Question

我想根据另一列在一列中查找信息。所以我在一列中有一些单词，在另一列中有完整的句子。我想知道它是否找到了那些句子中的单词。但有时单词不一样，所以我不能使用SQL like函数。因此，我认为模糊匹配+某种类似的＆＃39;喜欢＆＃39;函数会很有用，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.

这些数据有大约2,000行需要一个逻辑来判断Airplanes Sarl是否确实在句子中，而且它也适用于Kidco有限公司，它在句子中是＆＃39; Kidco.Ltd＆＃39; 。

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

我已经在Python中尝试过： df.apply（lambda s：fuzz.ratio（s [＆＃39; Names＆＃39;]，s [＆＃39; Sentences＆＃39;]），axis = 1）

但是我收到了很多unicode / ascii错误所以我放弃了并想尝试使用R. 关于如何在R中进行此操作的任何建议？我在Stackoverflow上看到了模糊匹配列中所有句子的答案，这与我想要的不同。有什么建议吗？

Answer 1

也许尝试标记化+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

以下是使用我在评论中建议的方法的解决方案，在此示例中效果很好：

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

模糊匹配下一列

2 个答案: