如何基于键合并字符串？R语言分段序列拼接技术问询

要实现将每个以TP<开头的分段对应的子行序列合并为完整字符串，并生成目标数据框out，可以通过分组标识+序列合并的思路来处理，以下提供两种实现方式（dplyr和base R）：

方法1：使用dplyr包（简洁直观） #

首先加载dplyr，然后通过分组标识将数据按TP<开头的行分组，再对每个组内的子序列进行合并，最后赋值到join列：

library(dplyr)

# 原始数据框
df <- structure(list(position = structure(c(6L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 7L, 1L, 2L, 3L, 4L, 8L, 1L, 2L, 3L, 4L), .Label = c("1,2,3,4,5,6,7,8,9,10,11,12,13,14,15", "2,3,4,5,6,7,8,9,10,11,12,13,14,15,16", "3,4,5,6,7,8,9,10,11,12,13,14,15,16,17", "4,5,6,7,8,9,10,11,12,13,14,15,16,17,18", "TP&lt;AMB88", "TP&lt;AMT55", "TP&lt;ELANE", "TP&lt;RACK1"), class = "factor"), col = structure(c(15L, 6L, 3L, 11L, 5L, 14L, 9L, 18L, 16L, 8L, 13L, 4L, 2L, 17L, 7L, 12L, 1L, 10L), .Label = c("EQMTLRGTLKGHNGW", "GRRLACLFLACVLPA", "GSLSNYALLQLTLTA", "LGRRLACLFLACVLP", "LSNYALLQLTLTAFL", "MGSLSNYALLQLTLT", "MTEQMTLRGTLKGHN", "MTLGRRLACLFLACV", "MVKETTYYDVLGVKP", "QMTLRGTLKGHNGWV", "SLSNYALLQLTLTAF", "TEQMTLRGTLKGHNG", "TLGRRLACLFLACVL", "TP&lt;AMB88", "TP&lt;AMT55", "TP&lt;ELANE", "TP&lt;RACK1", "VKETTYYDVLGVKPN"), class = "factor")), class = "data.frame", row.names = c(NA, -18L))

# 处理流程
out <- df %>%
  # 生成分组ID：每个TP<开头的行为新组的起始
  mutate(group_id = cumsum(grepl("^TP<", position))) %>%
  group_by(group_id) %>%
  mutate(
    # 合并组内子序列：第一个子序列 + 后续每个子序列的最后一个字符
    full_seq = if (n() > 1) {
      sub_seqs <- as.character(col[-1])
      paste0(sub_seqs[1], paste0(sapply(sub_seqs[-1], function(x) substr(x, nchar(x), nchar(x))), collapse = ""))
    } else "",
    # 赋值join列：标题行取自身position，第一个子行取合并后的序列，其余为空
    join = case_when(
      row_number() == 1 ~ as.character(position),
      row_number() == 2 ~ full_seq,
      TRUE ~ ""
    )
  ) %>%
  ungroup() %>%
  # 移除中间辅助列
  select(-group_id, -full_seq)

# 可选：将join列转为factor（与目标out格式匹配）
out$join <- factor(out$join, levels = unique(out$join)[nchar(unique(out$join)) > 0])

方法2：使用base R（无需额外包） #

如果不想加载dplyr，可以用base R的循环和分组处理实现：

# 原始数据框
df <- structure(list(position = structure(c(6L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 7L, 1L, 2L, 3L, 4L, 8L, 1L, 2L, 3L, 4L), .Label = c("1,2,3,4,5,6,7,8,9,10,11,12,13,14,15", "2,3,4,5,6,7,8,9,10,11,12,13,14,15,16", "3,4,5,6,7,8,9,10,11,12,13,14,15,16,17", "4,5,6,7,8,9,10,11,12,13,14,15,16,17,18", "TP&lt;AMB88", "TP&lt;AMT55", "TP&lt;ELANE", "TP&lt;RACK1"), class = "factor"), col = structure(c(15L, 6L, 3L, 11L, 5L, 14L, 9L, 18L, 16L, 8L, 13L, 4L, 2L, 17L, 7L, 12L, 1L, 10L), .Label = c("EQMTLRGTLKGHNGW", "GRRLACLFLACVLPA", "GSLSNYALLQLTLTA", "LGRRLACLFLACVLP", "LSNYALLQLTLTAFL", "MGSLSNYALLQLTLT", "MTEQMTLRGTLKGHN", "MTLGRRLACLFLACV", "MVKETTYYDVLGVKP", "QMTLRGTLKGHNGWV", "SLSNYALLQLTLTAF", "TEQMTLRGTLKGHNG", "TLGRRLACLFLACVL", "TP&lt;AMB88", "TP&lt;AMT55", "TP&lt;ELANE", "TP&lt;RACK1", "VKETTYYDVLGVKPN"), class = "factor")), class = "data.frame", row.names = c(NA, -18L))

# 初始化分组ID和join列
df$group_id <- cumsum(grepl("^TP<", df$position))
df$join <- ""

# 遍历每个分组处理
for (g in unique(df$group_id)) {
  group_rows <- which(df$group_id == g)
  # 标题行赋值
  df$join[group_rows[1]] <- as.character(df$position[group_rows[1]])
  # 有子行时合并序列
  if (length(group_rows) > 1) {
    sub_cols <- as.character(df$col[group_rows[-1]])
    full_seq <- sub_cols[1]
    if (length(sub_cols) > 1) {
      full_seq <- paste0(full_seq, paste0(sapply(sub_cols[-1], function(x) substr(x, nchar(x), nchar(x))), collapse = ""))
    }
    # 赋值给第一个子行
    df$join[group_rows[2]] <- full_seq
  }
}

# 生成最终输出数据框，移除辅助列
out <- df[, c("position", "col", "join")]

# 可选：转为factor格式匹配目标
out$join <- factor(out$join, levels = unique(out$join)[nchar(unique(out$join)) > 0])

关键逻辑说明 #

1. 分组标识：通过cumsum(grepl("^TP<", position))为每个以TP<开头的行创建独立分组ID，后续子行自动归为同一组。
2. 序列合并：每个子行的字符串是前一个子行去掉首字符、新增末尾字符，因此合并时只需取第一个子行的完整字符串，加上后续每个子行的最后一个字符即可得到完整序列。
3. join列赋值：标题行的join取自身position值，第一个子行取合并后的完整序列，其余子行留空，与目标格式完全匹配。

内容的提问来源于stack exchange，提问作者Learner