Is there a way to obtain a list separated by comma as the output of str_extract_all instead of the default output in R?

Question

I have searched high and low and nobody seems to have asked that exact question, so I'm at loss.

I have a data frame with a couple columns. One of this column contains various sentences that don't have a specific format of pattern, which limit how I can extract words from this column because I can't use position. My goal is to search this column and extract species name from the sentences. I need to be able to extract multiple species at once because sometimes the sample has tested + for more than one species and I need that information. My method works fairly well and the output works when there is a single species. The problem is when it identifies more than one species. I would want the output to be something like : sp1,sp2,sp3, but instead I get written: c("sp1","sp2"). I have no clue how I could change that. I tried using toString to no avail. Also, everything is written lower case, so there is no case issue

I tried

df = df %>% mutate(Species=str_extract_all(df$RESULT,"sp1|sp2|sp3|sp4"))

where RESULT is the column that I said contains different sentences. This is what I get as output:

Result	Species
Bla-bla-bla-sp1-bla-sp2	c("sp1","sp2")
Bla-bla-bla-sp3-bla-bla	sp3

But I would want:

Result	Species
Bla-bla-bla-sp1-bla-sp2	sp1, sp2
Bla-bla-bla-sp3-bla-bla	sp3

I tried:

df = df %>% mutate(Species=str_extract_all(toString(df$RESULT,"sp1|sp2|sp3|sp4")))

but the output ended up being the same

Thanks in advance for your help! I now this is not the most clear example, but I can't use my real data as it contains sensitive info Also just so you know, I wrote sp1, sp2 just to mimick species name but my real data doesn't have species all starting with the same letter which really limits the method of extraction I can use. For example it's cat,dog,bird, so methods with sp\d+ won't work, because it's not really species1, species2

In general your approach works, but don't use the data.frame name within mutate, rather try something like this df %>% mutate(Species = str_extract_all(Result, "sp\\d+")). "sp1|sp2|sp3|sp4" should also work. — Andre Wildberg
– Andre Wildberg, Commented Jul 20, 2024 at 18:00
If you want it to be a string try df %>% rowwise() %>% mutate(Species = toString(unlist(str_extract_all(Result, "sp\\d+")))) — Andre Wildberg
– Andre Wildberg, Commented Jul 20, 2024 at 18:06
Seems like paste with collapse=‘,’ should have been the first attempt. And you don’t want a list. You want a character value. — IRTFM
– IRTFM, Commented Jul 20, 2024 at 20:04

mmd356 · Accepted Answer · 2024-07-20 17:48:25Z

1

Try this

df %>% 
  mutate(Species = str_extract_all(Result, "sp1|sp2|sp3|sp4") %>%
           purrr::map_chr(~str_c(.x, collapse = ", ")))

answered Jul 20, 2024 at 17:48

mmd356

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Edward · Accepted Answer · 2024-07-22 03:29:31Z

For a pure base R approach, you can extract all species abbreviations using gregexpr and then regmatches (str_extract_all from stringr simplifies this approach). To collapse multiple extracted strings into one, use paste with collapse=",". Lastly, we wrap this all into lapply to return the results for each row of the data frame.

df$Species <- lapply(regmatches(df$Result, 
                         gregexpr("sp\\d+", df$Result)),
                     paste, collapse=",")
df

df
  id                  Result      Species
1  1 Bla-bla-bla-sp1-bla-sp2      sp1,sp2
2  2 Bla-bla-bla-sp3-bla-bla          sp3
3  3        sp7-bla-sp1-sp11 sp7,sp1,sp11

To sort the vectors in the result, we need to remove the "sp" using gsub, convert to numeric using as.numeric, and then sort the elements using sort.

df$Species <- lapply(
       regmatches(df$Result, 
             gregexpr("sp\\d+", df$Result)),
  \(x) paste0("sp", sort(as.numeric(
                    gsub(x, pattern='sp', replacement=''))),
              collapse=","))
df

  id                  Result      Species
1  1 Bla-bla-bla-sp1-bla-sp2      sp1,sp2
2  2 Bla-bla-bla-sp3-bla-bla          sp3
3  3        sp7-Bla-sp11-sp1 sp1,sp7,sp11

df <- structure(list(id = 1:3, Result = c("Bla-bla-bla-sp1-bla-sp2", 
"Bla-bla-bla-sp3-bla-bla", "sp7-bla-sp1-sp11")), class = "data.frame", row.names = c(NA, 
-3L))

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