How can I do an overlapping sequence count in Biopython?

Question

Biopython's .count() methods, like Python's str.count(), perform a non-overlapping count, how can I do an overlapping one?

For example, these code snippets return 2, but I want the answer 3:

>>> from Bio.Seq import Seq
>>> Seq('AAAA').count('AA')
2
>>> 'AAAA'.count('AA')
2

Answer 1

For Biopython 1.70, there is a new Seq.count_overlap() method, which includes optional start and end arguments:

>>> from Bio.Seq import Seq
>>> Seq('AAAA').count_overlap('AA')
3
>>> Seq('AAAA').count_overlap('AA', 1, 4)
2

This method is also implemented for the MutableSeq and UnknownSeq classes:

>>> from Bio.Seq import MutableSeq, UnknownSeq
>>> MutableSeq('AAAA').count_overlap('AA')
3
>>> UnknownSeq(4, character='A').count_overlap('AA')
3

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Disclaimer: I co-contributed the .count_overlap() methods with Peter Cock, see 97709cc

Answer 2

I've encountered this problem before, and used python re module to solve this problem.

import re
all = re.findall(r'(?=(AA))','AAAA')
counts = len(all)

You can get more details in this thread

Answer 3

You can use finditer from python's re module. The advantage of this approach is it allows for getting the indices of those matches, which could be handy down the track.

>>> import re
>>> matches = re.finditer(r'(?=(AA))', 'AAAA')
>>> indices = [match.span(1) for match in matches]
>>> indices
[(0, 2), (1, 3), (2, 4)]
>>> num_matches = len(indices)
>>> num_matches
3