Question

我的数据框如下所示：

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))

我想迭代df$epitope，对于这个具体的例子，只保留模式“E”连续出现至少五次的行，这样输出应该如下所示： / p>

  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

我们可以使用rleid中的data.table。转换＆＃39; data.frame＆＃39;到＆＃39; data.table＆＃39; （setDT(df)，根据＆＃39;表位＆＃39;的游程长度ID分组，创建逻辑向量（.N >=5 & epitope == "E"），提取行索引（.I）和子集数据

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

或者我们可以对rle

中的base R使用类似的方法

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

仅保留数据框中与特定模式长度匹配的行

1 个答案: