替换类似的拼写错误的单词答案

【问题标题】：Replace similar incorrectly spelled words替换类似的拼写错误的单词
【发布时间】：2018-06-04 19:52:17
【问题描述】：

我正在尝试修复我的调查数据。我的数据框包含多个应该相同的值；但是，拼写、间距和大小写存在差异，导致级别数超出预期。

str(data.frame$race)
"American Indian and Alaska Native" 
"Asian"                             
"Black of African American"        
"Black or African American"         
"Other"                            
"Unknown"                          
"white or Caucasian"                
"White or Caucasian"                
"White or Caucasion"

我如何“查找和替换”以创建一个统一的拼写并将其转换回具有适当级别数的因子？

【问题讨论】：

能否请您输入您的数据集？
你可能想看看 OpenRefine ...
stackoverflow.com/questions/40203830/…
你也可以看看phonicssoundex(c("Black of African American", "Black or African American", "white or Caucasian", "White or Caucasian", "White or Caucasion"))# [1] "B420" "B420" "W300" "W300" "W300"
这个答案非常接近你想要的...stackoverflow.com/questions/38956241/…...不确定是否值得一个新的答案...？

标签： r

【解决方案1】：

很难找到一种万能的解决方案。这是因为看起来相似的字符串可能描述了非常不同的事物（例如格拉纳达与格林纳达）。原帖下的cmets值得研究。

参见"Approximate string matching" on Wikipedia（有时也称为“模糊匹配”）。如您所见，有很多方法可以在字符串上定义“相似”。

最基本的工具是R函数adist。它计算所谓的编辑距离。

x <- c("American Indian and Alaska Native" ,
   "Asian"                             ,
   "Black of African American"        ,
   "Black or African American"         ,
   "Other"                            ,
   "Unknown"                          ,
   "white or Caucasian"                ,
   "White or Caucasian"                ,
   "White or Caucasion" )
u <- unique(x)
# compare all strings against each other
d <- adist(u)
# Do not list combinations of similar words twice
d[lower.tri(d)] <- NA
# Say your threshold below which you want to consider strings similar is 
# 2 edits:
a <- which(d > 0 & d < 2, arr.ind = TRUE)
a
##      row col
## [1,]   3   4
## [2,]   7   8
## [3,]   8   9
pairs <- cbind(u[a[,1]], u[a[,2]])
pairs
##      [,1]                        [,2]                       
## [1,] "Black of African American" "Black or African American"
## [2,] "white or Caucasian"        "White or Caucasian"       
## [3,] "White or Caucasian"        "White or Caucasion"

但最终，您必须自己策划结果，以避免不公平因素的意外均衡。

您可以通过使用命名向量作为翻译字典来重复执行此操作。例如，通过查看上面的示例，我可以创建以下字典：

dict <- c(
   # incorrect spellings          correct spellings
   # -------------------------    ----------------------------
   "Black of African American" =  "Black or African American",
   "white or Caucasian"        =  "white or Caucasian"       ,
   "White or Caucasion"        =  "White or Caucasian" 
)
# The correct levels need to be included, to
dict <- c(dict, setNames(u,u)

然后使用as.character 将您的因子列转换为字符并应用上面的字典就像我在这里用原始字符向量x：

xcorrected <- dict[x]
# show without names, but the result is also correct if you just use
# xcorrected alone (remove as.character here to see the difference).
as.character(xcorrected)
[1] "American Indian and Alaska Native" "Asian"                            
[3] "Black or African American"         "Black or African American"        
[5] "Other"                             "Unknown"                          
[7] "white or Caucasian"                "White or Caucasian"               
[9] "White or Caucasian"

【讨论】：