在 C++ 中创建稀疏数组的最佳方法是什么?

问题描述

我正在处理一个需要处理巨大矩阵的项目，特别是用于 copula 计算的金字塔求和.

I am working on a project that requires the manipulation of enormous matrices, specifically pyramidal summation for a copula calculation.

简而言之，我需要在矩阵(多维数组)的零海中跟踪相对较少的值(通常为 1，在极少数情况下大于 1).

In short, I need to keep track of a relatively small number of values (usually a value of 1, and in rare cases more than 1) in a sea of zeros in the matrix (multidimensional array).

稀疏数组允许用户存储少量值，并假设所有未定义的记录都是预设值.由于物理上不可能将所有值存储在内存中，因此我只需要存储少数非零元素.这可能是几百万个条目.

A sparse array allows the user to store a small number of values, and assume all undefined records to be a preset value. Since it is not physically possibly to store all values in memory, I need to store only the few non-zero elements. This could be several million entries.

速度是重中之重，我还想在运行时动态选择类中的变量数量.

Speed is a huge priority, and I would also like to dynamically choose the number of variables in the class at runtime.

我目前在一个使用二叉搜索树 (b-tree) 存储条目的系统上工作.有人知道更好的系统吗?

I currently work on a system that uses a binary search tree (b-tree) to store entries. Does anyone know of a better system?

推荐答案

对于 C++，地图效果很好.数百万个对象不会有问题.在我的电脑上处理 1000 万个项目需要大约 4.4 秒和大约 57 meg.

For C++, a map works well. Several million objects won't be a problem. 10 million items took about 4.4 seconds and about 57 meg on my computer.

我的测试应用如下:

#include <stdio.h>
#include <stdlib.h>
#include <map>

class triple {
public:
    int x;
    int y;
    int z;
    bool operator<(const triple &other) const {
        if (x < other.x) return true;
        if (other.x < x) return false;
        if (y < other.y) return true;
        if (other.y < y) return false;
        return z < other.z;
    }
};

int main(int, char**)
{
    std::map<triple,int> data;
    triple point;
    int i;

    for (i = 0; i < 10000000; ++i) {
        point.x = rand();
        point.y = rand();
        point.z = rand();
        //printf("%d %d %d %d
", i, point.x, point.y, point.z);
        data[point] = i;
    }
    return 0;
}

现在要动态选择变量的个数，最简单的办法就是将index表示为字符串，然后用字符串作为map的key.例如，位于 [23][55] 的项目可以通过23,55"字符串表示.我们还可以将这个解决方案扩展到更高的维度；例如对于三个维度，任意索引看起来像34,45,56".这种技术的简单实现如下:

Now to dynamically choose the number of variables, the easiest solution is to represent index as a string, and then use string as a key for the map. For instance, an item located at [23][55] can be represented via "23,55" string. We can also extend this solution for higher dimensions; such as for three dimensions an arbitrary index will look like "34,45,56". A simple implementation of this technique is as follows:

std::map data<string,int> data;
char ix[100];

sprintf(ix, "%d,%d", x, y); // 2 vars
data[ix] = i;

sprintf(ix, "%d,%d,%d", x, y, z); // 3 vars
data[ix] = i;

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