How can I extract a string rowwise using regex?

Question

I have a column filename in a dataframe that looks like this:

/testData/THQ/TAIRATE.20030314.190000.tif
/testData/THQ/TAIRATE.20030314.200000.tif
/testData/THQ/TAIRATE.20030314.210000.tif
/testData/THQ/TAIRATE.20030314.220000.tif

And I want to extract the timestamp from this and store it as another column. But I am not familiar with Regex. So far I have gotten to this:

tdat %>%
  dplyr::rowwise() %>% 
  dplyr::mutate(timestamp = str_extract(as.character(filename), "[^//TAIRATE]+$")) %>% 
  glimpse()

Result

.20030314.190000.tif
.20030314.200000.tif
.20030314.210000.tif
.20030314.220000.tif

Expected result

20030314190000
20030314200000
20030314210000
20030314220000

Question: How can I write the correct regex or is there a better way?

Try `str_extract(as.character(filename), "(?<=TAIRATE\\.)\\d+")` — Wiktor Stribiżew, Apr 24 '20 at 09:49
@WiktorStribiżew Thank you! But this is removing everything after the dot. Getting `20030314` — maximusdooku, Apr 24 '20 at 09:51
True, then what about `str_replace(as.character(filename), ".*TAIRATE\\.(\\d+)\.(\\d+).*", "\\1\\2")`? This will have a side effect: if there is no match found, you will end up with the whole file name unchanged. — Wiktor Stribiżew, Apr 24 '20 at 09:52

score 1 · Answer 1 · answered Apr 24 '20 at 09:52

1

str_extract and other such functions are vectorized you don't need row-wise.

In this case, you can do this in base R using sub.

sub('.*TAIRATE\\.(\\d+)\\.(\\d+).*', '\\1\\2', df$filename)
#[1] "20030314190000" "20030314200000" "20030314210000" "20030314220000"

answered Apr 24 '20 at 09:52

Ronak Shah

377,200
20
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@maximusdooku Same note as in [my comment](https://stackoverflow.com/questions/61405549/how-can-i-extract-a-string-rowwise-using-regex#comment108626908_61405549): This will have a side effect: if there is no match found, you will end up with the whole file name unchanged – Wiktor Stribiżew Apr 24 '20 at 09:55
@WiktorStribiżew Thank you for that cautionary note! Also, as a sidenote - what's a good resource to learn just enough about regex to make it work? I always avoided learning it. – maximusdooku Apr 24 '20 at 09:57
@maximusdooku I do not know your level of regex knowledge :) so that I can only suggest doing all lessons at [regexone.com](http://regexone.com/), reading through [regular-expressions.info](http://www.regular-expressions.info), [regex SO tag description](http://stackoverflow.com/tags/regex/info) (with many other links to great online resources), and the community SO post called [What does the regex mean](http://stackoverflow.com/questions/22937618/reference-what-does-this-regex-mean). Also, [rexegg.com](http://rexegg.com) is worth having a look at. – Wiktor Stribiżew Apr 24 '20 at 10:03
@maximusdooku For R, also see [this answer of mine](https://stackoverflow.com/a/47251004/3832970). – Wiktor Stribiżew Apr 24 '20 at 10:03

score 1 · Answer 2 · answered Apr 24 '20 at 10:17

1

Certainly less elegant than @akrun's solution but this one works too:

paste0(unlist(str_extract_all(filename, "[0-9]+")), collapse = "")

Data:

filename <- "/testData/THQ/TAIRATE.20030314.190000.tif"

answered Apr 24 '20 at 10:17

Chris Ruehlemann

20,321
4
12
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How can I extract a string rowwise using regex?

Result

Expected result

2 Answers2