R：可以将提取的字符串作为分隔字符保存到一列中吗？

Question

假设我需要根据评论行中的句子为人们分配课程。 （实际数据比这个复杂，我简化了） 因此，我使用带有 remathces()、gsub() 和 gregexpr() 的正则表达式从数据中的注释句子中提取字符串。然后将列表保存到列并将它们组合为字符，如下所示。

>cbind.data.frame(level,software,month,stringsAsFactors = FALSE) 

   level                         software             month
1  c("beginner1","beginner2")    c++                  Dec       
2                      NA        Java                 Jan       
3             "beginner3"        NA                   May   
4         "intermediate2"        NA                   NA      
5                      NA        Matlab               Mar    
6             "advanced1"        c("java","c++")      Jul     

我想用

将所有字符分成一列

-将列表 c("beginner1","beginner2") 分解为 "beginner1","beginner2"

-丢弃不适用

-保留为如下字符

  newcol
 "beginner1","beginner2","c++","Dec" 
 "Java","Jan" 
 "beginner3", "May"
 "intermediate2" 
 "Matlab", "Mar"    
 "advanced1","java","c++","Jul"  

但是，当我合并时，它被合并为一个字符。

> newcol<-unite(combined, newcol, 1:ncol(combined), remove=TRUE, sep = ",")

 "beginner1,beginner2,c++,Dec"  
 "Java,Jan" 
 "beginner3, May"
 "intermediate2" 
 "Matlab, Mar"    
 "advanced1,java,c++,Jul"  

是否可以将多个字符作为分隔字符保存到一列中？

Answer 1

这是一个基本的 R 解决方案，使用

f <- Vectorize(function(u) {
  z <- unlist(regmatches(u,gregexpr('\".*?\"',u,perl = T)))
  if (length(z)> 0) {
    r <- gsub('\"',"",z)
  } else {
    r <- u
  }
  r
})

df$newcol <- apply(df,1,function(x) f(na.omit(x)))

这样

> df
                       level        software month                         newcol
1 c("beginner1","beginner2")             c++   Dec beginner1, beginner2, c++, Dec
2                       <NA>            Java   Jan                      Java, Jan
3                  beginner3            <NA>   May                 beginner3, May
4              intermediate2            <NA>  <NA>                  intermediate2
5                       <NA>          Matlab   Mar                    Matlab, Mar
6                  advanced1 c("java","c++")   Jul      advanced1, java, c++, Jul

在哪里

> df$newcol
$`1`
$`1`$level
[1] "beginner1" "beginner2"

$`1`$software
[1] "c++"

$`1`$month
[1] "Dec"


$`2`
$`2`$software
[1] "Java"

$`2`$month
[1] "Jan"


$`3`
$`3`$level
[1] "beginner3"

$`3`$month
[1] "May"


$`4`
$`4`$level
[1] "intermediate2"


$`5`
$`5`$software
[1] "Matlab"

$`5`$month
[1] "Mar"


$`6`
$`6`$level
[1] "advanced1"

$`6`$software
[1] "java" "c++" 

$`6`$month
[1] "Jul"

数据

df <- structure(list(level = c("c(\"beginner1\",\"beginner2\")", NA, 
"beginner3", "intermediate2", NA, "advanced1"), software = c("c++", 
"Java", NA, NA, "Matlab", "c(\"java\",\"c++\")"), month = c("Dec", 
"Jan", "May", NA, "Mar", "Jul")), class = "data.frame", row.names = c("1", 
"2", "3", "4", "5", "6"))

Answer 2

这有帮助吗？

A<-data.frame(a=c("a","b","c"),b=c("a","b","c"),c=c("a","b","c"))

apply(A,2,paste,collapse=",")