Question

Here is my problem:

data

set.seed(123) 
randDNA = function(n) paste(sample(c("A", "C", "T", "G"), n, replace = TRUE), collapse = "")            
bigse <- randDNA(50000000) 

I want to see how many times the following small five letters match ("ATTGG", "TTTTT", "CCCCC", "ATATG"), in this large single unit vector bigse. Randomly generated just for an example

I tried Biostrings, unsuccessful but base R solution are also welcome.

require(Biostrings)            
pmatch (c("ATTGG", "TTTTT", "CCCCC", "ATATG"), bigse)
[1] NA NA NA NA

I believe that it is matching from one end, return NA. What I want is to these character can match to anywhere but need in the same sequence. If they are repeted twice, I need that information too…thanks…

Answer 1

I would be surprised if pkg:Biostrings had masked the base function pmatch, and I am guessing you are seeing the expected behavior of base::pmatch. Like joran, I thought your 50 million length test was excessive to I paired it down to something where the output of my successful use of the matchPattern “Biostrings” call would fit on this page, but do scroll past it for the full test. It was surprisingly fast on the full test. Much faster than constructing the string in fact.

I looked at the differences with stringr and see that it relates to how one counts matches with the highly duplicated segments. Given that you are working on biological data I think I would accept the Biostrings conventions unless you have specific reasons not to. In that case you should be looking further at the details of the functions and the more complete output

set.seed(123) 
randDNA = function(n) paste(sample(c("A", "C", "T", "G"), 
          n, replace = TRUE), collapse = "")            
bigse <- randDNA(10000)
# There is  a countPattern function that might  narrowly give you what you wanted.
 sapply(c("ATTGG", "TTTTT", "CCCCC", "ATATG"), countPattern, subject=bigse)

#ATTGG TTTTT CCCCC ATATG 
#    2    11     7     6 

Output of the more informative matchPattern with length= 5,000

sapply(c("ATTGG", "TTTTT", "CCCCC", "ATATG"), matchPattern, subject=bigse)
#
$ATTGG
  Views on a 5000-letter BString subject
subject: CGCGGATGTCGCTTAGAACGGTTGTTTTCAG...AGCGGCGAGAAGATTCCGCACGAGTCAGATA
views:
    start  end width
[1]  2035 2039     5 [ATTGG]

$TTTTT
  Views on a 5000-letter BString subject
subject: CGCGGATGTCGCTTAGAACGGTTGTTTTCAG...AGCGGCGAGAAGATTCCGCACGAGTCAGATA
views:
    start  end width
[1]   555  559     5 [TTTTT]
[2]   834  838     5 [TTTTT]
[3]  1905 1909     5 [TTTTT]
[4]  1906 1910     5 [TTTTT]
[5]  4419 4423     5 [TTTTT]

$CCCCC
  Views on a 5000-letter BString subject
subject: CGCGGATGTCGCTTAGAACGGTTGTTTTCAG...AGCGGCGAGAAGATTCCGCACGAGTCAGATA
views:
    start  end width
[1]   798  802     5 [CCCCC]
[2]  3268 3272     5 [CCCCC]
[3]  3629 3633     5 [CCCCC]

$ATATG
  Views on a 5000-letter BString subject
subject: CGCGGATGTCGCTTAGAACGGTTGTTTTCAG...AGCGGCGAGAAGATTCCGCACGAGTCAGATA
views:
    start  end width
[1]  1264 1268     5 [ATATG]
[2]  2924 2928     5 [ATATG]
[3]  3103 3107     5 [ATATG]

But just for kicks I ran it with your big string:

 sapply(c("ATTGG", "TTTTT", "CCCCC", "ATATG"), countPattern, subject=bigse)
# ATTGG TTTTT CCCCC ATATG 
# 48850 48933 49111 49073 

Here are the speed comparisons:

> system.time( sapply(c("ATTGG", "TTTTT", "CCCCC", "ATATG"), 
                   countPattern, subject=bigse) )
   user  system elapsed 
  1.507   0.119   1.618 

> system.time(str_count(bigse,c("ATTGG", "TTTTT", "CCCCC", "ATATG")))
   user  system elapsed 
  6.332   0.017   6.337 

# Added the gregexpr solution timing (not surprising to see similarity with stingr times)
> system.time( sapply(motif,function(x) length(gregexpr(x,bigse)[[1]])) )
   user  system elapsed 
  6.768   0.046   6.794