通过至少两列中的一个匹配合并数据帧

Question

我一直在寻找解决方案并一直在尝试，但似乎无法执行我应该做的简单任务。

我有两个数据帧格式化类似于下面玩具实例

DF1 = data.frame(A=c("cats","dogs",NA,"dogs"), B=c("kittens","puppies","kittens",NA), C=c(88,99,101,110)) 

    A  B   C 
1 cats kittens  88 
2 dogs puppies  99 
3 NA  kittens  101 
4 dogs NA   110 


DF2 = data.frame(D=c(1,2), A=c("cats","dogs"), B=c("kittens","puppies")) 

    D A  B 
1 1 cats kittens 
2 2 dogs puppies 

我希望合并这两个数据集，使得输出是：

 A  B   C  D 
1 cats kittens 88 1 
2 dogs puppies 99 2 
3 dogs NA  110 2 
4  NA kittens 101 1 

换句话说，任何行带有标签A ==“猫”或B ==“小猫”将被映射到列D中的1，任何具有A ==“狗”或B ==“小狗”的行将被映射到2.

我已经使用命令

merge(DF1, DF2, by=c("A","B"), all.x=TRUE) 

然而，这并不正确匹配的行3和第4，仅行1和2。我得到的输出

 A  B   C  D 
1 cats kittens 88 1 
2 dogs puppies 99 2 
3 dogs NA  110 NA 
4  NA kittens 101 NA 

请注意我的工作实际数据集是很长。事实上DF1超过1,000,000行，DF2超过300,000行每行数千行，所以可以缩放的解决方案就是我真正需要的。

Answer 1

这里有一个不同的方法：

library(functional) 

partial.merge <- function(DF1, DF2) { 
    common.cols <- intersect(names(DF1), names(DF2)) 
    result.col <- names(DF2)[!(names(DF2) %in% common.cols)] 

    # This can only handle one result column: 
    stopifnot(length(result.col) == 1) 

    # Merge in each common column, one at a time. 
    # The identical operation is done for each common column, so Reduce is useful: 
    r <- Reduce(function(D, C) merge(D, DF2[c(C, result.col)], by=c(C), all.x=TRUE), x=common.cols, init=DF1) 

    # The merge created cols like c('D.x', 'D.y'). These are the columns: 
    merge.cols <- paste(result.col, c('x', 'y'), sep='.') 

    # The .x and .y columns are partial, put them together: 
    r[[result.col]] <- rowMeans(r[merge.cols], na.rm=TRUE) 

    # Remove the temporaries: 
    for (i in merge.cols) { 
    r[[i]] <- NULL 
    } 
    return(r) 
} 

partial.merge(DF1, DF2) 
##   B A C D 
## 1 kittens cats 88 1 
## 2 kittens <NA> 101 1 
## 3 puppies dogs 99 2 
## 4 <NA> dogs 110 2 

Answer 2

也许你可以尝试的东西沿着这些路线：

temp <- merge(DF1, DF2, by=c("A","B"), all.x=TRUE) 

within(temp, { 
    M1 <- c("cats", "kittens") 
    D <- ifelse(A %in% M1 | B %in% M1, 1, 2) 
    rm(M1) 
}) 
#  A  B C D 
# 1 cats kittens 88 1 
# 2 dogs puppies 99 2 
# 3 dogs <NA> 110 2 
# 4 <NA> kittens 101 1 

可以嵌套如果ifelse语句您需要的不仅仅是这两个选项的更多。

Answer 3

DF1[which(DF1$A=="cats"|DF1$B=="kittens"), "D"] <- DF2[which(DF2$A=="cats"|DF2$B=="kittens"), "D"] 
DF1[which(DF1$A=="dogs"|DF1$B=="puppies"), "D"] <- DF2[which(DF2$A=="dogs"|DF2$B=="puppies"), "D"] 
DF1 
#------- 
    A  B C D 
1 cats kittens 88 1 
2 dogs puppies 99 2 
3 <NA> kittens 101 1 
4 dogs <NA> 110 2 

功能化：

idxpick <- function(a,b) DF1[which(DF1$A==a|DF1$B==b), "D"] <<- # Yes, I feel guilty. 
            DF2[which(DF2$A==a|DF2$B==b), "D"] 
DF1 = data.frame(A=c("cats","dogs",NA,"dogs"), 
       B=c("kittens","puppies","kittens",NA), 
       C=c(88,99,101,110)) 
DF2 = data.frame(D=c(1,2), A=c("cats","dogs"), B=c("kittens","puppies")) 
apply(DF2, 1, function(rr) idxpick(rr["A"], rr["B"])) 
#------------ 
[1] 1 2 

DF1 
    A  B C D 
1 cats kittens 88 1 
2 dogs puppies 99 2 
3 <NA> kittens 101 1 
4 dogs <NA> 110 2