javascript - Algoritmos LogLog y HyperLogLog para contar grandes cardinalidades

algorithm counting

Question

¿Dónde puedo encontrar una implementación válida del algoritmo LogLog ? He tratado de implementarlo por mí mismo, pero mi proyecto de implementación arroja resultados extraños.

Here está:

function LogLog(max_error, max_count)
{
function log2(x)
{
return Math.log(x) / Math.LN2;
}
var m = 1.30 / max_error;
var k = Math.ceil(log2(m * m));
m = Math.pow(2, k);
var k_comp = 32 - k;
var l = log2(log2(max_count / m));
if (isNaN(l)) l = 1; else l = Math.ceil(l);
var l_mask = ((1 << l) - 1) >>> 0;
var M = [];
for (var i = 0; i < m; ++i) M[i] = 0;
function count(hash)
{
if (hash !== undefined)
{
var j = hash >>> k_comp;
var rank = 0;
for (var i = 0; i < k_comp; ++i)
{
if ((hash >>> i) & 1)
{
rank = i + 1;
break;
}
}
M[j] = Math.max(M[j], rank & l_mask);
}
else
{
var c = 0;
for (var i = 0; i < m; ++i) c += M[i];
return 0.79402 * m * Math.pow(2, c / m);
}
}
return {count: count};
}
function fnv1a(text)
{
var hash = 2166136261;
for (var i = 0; i < text.length; ++i)
{
hash ^= text.charCodeAt(i);
hash += (hash << 1) + (hash << 4) + (hash << 7) +
(hash << 8) + (hash << 24);
}
return hash >>> 0;
}
var words = [''aardvark'', ''abyssinian'', ... ,''zoology'']; // about 2 300 words
var log_log = LogLog(0.01, 100000);
for (var i = 0; i < words.length; ++i) log_log.count(fnv1a(words[i]));
alert(log_log.count());

Por razones desconocidas, la implementación es muy sensible al parámetro max_error , es el factor principal que determina la magnitud del resultado. Estoy seguro, hay un error estúpido :)

ACTUALIZACIÓN: este problema se resuelve en la versión más nueva de algoritmo. Publicaré su implementación más tarde.

Answer 1

Aquí hay una versión ligeramente modificada que agrega la operación de fusión.

Merge le permite tomar los contadores de varias instancias de HyperLogLog y determinar los contadores únicos en general.

Por ejemplo, si tiene visitantes únicos que se recogen los lunes, martes y miércoles, puede combinar los segmentos y contar el número de visitantes únicos en el lapso de tres días:

var pow_2_32 = 0xFFFFFFFF + 1;
function HyperLogLog(std_error)
{
function log2(x)
{
return Math.log(x) / Math.LN2;
}
function rank(hash, max)
{
var r = 1;
while ((hash & 1) == 0 && r <= max) { ++r; hash >>>= 1; }
return r;
}
var m = 1.04 / std_error;
var k = Math.ceil(log2(m * m)), k_comp = 32 - k;
m = Math.pow(2, k);
var alpha_m = m == 16 ? 0.673
: m == 32 ? 0.697
: m == 64 ? 0.709
: 0.7213 / (1 + 1.079 / m);
var M = []; for (var i = 0; i < m; ++i) M[i] = 0;
function merge(other)
{
for (var i = 0; i < m; i++)
M[i] = Math.max(M[i], other.buckets[i]);
}
function count(hash)
{
if (hash !== undefined)
{
var j = hash >>> k_comp;
M[j] = Math.max(M[j], rank(hash, k_comp));
}
else
{
var c = 0.0;
for (var i = 0; i < m; ++i) c += 1 / Math.pow(2, M[i]);
var E = alpha_m * m * m / c;
// -- make corrections
if (E <= 5/2 * m)
{
var V = 0;
for (var i = 0; i < m; ++i) if (M[i] == 0) ++V;
if (V > 0) E = m * Math.log(m / V);
}
else if (E > 1/30 * pow_2_32)
E = -pow_2_32 * Math.log(1 - E / pow_2_32);
// --
return E;
}
}
return {count: count, merge: merge, buckets: M};
}
function fnv1a(text)
{
var hash = 2166136261;
for (var i = 0; i < text.length; ++i)
{
hash ^= text.charCodeAt(i);
hash += (hash << 1) + (hash << 4) + (hash << 7) +
(hash << 8) + (hash << 24);
}
return hash >>> 0;
}

Entonces puedes hacer algo como esto:

// initialize one counter per day
var ll_monday = HyperLogLog(0.01);
var ll_tuesday = HyperLogLog(0.01);
var ll_wednesday = HyperLogLog(0.01);
// add 5000 unique values in each day
for(var i=0; i<5000; i++) ll_monday.count(fnv1a('''' + Math.random()));
for(var i=0; i<5000; i++) ll_tuesday.count(fnv1a('''' + Math.random()));
for(var i=0; i<5000; i++) ll_wednesday.count(fnv1a('''' + Math.random()));
// add 5000 values which appear every day
for(var i=0; i<5000; i++) {ll_monday.count(fnv1a(''''+i)); ll_tuesday.count(fnv1a('''' + i)); ll_wednesday.count(fnv1a('''' + i));}
// merge three days together
together = HyperLogLog(0.01);
together.merge(ll_monday);
together.merge(ll_tuesday);
together.merge(ll_wednesday);
// report
console.log(''unique per day: '' + Math.round(ll_monday.count()) + '' '' + Math.round(ll_tuesday.count()) + '' '' + Math.round(ll_wednesday.count()));
console.log(''unique numbers overall: '' + Math.round(together.count()));

Answer 2

Hemos abierto un proyecto llamado Stream-Lib que tiene una implementación LogLog . El trabajo se basó en este documento .

Answer 3

Implementé loglog e hyperloglog en JS y PHP y el código bien comentado https://github.com/buryat/loglog

Answer 4

Sé que esta es una publicación anterior, pero la implementación de @buryat se ha movido, y en cualquier caso es incompleta, y un poco lenta (lo siento o_o).

Tomé la implementación utilizada por la nueva versión de Redis que se puede encontrar here y se la transfirió a PHP. El informe está aquí https://github.com/joegreen0991/HyperLogLog

<?php
class HyperLogLog {
private $HLL_P_MASK;
private $HLL_REGISTERS;
private $ALPHA;
private $registers;
public function __construct($HLL_P = 14)
{
$this->HLL_REGISTERS = (1 << $HLL_P); /* With P=14, 16384 registers. */
$this->HLL_P_MASK = ($this->HLL_REGISTERS - 1); /* Mask to index register. */
$this->ALPHA = 0.7213 / (1 + 1.079 / $this->HLL_REGISTERS);
$this->registers = new SplFixedArray($this->HLL_REGISTERS);
for ($i = 0; $i < $this->HLL_REGISTERS; $i++) {
$this->registers[$i] = 0;
}
}
public function add($v)
{
$h = crc32(md5($v));
$h |= 1 << 63; /* Make sure the loop terminates. */
$bit = $this->HLL_REGISTERS; /* First bit not used to address the register. */
$count = 1; /* Initialized to 1 since we count the "00000...1" pattern. */
while(($h & $bit) == 0) {
$count++;
$bit <<= 1;
}
/* Update the register if this element produced a longer run of zeroes. */
$index = $h & $this->HLL_P_MASK; /* Index a register inside registers. */
if ($this->registers[$index] < $count) {
$this->registers[$index] = $count;
}
}
public function export()
{
$str = '''';
for ($i = 0; $i < $this->HLL_REGISTERS; $i++) {
$str .= chr($this->registers[$i]);
}
return $str;
}
public function import($str)
{
for ($i = 0; $i < $this->HLL_REGISTERS; $i++) {
$this->registers[$i] = isset($str[$i]) ? ord($str[$i]) : 0;
}
}
public function merge($str)
{
for ($i = 0; $i < $this->HLL_REGISTERS; $i++) {
if(isset($str[$i]))
{
$ord = ord($str[$i]);
if ($this->registers[$i] < $ord) {
$this->registers[$i] = $ord;
}
}
}
}
/**
* @static
* @param $arr
* @return int Number of unique items in $arr
*/
public function count() {
$E = 0;
$ez = 0;
for ($i = 0; $i < $this->HLL_REGISTERS; $i++) {
if ($this->registers[$i] !== 0) {
$E += (1.0 / pow(2, $this->registers[$i]));
} else {
$ez++;
$E += 1.0;
}
}
$E = (1 / $E) * $this->ALPHA * $this->HLL_REGISTERS * $this->HLL_REGISTERS;
/* Use the LINEARCOUNTING algorithm for small cardinalities.
* For larger values but up to 72000 HyperLogLog raw approximation is
* used since linear counting error starts to increase. However HyperLogLog
* shows a strong bias in the range 2.5*16384 - 72000, so we try to
* compensate for it. */
if ($E < $this->HLL_REGISTERS * 2.5 && $ez != 0) {
$E = $this->HLL_REGISTERS * log($this->HLL_REGISTERS / $ez);
}
else if ($this->HLL_REGISTERS == 16384 && $E < 72000) {
// We did polynomial regression of the bias for this range, this
// way we can compute the bias for a given cardinality and correct
// according to it. Only apply the correction for P=14 that''s what
// we use and the value the correction was verified with.
$bias = 5.9119 * 1.0e-18 * ($E*$E*$E*$E)
-1.4253 * 1.0e-12 * ($E*$E*$E)+
1.2940 * 1.0e-7 * ($E*$E)
-5.2921 * 1.0e-3 * $E+
83.3216;
$E -= $E * ($bias/100);
}
return floor($E);
}
}

Answer 5

Utilizando la versión js @actual proporcionada, traté de implementar lo mismo en C #, que parece lo suficientemente cerca. Acabo de cambiar la función fnv1a un poco y le cambié el nombre a getHashCode. (El crédito va a la función hash de Jenkins, http://en.wikipedia.org/wiki/Jenkins_hash_function )

public class HyperLogLog
{
private double mapSize, alpha_m, k;
private int kComplement;
private Dictionary<int, int> Lookup = new Dictionary<int, int>();
private const double pow_2_32 = 4294967297;
public HyperLogLog(double stdError)
{
mapSize = (double)1.04 / stdError;
k = (long)Math.Ceiling(log2(mapSize * mapSize));
kComplement = 32 - (int)k;
mapSize = (long)Math.Pow(2, k);
alpha_m = mapSize == 16 ? (double)0.673
: mapSize == 32 ? (double)0.697
: mapSize == 64 ? (double)0.709
: (double)0.7213 / (double)(1 + 1.079 / mapSize);
for (int i = 0; i < mapSize; i++)
Lookup[i] = 0;
}
private static double log2(double x)
{
return Math.Log(x) / 0.69314718055994530941723212145818;//Ln2
}
private static int getRank(uint hash, int max)
{
int r = 1;
uint one = 1;
while ((hash & one) == 0 && r <= max)
{
++r;
hash >>= 1;
}
return r;
}
public static uint getHashCode(string text)
{
uint hash = 0;
for (int i = 0, l = text.Length; i < l; i++)
{
hash += (uint)text[i];
hash += hash << 10;
hash ^= hash >> 6;
}
hash += hash << 3;
hash ^= hash >> 6;
hash += hash << 16;
return hash;
}
public int Count()
{
double c = 0, E;
for (var i = 0; i < mapSize; i++)
c += 1d / Math.Pow(2, (double)Lookup[i]);
E = alpha_m * mapSize * mapSize / c;
// Make corrections & smoothen things.
if (E <= (5 / 2) * mapSize)
{
double V = 0;
for (var i = 0; i < mapSize; i++)
if (Lookup[i] == 0) V++;
if (V > 0)
E = mapSize * Math.Log(mapSize / V);
}
else
if (E > (1 / 30) * pow_2_32)
E = -pow_2_32 * Math.Log(1 - E / pow_2_32);
// Made corrections & smoothen things, or not.
return (int)E;
}
public void Add(object val)
{
uint hashCode = getHashCode(val.ToString());
int j = (int)(hashCode >> kComplement);
Lookup[j] = Math.Max(Lookup[j], getRank(hashCode, kComplement));
}
}

Answer 6

Here está la versión actualizada del algoritmo basada en el documento más nuevo :

var pow_2_32 = 0xFFFFFFFF + 1;
function HyperLogLog(std_error)
{
function log2(x)
{
return Math.log(x) / Math.LN2;
}
function rank(hash, max)
{
var r = 1;
while ((hash & 1) == 0 && r <= max) { ++r; hash >>>= 1; }
return r;
}
var m = 1.04 / std_error;
var k = Math.ceil(log2(m * m)), k_comp = 32 - k;
m = Math.pow(2, k);
var alpha_m = m == 16 ? 0.673
: m == 32 ? 0.697
: m == 64 ? 0.709
: 0.7213 / (1 + 1.079 / m);
var M = []; for (var i = 0; i < m; ++i) M[i] = 0;
function count(hash)
{
if (hash !== undefined)
{
var j = hash >>> k_comp;
M[j] = Math.max(M[j], rank(hash, k_comp));
}
else
{
var c = 0.0;
for (var i = 0; i < m; ++i) c += 1 / Math.pow(2, M[i]);
var E = alpha_m * m * m / c;
// -- make corrections
if (E <= 5/2 * m)
{
var V = 0;
for (var i = 0; i < m; ++i) if (M[i] == 0) ++V;
if (V > 0) E = m * Math.log(m / V);
}
else if (E > 1/30 * pow_2_32)
E = -pow_2_32 * Math.log(1 - E / pow_2_32);
// --
return E;
}
}
return {count: count};
}
function fnv1a(text)
{
var hash = 2166136261;
for (var i = 0; i < text.length; ++i)
{
hash ^= text.charCodeAt(i);
hash += (hash << 1) + (hash << 4) + (hash << 7) +
(hash << 8) + (hash << 24);
}
return hash >>> 0;
}
var words = [''aardvark'', ''abyssinian'', ..., ''zoology'']; // 2336 words
var seed = Math.floor(Math.random() * pow_2_32); // make more fun
var log_log = HyperLogLog(0.065);
for (var i = 0; i < words.length; ++i) log_log.count(fnv1a(words[i]) ^ seed);
var count = log_log.count();
alert(count + '', error '' +
(count - words.length) / (words.length / 100.0) + ''%'');