计数+识别两个字符串向量中的常用词 [R]

Question

我如何编写一个 R 函数，它可以接受两个字符串向量和 returns 常用词的数量以及哪些常用词比较元素 1 从 stringvec1 到元素 1 的 stringvec2，元素 2 的 strinvec1 到stringvec2 的元素 2 等

假设我有这些数据：

#string vector 1
strvec1 <- c("Griffin Rahea Petersen Deana Franks Morgan","Story Keisha","Douglas Landon Lark","Kinsman Megan Thrall Michael Michels Breann","Gutierrez Mccoy Tyler Westbrook Grayson Swank Shirley Didas Moriah")

#string vector 2
strvec2 <- c("Griffin Morgan Rose Manuel","Van De Grift Sarah Sell William","Mark Landon Lark","Beerman Carlee Megan Thrall Michels","Mcmillan Tyler Jonathan Westbrook Grayson Didas Lloyd Connor")

理想情况下，我有一个函数可以 return 常用词的数量以及常用词是什么：

#Non working sample of how functions would ideally work
desiredfunction_numwords(strvec1,strvec2)
[1] 2 0 2 3 4

desiredfunction_matchwords(strvec1,strvec2)
[1] "Griffin Morgan" "" "Landon Lark" "Megan Thrall Michels" "Tyler Westbrook Grayson Didas"

Answer 1

您可以在每个单词处拆分字符串并执行操作。

在基数 R 中：

numwords <- function(str1, str2) {
  mapply(function(x, y) length(intersect(x, y)), 
         strsplit(str1, ' '), strsplit(str2, ' '))
}

matchwords <- function(str1, str2) {
  mapply(function(x, y) paste0(intersect(x, y),collapse = " "), 
         strsplit(str1, ' '), strsplit(str2, ' '))
}

numwords(strvec1, strvec2)
#[1] 2 0 2 3 4

matchwords(strvec1, strvec2)
#[1] "Griffin Morgan"          ""                "Landon Lark"                  
#[4] "Megan Thrall Michels"          "Tyler Westbrook Grayson Didas"

Answer 2

您可以将 strvec1 用作 正则表达式模式 ，方法是 strsplit 将其分成单独的词，然后 paste 将这些词与交替标记 |:

pattern <- paste0(unlist(strsplit(strvec1, " ")), collapse = "|")

您可以将此模式与 str_count 和 str_extract_all 一起使用：

library(stringr) 
# counts:
str_count(strvec2, pattern)
[1] 2 0 2 3 4

# matches:
str_extract_all(strvec2, pattern)
[[1]]
[1] "Griffin" "Morgan" 

[[2]]
character(0)

[[3]]
[1] "Landon" "Lark"  

[[4]]
[1] "Megan"   "Thrall"  "Michels"

[[5]]
[1] "Tyler"     "Westbrook" "Grayson"   "Didas"

计数+识别两个字符串向量中的常用词 [R]

Count+Identify common words in two string vectors [R]

r

string

substring

string-matching

longest-substring