1

我有一张拼错单词的桌子。我需要更正那些使用与那个词更相似的词，即频率更高的词。

例如，在我运行之后

aggregate(CustomerID ~ Province, ventas2, length)

我明白了

1                             
2                     AMBA         29
    3                   BAIRES          1
    4              BENOS AIRES          1

    12            BUENAS AIRES          1

    17           BUENOS  AIRES          4
    18            buenos aires          7
    19            Buenos Aires          3
    20            BUENOS AIRES      11337
    35                 CORDOBA       2297
    36                cordoba           1
    38               CORDOBESA          1
    39              CORRIENTES        424

所以我需要用 BUENOS AIRES 替换 buenos aires, Buenos Aires, Baires, BUENOS AIRES，但不应该替换 AMBA。CORDOBESA 和 cordoba 也应替换为 CORDOBA，而不是 CORRIENTES。

我怎样才能在 R 中做到这一点？

谢谢！

4

2 回答 2

3

这是一个可能的解决方案。

免责声明：
此代码似乎适用于您当前的示例。我不保证当前参数（例如切割高度、集群聚集方法、距离方法等）对您的真实（完整）数据有效。

# recreating your data
data <- 
read.csv(text=
'City,Occurr
AMBA,29
BAIRES,1
BENOS AIRES,1
BUENAS AIRES,1
BUENOS  AIRES,4
buenos aires,7
Buenos Aires,3
BUENOS AIRES,11337
CORDOBA,2297
cordoba,1
CORDOBESA,1
CORRIENTES,424',stringsAsFactors=F)


# simple pre-processing to city strings:
# - removing spaces
# - turning strings to uppercase
cities <- gsub('\\s+','',toupper(data$City))

# string distance computation
# N.B. here you can play with single components of distance costs 
d <- adist(cities, costs=list(insertions=1, deletions=1, substitutions=1))
# assign original cities names to distance matrix
rownames(d) <- data$City
# clustering cities
hc <- hclust(as.dist(d),method='single')

# plot the cluster dendrogram
plot(hc)
# add the cluster rectangles (just to see the clusters) 
# N.B. I decided to cut at distance height < 5
#      (read it as: "I consider equal 2 strings needing
#       less than 5 modifications to pass from one to the other")
#      Obviously you can use another value.
rect.hclust(hc,h=4.9)

# get the clusters ids
clusters <- cutree(hc,h=4.9) 
# turn into data.frame
clusters <- data.frame(City=names(clusters),ClusterId=clusters)

# merge with frequencies
merged <- merge(data,clusters,all.x=T,by='City') 

# add CityCorrected column to the merged data.frame
ret <- by(merged, 
          merged$ClusterId,
          FUN=function(grp){
                idx <- which.max(grp$Occur)
                grp$CityCorrected <- grp[idx,'City']
                return(grp)
              })

fixed <- do.call(rbind,ret)

结果 ：

> fixed
              City Occurr ClusterId CityCorrected
1             AMBA     29         1          AMBA
2.2         BAIRES      1         2  BUENOS AIRES
2.3    BENOS AIRES      1         2  BUENOS AIRES
2.4   BUENAS AIRES      1         2  BUENOS AIRES
2.5  BUENOS  AIRES      4         2  BUENOS AIRES
2.6   buenos aires      7         2  BUENOS AIRES
2.7   Buenos Aires      3         2  BUENOS AIRES
2.8   BUENOS AIRES  11337         2  BUENOS AIRES
3.9        cordoba      1         3       CORDOBA
3.10       CORDOBA   2297         3       CORDOBA
3.11     CORDOBESA      1         3       CORDOBA
4       CORRIENTES    424         4    CORRIENTES

聚类图：

在此处输入图像描述

于 2014-09-09T21:56:40.443 回答
0

这是我对您的聚合结果的小型复制您需要更改对数据框的所有调用以适应您的数据结构。

df
#output
#       word freq
#1         a    1
#2         b    2
#3         c    3

#find the max frequency
mostFrequent<-max(df[,2])  #doesn't handle ties

#find the word we will be replacing with
replaceString<-df[df[,2]==mostFrequent,1]
#[1] "c"

#find all the other words to be replaced
tobereplaced<-df[df[,2]!=mostFrequent,1]
#[1] "a" "b"

现在假设您有以下包含整个数据集的数据框，我将只复制一个带有单词的列。

totalData
 #    [,1]
 #[1,] "a" 
 #[2,] "c" 
 #[3,] "b" 
 #[4,] "d" 
 #[5,] "f" 
 #[6,] "a" 
 #[7,] "d" 
 #[8,] "b" 
 #[9,] "c"

我们可以通过以下调用将所有要替换的单词替换为要替换的字符串

totaldata[totaldata%in%tobereplaced]<-replaceString
 #    [,1]
 #[1,] "c" 
 #[2,] "c" 
 #[3,] "c" 
 #[4,] "d" 
 #[5,] "f" 
 #[6,] "c" 
 #[7,] "d" 
 #[8,] "c" 
 #[9,] "c"

可以看到，所有的a和b都被c替换了，其他的词都一样

于 2014-09-09T20:04:49.933 回答