从一个数据框的不同列创建一个新列，该条件以另一个数据框的另一列为条件

Question

Suppose I have two data frame

df1 <- data.frame(A = 1:6, B = 7:12, C = rep(1:2, 3))
df2 <- data.frame(C = 1:2, D = c("A", "B"))

I want to create a new column E in df1 whose value is based on the values of Column C, which can then be connected to Column D in df2. For example, the C value in the first row of df1 is "1". And value 1 of column C in df2 corresponds to "A" of Column D, so the value E created in df2 should from column "A", ie, 1.

As suggested by Select values from different columns based on a variable containing column names , I can achieve this by two steps:

setDT(df1)
setDT(df2)
df3 <- df1[df2, on = "C"] # step 1 combines the two data.tables
df3[, E := .SD[[.BY[[1]]]], by = D] # step 2

My question is: Could we do this in one step? Furthermore, as my data is relatively large, the first step in this original solution takes a lot time. Could we do this in a faster way? Any suggestions?

Answer 1

您可以尝试这样做，C列可以指示df1中的列值

setDT(df1) df1[, e := eval(parse(text = names(df1)[C])), by = 1:nrow(df1)] df1

ABC e 1: 1 7 1 1 2: 2 8 2 8 3: 3 9 1 3 4: 4 10 2 10 5: 5 11 1 5 6: 6 12 2 12

Answer 2

Here's how I would do it:

df1[df2, on=.(C), D := i.D][, E := .SD[[.BY$D]], by=D]

   A  B C D  E
1: 1  7 1 A  1
2: 2  8 2 B  8
3: 3  9 1 A  3
4: 4 10 2 B 10
5: 5 11 1 A  5
6: 6 12 2 B 12

This adds the columns to df1 by reference instead of making a new table and so I guess is more efficient than building df3 . Also, since they're added to df1 , the rows retain their original ordering.