Question

我正在努力＆＃34;合并＆＃34;如果列X和Y等于（我必须匹配dOne.X == dTwo.X & dOne.Y == dTwo.Y和dOne.X == dTwo.Y & dOne.Y == dTwo.X），则数据框的第五列在另一个数据框中我使用for循环解决了这个问题，但是当数据框dOne很大时它很慢（在我的机器中，如果length(dOne.X) == 500000需要25分钟）。我想知道是否有办法使用更快的＆＃34;矢量化＆＃34;来解决这个问题。操作。以上是我想要做的事例：

Data Frame ONE
X Y  V
a b  2
a c  3
a d  0
a e  0
b c  2
b d  3
b e  0
c d  2
c e  0
d e  0

Data Frame TWO
X Y  V
a b  1
a c  1
a d  1
b c  1
b d  1
c d  1
e d  1

Expected Data Frame after the columns are merged
X Y  V V2
a b  2  1
a c  3  1
a d  0  1
a e  0  0
b c  2  1
b d  3  1
b e  0  0
c d  2  1
c e  0  0
d e  0  1

这是我目前使用的代码，当dOne很大（数十万或几行）时，它很慢：

copyadjlistValueColumn <- function(dOne, dTwo) {
    dOne$V2 <- 0

    lv <- union(levels(dOne$Y), levels(dOne$X))

    dTwo$X <- factor(dTwo$X, levels = lv)
    dTwo$Y <- factor(dTwo$Y, levels = lv)
    dOne$X <- factor(dOne$X, levels = lv)
    dOne$Y <- factor(dOne$Y, levels = lv)

    for(i in 1:nrow(dTwo)) {
      row <- dTwo[i,]
      dOne$V2[dOne$X == row$X & dOne$Y == row$Y] <- row$V
      dOne$V2[dOne$X == row$Y & dOne$Y == row$X] <- row$V
    }
    dOne
}

这是一个测试案例，涵盖了我期望的内容（使用上面的数据框）：

test_that("Copy V column to another Data Frame", {
    dfOne <- data.frame(X=c("a", "a", "a", "a", "b", "b", "b", "c", "c", "d"),
                        Y=c("b", "c", "d", "e", "c", "d", "e", "d", "e", "e"),
                        V=c(2, 3, 0, 0, 2, 3, 0, 2, 0, 0))

    dfTwo <- data.frame(X=c("a", "a", "a", "b", "b", "c", "e"),
                        Y=c("b", "c", "d", "c", "d", "d", "d"),
                        V=c(1, 1, 1, 1, 1, 1, 1))

    lv <- union(levels(dfTwo$Y), levels(dfTwo$X))
    dfExpected <- data.frame(X=c("a", "a", "a", "a", "b", "b", "b", "c", "c", "d"),
                             Y=c("b", "c", "d", "e", "c", "d", "e", "d", "e", "e"),
                             V=c(2, 3, 0, 0, 2, 3, 0, 2, 0, 0),
                             V2=c(1, 1, 1, 0, 1, 1, 0, 1, 0, 1))
    dfExpected$X <- factor(dfExpected$X, levels = lv)
    dfExpected$Y <- factor(dfExpected$Y, levels = lv)

    dfMerged <- copyadjlistValueColumn(dfOne, dfTwo)

    expect_identical(dfMerged, dfExpected)
})

有什么建议吗？

非常感谢：）

Answer 1

尝试两个merge，其中匹配列的顺序在第二个中反转，以获得“双向”匹配。然后你可以使用例如rowSums将两个已创建的列折叠为一个。

d1 <- merge(dfOne, dfTwo, by.x = c("X", "Y"), by.y = c("X", "Y"), all.x = TRUE)
d2 <- merge(d1, dfTwo, by.x = c("X", "Y"), by.y = c("Y", "X"), all.x = TRUE)
cbind(dfOne, V2 = rowSums(cbind(d2$V.y, d2$V), na.rm = TRUE))


#    X Y V V2
# 1  a b 2  1
# 2  a c 3  1
# 3  a d 0  1
# 4  a e 0  0
# 5  b c 2  1
# 6  b d 3  1
# 7  b e 0  0
# 8  c d 2  1
# 9  c e 0  0
# 10 d e 0  1

要更快地替换merge，请在此处检查data.table和dplyr替代方案：stackoverflow.com/questions/1299871/how-to-join-data-frames-in-r-内外 - 左 - 右/

Answer 2

这是一种可能的data.table套餐方法。对于像您这样的大数据集，这种方法应该特别有效：

首先转换为data.table对象并添加键

library(data.table)
setkey(setDT(dfOne), X, Y)
setkey(setDT(dfTwo), X, Y)

然后在X & Y组合上执行联接 - 通过将X,Y的关键列dfOne与X,Y的关键列dfTwo分别匹配来执行联接。

dfOne[dfTwo, V2 := i.V]

现在在Y & X组合上执行联接 - 通过将X,Y的关键列dfOne与Y,X的关键列dfTwo分别匹配来执行联接。

setkey(dfTwo, Y, X)
dfOne[dfTwo, V2 := i.V][]

结果（我将保持不匹配的NA而不是零，因为这样更有意义）：

#     X Y V V2
#  1: a b 2  1
#  2: a c 3  1
#  3: a d 0  1
#  4: a e 0 NA
#  5: b c 2  1
#  6: b d 3  1
#  7: b e 0 NA
#  8: c d 2  1
#  9: c e 0 NA
# 10: d e 0  1

Answer 3

使用dplyr：

library(dplyr)

left_join(dfOne, dfTwo, by = c("X", "Y")) %>% 
  left_join(dfTwo, by = c("X" = "Y", "Y" = "X")) %>% 
  mutate(V2 = ifelse(is.na(V.y), V, V.y)) %>% 
  select(X, Y, V = V.x, V2) %>% 
  do(replace(., is.na(.), 0))

通过匹配列来合并具有不同大小的两个数据帧

3 个答案: