Question

我有一个非常奇特的问题，我无法处理。我有一个字符串向量，每个元素代表一个小说的句子。

我需要做的只是折叠同一对话内的那些线。例如，采取以下行：

snap <- c("It was a few seconds before Mr Dursley realised that the man was wearing a violet cloak.",
      "He didn't seem at all upset at being almost knocked to the ground.",
      "On the contrary, his face split into a wide smile and he said in a squeaky voice that made passers-by stare: \"Don't be sorry, my dear sir, for nothing could upset me today!",
      "Rejoice, for You-Know-Who has gone at last!",
      "Even Muggles like yourself should be celebrating, this happy, happy day!\"",
      "And the old man hugged Mr Dursley around the middle and walked off."
      )

第3行到第5行属于同一个对话框，因此它们必须折叠，结果向量为：

snap.2 <- c("It was a few seconds before Mr Dursley realised that the man was wearing a violet cloak.",
      "He didn't seem at all upset at being almost knocked to the ground.",
      "On the contrary, his face split into a wide smile and he said in a squeaky voice that made passers-by stare: \"Don't be sorry, my dear sir, for nothing could upset me today! Rejoice, for You-Know-Who has gone at last! Even Muggles like yourself should be celebrating, this happy, happy day!\"",
      "And the old man hugged Mr Dursley around the middle and walked off."
      )

我可以用以下方法检测不平衡的双引号：

which((str_count(snap, "\"") %% 2) != 0)
[3 5]

但后来我不知道如何合并，如上面的例子，第3,4和5行

关于如何做到这一点的任何想法？

Answer 1

我们可以paste将它们放在一起，然后根据正则表达式进行拆分

out <- strsplit(paste(snap, collapse=' '), '(?<=\\.)\\s*|(?<=["])\\s', perl = TRUE)[[1]]
identical(out, snap.2)
#[1] TRUE

注意：模式不清楚。

Answer 2

它可能不是最好的方法（非常难看的代码），但它有效。基本上是：

成对分割which的输出（情侣将代表对话开始和结束偏移）
使用dplyr
填充虚拟夫妇的空白，其中dialog.start = dialog.end for not dialogue lines
使用输出数据集作为粘贴的索引

在代码中：

dialogue.start <- which((str_count(snap, "\"") %% 2) != 0)

quotes.fill <- data.frame(dialogue.start) %>%
  mutate(n = row_number())

quotes.fill$dialogue.end <- ifelse((quotes.fill$n %% 2) != 0, lead(quotes.fill$dialogue.start, 1), NA)
quotes.fill$dialogue.next <- ifelse((quotes.fill$n %% 2) != 0, lead(quotes.fill$dialogue.start, 2, default = NROW(snap)), NA)
quotes.fill$dialogue.before <- ifelse((quotes.fill$n %% 2) != 0, lag(quotes.fill$dialogue.start, 2, default = 0), NA)


quotes.fill <- quotes.fill %>% filter(!is.na(dialogue.end)) %>%
  select(-n)

quotes.gaps <- do.call(rbind, lapply(split(quotes.fill, seq(nrow(quotes.fill))), function(x) { 

  prologue <- NULL

  dialogue.hold <- seq(to = (x$dialogue.next - 1), from = (x$dialogue.end + 1))
  dialogue.prologue <- seq(to = (x$dialogue.start - 1), from = (x$dialogue.before + 1))

  if(x$dialogue.before == 0 & x$dialogue.start > 0) prologue <- data.frame(dialogue.start = dialogue.prologue, dialogue.end = dialogue.prologue, stringsAsFactors = FALSE)

  if((x$dialogue.end + 1) >= x$dialogue.next) return(rbind(prologue, x[,c("dialogue.start", "dialogue.end")]))


  return(rbind(prologue, x[,c("dialogue.start", "dialogue.end")], data.frame(dialogue.start = dialogue.hold, dialogue.end = dialogue.hold, stringsAsFactors = FALSE)))
})
)

snap.2 <- do.call(c, lapply(split(quotes.gaps, seq(nrow(quotes.gaps))), function(c, novel) {
  paste(novel[c$dialogue.start:c$dialogue.end], collapse = " ")
}, novel = snap))

合并（折叠）字符串向量的一些元素但不是全部在R中

2 个答案: