は、特定のパターン長

Question

に一致するデータフレームの行は、私は次のようなデータフレームを持っているだけでください：

df <- structure(list(Sequence = c("Sequence", "Sequence", "Sequence", 
"Sequence", "Sequence", "Sequence", "Sequence", "Sequence", "Sequence", 
"Sequence", "Sequence", "Sequence", "Sequence", "Sequence", "Sequence", 
"Sequence", "Sequence", "Sequence", "Sequence", "Sequence", "Sequence", 
"Sequence", "Sequence", "Sequence", "Sequence"), start = c(1, 
2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 
20, 21, 22, 23, 24, 25), end = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 
11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25), 
    score = c(-0.205, -0.229, -0.115, -0.427, -0.327, -0.543, 
    -0.717, -0.923, -1.241, -1.471, -1.737, -1.717, -1.247, -1.137, 
    -0.689, -0.731, -0.337, 0.091, 0.579, 0.93, 0.575, 0.128, 
    -0.036, -0.186, -0.259), residue = c("M", "D", "A", "R", 
    "M", "R", "E", "L", "S", "F", "K", "V", "V", "L", "L", "G", 
    "E", "G", "R", "V", "G", "K", "T", "S", "L"), epitope = c("E", 
    "E", "E", ".", ".", ".", ".", ".", "E", ".", ".", "E", "E", 
    ".", ".", ".", ".", ".", "E", "E", "E", "E", "E", "E", "." 
    )), .Names = c("Sequence", "start", "end", "score", "residue", 
"epitope"), class = c("data.table", "data.frame"), row.names = c(NA, 
-25L)) 

私はdf$epitopeを反復処理したいと、この特定の例のために、唯一保ちます出力は次のようになりますようにパターンが「E」は、行に少なくとも5回出現する行：

Sequence start end score residue epitope 
1 Sequence 19 19 0.579  R  E 
2 Sequence 20 20 0.930  V  E 
3 Sequence 21 21 0.575  G  E 
4 Sequence 22 22 0.128  K  E 
5 Sequence 23 23 -0.036  T  E 
6 Sequence 24 24 -0.186  S  E 

Answer 1

我々はdata.tableからrleidを使用することができます。 'data.table'（setDT(df)）を 'エピトープ'のランレングスIDでグループ化し、論理ベクトル（.N >=5 & epitope == "E"）を作成し、行インデックス（.I）を抽出し、データをサブセット化

library(data.table) 
setDT(df)[df[, .I[.N >=5 & epitope == "E"], rleid(epitope)]$V1] 
#  Sequence start end score residue epitope 
#1: Sequence 19 19 0.579  R  E 
#2: Sequence 20 20 0.930  V  E 
#3: Sequence 21 21 0.575  G  E 
#4: Sequence 22 22 0.128  K  E 
#5: Sequence 23 23 -0.036  T  E 
#6: Sequence 24 24 -0.186  S  E 

---

それともbase R

df[inverse.rle(within.list(rl, values[!(lengths >=5 & values)] <- FALSE)),] 

から rleと同様のアプローチを使用することができます