Question

我有一个功能，我正在尝试对无线电格式进行分类，但是，它无法正常工作。

findFormat <- function(format) {
  currentFormat <- c(strsplit(tolower(format), " "))
  if ("christian" %in% currentFormat || "gospel" %in% currentFormat || "religious" %in% currentFormat || "religion" %in% currentFormat) {
    return("Religious")
  }
  if ("pop" %in% currentFormat || "contemporary" %in% currentFormat || "mainstream" %in% currentFormat || "top" %in% currentFormat || "hot" %in% currentFormat || "hit" %in% currentFormat) {
    return("Pop or Contemporary")
  }
  if ("rock" %in% currentFormat || "alternative" %in% currentFormat || "indie" %in% currentFormat) {
    return("Rock, Alternative, or Indie")
  }
  if ("country" %in% currentFormat || "southern" %in% currentFormat) { 
    return("Country")
  }
  if ("urban" %in% currentFormat || "hip" %in% currentFormat || "rap" %in% currentFormat || "hip-hop" %in% currentFormat) {
    return("Hip-hop")
  }
  if ("jazz" %in% currentFormat || "blues" %in% currentFormat) {
    return("Jazz or Blues")
  }
  if ("latin" %in% currentFormat || "mexican" %in% currentFormat || "international" %in% currentFormat) {
    return("International")
  }
  if ("oldies" %in% currentFormat) { 
    return("Oldies")
  }
  if ("news/talk" %in% currentFormat || "news" %in% currentFormat || "talk" %in% currentFormat || "public" %in% currentFormat) { 
    return("News and Talk")
  }
  # Default
  return("Other")
}

现在，如果我运行findFormat("rap")，它将按预期返回[1] "Hip-hop"，但如果我运行findFormat("rap and rhythm")，则会获得[1] Other。我不知道为什么，因为currentFormat应该只是一个字符串向量而"rap" %in% c("rap", "and", "rhythm")返回[1] TRUE。任何帮助将不胜感激！

Answer 1

我认为通过查找表解决这个问题可以更好地服务，或许是这样：

findFormat = function(fmt) {
  matchDT = 
    data.table(keyword = c('christian', 'gospel', 'religious', 'religion',
                           'pop', 'contemporary', 'mainstream', 'top',
                           'hot', 'hit', 'rock', 'alternative',' indie',
                           'country', 'southern', 'urban', 'hip', 
                           'hop', 'rap', 'jazz', 'blues', 'latin',
                           'mexican', 'international', 'oldies',
                           'news', 'talk'),
               format = c(rep('Religious', 4L), 
                          rep('Pop or Contemporary', 6L),
                          rep('Rock, Alternative, or Indie', 3L),
                          rep('Country', 2L), rep('Hip-Hop', 4L),
                          rep('Jazz or Blues', 2L),
                          rep('International', 3L), 'Oldies',
                          rep('News and Talk', 2L)),
               key = 'keyword')
  out = matchDT[strsplit(fmt, '[ [:punct:]]')[[1L]], nomatch = 0L]
  if (!length(out)) return('Other') else return(out$format)
}

findFormat('rap and rhythm')
# [1] "Hip-Hop"

％in％不正确匹配字符串？

1 个答案: