Replace numeric part of string with addition

Question

I have a large list with more than 100 elements. I would like to add 200 to the ID.

var <- c("ID66Session1_Date-2017-02-28__10-38.csv","ID66Session2_Date-2017-03-04__21-31.csv","ID66Session3_Date-2017-03-07__19-17.csv",
         "ID66Session4_Date-2017-03-09__18-08.csv","ID66Session5_Date-2017-03-15__16-26.csv","ID66Session6_Date-2017-04-15__23-39.csv",
         "ID6Session1_Date-2017-01-20__11-06.csv")

aim <- c("ID266Session1_Date-2017-02-28__10-38.csv","ID266Session2_Date-2017-03-04__21-31.csv","ID266Session3_Date-2017-03-07__19-17.csv",
         "ID266Session4_Date-2017-03-09__18-08.csv","ID266Session5_Date-2017-03-15__16-26.csv","ID266Session6_Date-2017-04-15__23-39.csv",
         "ID206Session1_Date-2017-01-20__11-06.csv")

I believe it's possible to solve this with grepl. But other solutions are also welcomed. I am looking for a handy and short code.

What is the range of the original ID's? Maybe from 1 to 99?

Julius Vainora
– Julius Vainora

2018-12-04 12:51:05 +00:00
Commented Dec 4, 2018 at 12:51 — Julius Vainora
– Julius Vainora, Commented Dec 4, 2018 at 12:51
The range is 1 to 150

SDahm
– SDahm

2018-12-04 12:52:50 +00:00
Commented Dec 4, 2018 at 12:52 — SDahm
– SDahm, Commented Dec 4, 2018 at 12:52

Julius Vainora · Accepted Answer · 2018-12-04 13:06:02Z

2

gsubfn is exactly for that:

library(gsubfn)
gsubfn("((?<=ID)\\d+)", function(x, ...) as.numeric(x) + 200, var, perl = TRUE)
# [1] "ID266Session1_Date-2017-02-28__10-38.csv" "ID266Session2_Date-2017-03-04__21-31.csv"
# [3] "ID266Session3_Date-2017-03-07__19-17.csv" "ID266Session4_Date-2017-03-09__18-08.csv"
# [5] "ID266Session5_Date-2017-03-15__16-26.csv" "ID266Session6_Date-2017-04-15__23-39.csv"
# [7] "ID206Session1_Date-2017-01-20__11-06.csv"

As to avoid constructing a replacement value by "ID" + (200 + ID value) + "Session...", we use a positive lookbehind: If there is "ID" right behind us, we take all the subsequent digits. Those are then converted to a number and 200 is added.

answered Dec 4, 2018 at 13:06

Julius Vainora

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Maurits Evers · Accepted Answer · 2018-12-04 13:10:57Z

Here is a solution using stringr and purrr

library(stringr)
library(purrr)
map_chr(map(str_split(var, "(?<=ID)"), function(x)
    unlist(str_split(x, "(?=Session)"))), function(y)
        paste0(y[1], as.numeric(y[2]) + 200, y[3]))
#[1] "ID266Session1_Date-2017-02-28__10-38.csv"
#[2] "ID266Session2_Date-2017-03-04__21-31.csv"
#[3] "ID266Session3_Date-2017-03-07__19-17.csv"
#[4] "ID266Session4_Date-2017-03-09__18-08.csv"
#[5] "ID266Session5_Date-2017-03-15__16-26.csv"
#[6] "ID266Session6_Date-2017-04-15__23-39.csv"
#[7] "ID206Session1_Date-2017-01-20__11-06.csv"

Explanation: We split var into three parts using a positive look-behind and look-ahead: part 1 ending in ID, part 2 between (and excluding) ID and Session (i.e. the relevant part containing the ID number), and part 3 starting with Session. We then convert the middle part as.numeric and add 200 before concatenating all three parts.

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