在R中迭代拼写检查单词的向量

Question

我有一个充满不适当间隔句子的数据集.我想找到一种方法来删除一些空格.

我从一个我转换为单词数据框的句子开始:

> word5 <- "hotter the doghou se would be bec ause the co lor was diffe rent"
> abc1 <- data.frame(filler1 = 1,words1=factor(unlist(strsplit(word5, split=" "))))
> abc1
   filler1 words1
1        1 hotter
2        1    the
3        1 doghou
4        1     se
5        1  would
6        1     be
7        1    bec
8        1   ause
9        1    the
10       1     co
11       1    lor
12       1    was
13       1  diffe
14       1   rent

接下来,我使用以下代码尝试拼写检查并组合单词之前或之后组合的单词:

abc2 <- abc1
i <- 1
while(i < nrow(abc1)){
  print(abc2)
  if(nrow(aspell(abc1$words1[i])) == 0){
    print(paste(i,"Words OK",sep=" | "));flush.console() 
    i <- i + 1
  }
 else{
  if(nrow(aspell(abc1$words1[i])) > 0 & i != 1){
    preWord1 <- abc1$words1[i-1]
    postWord1 <- abc1$words1[i+1]
    badWord1 <- abc1$words1[i]
    newWord1 <- factor(paste(preWord1,badWord1,sep=""))
    newWord2 <- factor(paste(badWord1,postWord1,sep=""))

    if(nrow(aspell(newWord1)) == 0 & nrow(aspell(newWord2)) != 0){
      abc2[i,"words1"] <-as.character(newWord1)
      abc2 <- abc2[-c(i+1),]
      print(paste(i,"word1",sep=" | "));flush.console()
      i <- i + 1
    }

    if(nrow(aspell(newWord1)) != 0 & nrow(aspell(newWord2)) == 0){
      abc2[i ,"words1"] <-as.character(newWord2)
      abc2 <- abc2[-c(i-1),]
      print(paste(i,"word2",sep=" | "));flush.console()
      i <- i + 1
    }

  }
}
}

在玩了一段时间之后,我得出的结论是我需要某种类型的迭代器,但不确定如何在R中实现它.任何建议？

Answer 1

注意:我提出了一个完全不同的,更好的解决方案,因为它避开了以前解决方案的所有缺点.但我仍然希望保留原有的解决方案.因此,我将其添加为新答案,如果我做错了,请纠正我.

在这种方法中,我稍微重新格式化了数据集.基础是我称之为wordpair对象.例如:

> word5
[1] "hotter the doghou se would be bec ause the col or was diffe rent"

看起来像:

> abc1_pairs
    word1  word2
1  hotter    the
2     the doghou
3  doghou     se
4      se  would
5   would     be
6      be    bec
7     bec   ause
8    ause    the
9     the    col
10    col     or
11     or    was
12    was  diffe
13  diffe   rent

接下来,我们遍历wordpairs并查看它们是否是有效的单词本身,递归执行此操作直到找不到有效的新单词(请注意,此帖子的底部列出了一些其他函数):

# Recursively delete wordpairs which lead to a correct word
merge_wordpairs = function(wordpairs) {
  require(plyr)
  merged_pairs = as.character(mlply(wordpairs, merge_word))
  correct_words_idxs = which(sapply(merged_pairs, word_correct))
  if(length(correct_words_idxs) == 0) {
    return(wordpairs)
  } else {
    message(sprintf("Number of words about to be merged in this pass: %s", length(correct_words_idxs)))
    for(idx in correct_words_idxs) {
      wordpairs = merge_specific_pair(wordpairs, idx, delete_pair = FALSE)
    }
    return(merge_wordpairs(wordpairs[-correct_words_idxs,])) # recursive call
  }
}

应用于示例数据集,这将导致:

> word5 <- "hotter the doghou se would be bec ause the col or was diffe rent"
> abc1 = strsplit(word5, split = " ")[[1]]
> abc1_pairs = wordlist2wordpairs(abc1)
> abc1_pairs
    word1  word2
1  hotter    the
2     the doghou
3  doghou     se
4      se  would
5   would     be
6      be    bec
7     bec   ause
8    ause    the
9     the    col
10    col     or
11     or    was
12    was  diffe
13  diffe   rent
> abc1_merged_pairs = merge_wordpairs(abc1_pairs)
Number of words about to be merged in this pass: 4
> merged_sentence = paste(wordpairs2wordlist(abc1_merged_pairs), collapse = " ")
> c(word5, merged_sentence)
[1] "hotter the doghou se would be bec ause the col or was diffe rent"
[2] "hotter the doghouse would be because the color was different"    

所需的其他功能:

# A bunch of functions
# Data transformation
wordlist2wordpairs = function(word_list) {
  require(plyr)
  wordpairs = ldply(seq_len(length(word_list) - 1), 
                    function(idx) 
                      return(c(word_list[idx], 
                               word_list[idx+1])))
  names(wordpairs) = c("word1", "word2")
  return(wordpairs)
}
wordpairs2wordlist = function(wordpairs) {
  return(c(wordpairs[[1]], wordpairs[[2]][nrow(wordpairs)]))
}

# Some checking functions
# Is the word correct?
word_correct = function(word) return(nrow(aspell(factor(word))) == 0)
# Merge two words
merge_word = function(word1, word2) return(paste(word1, word2, sep = ""))

# Merge a specific pair, option to postpone deletion of pair
merge_specific_pair = function(wordpairs, idx, delete_pair = TRUE) {
  # merge pair into word
  merged_word = do.call("merge_word", wordpairs[idx,])
  # assign the pair to the idx above
  if(!(idx == 1)) wordpairs[idx - 1, "word2"] = merged_word
  if(!(idx == nrow(wordpairs))) wordpairs[idx + 1, "word1"] = merged_word
  # assign the pair to the index below (if not last one)
  if(delete_pair) wordpairs = wordpairs[-idx,]
  return(wordpairs)
}