to split the street column into two lists one with the streetnames and one for the house numbers including cases like "1-3","14a", so that in the end, the result could be assigned to the data frame and look like.
> shops
Name city Streetname HouseNumber
1 Something Fakecity New Street 3
2 SomethingOther Fakecity Some-Complicated-Casestreet 1-3
3 SomethingDifferent Fakecity Fake Street 14a
Example: Easyfakestreet 5 --> Easyfakestreet , 5
It gets slightly complicated by the fact that some of my street strings will have hyphenated street addresses and have non numerical components.
Examples: New Street 3 --> ['New Street', '3 ']
Some-Complicated-Casestreet 1-3 --> ['Some-Complicated-Casestreet','1-3']
Fake Street 14a --> ['Fake Street', '14a']
I would appreciate some help!

Answer 1

Here's a possible tidyr solution
library(tidyr)
extract(df, "street", c("Streetname", "HouseNumber"), "(\\D+)(\\d.*)")
# Name city Streetname HouseNumber
# 1 Something Fakecity New Street 3
# 2 SomethingOther Fakecity Some-Complicated-Casestreet 1-3
# 3 SomethingDifferent Fakecity Fake Street 14a

Answer 2

You can try:
shops$Streetname <- gsub("(.+)\\s[^ ]+$","\\1", shops$street)
shops$HousNumber <- gsub(".+\\s([^ ]+)$","\\1", shops$street)
data
shops$street
#[1] "New Street 3" "Some-Complicated-Casestreet 1-3" "Fake Street 14a"
results
shops$Streetname
#[1] "New Street" "Some-Complicated-Casestreet" "Fake` Street"
shops$HousNumber
#[1] "3" "1-3" "14a"

Answer 3

Create a pattern with back references that match both the street and the number and then using sub replace it by each backreference in turn. No packages are needed:
pat <- "(.*) (\\d.*)"
transform(shops,
street = sub(pat, "\\1", street),
HouseNumber = sub(pat, "\\2", street)
)
giving:
Name city street HouseNumber
1 Something Fakecity New Street 3
2 SomethingOther Fakecity Some-Complicated-Casestreet 1-3
3 SomethingDifferent Fakecity Fake Street 14a
Here is a visualization of pat:
(.*) (\d.*)
Debuggex Demo
Note:
1) We used this for shops:
shops <-
structure(list(Name = c("Something", "SomethingOther", "SomethingDifferent"
), city = c("Fakecity", "Fakecity", "Fakecity"), street = c("New Street 3",
"Some-Complicated-Casestreet 1-3", "Fake Street 14a")), .Names = c("Name",
"city", "street"), class = "data.frame", row.names = c(NA, -3L))
2) David Arenburg's pattern could alternately be used here. Just set pat to it. The pattern above has the advantage that it allows street names that have embedded numbers in them but David's has the advantage that the space may be missing before the street number.

Answer 4

You could use the package unglue
library(unglue)
unglue_unnest(shops, street, "{street} {value=\\d.*}")
#> Name city street value
#> 1 Something Fakecity New Street 3
#> 2 SomethingOther Fakecity Some-Complicated-Casestreet 1-3
#> 3 SomethingDifferent Fakecity Fake Street 14a
Created on 2019-10-08 by the reprex package (v0.3.0)

Answer 5

Very complex issue for international addresses
$re = '/(\d+[\d\/\-\. ,]*[ ,\d\-\w]{0,2} )/m';
$str = '234 Test Road, Testville
456b Tester Road, Testville
789 c Tester Road, Testville
Mystreet 14a
123/3 dsdsdfs
Roobertinkatu 36-40
Flats 1-24 Acacia Avenue
Apartment 9D, 1 Acacia Avenue
Flat 24, 1 Acacia Avenue
Moscow Street, plot,23 building 2
Apartment 5005 no. 7 lane 31 Wuming Rd
Quinta da Redonda Lote 3 - 1 º
102 - 3 Esq
Av 1 Maio 16,2 dt,
Rua de Ceuta Lote 1 Loja 5
11334 Nc Highway 72 E ';
preg_match_all($re, $str, $matches, PREG_SET_ORDER, 0);
// Print the entire match result
var_dump($matches);
Output example
https://regex101.com/r/WVPBji/1