Question

我有两个数据框。第一个 - 保存在名为b的对象中：

structure(list(CONTENT = c("@myntra beautiful teamä»ç where is the winners list?", 
"The best ever Puma wishlist for Workout freaks, Head over to @myntra  https://t.co/V58Gk3EblW #MyPUMACollection Hit Like if you Find it good", 
"I finalised on buy a top from Myntra, and then I found the same top at 20% off in jabong. I feel like I've achieved so much in life!", 
"Check out #myPUMAcollection on @Myntra. Its perfect for a day at gym.  https://t.co/VeRy4G3c7X https://t.co/fOpBRWCdSh", 
"Check out #myPUMAcollection on @Myntra. Its perfect for a day at gym.  https://t.co/VeRy4G3c7X.....", 
"@DrDrupad @myntra #myPUMAcollection superb :)", "Super exclusive collection @myntra #myPUMAcollection   https://t.co/Qm9dZzJdms", 
"@myntra gave my best Love playing wid u Hope to win  #myPUMAcollection", 
"Check out PUMA Unisex Black Running Performance Gloves on Myntra!  https://t.co/YD6IcvuG98 @myntra  #myPUMAcollection", 
"@myntra i have been mailing my issue daily since past week.All i get in reply is an auto generated assurance mail. 1st time pissed wd myntra"
), score = c(7.129, 7.08, 6.676, 5.572, 5.572, 5.535, 5.424, 
5.205, 4.464, 4.245)), .Names = c("CONTENT", "score"), row.names = c(25L, 
103L, 95L, 66L, 90L, 75L, 107L, 32L, 184L, 2L), class = "data.frame")

第二个数据库 - 保存在名为c：

的对象中

structure(list(CONTENT = c("The best ever for workout  over to myntra like if you find it good", 
"i finalised buy a top  myntra and found the at in feel like i so in life"
)), .Names = "CONTENT", row.names = c(103L, 95L), class = "data.frame")

我想找到第二个数据帧（c）中的每个语句，第一个数据帧（b）中最接近的匹配，并从第一个数据帧（b）返回分数。

例如，语句The best ever for workout over to myntra like if you find it good与数据框1中的第二个语句紧密匹配，因此我应该返回分数7.080。

我尝试使用堆栈溢出代码进行一些调整：

cp <- str_split(c$CONTENT, " ")
library(data.table)
nn <- lengths(cp)  ## Or, for < R-3.2.0, `nn <- sapply(wordList, length)` 
dt <- data.table(grp=rep(seq_along(nn), times=nn), X = unlist(cp), key="grp")
dt[,Score:=b$score[pmatch(X,b$CONTENT)]]
dt[!is.na(Score), list(avgScore=sum(Score)), by="grp"]

这将返回df c中只有一个语句的值。有人可以帮忙吗？

Answer 1

这是使用stringsim包中的stringdist的一种方法。有几种method s（算法）可供选择 - 我决定使用Jaro distance指标来计算相似性，因为它似乎可以为您的数据产生合理的结果。话虽如此，我对这个主题的体验充其量是偶然的，所以你可能想花点时间阅读并试验stringdist提供的各种算法。

为了减少混乱，我使用这个包装函数返回给定字符串最相似（最高相似度值）元素的索引，

library(stringdist)
library(data.table)

best_match <- function(x, y, method = "jw", ...) {
    which.max(stringsim(x, y, method, ...))
}

并使data.table包含要匹配的字符串，为行方式操作添加虚拟索引：

Dt <- data.table(
    MatchPhrase = df_c$CONTENT,
    Idx = 1:nrow(df_c)
)

使用best_match，添加一个包含最佳匹配索引的列（之后删除虚拟Idx列），

Dt[, MatchIdx := best_match(df_b$CONTENT, MatchPhrase), 
    by = "Idx"][,Idx := NULL]

并从df_b中提取相应的元素（我将您的数据分别从b和c重命名为df_b和df_c）：

Dt[, .(Score = df_b$score[MatchIdx],
       BestMatch = df_b$CONTENT[MatchIdx]),
   by = "MatchPhrase"]
#                                                                MatchPhrase Score
#1:       The best ever for workout  over to myntra like if you find it good 7.080
#2: i finalised buy a top  myntra and found the at in feel like i so in life 6.676

#                                                                                                                                      BestMatch
#1: The best ever Puma wishlist for Workout freaks, Head over to @myntra  https://t.co/V58Gk3EblW #MyPUMACollection Hit Like if you Find it good
#2:         I finalised on buy a top from Myntra, and then I found the same top at 20% off in jabong. I feel like I've achieved so much in life!

R

1 个答案: