添加到boost：：累加器的值的数量是否有限制？

Question

对于一个boost：：累加器可以添加多少个值有限制吗？如果添加了大量条目，是否存在累加器停止正常工作的点，或者内部算法是否足够健壮，足以考虑一组接近无穷大的值？

#include <iostream>
#include <boost/accumulators/accumulators.hpp>
#include <boost/accumulators/statistics/stats.hpp>
#include <boost/accumulators/statistics/mean.hpp>
#include <boost/accumulators/statistics/moment.hpp>
using namespace boost::accumulators;

int main()
{
    // Define an accumulator set for calculating the mean and the
    // 2nd moment ...
    accumulator_set<double, stats<tag::mean, tag::moment<2> > > acc;

    // push in some data ...
    for (std::size_t i=0; i<VERY_LARGE_NUMBER; i++)
    {
      acc(i);
    }


    // Display the results ...
    std::cout << "Mean:   " << mean(acc) << std::endl;
    std::cout << "Moment: " << moment<2>(acc) << std::endl;

    return 0;
}

Answer 1

累加器统计信息不考虑溢出，因此需要仔细选择累加器类型。它不需要匹配您要添加的对象的初始类型--您可以在积累时转换它，然后获取统计数据并将其转换回原始类型。

你可以在这个简单例子上看到它

#include <bits/stdc++.h>
#include <boost/accumulators/accumulators.hpp>
#include <boost/accumulators/statistics.hpp>

using namespace boost::accumulators;

int main(void) {
    accumulator_set<int8_t, features<tag::mean>> accInt8;
    accumulator_set<double, features<tag::mean>> accDouble;
    
    int8_t sum = 0; // range of int8_t: -128 to 127
    for (int8_t i = 1; i <= 100; i++) {
        sum += i; // this will overflow!
        accInt8(i); // this will also overflow
        accDouble((double)i);
    }
    
    std::cout << "sum from 1 to 100: " << (int)sum << " (overflow)\n";
    std::cout << "mean(<int8_t>): " << extract::mean(accInt8) << " (overflow)\n";
    std::cout << "mean(<double>): " << (int)extract::mean(accDouble) << "\n";
    
    return 0;
}

我使用了int8_t，它的范围很小(-128到127)，它演示了如果使用int8_t作为accumulator_set的内部类型，从1到100 (应该是50)的平均值就会溢出。

产出如下：

sum from 1 to 100: -70 (overflow)
mean(<int8_t>): -7 (overflow)
mean(<double>): 50