Question

我一直在寻找解决方案并且一直在尝试，但我似乎无法执行我应该做的简单任务。

我有两个数据框，其格式类似于以下玩具示例

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

我希望合并两个数据集，使输出为：

      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，任何带有标签A ==“cats”或B ==“kittens”的行都将映射到D列中的1，任何带有A ==“dogs”或B ==“puppies”的行将被映射到2.

我使用了命令

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

然而，这不正确地匹配第3行和第4行，只有第1行和第2行。我得到了输出

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

请注意我正在使用的实际数据集非常长。实际上DF1超过1,000,000行，而DF2每行超过300,000行数千行，因此可以扩展的解决方案是我真正需要的。

Answer 1

也许你可以尝试这些方面：

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

如果您需要的不仅仅是这两个选项，您可以嵌套ifelse语句。

Answer 2

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

官能化：

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

Answer 3

这是一种不同的方法：

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

在两列中的至少一列中按匹配合并数据帧

3 个答案: