Ah! That's perfect. And for sorted non-overlapping inputs (separated by
gap_length 0's), I can just bm.appendFill(0, gap_length); bm.appendFill(1,
length); and be done (no OR-ing), right?
Thanks,
Andrew
On Sep 27, 2012, at 12:23 PM, K. John Wu wrote:
> Hi, Andrew,
>
> The statement "ibis::bitvector bv(from,from+length);" could be
> implemented as follows
>
> ibis::bitvector bm;
> bm.appendFill(0, from);
> bm.appendFill(1, length);
> bm.adjustSize(0, max); // bm to represent a total of max bits
>
>
>
> On 9/27/12 8:16 AM, Olson, Andrew wrote:
>> I've used ibis::bitvector::setBit() to set individual bits in a (compressed)
>> bitvector, but I'd like to build a bitvector corresponding to a set of
>> continuous ranges. Is there a convenient way to do this already? I was
>> thinking of creating a bitvector for each continuous range and OR-ing them
>> together as follows:
>>
>> ibis::bitvector mask;
>>
>> int intervals = 10000;
>> int min_span = 100;
>> int max_span = 10000 - min_span;
>> int max=100000000;
>>
>> for (int i=0; i<intervals; i++) {
>> int from = rand() % max;
>> int length = rand() % max_span + min_span;
>> ibis::bitvector bv(from,from+length); // the constructor I'm looking for
>> mask |= bv;
>> }
>>
>> In practice, the list of intervals could be sorted by start position and
>> would usually be non-overlapping. Would this make the OR-ing faster?
>>
>> Andrew
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>>
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