Question

我有两个字符串：

a <- "Roy lives in Japan and travels to Africa"
b <- "Roy travels Africa with this wife"

我希望得到这些字符串之间常用词的数量。

答案应该是3。

＆＃34;罗伊＆＃34;
＆＃34;行进＆＃34;
＆＃34;非洲＆＃34;

是常用词

这就是我的尝试：

stra <- as.data.frame(t(read.table(textConnection(a), sep = " ")))
strb <- as.data.frame(t(read.table(textConnection(b), sep = " ")))

采取独特的措施以避免重复计算

stra_unique <-as.data.frame(unique(stra$V1))
strb_unique <- as.data.frame(unique(strb$V1))
colnames(stra_unique) <- c("V1")
colnames(strb_unique) <- c("V1")

common_words <-length(merge(stra_unique,strb_unique, by = "V1")$V1)

我需要这个用于超过2000和1200字符串的数据集。我必须评估字符串的总时间是2000 X 1200.任何快速方式，不使用循环。

Answer 1

您可以使用base库中的strsplit和intersect：

> a <- "Roy lives in Japan and travels to Africa"
> b <- "Roy travels Africa with this wife"
> a_split <- unlist(strsplit(a, sep=" "))
> b_split <- unlist(strsplit(b, sep=" "))
> length(intersect(a_split, b_split))
[1] 3

Answer 2

也许，使用intersect和str_extract 对于multiple strings，您可以将它们设为list或vector

 vec1 <- c(a,b)
 Reduce(`intersect`,str_extract_all(vec1, "\\w+"))
 #[1] "Roy"     "travels" "Africa"

对于faster选项，请考虑stringi

 library(stringi)
 Reduce(`intersect`,stri_extract_all_regex(vec1,"\\w+"))
 #[1] "Roy"     "travels" "Africa"

计算：

 length(Reduce(`intersect`,stri_extract_all_regex(vec1,"\\w+")))
 #[1] 3

或使用base R

  Reduce(`intersect`,regmatches(vec1,gregexpr("\\w+", vec1)))
  #[1] "Roy"     "travels" "Africa"

Answer 3

这种方法可推广到n个向量：

a <- "Roy lives in Japan and travels to Africa"
b <- "Roy travels Africa with this wife"
c <- "Bob also travels Africa for trips but lives in the US unlike Roy."

library(stringi);library(qdapTools)
X <- stri_extract_all_words(list(a, b, c))
X <- mtabulate(X) > 0
Y <- colSums(X) == nrow(X); names(Y)[Y]

[1] "Africa"  "Roy"     "travels"

计算两个字符串中的常用单词

3 个答案: