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:
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......
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)
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"))