在R中的两个数据表之间按行计算匹配元素

Question

我在R中有两个数据框,我需要逐行计算元素匹配,最后获得一个列,其中包含两个表的笛卡尔积的长度和两行的ID.此外,表格非常大,行数不同,但列数相同.

我有以下代码,但多次运行时速度很慢.

library(data.table)

table_1<-data.table(matrix(c(1:24),nrow = 4))
table_2<-data.table(matrix(c(11:34),nrow = 4))

names(table_1)<-c("s1", "s2","s3","s4","s5","s6")
names(table_2)<-c("a1","a2","a3","a4","a5","a6")

table_1$ID<-seq.int(nrow(table_1))
table_2$ID_ap<-seq.int(nrow(table_2))

setcolorder(table_1, c("ID", "s1", "s2","s3","s4","s5","s6"))
setcolorder(table_2, c("ID_ap","a1","a2","a3","a4","a5","a6"))

CJ.table<-function(X,Y) setkey(X[,c(k=1,.SD)],k)[Y[,c(k=1,.SD)],allow.cartesian=TRUE][,k:=NULL] 

join<-CJ.table(table_1,table_2)

R<-subset(join, select=c("ID_ap","ID"))

R$Ac<- (join$s1 == join$a1) + (join$s1 ==join$a2) + (join$s1 ==join$a3) + (join$s1 ==join$a4) + (join$s1 ==join$a5) + (join$s1 ==join$a6)+ 
(join$s2 == join$a1) + (join$s2 ==join$a2) + (join$s2 ==join$a3) + (join$s2 ==join$a4) + (join$s2 ==join$a5) + (join$s2 ==join$a6)+ 
(join$s3 == join$a1) + (join$s3 ==join$a2) + (join$s3 ==join$a3) + (join$s3 ==join$a4) + (join$s3 ==join$a5) + (join$s3 ==join$a6)+ 
(join$s4 == join$a1) + (join$s4 ==join$a2) + (join$s4 ==join$a3) + (join$s4 ==join$a4) + (join$s4 ==join$a5) + (join$s4 ==join$a6)+ 
(join$s5 == join$a1) + (join$s5 ==join$a2) + (join$s5 ==join$a3) + (join$s5 ==join$a4) + (join$s5 ==join$a5) + (join$s5 ==join$a6)+ 
(join$s6 == join$a1) + (join$s6 ==join$a2) + (join$s6 ==join$a3) + (join$s6 ==join$a4) + (join$s6 ==join$a5) + (join$s6 ==join$a6)

这使

   R
   ID_ap ID Ac
 1:     1  1  0
 2:     1  2  0
 3:     1  3  4
 4:     1  4  0
 5:     2  1  0
 6:     2  2  0
 7:     2  3  0
 8:     2  4  4
 9:     3  1  3
10:     3  2  0
11:     3  3  0
12:     3  4  0
13:     4  1  0
14:     4  2  3
15:     4  3  0
16:     4  4  0

Answer 1

以长格式放置数据，因为列顺序无关紧要：

setnames(table_2, "ID_ap", "ID")
tabs = rbind(
  melt(table_1, id="ID")[, variable := NULL],
  melt(table_2, id="ID")[, variable := NULL],
  idcol = TRUE)

(1) 对于每个值，确定相关对；和

(2) 对于成对，计数值：

tabs[, 
  if (uniqueN(.id) > 1L) CJ(ID1 = ID[.id == 1L], ID2 = ID[.id == 2L])
, by=value][,
   .N
, by=.(ID1, ID2)]


   ID1 ID2 N
1:   3   1 4
2:   4   2 4
3:   1   3 3
4:   2   4 3

(ID1, ID2)我认为所有其他组合都是零，不需要明确列举。

如果每个 table 中的值不同，如在 OP 的示例中，那么我们可以简化：

tabs[, if (.N==2L) .(ID1 = ID[1L], ID2 = ID[2L]), by=value][, .N, by=.(ID1, ID2)]