匹配两个列表,一个包含部分字符串,另一个包含完整字符串,如果匹配则返回整个字符串

Question

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(df14vs df12)上有唯一匹配,则返回整个字符串df12.

这是df12(长串列表)

    [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?

Answer 1

这是一种将 df14 中的每个短字符串与 df12 中所有可能的匹配项进行匹配并输出它们的方法，包括将短字符串作为列表中的索引，以了解哪一个与 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?"