R:如何从一个非常大的表中快速选择两列中的常用词或相同数字?

问题描述 投票:3回答:1

我有一个非常大的表(1,000,000 X 20)来处理并且需要以快速的方式完成。

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个包含的新列

1)常用词或相同数字

例如:[1] "100020003001" "100020003002"

2)常用词的数量或相同的数字

例如:[1] 2

我已经从下面的线程尝试了这个方法,但是,因为我用for循环执行它所以处理时间很慢:

Count common words in two strings

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

谢谢您的帮助!我真的在这里挣扎......

r dataframe data-science
1个回答
1
投票

我们可以用,分割'X2','X3'列,用intersect得到相应的list元素的map2,并使用lengths'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)

data

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"))
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