Simple counts() method for pairwise alignments

[peterjc]

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Closes #3538, while leaving scope for future enhancement.

[mdehoon]

Just two suggestions:

• It may be better to return a namedtuple instead of a plain tuple, as future versions of this function may return additional numbers (e.g. number of gap opens, number of gap extensions, number of upper-vs-lower matches, etc.). Then users can use the namedtuple keywords to ensure that they are getting the number they think they are getting.
• For multiple (> 2) alignments, it is trivial to extend this function to include all pairs, as in

for i, seq1 in alignment:
    for j, seq2 in alignment:
        if i==j: break
        for a, b in zip(seq1, seq2):
           if a == "-" or b == "-":
                gaps += 1
            elif a == b:
                identities += 1
            else:
                mismatches += 1

[peterjc]

Good idea on using a named tuple, although it is possible a future optional argument might change "identities" into something else like "close matches". Go with the current variable names?

Do you think the counts defined that way would be useful for n>2?

If so, should we exclude double counting - i.e. something equivalent to this:

for i, seq1 in alignment:
    for j, seq2 in alignment:
        # Don't count seq1 vs seq2 and seq2 vs seq1
        if i >= j: break
        for a, b in zip(seq1, seq2):
           if a == "-" or b == "-":
                gaps += 1
            elif a == b:
                identities += 1
            else:
                mismatches += 1

[mdehoon]

Good idea on using a named tuple, although it is possible a future optional argument might change "identities" into something else like "close matches". Go with the current variable names?

Yes; identities is not ambiguous.

Do you think the counts defined that way would be useful for n>2?

I have seen this used before in Biopython, though I can't remember now where. In any case, it does not hurt to allow that possibility.

If so, should we exclude double counting - i.e. something equivalent to this:

I believe that works out to be the same as breaking on i == j.

[peterjc]

You're probably right about the loop. I'll update this shortly...

[peterjc]

Thank you - that looks better than the first iteration.