您好我正在尝试计算一系列整数或浮点数的加权方差和加权标准差。我找到了这些链接：
http://math.tutorvista.com/statistics/standard-deviation.html#weighted-standard-deviation
http://www.itl.nist.gov/div898/software/dataplot/refman2/ch2/weightsd.pdf（警告pdf）
到目前为止，这是我的模板功能。方差和标准差工作正常，但对于我的生活，我无法得到加权版本以匹配pdf底部的测试用例：
template <class T>
inline float Mean( T samples[], int count )
{
float mean = 0.0f;
if( count >= 1 )
{
for( int i = 0; i < count; i++ )
mean += samples[i];
mean /= (float) count;
}
return mean;
}
template <class T>
inline float Variance( T samples[], int count )
{
float variance = 0.0f;
if( count > 1 )
{
float mean = 0.0f;
for( int i = 0; i < count; i++ )
mean += samples[i];
mean /= (float) count;
for( int i = 0; i < count; i++ )
{
float sum = (float) samples[i] - mean;
variance += sum*sum;
}
variance /= (float) count - 1.0f;
}
return variance;
}
template <class T>
inline float StdDev( T samples[], int count )
{
return sqrtf( Variance( samples, count ) );
}
template <class T>
inline float VarianceWeighted( T samples[], T weights[], int count )
{
float varianceWeighted = 0.0f;
if( count > 1 )
{
float sumWeights = 0.0f, meanWeighted = 0.0f;
int numNonzero = 0;
for( int i = 0; i < count; i++ )
{
meanWeighted += samples[i]*weights[i];
sumWeights += weights[i];
if( ((float) weights[i]) != 0.0f ) numNonzero++;
}
if( sumWeights != 0.0f && numNonzero > 1 )
{
meanWeighted /= sumWeights;
for( int i = 0; i < count; i++ )
{
float sum = samples[i] - meanWeighted;
varianceWeighted += weights[i]*sum*sum;
}
varianceWeighted *= ((float) numNonzero)/((float) count*(numNonzero - 1.0f)*sumWeights); // this should be right but isn't?!
}
}
return varianceWeighted;
}
template <class T>
inline float StdDevWeighted( T samples[], T weights[], int count )
{
return sqrtf( VarianceWeighted( samples, weights, count ) );
}
测试用例：
int samples[] = { 2, 3, 5, 7, 11, 13, 17, 19, 23 };
printf( "%.2f\n", StdDev( samples, 9 ) );
int weights[] = { 1, 1, 0, 0, 4, 1, 2, 1, 0 };
printf( "%.2f\n", StdDevWeighted( samples, weights, 9 ) );
结果：
7.46
1.94
应该是：
7.46
5.82
我认为问题在于加权方差有一些不同的解释，我不知道哪一个是标准的。我发现了这种变化：
http://en.wikipedia.org/wiki/Algorithms_for_calculating_variance#Weighted_incremental_algorithm
template <class T>
inline float VarianceWeighted( T samples[], T weights[], int count )
{
float varianceWeighted = 0.0f;
if( count > 1 )
{
float sumWeights = 0.0f, meanWeighted = 0.0f, m2 = 0.0f;
for( int i = 0; i < count; i++ )
{
float temp = weights[i] + sumWeights,
delta = samples[i] - meanWeighted,
r = delta*weights[i]/temp;
meanWeighted += r;
m2 += sumWeights*delta*r; // Alternatively, m2 += weights[i] * delta * (samples[i]−meanWeighted)
sumWeights = temp;
}
varianceWeighted = (m2/sumWeights)*((float) count/(count - 1));
}
return varianceWeighted;
}
结果：
7.46
5.64
我也尝试过看看boost和esutil但是他们没有多大帮助：
http://www.boost.org/doc/libs/1_48_0/boost/accumulators/statistics/weighted_variance.hpp http://esutil.googlecode.com/svn-history/r269/trunk/esutil/stat/util.py
我不需要整个统计库，更重要的是，我想了解实现。
有人可以发布功能来正确计算这些吗？
奖励积分，如果您的功能可以一次性完成。
此外，是否有人知道加权方差是否与重复值的普通方差相同？例如，样本[] = {1,2,3,3}的方差是否与样本的加权方差相同[] = {1,2,3}，权重[] = {1,1,2}吗
更新：这是我设置的谷歌文档电子表格，用于探索问题。不幸的是，我的答案与NIST pdf无关。我认为问题出在unbias步骤，但我看不出如何解决它。
https://docs.google.com/spreadsheet/ccc?key=0ApzPh5nRin0ldGNNYjhCUTlWTks2TGJrZW4wQUcyZnc&usp=sharing
结果的加权方差为3.77，这是我在c ++代码中得到的加权标准差为1.94的平方。
我正在尝试在我的Mac OS X设置上安装八度音，这样我就可以使用权重运行他们的var（）函数，但是它需要永远用brew安装它。我现在已深入yak shaving。
答案 0 :(得分：3)
float mean(uint16_t* x, uint16_t n) {
uint16_t sum_xi = 0;
int i;
for (i = 0; i < n; i++) {
sum_xi += x[i];
}
return (float) sum_xi / n;
}
/**
* http://www.itl.nist.gov/div898/software/dataplot/refman2/ch2/weigmean.pdf
*/
float weighted_mean(uint16_t* x, uint16_t* w, uint16_t n) {
int sum_wixi = 0;
int sum_wi = 0;
int i;
for (i = 0; i < n; i++) {
sum_wixi += w[i] * x[i];
sum_wi += w[i];
}
return (float) sum_wixi / (float) sum_wi;
}
float variance(uint16_t* x, uint16_t n) {
float mean_x = mean(x, n);
float dist, dist2;
float sum_dist2 = 0;
int i;
for (i = 0; i < n; i++) {
dist = x[i] - mean_x;
dist2 = dist * dist;
sum_dist2 += dist2;
}
return sum_dist2 / (n - 1);
}
/**
* http://www.itl.nist.gov/div898/software/dataplot/refman2/ch2/weighvar.pdf
*/
float weighted_variance(uint16_t* x, uint16_t* w, uint16_t n) {
float xw = weighted_mean(x, w, n);
float dist, dist2;
float sum_wi_times_dist2 = 0;
int sum_wi = 0;
int n_prime = 0;
int i;
for (i = 0; i < n; i++) {
dist = x[i] - xw;
dist2 = dist * dist;
sum_wi_times_dist2 += w[i] * dist2;
sum_wi += w[i];
if (w[i] > 0)
n_prime++;
}
if (n_prime > 1) {
return sum_wi_times_dist2 / ((float) ((n_prime - 1) * sum_wi) / n_prime);
} else {
return 0.0f;
}
}
/**
* http://en.wikipedia.org/wiki/Algorithms_for_calculating_variance#Weighted_incremental_algorithm
*/
float weighted_incremental_variance(uint16_t* x, uint16_t* w, uint16_t n) {
uint16_t sumweight = 0;
float mean = 0;
float M2 = 0;
int n_prime = 0;
uint16_t temp;
float delta;
float R;
int i;
for (i = 0; i < n; i++) {
if (w[i] == 0)
continue;
temp = w[i] + sumweight;
delta = x[i] - mean;
R = delta * w[i] / temp;
mean += R;
M2 += sumweight * delta * R;
sumweight = temp;
n_prime++;
}
if (n_prime > 1) {
float variance_n = M2 / sumweight;
return variance_n * n_prime / (n_prime - 1);
} else {
return 0.0f;
}
}
void main(void) {
uint16_t n = 9;
uint16_t x[] = { 2, 3, 5, 7, 11, 13, 17, 19, 23 };
uint16_t w[] = { 1, 1, 0, 0, 4, 1, 2, 1, 0 };
printf("%f\n", weighted_variance(x, w, n)); /* outputs: 33.900002 */
printf("%f\n", weighted_incremental_variance(x, w, n)); /* outputs: 33.900005 */
}
答案 1 :(得分：0)
您在方差更新字词的分母中意外添加了一个额外的术语“计数”。
使用下面的更正时，我会得到您预期的答案
5.82
仅供参考，在进行代码审查时，采用这种方法的一种方法是进行“维度分析”。等式的“单位”是错误的。当你应该是一个N阶项时，你实际上除以了一个N阶平方项。
template <class T>
inline float VarianceWeighted( T samples[], T weights[], int count )
{
...
varianceWeighted *= ((float) numNonzero)/((float) count*(numNonzero - 1.0f)*sumWeights); // this should be right but isn't?!
...
}
删除“计数”此行应替换为
template <class T>
inline float VarianceWeighted( T samples[], T weights[], int count )
{
...
varianceWeighted *= ((float) numNonzero)/((float) (numNonzero - 1.0f)*sumWeights); // removed count term
...
}
答案 2 :(得分：0)
这是一个简短得多的版本，具有有效的 Demo ：
#include <iostream>
#include <vector>
#include <boost/accumulators/accumulators.hpp>
#include <boost/accumulators/statistics/stats.hpp>
#include <boost/accumulators/statistics/weighted_variance.hpp>
#include <boost/accumulators/statistics/variance.hpp>
namespace ba = boost::accumulators;
int main() {
std::vector<double> numbers{2, 3, 5, 7, 11, 13, 17, 19, 23};
std::vector<double> weights{1, 1, 0, 0, 4, 1, 2, 1, 0 };
ba::accumulator_set<double, ba::stats<ba::tag::variance > > acc;
ba::accumulator_set<double, ba::stats<ba::tag::weighted_variance > , double > acc_weighted;
double n = numbers.size();
double N = n;
for(size_t i = 0 ; i<numbers.size() ; i++ ) {
acc ( numbers[i] );
acc_weighted( numbers[i] , ba::weight = weights[i] );
if(weights[i] == 0) {
n=n-1;
}
};
std::cout << "Sample Standard Deviation, s: " << std::sqrt(ba::variance(acc) *N/(N-1)) << std::endl;
std::cout << "Weighted Sample Standard Deviation, s: " << std::sqrt(ba::weighted_variance(acc_weighted)*n/(n-1)) << std::endl;
}
请注意，n必须反映权重非零的样本数，因此要额外添加n=n-1;行。
Sample Standard Deviation, s: 7.45729
Weighted Sample Standard Deviation, s: 5.82237