R中Hunspell无拼写建议时越界错误的解决方法

我尝试自动检查data.table/data.frame的字符串列拼写，发现当hunspell_suggest返回无建议的空列表（比如像“pippasnjfjsfiadjg”这种完全无意义的词汇）时，多种处理方法都会触发“越界”错误。我知道需要识别这些空建议，将其排除在“选取首个建议”的代码逻辑之外，但不知道具体怎么实现。

原测试代码如下：

library(dplyr)
library(stringi)
library(hunspell)

df1 <- data.frame("Index" = 1:7, "Text" = c("pippasnjfjsfiadjg came to dinner with us tonigh.",
                                            "Wuld you like to trave with me?",
                                            "There is so muh to undestand.",
                                            "Sentences cone in many shaes and sizes.",
                                            "Learnin R is fun",
                                            "yesterday was Friday",
                                            "bing search engine"),
                  stringsAsFactors = FALSE)

# 获取拼写错误的词汇
badwords <- hunspell(df1$Text) %>% unlist

# 提取每个错误词汇的首个拼写建议
suggestions <- sapply(hunspell_suggest(badwords), "[[", 1)

# 替换错误词汇
mutate(df1, Text = stri_replace_all_fixed(str = Text,
                                          pattern = badwords,
                                          replacement = suggestions,
                                          vectorize_all = FALSE)) -> out

解决方案 #

核心是在提取拼写建议时，先判断每个建议列表是否为空：为空的情况直接保留原词（或返回NA，根据需求调整），避免直接取[[1]]触发越界错误。修改后的代码如下：

library(dplyr)
library(stringi)
library(hunspell)

df1 <- data.frame("Index" = 1:7, "Text" = c("pippasnjfjsfiadjg came to dinner with us tonigh.",
                                            "Wuld you like to trave with me?",
                                            "There is so muh to undestand.",
                                            "Sentences cone in many shaes and sizes.",
                                            "Learnin R is fun",
                                            "yesterday was Friday",
                                            "bing search engine"),
                  stringsAsFactors = FALSE)

# 获取拼写错误的词汇
badwords <- hunspell(df1$Text) %>% unlist

# 提取首个建议，空建议时返回原词
suggestions <- sapply(badwords, function(word) {
  sugg <- hunspell_suggest(word)[[1]]
  if(length(sugg) > 0) sugg[[1]] else word
})

# 替换错误词汇
out <- mutate(df1, Text = stri_replace_all_fixed(str = Text,
                                          pattern = badwords,
                                          replacement = suggestions,
                                          vectorize_all = FALSE))

关键说明 #

• 替换了原代码中直接用sapply(hunspell_suggest(badwords), "[[", 1)的逻辑，改为逐个处理每个错误词汇：先获取它的建议列表，若列表长度大于0则取首个建议，否则返回原词。
• 这样处理后，像“pippasnjfjsfiadjg”这类无拼写建议的词汇会被原样保留，不会触发越界错误，同时其他有建议的错误词汇会正常替换。

内容的提问来源于stack exchange，提问作者Magasinus