Question

让我采用如下数据框：

>df : Col_1   Col_2  Col_3
       A       B       z
       C       D       x
       A       D       c
       B       A       g
       D       C       f

现在，我需要订购这样的df：

>df1 : Col_1   Col_2  Col_3
        A       B      z
        C       D      x
        A       D      c
        A       B      g
        C       D      f

所以，我需要在第一次尝试时按照它们出现的顺序排列这个数据框（这里df我们有A，B在行-1和B，A在第4行）。我想把它们安排为A，B在第1行和A，B在第4行，如df1所示

Answer 1

重新阅读问题后（在@ DavidArenburg的评论之后），一个可能的解决方案：

# order the letters rowwise in alphabetical order and paste them together in a character vector
x <- do.call(paste0, data.frame(t(apply(mydf[1:2], 1, function(x) x[order(x)]))))
# create an index for the first occurance
idx <- as.integer(factor(x, levels = unique(x)))
# replace with the first occurance
mydf[,-3] <- mydf[idx,-3]

给出：

> mydf
  Col_1 Col_2 Col_3
1     B     A     z
2     C     D     x
3     A     D     c
4     B     A     g
5     C     D     f

使用过的数据：

mydf <- structure(list(Col_1 = structure(c(2L,3L,1L,2L,4L), .Label = c("A","B","C","D"), class = "factor"), 
                       Col_2 = structure(c(1L,4L,4L,1L,3L), .Label = c("A","B","C","D"), class = "factor"), 
                       Col_3 = structure(c(5L,4L,1L,3L,2L), .Label = c("c","f","g","x","z"), class = "factor")), 
                  .Names = c("Col_1","Col_2","Col_3"), row.names = c(NA, -5L), class = "data.frame")

旧答案：您可以创建一个行的索引，其中Col_1中的字母的序数高于Col_2中的字母，然后只需交换这两个值：

# create a match vector
ltrs <- setNames(LETTERS,1:26)

# create an index for which rows the letter in 'Col_1' is of a higher order than in 'Col_2'
idx <- match(mydf$Col_1, ltrs) > match(mydf$Col_2, ltrs)

# swap the two values
mydf[idx,-3] <- mydf[idx, c(2,1)]

给出：

> mydf
  Col_1 Col_2 Col_3
1     A     B     z
2     C     D     x
3     A     D     c
4     A     B     g
5     C     D     f

根据数据框的外观对数据框中的元素进行排序

1 个答案: