模糊匹配一列中的行与下一列中的相同行

Question

我想在一个栏目的基础上找到另一个栏目的信息。因此，我在一个专栏中有一些单词，在另一个专栏中有完整的句子。我想知道它是否找到了这些句子中的单词。但是有时候单词不一样，所以我不能使用SQL like函数。因此，我认为模糊匹配+某种“喜欢”函数会很有帮助，因为数据看起来像这样：

Names                    Sentences
Airplanes Sarl           Airplanes-Sàrl is part of Airplanes-Group Sarl. 
Kidco Ltd.               100% ownership of Kidco.Ltd. is the mother company.
Popsi Co.                Cola Inc. is 50% share of PopsiCo which is part of LaLo.

这些数据大约有2000行，需要一个逻辑来判断Airplanes Sarl是否真的在句子中，它也适用于Kidco Ltd.，它在句子中被称为“Kidco.Ltd”。

为了简化问题，我不需要它搜索列中的所有句子，它只需要查找单词Kidco Ltd.并在数据帧的同一行中搜索它。

我已经用df.apply(lambda s: fuzz.ratio(s‘’Names‘，s’‘Sentences’)，axis=1)在Python中尝试过了。

但是我得到了很多unicode /ascii错误，所以我放弃了，并想在R中尝试一下。有什么建议可以在R中进行吗？我在Stackoverflow上看到的答案会模糊地匹配专栏中的所有句子，这与我想要的不同。有什么建议吗？

Answer 1

也许可以尝试标记化+语音匹配：

library(RecordLinkage)
library(quanteda)
df <- read.table(header=T, sep=";", text="
Names                    ;Sentences
Airplanes Sarl           ;Airplanes-Sàrl is part of Airplanes-Group Sarl. 
Kidco Ltd.               ;Airplanes-Sàrl is part of Airplanes-Group Sarl. 
Kidco Ltd.               ;100% ownership of Kidco.Ltd. is the mother company.
Popsi Co.                ;Cola Inc. is 50% share of PopsiCo which is part of LaLo.
Popsi Co.                ;Cola Inc. is 50% share of Popsi Co which is part of LaLo.")
f <- soundex
tokens <- tokenize(as.character(df$Sentences), ngrams = 1:2) # 2-grams to catch "Popsi Co"
tokens <- lapply(tokens, f)
mapply(is.element, soundex(df$Names), tokens)
 # A614  K324  K324  P122  P122 
 # TRUE FALSE  TRUE  TRUE  TRUE 

Answer 2

这是一个使用我在评论中建议的方法的解决方案，在这个例子中它工作得很好：

library("stringdist")

df <- as.data.frame(matrix(c("Airplanes Sarl","Airplanes-Sàrl is part of Airplanes-Group Sarl.",
                             "Kidco Ltd.","100% ownership of Kidco.Ltd. is the mother company.",
                             "Popsi Co.","Cola Inc. is 50% share of PopsiCo which is part of LaLo.",
                             "some company","It is a truth universally acknowledged...",
                             "Hello world",list(NULL)),
                     ncol=2,byrow=TRUE,dimnames=list(NULL,c("Names","Sentences"))),stringsAsFactors=FALSE)

null_elements <- which(sapply(df$Sentences,is.null))
df$Sentences[null_elements] <- "" # replacing NULLs to avoid errors
df$dist <- mapply(stringdist,df$Names,df$Sentences)
df$n2 <- nchar(df$Sentences)
df$n1 <- nchar(df$Names)
df$match_quality <- df$dist-(df$n2-df$n1)
cutoff <- 2
df$match <- df$match_quality <= cutoff
df$Sentences[null_elements] <- list(NULL) # setting null elements back to initial value
df$match[null_elements] <- NA # optional, set to FALSE otherwise, as it will prevent some false positives if Names is shorter than cutoff

# Names                                                Sentences dist n2 n1 match_quality match
# 1 Airplanes Sarl          Airplanes-Sàrl is part of Airplanes-Group Sarl.   33 47 14             0  TRUE
# 2     Kidco Ltd.      100% ownership of Kidco.Ltd. is the mother company.   42 51 10             1  TRUE
# 3      Popsi Co. Cola Inc. is 50% share of PopsiCo which is part of LaLo.   48 56  9             1  TRUE
# 4   some company                It is a truth universally acknowledged...   36 41 12             7 FALSE
# 5    Hello world                                                     NULL   11  0 11            22    NA