Match household partners without loop - r

Question

I am trying to avoid using a loop in my attempt to recode household partners' variable identification.
hldid denotes the household while persid the person in the household.
The variable partner indicates the persid of the partner and child indicate if the row is a child.
What is missing from the variable partner is the persid for both partners.
For instance for hldid == 1, the persid == 1 has a value of 0 for partner while it should be 2.
This is how the data looks like:
> test
hldid persid age sex relresp partner child
1 1 1 26 2 0 0 0
2 1 2 26 1 1 1 0
3 2 1 59 2 0 0 0
4 2 2 64 1 1 1 0
5 3 1 76 2 0 0 0
6 4 1 65 2 0 0 0
7 4 2 64 1 1 1 0
8 5 1 52 2 0 0 0
9 5 2 51 1 1 1 0
10 5 3 20 2 21 0 1
11 5 4 14 2 21 0 1
12 7 1 69 1 0 0 0
13 7 2 70 2 1 1 0
I managed to create a quite ugly loop, however it is too slow for the entire dataset.
test$partnerREC = test$partner
for(i in 1:13){
for(j in 1:13){
if(
test$hldid[i] == test$hldid[i+1] & # verify if household is the same
(test$persid[i] == test$partner[j])
)
{
test$partnerREC[i] = test$persid[j] # put the persid for each partner
}
}
}
> test
hldid persid age sex relresp partner child partnerREC
1 1 1 26 2 0 0 0 2
2 1 2 26 1 1 1 0 1
3 2 1 59 2 0 0 0 2
4 2 2 64 1 1 1 0 1
5 3 1 76 2 0 0 0 0
6 4 1 65 2 0 0 0 2
7 4 2 64 1 1 1 0 1
8 5 1 52 2 0 0 0 2
9 5 2 51 1 1 1 0 1
10 5 3 20 2 21 0 1 0
11 5 4 14 2 21 0 1 0
12 7 1 69 1 0 0 0 2
13 7 2 70 2 1 1 0 1
Any idea how I could use data.table for solving this?
test = structure(list(hldid = c(1, 1, 2, 2, 3, 4, 4, 5, 5, 5, 5, 7,
7), persid = c(1, 2, 1, 2, 1, 1, 2, 1, 2, 3, 4, 1, 2), age = c(26,
26, 59, 64, 76, 65, 64, 52, 51, 20, 14, 69, 70), sex = c(2, 1,
2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2), relresp = c(0, 1, 0, 1, 0,
0, 1, 0, 1, 21, 21, 0, 1), partner = c(0, 1, 0, 1, 0, 0, 1, 0,
1, 0, 0, 0, 1), child = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0,
0)), class = "data.frame", row.names = c(NA, -13L))

Answer 1

A possible solution:
library(data.table) # load the package
setDT(test) # convert 'test' to a 'data.table'
test[, partnerREC := persid[c(pmin(2,.N):1,rep(0,(pmax(.N,2)-2)))] *
(persid %in% 1:2) *
(.N != 1)
, by = hldid][]
which gives:
> test
hldid persid age sex relresp partner child partnerREC
1: 1 1 26 2 0 0 0 2
2: 1 2 26 1 1 1 0 1
3: 2 1 59 2 0 0 0 2
4: 2 2 64 1 1 1 0 1
5: 3 1 76 2 0 0 0 0
6: 4 1 65 2 0 0 0 2
7: 4 2 64 1 1 1 0 1
8: 5 1 52 2 0 0 0 2
9: 5 2 51 1 1 1 0 1
10: 5 3 20 2 21 0 1 0
11: 5 4 14 2 21 0 1 0
12: 7 1 69 1 0 0 0 2
13: 7 2 70 2 1 1 0 1
This solution is based on the assumption (derived from the example data) that only 'persid' 1 & 2 are partners, anything higher are children.
What this does:
Group by hldid
Change the order of the first two persons in a hould but only when there are more people in a household with
persid[c(pmin(2,.N):1,rep(0,(pmax(.N,2)-2)))] where pmin is used to assure that a vector of length 1 is constructed when a household has only one person.
Multiply that with (persid %in% 1:2)to get zores for children.
Multiply that with (.N != 1) to get a value of zero for households of one person.

Answer 2

df <- data.frame(matrix(data = NA, ncol = 7))
names(df) <- names(test)
for(id in unique(test$hldid)){
t <- test[test$hldid==id,]
t$partner[t$persid == t$partner[t$partner!=0]] <- t$persid[which(t$partner!=0)]
df <- rbind(df, t)
}
df <- df[-1,]

Answer 3

A base R solution is more complicated than Jaap's data.table solution.
I work with a copy.
test2 <- test
After running the code in the question, run the following.
test2$partnerREC <- test2$partner
sp <- split(test2, test2$hldid)
test2 <- lapply(sp, function(DF){
i <- with(DF, which(persid %in% partner))
j <- with(DF, which(partner %in% persid))
#cat("i:", i, "\tj:", j, "\n")
DF$partnerREC[i] <- DF$persid[j]
DF
})
test2 <- do.call(rbind, test2)
row.names(test2) <- NULL
Now compare both results.
identical(test, test2)
#[1] TRUE

Answer 4

You can get there through some dplyr steps to join the data on itself and update the value of partner when persid == partner.
test2 <- left_join(test, test %>% select(hldid, persid, partner) %>% filter(partner != 0), by=c("hldid")) %>%
filter(persid.x == partner.y) %>%
mutate(partner.x = persid.y)
This will give you the head of household with their partner id matched up but you'd have to rejoin this to the original data (I'm not sure the equivalent of SQL update in dplyr lingo).

Answer 5

1. Create your test data.frame
library(tidyverse)
test <- tribble(
~hldid, ~persid, ~age, ~sex, ~relresp, ~partner, ~child,
1, 1, 26, 2, 0, 0, 0,
1, 2, 26, 1, 1, 1, 0,
2, 1, 59, 2, 0, 0, 0,
2, 2, 64, 1, 1, 1, 0,
3, 1, 76, 2, 0, 0, 0,
4, 1, 65, 2, 0, 0, 0,
4, 2, 64, 1, 1, 1, 0,
5, 1, 52, 2, 0, 0, 0,
5, 2, 51, 1, 1, 1, 0,
5, 3, 20, 2, 21, 0, 1,
5, 4, 14, 2, 21, 0, 1,
7, 1, 69, 1, 0, 0, 0,
7, 2, 70, 2, 1, 1, 0)
2. arrange(), group_by(), and mutate() come to rescue
test %>%
# arrange the data in case the raw data did not
arrange(hldid, child, persid) %>%
# group each household
group_by(hldid) %>%
# match first and second household person as each other's partner
mutate(partnerREC = ifelse(persid == first(persid), nth(persid, 2), first(persid))) %>%
# correct partnerREC for child and single
mutate(partnerREC = ifelse(child == 1 | is.na(partnerREC), 0, partnerREC))
# un-group it
ungroup()

Answer 6

A loop solution using rcpp
Source the rcpp script
#include <Rcpp.h>
using namespace Rcpp;
// [[Rcpp::export]]
NumericVector HHgrid(CharacterVector hid, NumericVector persid, NumericVector partner,
NumericVector partnerRec) {
int nrows = hid.size();
for (int i = 1; i < nrows - 1; i ++){
for (int j = 0; j < nrows - 1; j++){
if( (hid(i) == hid(i+1)) & ( persid(i) == partner(j) ) ){
partnerRec(i) = persid(j);
}
}
Rcout << i << std::endl;
}
return(partnerRec);
}
and run the function
HHgrid(hid = test$hldid, persid = test$persid, partner = test$partner, partnerRec = test$partnerRec)
There is just a slight issue with the first case (if you know how to fix it)