我自己实现了HyperLogLog algorithm. 它工作得很好，但有时我必须获取很多（大约 10k-100k）的 HLL 结构并将它们合并。
我将它们中的每一个都存储为一个位字符串，所以首先我必须将每个位字符串转换为存储桶。由于有很多 HLL，它需要的时间比我想要的要多。
目前，大约 80% 的运行时会为每个 HLL 调用一次这行代码：
my @buckets = map { oct '0b'.$_ } unpack('(a5)1024', $bitstring);
有什么方法可以更快地做到这一点？
如果我们将 HyperLogLog 的定义抛在脑后，任务可以这样解释：假设它$bitstring由 1024 个 5 位计数器组成（因此每个计数器的值最高可达 32），我们必须将其转换为 1024 个整数的数组。
The a denotes arbitrary, zero-padded binary data. Here, you treat that data as ASCII text but it can only contain 1 and 0! This is inefficient as a5 ends up using five bytes. The easiest and most efficient solution would be to store a 8-bit number for each counter, then: my @buckets = unpack 'C1024', $bitstring.
If you only want to store five bits per counter, you end up saving very little memory for very much hassle. You'd have to use something insane like this for a round-trip conversion:
my $bitstring = pack "(b5)1024", map { sprintf "%b", $_ } @buckets;
@buckets = map { oct "0b$_" } unpack "(b5)1024", $bitstring;