> On 18 Mar 2020, at 14:39, Claude Warren <cla...@xenei.com> wrote:
>
>>> Shape Discussion:
>>>
>>> as for getNumberOfBytes() it should return the maximum number of bytes
>>> returned by a getBits() call to a filter with this shape. So yes, if
> there
>>> is a compressed internal representation, no it won't be that. It is a
>>> method on Shape so it should literally be Math.ceil( getNumberOfBits() /
>>> 8.0 )
>>>
>>> Basically, if you want to create an array that will fit all the bits
>>> returned by BloomFilter.iterator() you need an array of
>>> Shape.getNumberOfBytes(). And that is actually what I use it for.
>
>> Then you are also mapping the index to a byte index and a bit within the
> byte. So if you are doing these two actions then this is something that you
> should control.
>
> BloomFilter.getBits returns a long[]. that long[] may be shorter than the
> absolute number of bytes specified by Shape. It also may be longer.
>
> If you want to create a copy of the byte[] you have to know how long it
> should be. The only way to determine that is from Shape, and currently
> only if you do the Ceil() method noted above. There is a convenience in
> knowing how long (in bytes) the buffer can be.
Copy of what byte[]?
There is no method to create a byte[] for a BloomFilter. So no need for
getNumberOfBytes().
Are you talking about compressing the long[] to a byte[] by truncating the
final long into 1-8 bytes?
BloomFilter bf;
long[] bits = bf.getBits();
ByteBuffer bb = ByteBuffer.allocate(bits.length *
Long.BYTES).order(ByteOrder.LITTLE_ENDIAN);
Arrays.stream(bits).forEachOrdered(bb::putLong);
byte[] bytes = bb.array();
int expected = (int) Math.ceil(bf.getShape().getNumberOfBits() / 8.0);
if (bytes.length != expected) {
bytes = Arrays.copyOf(bytes, expected);
}
For a BloomFilter of any reasonable number of bits the storage saving will be
small.
Is this for serialisation? This is outside of the scope of the library.
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