Question: I am new to R and have a very large irregular column in a data frame like this:

```
x <- data.frame(section = c("BOOK I: Introduction", "Page one: presentation", "Page two: acknowledgments", "MAGAZINE II: Considerations", "Page one: characters", "Page two: index", "BOOK III: General Principles", "BOOK III: General Principles", "Page one: invitation"))

section
BOOK I: Introduction
Page one: presentation
Page two: acknowledgments
MAGAZINE II: Considerations 
Page one: characters
Page two: index
BOOK III: General principles
BOOK III: General principles
Page one: invitation

```

I need to concatenate this column to look like this:

```
section
BOOK I: Introduction 
BOOK I: Introduction / Page one: presentation
BOOK I: Introduction / Page two: acknowledgments
MAGAZINE II: Considerations
MAGAZINE II: Considerations / Page one: characters
MAGAZINE II: Considerations / Page two: index
BOOK III: General Principles
BOOK III: General Principles
BOOK III: General Principles / Page one: invitation

```

Basically the goal is to extract the value of the upper string based in a condition and then concatenate with the lower actualizing the value with a regex expression, but I really don't know how to do it.

Thanks in advance.

Answer: 
You can do:

```r
unlist(lapply(split(x$section, cumsum(grepl('^[A-Z]{3}', x$section))), 
              function(y) {
                  if(length(y) == 1) return(y)
                  else c(y[1], paste(y[1], y[-1], sep = " / "))
                }), use.names = FALSE)
#> [1] "BOOK I: Introduction"                               
#> [2] "BOOK I: Introduction / Page one: presentation"      
#> [3] "BOOK I: Introduction / Page two: acknowledgments"   
#> [4] "MAGAZINE II: Considerations"                        
#> [5] "MAGAZINE II: Considerations / Page one: characters" 
#> [6] "MAGAZINE II: Considerations / Page two: index"      
#> [7] "BOOK III: General Principles"                       
#> [8] "BOOK III: General Principles"                       
#> [9] "BOOK III: General Principles / Page one: invitation"

```
An slightly simpler `data.table` approach:

```
library(data.table)
setDT(x)

x[, g := cumsum(grepl('(BOOK|MAGAZINE)', section))]
x[, section := ifelse(seq_along(section) == 1,
    section, paste(section[1], section, sep = ' / ')), by = .(g)]
x[, g := NULL]

```

The output is:

```
> x
                                               section
1:                                BOOK I: Introduction
2:       BOOK I: Introduction / Page one: presentation
3:    BOOK I: Introduction / Page two: acknowledgments
4:                         MAGAZINE II: Considerations
5:  MAGAZINE II: Considerations / Page one: characters
6:       MAGAZINE II: Considerations / Page two: index
7:                        BOOK III: General Principles
8:                        BOOK III: General Principles
9: BOOK III: General Principles / Page one: invitation

```