Extract a certain part of a string separated with "/"

Question

I have a list of file names

fastq/SRR17405580 
fastq/SRR17405578 
fastq/SRR17405579

I want to change their name to look like

SRR17405580 
SRR17405578 
SRR17405579

Does this answer your question? [How do I specify a dynamic position for the start of substring?](https://stackoverflow.com/questions/3003527/how-do-i-specify-a-dynamic-position-for-the-start-of-substring) — user438383, Aug 30 '22 at 10:30

Darren Tsai · Accepted Answer · 2022-08-27T05:56:34.977

2

You could use basename() to remove all of the path up to and including the last path separator.

x <- c("fastq/SRR17405580", "fastq/SRR17405578", "fastq/SRR17405579")

basename(x)

# [1] "SRR17405580" "SRR17405578" "SRR17405579"

You could also use regex:

sub('.+/(.+)', '\\1', x)

# [1] "SRR17405580" "SRR17405578" "SRR17405579"

edited Aug 27 '22 at 05:56

answered Aug 27 '22 at 05:51

Darren Tsai

score 1 · Answer 2 · answered Aug 27 '22 at 06:10

1

Another option is to use word from stringr package:

library(stringr)
word(x, 2, sep="/")

[1] "SRR17405580" "SRR17405578" "SRR17405579"

answered Aug 27 '22 at 06:10

TarJae

score 1 · Answer 3 · answered Aug 27 '22 at 08:25

1

Another option using str_split from stringr like this:

library(stringr)
str_split(x, '/', simplify = TRUE)[,2]
#> [1] "SRR17405580" "SRR17405578" "SRR17405579"

^{Created on 2022-08-27 with reprex v2.0.2}

answered Aug 27 '22 at 08:25

Quinten

score 1 · Answer 4 · answered Aug 27 '22 at 17:30

1

Using trimws from base R

trimws(x, whitespace = ".*/")
[1] "SRR17405580" "SRR17405578" "SRR17405579"

answered Aug 27 '22 at 17:30

akrun

4 Answers4