Open scott-griffiths opened 7 months ago
So instead of just counting how many zeros or ones, allow bs.count(x) which is roughly equivalent to len(bs.findall(x)).
bs.count(x)
len(bs.findall(x))
There is now a count in bitarray that can be efficiently used. It probably makes sense to allow a bytealigned parameter too.
count
bytealigned
So instead of just counting how many zeros or ones, allow
bs.count(x)which is roughly equivalent tolen(bs.findall(x)).There is now a
countin bitarray that can be efficiently used. It probably makes sense to allow abytealignedparameter too.