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.
CodePudding user response:
Here is one method:
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"))
x <- dplyr::mutate(x,
isSection = stringr::str_starts(section, "Page", negate = TRUE),
sectionNum = cumsum(isSection)
) |>
dplyr::group_by(sectionNum) |>
dplyr::mutate(newSection = dplyr::if_else(
condition = isSection,
true = section,
false = paste(dplyr::first(section), section, sep = " / ")
)) |>
ungroup()
x
#> # A tibble: 9 × 4
#> section isSection sectionNum newSection
#> <chr> <lgl> <int> <chr>
#> 1 BOOK I: Introduction TRUE 1 BOOK I: Introduction
#> 2 Page one: presentation FALSE 1 BOOK I: Introduction / Page…
#> 3 Page two: acknowledgments FALSE 1 BOOK I: Introduction / Page…
#> 4 MAGAZINE II: Considerations TRUE 2 MAGAZINE II: Considerations
#> 5 Page one: characters FALSE 2 MAGAZINE II: Considerations…
#> 6 Page two: index FALSE 2 MAGAZINE II: Considerations…
#> 7 BOOK III: General Principles TRUE 3 BOOK III: General Principles
#> 8 BOOK III: General Principles TRUE 4 BOOK III: General Principles
#> 9 Page one: invitation FALSE 4 BOOK III: General Principle…
Created on 2022-03-25 by the reprex package (v2.0.1)
Here, we first determine if the section
is a section title or a page title and save that as TRUE
or FALSE
.
Then, we label the pages belonging to a section by using cumsum()
(cumulative sum). When we add up TRUE
and FALSE
values, TRUE
(here, sections) become 1
and increment the cumulative sum, but FALSE
(here, pages) become 0
and don't increment the cumulative sum, so all of the pages within a specific section receive the same value.
Lastly, we make a new section variable, this time using group_by()
and if_else()
to conditionally set the value. If isSection
is TRUE
, we just keep the existing value of section
(the section title). If isSection
is FALSE
, we concatenate the first value of section
from the group with the existing value of section
, separated by " / "
.
CodePudding user response:
using data.table:
library(data.table)
setDT(x)[grepl("^Page.",section)==F, header:=section] %>%
.[,header:=zoo::na.locf(header)] %>%
.[section!=header,header:=paste0(header, " / ",section)] %>%
.[,.(section = header)] %>%
.[]
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
CodePudding user response:
A rolling join could achieve this. In data.table:
library( data.table )
# add a row column for joining by reference
x[ , row := .I ]
# pick out just the title rows. It looks like these start with either "BOOK" or "MAGAZINE"
books_magazines <- x[ grepl("^BOOK|^MAGAZINE", section),
.(row, book_magazine = section) ]
# join the 2 tables, using a rolling join to add the title row to subsequent rows
both_cols <- books_magazines[ x, on = .(row), roll = TRUE ]
# concatenate the 2 columns together where necessary, leave it alone if it's the title row
result <- both_cols[ , .(
section_string = fifelse( book_magazine == section,
book_magazine,
sprintf("%s / %s", book_magazine, section) )
) ]
This gives:
> result$section_string
[1] "BOOK I: Introduction"
[2] "BOOK I: Introduction / Page one: presentation"
[3] "BOOK I: Introduction / Page two: acknowledgments"
[4] "MAGAZINE II: Considerations"
[5] "MAGAZINE II: Considerations / Page one: characters"
[6] "MAGAZINE II: Considerations / Page two: index"
[7] "BOOK III: General Principles"
[8] "BOOK III: General Principles"
[9] "BOOK III: General Principles / Page one: invitation"
CodePudding user response:
You can do:
unlist(lapply(split(x$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"
CodePudding user response:
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