连接空data.table并赋值时报错：多列赋值失败，单列正常

Hey there, let's break down why you're hitting this error and how to fix it.

The Root Cause #

When you're doing a join with an empty right table (D2 here) and trying to assign both an existing column and a new column in one := call, data.table throws this error because of its strict validation rules for existing columns.

Here's your code again for context:

D1 <- data.table(l = "a", x = 0)
D2 <- data.table(l = character(0), z = numeric(0))
# This throws an error
D1[D2, `:=`(x = x + z, u = x * z), on = "l"]

Since D2 is empty, there are no matching rows in D1 to update. The expression x + z evaluates to a zero-length vector, and data.table doesn't allow assigning a zero-length (non-NULL) vector to an existing column—it can't tell if you meant to wipe the column or if this was an accident. When you assign columns one by one, though, data.table skips the update entirely (since there's nothing to change), so no error gets triggered.

Solutions #

1. Assign Columns One at a Time (Your Working Approach) #

This is the simplest fix since you already know it works. Split the assignment into two separate calls:

# Update existing column x
D1[D2, x := x + z, on = "l"]
# Create new column u
D1[D2, u := x * z, on = "l"]

Since there are no matching rows, neither call will modify D1, but they won't throw errors either.

2. Use NULL for Existing Columns When No Matches #

You can check if there are any matching rows with .N (the number of rows in the current join group) and return NULL for existing columns when there's nothing to update. data.table accepts NULL as a valid value for skipping updates to existing columns:

D1[D2, `:=`(
  x = if (.N == 0) NULL else x + z,
  u = x * z
), on = "l"]

When .N == 0 (i.e., D2 is empty), x gets assigned NULL (so no change to the column), and u will be created as an empty column (you can adjust this if you don't want an empty column at all).

3. Check if the Right Table is Empty First #

If empty right tables are a common case in your workflow, you can just skip the assignment entirely when there's nothing to join:

if (nrow(D2) > 0) {
  D1[D2, `:=`(x = x + z, u = x * z), on = "l"]
}

This is the most efficient approach because it avoids running the join logic altogether when it's unnecessary.

Why Single-Column Assignment Works #

When you assign one column at a time, data.table checks if there are any rows to update first. Since there are no matches with the empty D2, it doesn't attempt to evaluate the assignment for existing columns in a way that triggers the validation error. When you do multiple columns at once, it parses all the RHS expressions first, catches the zero-length vector for the existing column, and throws the error immediately.

内容的提问来源于stack exchange，提问作者eaglefreeman