大基数计数的LogLog和HyperLogLog算法

Question

在哪里可以找到有效的LogLog algorithm实现？我试着自己实现它，但我的草案实现产生了奇怪的结果。

Here是这样的：

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());

由于未知的原因，实现对max_error参数非常敏感，它是决定结果大小的主要因素。我敢肯定，肯定有一些愚蠢的错误:)

更新：这个问题在算法的newer version中得到了解决。我将在稍后发布它的实现。

Answer 1

Here它是基于newer paper的算法的更新版本

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) + '%');

Answer 2

这是一个稍微修改过的版本，其中添加了合并操作。

Merge允许您从多个HyperLogLog实例中获取计数器，并确定总体上唯一的计数器。

例如，如果您在周一、周二和周三收集了独立访客，则可以将这些存储桶合并在一起，并计算这三天内的独立访客数量：

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;
}

然后你可以这样做：

// 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 3

我们已经开源了一个名为Stream-Lib的项目，它有一个LogLog implementation。这项工作基于this paper。