Loops: How can I loop case_when function in R? - r

Question

Here's the code, where I am trying to create a variable by detecting the words and matching them. Here I use dplyr package and its function mutate in combination with case_when. The problem is I am adding each one of the values manually as you see. How can I automate it by applying some loop functions to match the two?
city <- LETTERS #26 cities
district <- letters[10:20] #11 districts
streets <- paste0(district, district)
streets <- streets[-c(5:26)] #4 streets
df <- data.frame(x = c(1:5),
address = c("A, b, cc,", "B, dd", "a, dd", "C", "D, a, cc"))
library(dplyr)
library(stringi)
df2 <- df %>%
mutate(districts = case_when(
stri_detect_fixed(address, "b") ~ "b", #address[1]
#address[2]
stri_detect_fixed(address, "a") ~ "a", #address[3]
#address[4]
stri_detect_fixed(address, "cc") ~ "cc" #address[5]
))
The code scans through address for the value in district vector. I would love to do the same for city and street variables. So I used the modified version of the code from another question in Stack Overflow. It produces an error.
for (j in town_village2) {
trn_house3[,93] <- case_when(
stri_detect_fixed(trn_house3[1:6469, 4], j) ~ j)
}
I seek to produce this result:
x address city district street
1 A, b, cc, A b cc
2 B, dd B NA dd
3 a, dd NA a dd
4 C C NA NA
5 D, a, cc D a cc

Answer 1

If you are going to add a loop, it makes no sense to use case_when(); you don't have to add all options into it if you can loop over them.
You can solve it with a for-loop:
library(stringi)
df2 <- df
for(c in city) df2$city[stri_detect_fixed(df2$address, c)] <- c
for(d in district) df2$district[stri_detect_fixed(df2$address, d)] <- d
for(s in streets) df2$street[stri_detect_fixed(df2$address, s)] <- s
Note that your example code didn't work; the district names are 'a' and 'b' in your example dataset, but you generate names 'j' through 't'. I fixed that in my code above.
And it will cause an error if names of cities, districts and/or streets overlap. For instance, if one row is in the district 'b', and in the street 'cc', stri_detect_fixed will also see the 'c' and think it is in 'c'. I propose a completely different method to overcome this:
Alternative method
Given your example data, it makes most sense to first split the given address by ,, then look for exact matches with your reference city/district/street names. We can look for those exact matches with intersect().
# example reference address parts
cities <- c("A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L",
"M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y",
"Z")
districts <- c("a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k")
streets <- c("aa", "bb", "cc", "dd")
# example dataset
df <- data.frame(x = c(1:5),
address = c("A, b, cc,", "B, dd", "a, dd", "C", "D, a, cc"))
# vectorize address into elements
address_elems = strsplit(df$address, ',') # split by comma
address_elems = sapply(address_elems, trimws) # trim whitespace
Compare df$address and the newly created address_elems:
> df$address
[1] "A, b, cc," "B, dd" "a, dd" "C" "D, a, cc"
> address_elems
[[1]]
[1] "A" "b" "cc"
[[2]]
[1] "B" "dd"
[[3]]
[1] "a" "dd"
[[4]]
[1] "C"
[[5]]
[1] "D" "a" "cc"
We could find matching cities for just the first vector in address_elems in with intersect(cities, address_elems[[1]]).
Because we might get multiple matches, we only take the first element, with intersect(cities, address_elems[[1]])[[1]].
To apply this to every vector in address_elems, we can use sapply() or lapply():
# intersect the respective reference lists with each list of
# address items, taking only the first element
df$cities = sapply(address_elems, function(x) intersect(cities, x)[1])
df$district = sapply(address_elems, function(x) intersect(districts, x)[1])
df$street = sapply(address_elems, function(x) intersect(streets, x)[1])
PIAT
Putting it all together we get:
# example reference address parts
cities <- c("A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L",
"M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y",
"Z")
districts <- c("a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k")
streets <- c("aa", "bb", "cc", "dd")
# example dataset
df <- data.frame(x = c(1:5),
address = c("A, b, cc,", "B, dd", "a, dd", "C", "D, a, cc"))
# create vector of address elements
address_elems = strsplit(df$address, ',') # split by comma
address_elems = sapply(address_elems, trimws) # trim whitespace
# intersect the respecitve reference lists with each list of
# address items, take only the first element
df$cities = lapply(address_elems, function(x) intersect(cities, x)[1])
df$district = sapply(address_elems, function(x) intersect(districts, x)[1])
df$street = sapply(address_elems, function(x) intersect(streets, x)[1])
# cleanup
rm(address_elems)

Answer 2

This will separate the elements into vectors:
library(tidyverse)
df <- data.frame(
x = c(1:5),
address = c("A, b, cc,", "B, dd", "a, dd", "C", "D, a, cc")
)
df3 <-
df %>%
separate_rows(address, sep = "[, ]+") %>%
filter(nchar(address) > 0) %>%
nest(address) %>%
transmute(x, districts = data %>% map(~ .x[[1]]))
#> Warning: All elements of `...` must be named.
#> Did you want `data = address`?
df3
#> # A tibble: 5 × 2
#> x districts
#> <int> <list>
#> 1 1 <chr [3]>
#> 2 2 <chr [2]>
#> 3 3 <chr [2]>
#> 4 4 <chr [1]>
#> 5 5 <chr [3]>
df3$districts[[1]]
#> [1] "A" "b" "cc"
Created on 2022-04-14 by the reprex package (v2.0.0)

Answer 3

a data.table approach
library(data.table)
DT <- data.table(city, streets, district)
# create a lookup table with all elements
lookup <- melt(DT, measure.vars = names(DT))
# set df to data.table format
setDT(df)
final <- df[, .(address = unlist(tstrsplit(address, ",[ ]*", perl = TRUE))), by = .(x)]
# now add elements
final[lookup, type := i.variable, on = .(address = value)]
# and dcast to wide
dcast(final, x ~ type, value.var = "address")
# x city streets district
# 1: 1 A cc b
# 2: 2 B dd <NA>
# 3: 3 <NA> dd a
# 4: 4 C <NA> <NA>
# 5: 5 D cc a