在 R 中迭代以拼写检查单词向量

Question

我有一个数据集，其中包含不适当间隔的句子。我正在尝试想出一种方法来删除一些空格。

我从一个句子开始，然后将其转换为单词的数据框：

> 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

接下来我使用下面的代码来尝试拼写检查和组合之前或之后的单词组合的单词：

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
    }

  }
}
}

玩了一段时间后，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。有什么建议吗？

Answer 1

你可以做的是使用递归。下面的代码对您的示例进行了稍微修改的版本。它检查所有单词是否正确，如果正确，则返回单词列表。如果不是，它会尝试将该单词与前面的单词以及后面的单词结合起来。如果前一个单词的合并是正确的，这将导致一个看起来像paste(word_before, word, word_after) 的合并。在尝试合并之后，在新单词列表上调用合并单词的函数。这种递归一直持续到没有留下错误的单词为止。

# Wrap the spell checking in a function, makes your code much more readable
word_correct = function(word) return(nrow(aspell(factor(word))) == 0)
# Merge two words
merge_word = function(word1, word2) return(paste(word1, word2, sep = ""))
# Merge two words and replace in list
merge_words_in_list = function(word_list, idx1, idx2) {
  word_list[idx1] = merge_word(word_list[idx1], word_list[idx2])
  return(word_list[-idx2])
}
# Function that recursively combines words 
combine_words = function(word_list) {
  message("Current sentence: ", paste(word_list, collapse = " "))
  words_ok = sapply(word_list, word_correct)
  if(all(words_ok)) {
    return(word_list) 
  } else {
    first_wrong_word = which(!words_ok)[1]
    combination_before = merge_word(word_list[first_wrong_word], 
                                    word_list[first_wrong_word-1])
    if(word_correct(combination_before)) {
      word_list = merge_words_in_list(word_list, first_wrong_word-1, 
                                      first_wrong_word)
    }
    combination_after = merge_word(word_list[first_wrong_word], 
                                   word_list[first_wrong_word+1])
    if(word_correct(combination_after)) {
      word_list = merge_words_in_list(word_list, first_wrong_word, 
                                      first_wrong_word+1)
    }
    return(combine_words(word_list))  # Recursive call
  }
}

将这组函数应用于（稍作修改的）您的句子版本：

word5 <- "hotter the doghou se would be bec ause the col or was diffe rent"
abc1 = strsplit(word5, split = " ")[[1]]
combine_words(abc1)
Current sentence: hotter the doghou se would be bec ause the col or was diffe rent
Current sentence: hotter the doghouse would be bec ause the col or was diffe rent
Current sentence: hotter the doghouse would be because the col or was diffe rent
Current sentence: hotter the doghouse would be because the col or was different

一些问题：

• 仍然存在如果combination_before 和combination_after 都无效，程序会陷入无限递归的问题。程序仅在所有单词都有效时停止。
• 如果两者都与前一个词合并，并且下一个词有效，我们该怎么办？
• 代码只合并错误的单词，例如'col' 和 'or' 被 aspell 判断为好词，而您可能想要合并。这带来了一个新的挑战：在这种情况下，合并是显而易见的，但在大型数据集中，如何组合一组本身正确的单词可能并不明显。

不过，我认为这个例子很好地说明了递归方法。

Answer 2

注意：我想出了一个完全不同的更好的解决方案，因为它规避了之前解决方案的所有缺点。但我仍然想保留旧的解决方案。因此，我将其添加为新答案，如果我这样做错了，请纠正我。

在这种方法中，我稍微重新格式化了数据集。基础就是我所说的 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

接下来我们遍历单词对并查看它们本身是否是有效单词，递归执行此操作直到找不到有效的新单词（请注意，本文底部列出了一些附加函数）：

# 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"    

需要额外的功能：

# 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)
}