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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
The text was updated successfully, but these errors were encountered:
scott-griffiths
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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
count
in bitarray that can be efficiently used. It probably makes sense to allow abytealigned
parameter too.The text was updated successfully, but these errors were encountered: