我们可以计算行之间的距离矩阵,形成簇并选择簇成员 作为相似行的候选。
使用R
和stringdistmatrix
函数从stringdist
包允许 字符串输入之间的距离计算。
stringdist支持的距离方法如下。见package manual 更多细节
#Method name; Description
#osa ; Optimal string aligment, (restricted Damerau-Levenshtein distance).
#lv ; Levenshtein distance (as in R's native adist).
#dl ; Full Damerau-Levenshtein distance.
#hamming ; Hamming distance (a and b must have same nr of characters).
#lcs ; Longest common substring distance.
#qgram ;q-gram distance.
#cosine ; cosine distance between q-gram profiles
#jaccard ; Jaccard distance between q-gram profiles
#jw ; Jaro, or Jaro-Winker distance.
#soundex ; Distance based on soundex encoding (see below)
数据:
library("stringdist")
#have modified the data slightly to include dissimilar datapoints
Date = c("07-Jan-17","06-Feb-17","03-Mar-17")
name = c("Game of ThOnes Books for selling","Selling Game of Thrones books","Harry Potter BlueRay")
address = c("George Washington street","George Washington st.","Central Avenue")
phone = c("555-55-55","0(555)-55-55","111-222-333")
DF = data.frame(Date,name,address,phone,stringsAsFactors=FALSE)
DF
# Date name address phone
#1 07-Jan-17 Game of ThOnes Books for selling George Washington street 555-55-55
#2 06-Feb-17 Selling Game of Thrones books George Washington st. 0(555)-55-55
#3 03-Mar-17 Harry Potter BlueRay Central Avenue 111-222-333
层次聚类:
rowLabels = sapply(DF[,"name"],function(x) paste0(head(unlist(strsplit(x," ")),2),collapse="_"))
#create string distance matrix, hierarchical cluter object and corresponding plot
nameDist = stringdistmatrix(DF[,"name"])
nameHC = hclust(nameDist)
plot(nameHC,labels = rowLabels ,main="HC plot : name")
addressDist = stringdistmatrix(DF[,"address"])
addressDistHC = hclust(addressDist)
plot(addressDistHC ,labels = rowLabels, main="HC plot : address")
phoneDist = stringdistmatrix(DF[,"phone"])
phoneHC = hclust(phoneDist)
plot(phoneHC ,labels = rowLabels, main="HC plot : phone")
类似的行:
该行始终形成该数据集两个集群,以识别集群的成员,我们可以做
clusterDF = data.frame(sapply(DF[,-1],function(x) cutree(hclust(stringdistmatrix(x)),2)))
clusterDF$rowSummary = rowSums(clusterDF)
clusterDF
# name address phone rowSummary
#1 1 1 1 3
#2 1 1 1 3
#3 2 2 2 6
#row frequency
rowFreq = table(clusterDF$rowSummary)
#3 6
#2 1
#we filter rows with frequency > 1
similarRowValues = as.numeric(names(which(rowFreq>1)))
DF[clusterDF$rowSummary == similarRowValues,]
# Date name address phone
#1 07-Jan-17 Game of ThOnes Books for selling George Washington street 555-55-55
#2 06-Feb-17 Selling Game of Thrones books George Washington st. 0(555)-55-55
这个演示对简单/玩具数据集运行良好,但是对于真正的数据集,你不得不用字符串距离计算方法,簇数等来修饰,但是我希望这可以让你指向正确的方向。
你想在R或Python的解决方案吗? –
@RonakShah其实没关系。我只想要合理的解决方案。 –