How to match different ids within a single data set in R? - r

Question

A sample of my data is
df1 <- read.table(text = " id1 time id2 gender id3 group id4 house
123 12 141 F 13 1 156 A
141 19 144 F 144 1 123 A
144 22 123 M 123 1 141 M
168 14 13 M 141 2 144 M
156 13 168 M 168 2 13 Q
13 11 156 F 156 2 168 Q
", header = TRUE)
I want to get the following outcome. For example, id123, time= 12, Gender=M, group=1, house= A, by looking at other ids
df1 <- read.table(text = " id time gender group house
123 12 M 1 A
141 19 F 2 M
144 22 F 1 M
168 14 M 2 Q
156 13 F 2 A
13 11 M 1 Q
", header = TRUE)
I have tried left_join, but I struggled to get the outcome of interest
df1 <- left_join(id2,id3,id4 by = "id1")

Answer 1

You've got the folks confused here because your table is in an unusual format. Typically in R, we expect one variable per column and one observation per row. What you have is effectively four tables stuck side-by-side, where id1, id2, id3 and id4 are all actually just "id". So effectively, you are looking to left join columns 3:4 to columns 1:2, then left join columns 5:6 to that, and so on.
I'll show one way of doing that, then maybe some of the smart folks here can show you a better way:
library(dplyr)
df_list <- lapply(list(1:2, 3:4, 5:6, 7:8), function(x) df1[x])
df_list <- lapply(df_list, function(x) {names(x)[1] <- "id"; x})
df2 <- df_list[[1]] %>%
left_join(df_list[[2]]) %>%
left_join(df_list[[3]]) %>%
left_join(df_list[[4]])
df2
#> id time gender group house
#> 1 123 12 M 1 A
#> 2 141 19 F 2 M
#> 3 144 22 F 1 M
#> 4 168 14 M 2 Q
#> 5 156 13 F 2 A
#> 6 13 11 M 1 Q
Created on 2020-07-01 by the reprex package (v0.3.0)

Answer 2

It seems like we need a match for different 'id' columns and corresponding 'group', 'gender' etc columns
nm1 <- c('id1', 'time', 'gender', 'group', 'house')
out1 <- transform(df1, gender = gender[match(id1, id2)],
group = group[match(id1, id3)],
house = house[match(id1, id4)])[nm1]
names(out1)[1] <- 'id'
out1
# id time gender group house
#1 123 12 M 1 A
#2 141 19 F 2 M
#3 144 22 F 1 M
#4 168 14 M 2 Q
#5 156 13 F 2 A
#6 13 11 M 1 Q
In addition to the above base R, an alternative option to #AllanCameron's solution would be to split subset of columns based on the occurrence of 'id' column (split.default), then change the first column name to 'id' and apply left_join within reduce
library(dplyr)
library(purrr)
df1 %>%
split.default(cumsum(startsWith(names(.), "id"))) %>%
map(~ rename_at(.x, 1, ~ 'id')) %>%
reduce(left_join, by = 'id')
# id time gender group house
#1 123 12 M 1 A
#2 141 19 F 2 M
#3 144 22 F 1 M
#4 168 14 M 2 Q
#5 156 13 F 2 A
#6 13 11 M 1 Q