匹配R中的两个列表,一个带有部分字符串,另一个带有完整字符串,如果匹配则返回整个字符串.仅返回唯一匹配(一次).
所以,假设我有一个CSV文件,每行都有一个长字符串(长列表).然后,我使用substr缩短字符串,然后使用unique删除任何重复的字符串.然后我想比较长字符串列表df12和唯一的短列表df14,如果在部分字符串搜索(df14 vs df12)上有唯一匹配,则从df12返回整个字符串.
这是df12(长字符串列表)
[1] I like stackoverflow very much today
[2] I like stackoverflow much today
[3] I dont like stackoverflow very much today
[4] I dont like you!
[5] What?
df13<-substr(df12,start=0,stop=30)
这是df13(缩短的字符串 – 不是唯一的)
[1] I like stacko [2] I like stacko [3] I dont like s [4] I dont like y [5] What? df14<-unique(df13)
这是df14(缩短的字符串 – 应用唯一方法后的唯一字符串)
[1] I like stacko
[2] I dont like s
[3] I dont like y
[4] What?
这是我最终想要的结果
[1] I like stackoverflow very much today
[2] I dont like stackoverflow very much today
[3] I dont like you!
[4] What?
这是将df14中的每个短字符串与df12中的所有可能匹配进行匹配并输出它们的一种方法,包括短字符串作为列表中的索引,以便知道哪个与df12中的匹配：
df1 <- c('I like stackoverflow very much today','I like stackoverflow much today','I dont like stackoverflow very much today','I dont like you!','What?')
df2 <- c('I like stacko','I dont like s','I dont like y','What?')
sapply(df2,function(x) df1[grepl(x,df1)])
$`I like stacko`
[1] "I like stackoverflow very much today" "I like stackoverflow much today"
$`I dont like s`
[1] "I dont like stackoverflow very much today"
$`I dont like y`
[1] "I dont like you!"
$`What?`
[1] "What?"
