R：按组，检查是否对于一个var的每个唯一值，至少有一个观察值，其中var的值等于另一个var的值

Question

I think I am on the right direction with this code, but I am not quite there yet.

I tried finding something useful on Google and SE, but I did not seem to be able to formulate the question in a way that gets me the answer I am looking for.

I could write a for-loop for this, comparing for each id and for each unique value of a per row, but I strive to achieve a higher level of R-understanding and thus want to avoid loops.

id <- c(1,1,1,2,2,2,3,3,3,4,4,4,5,5,5)
a <- c(1,1,1,2,2,2,3,3,4,4,4,5,5,5,6)
b <- c(1,2,3,3,3,4,3,4,5,4,4,5,6,7,8)

require(data.table)
dt <- data.table(id, a, b)

dt
dt[,unique(a) %in% b, by=id]
tmp <- dt[,unique(a) %in% b, by=id]
tmp$id[tmp$V1 == FALSE]

In my example, IDs 2, 3 and 5 should be the result, the decision rule being: "By id , check if for each unique value of a if there is at least one observation where the value of b equals value of a ."

However, my code only outputs IDs 2 and 5, but not 3. This is because for ID 3, the 4 is matched with the 4 of the previous observation.

The result should either output the IDs for which the condition is not met, or add a dummy variable to the original table that indicated whether the condition is met for the ID.

Answer 1

How about

dt[, all(sapply(unique(a), function(i) any(a == i & b == i))), by = id]

#   id    V1
#1:  1  TRUE
#2:  2 FALSE
#3:  3 FALSE
#4:  4  TRUE
#5:  5 FALSE

If you want to add a dummy variable to the original table, you can modify it like

dt[, check:=all(sapply(unique(a), function(i) any(a == i & b == i))), by = id]

Answer 2

I wondered if I can find are more data.table-esk solution for this old question using the enhanced join capabilities which were introduced to data.table in version 1.9.6 (on CRAN 19 Sep 2015). With that version, data.table has gained the ability to join without having to set keys by using the on argument.

Variant 1

dt[a == b][dt[, unique(a), by = id], on = .(id, a == V1)][is.na(b), unique(id)]

 [1] 2 3 5 

First, the rows of dt where a and b are equal are selected. Only these rows are right joined with the unique values of a for each id . The result of the join is

dt[a == b][dt[, unique(a), by = id], on = .(id, a == V1)]

  id ab 1: 1 1 1 2: 2 2 NA 3: 3 3 3 4: 3 4 NA 5: 4 4 4 6: 4 4 4 7: 4 5 5 8: 5 5 NA 9: 5 6 NA 

The NA values in column b indicate that no match is found. Any id which has an NA value indicates that OP's condition is not met.

Variant 2

dt[dt[, unique(a), by = id], on = .(id, a == V1, b == V1), unique(id[is.na(x.a)])]

 [1] 2 3 5 

This variant right joins dt (unfiltered!) with the unique values of a for each id but the join conditions require matches in id as well as matches in a and b . (This resembles the a == i & b == i expression in konvas' accepted answer . Finally, those id s are returned which have at least one NA value in the join result indicating a missing match.