R:如何快速地从一个非常大的表格中选择两列中的常用单词或相同数字？

Question

我有一个非常大的表 (1,000,000 X 20) 需要处理并且需要快速处理。

例如，我的表中有 2 列 X2 和 X3:

enter image description here

    X1  X2                                          X3
c1  1   100020003001, 100020003002, 100020003003    100020003001, 100020003002, 100020003004
c2  2   100020003001, 100020004002, 100020004003    100020003001, 100020004007, 100020004009
c3  3   100050006003, 100050006001, 100050006001    100050006011, 100050006013, 100050006021

现在我想创建 2 个新列，其中包含

1)常用词或相同数字

例如:[1] "100020003001""100020003002"

2) 常用词或相同数字的个数

例如:[1] 2

我试过下面线程的方法，但是，处理时间很慢，因为我是用 f​​or 循环做的:

Count common words in two strings

 library(stringi)
 Reduce(`intersect`,stri_extract_all_regex(vec1,"\\w+"))

感谢您的帮助! 我真的在这里挣扎......

Answer 1

我们可以通过,拆分'X2','X3'列，得到对应list元素的intersect与map2 并使用 lengths 来“计算”list

中的元素数量

library(tidyverse)
df1 %>%
   mutate(common_words = map2(strsplit(X2, ", "),
                              strsplit(X3, ", "),  
                                   intersect), 
          count = lengths(common_words))
# X1                                       X2                                       X3
#1  1 100020003001, 100020003002, 100020003003 100020003001, 100020003002, 100020003004
#2  2 100020003001, 100020004002, 100020004003 100020003001, 100020004007, 100020004009
#3  3 100050006003, 100050006001, 100050006001 100050006011, 100050006013, 100050006021
#                common_words count
#1 100020003001, 100020003002     2
#2               100020003001     1
#3                                0

---

或者使用base R

df1$common_words <- Map(intersect, strsplit(df1$X2, ", "), strsplit(df1$X3, ", "))
df1$count <- lengths(df1$common_words)

数据

df1 <- structure(list(X1 = 1:3, X2 = c("100020003001, 100020003002, 100020003003", 
"100020003001, 100020004002, 100020004003", "100050006003, 
 100050006001, 100050006001"
 ), X3 = c("100020003001, 100020003002, 100020003004", "100020003001, 
 100020004007, 100020004009", 
 "100050006011, 100050006013, 100050006021")), class = "data.frame", 
  row.names = c("c1", "c2", "c3"))