r - rle を使用した R の一意のタプル

Question

このような df でタプルを一意 (rle 一意) にする方法

structure(c("M01", "M01", "M01", "M01", "M01", "M02", "M02", 
"M02", "M02", "M03", "M03", "F04", "F04", "F02", "F02", "F04", 
"F10", "F10", NA, "F10", "F01", "F01"), .Dim = c(11L, 2L), .Dimnames = list(
    NULL, c("a", "b")))

> sample
      a     b    
 [1,] "M01" "F04"
 [2,] "M01" "F04"
 [3,] "M01" "F02"
 [4,] "M01" "F02"
 [5,] "M01" "F04"
 [6,] "M02" "F10"
 [7,] "M02" "F10"
 [8,] "M02" NA   
 [9,] "M02" "F10"
[10,] "M03" "F01"
[11,] "M03" "F01"

これを取得するには：

structure(c("M01", "M01", "M01", "M02", "M02", "M03", "F04", 
"F02", "F04", "F10", "F10", "F01"), .Dim = c(6L, 2L), .Dimnames = list(
    NULL, c("d", "c")))
> output
     d     c    
[1,] "M01" "F04"
[2,] "M01" "F02"
[3,] "M01" "F04"
[4,] "M02" "F10"
[5,] "M02" "F10"
[6,] "M03" "F01"

したがって、アイデアはタプルを使用してdfを取得することですが、行に基づいて、前の要素のみに基づいて一意であるため、 : unique(sample) 必要なものが得られません。タプルを考慮し、出力として df を保持する方法で、この df で rle を実行できますか? より良いアプローチはありますか？

rle(sample[,2]$values)

望ましい結果が得られますが、明らかに列 1 の貴重な情報が失われます。

Answer 1

これはどう？

# dd is the matrix structure you posted in the question
dd <- as.data.frame(dd)                     ## convert to data.frame
dd[] <- lapply(dd, as.character)            ## change columns to character
na.omit(dd[cumsum(rle(dd$b)$lengths), ])    ## get indices by cumsum'ing rle-lengths 
                                            ## wrap with na.omit to remove NA rows
#      a   b
# 2  M01 F04
# 4  M01 F02
# 5  M01 F04
# 7  M02 F10
# 9  M02 F10
# 11 M03 F01