CODEWITHSUNDEEP
rregex
Michaela
Michaela

Reputation: 39

Extract string between nth occurrence of character and another character

filelist <- c(
  "http://content.caiso.com/green/renewrpt/20171015_DailyRenewablesWatch.txt",
  "http://content.caiso.com/green/renewrpt/20171016_DailyRenewablesWatch.txt",
  "http://content.caiso.com/green/renewrpt/20171017_DailyRenewablesWatch.txt",
  "http://content.caiso.com/green/renewrpt/20171018_DailyRenewablesWatch.txt",
  "http://content.caiso.com/green/renewrpt/20171019_DailyRenewablesWatch.txt",
  "http://content.caiso.com/green/renewrpt/20171020_DailyRenewablesWatch.txt",
  "http://content.caiso.com/green/renewrpt/20171021_DailyRenewablesWatch.txt",
  "http://content.caiso.com/green/renewrpt/20171022_DailyRenewablesWatch.txt"
)

I am looking to extract the string between the 5th occurrence of / and _

Ex: From "http://content.caiso.com/green/renewrpt/20171015_DailyRenewablesWatch.txt" I would want 20171015. I have tried 

regmatches(filelist, regexpr("/{4}([^_]+)", filelist))

but it returns empty.

Upvotes: 1

Views: 2175

Answers (4)

ToWii
ToWii

Reputation: 660

There is a function to get rid of the url first:

filelist <- basename(filelist)

Then try removing all after "_" using str_remove from the stringr package:

library(stringr)
str_remove(filelist, "_.*")

Output:

[1] "20171015" "20171016" "20171017" "20171018" "20171019" "20171020" "20171021" "20171022"

Check the lubridate package's ymd function in case you would like to turn this into a date.

Upvotes: 0

G. Grothendieck
G. Grothendieck

Reputation: 269471

Here are a few approaches which use regular expressions:

sub(".*(\\d{8}).*", "\\1", filelist)

sub(".*/", "", sub("_.*", "", filelist))

sub("_.*", "", basename(filelist))

sapply(strsplit(filelist, "[/_]"), "[", 6)

gsub("\\D", "", filelist)

m <- gregexpr("\\d{8}", filelist)
unlist(regmatches(filelist, m))

strcapture("(\\d{8})", filelist, data.frame(character()))[[1]]

library(gsubfn)
strapplyc(filelist, "\\d{8}", simplify = TRUE)

These solutions do not use regular expressions at all:

substring(filelist, 41, 48)

substring(basename(filelist), 1, 8)

read.table(text = filelist, comment.char = "_", sep = "/")[[6]]

as.Date(basename(filelist), "%Y%m%d")  # returns Date class object

Update: Added a few more approaches.

Upvotes: 1

d.b
d.b

Reputation: 32548

substr(x = filelist,
       start = sapply(gregexpr(pattern = "/", filelist), function(x) x[5])+1,
       stop = sapply(gregexpr(pattern = "_", filelist), function(x) x[1])-1)
#[1] "20171015" "20171016" "20171017" "20171018" "20171019" "20171020" "20171021"
#[8] "20171022"

Upvotes: 0

MrFlick
MrFlick

Reputation: 206187

This should work

gsub("(?:.*/){4}([^_]+)_.*", "\\1", filelist)
# [1] "20171015" "20171016" "20171017" "20171018" "20171019" "20171020" "20171021"
# [8] "20171022"

We need to also match the stuff in front of each of the slashed in the capture.

Upvotes: 4

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