我需要您的帮助来解决以下问题:
c++/opencv 中是否有等效于以下代码的函数:
np.random.choice(len(vec), samples, p=probabilities[:,0], replace=True)
提前致谢。
我需要您的帮助来解决以下问题:
c++/opencv 中是否有等效于以下代码的函数:
np.random.choice(len(vec), samples, p=probabilities[:,0], replace=True)
提前致谢。
好吧,让我们看看:(numpy.random.choice(a, size=None, replace=True, p=None)
看我的评论,我猜你混淆了一些函数的参数。)
对于输入a
,您使用的是一组样本。作为您想要的输出大小len(vec)
,您希望进行替换采样并具有自定义的非均匀分布。
首先使用随机分布生成索引数组,然后使用索引数组生成选定元素的数组可能就足够了。
C++ 提供了生成非均匀分布数的帮助,是std::discrete_distribution
例子:
#include <random>
#include <vector>
#include <algorithm>
#include <iostream>
int main()
{
auto const samples = { 1, 2, 3, 4, 5, 6 }; // deducts to std::initializer_list<int>
auto const probabilities = { 0.1, 0.2, 0.1, 0.5, 0.0, 1.0 }; // deducts to std::initializer_list<double>
if (samples.size() < probabilities.size()) {
std::cerr << "If there are more probabilities then samples, you will get out-of-bounds indices = UB!\n";
return -1;
}
// generate non-uniform distribution (default result_type is int)
std::discrete_distribution const distribution{probabilities};
// note, for std::vector or std::array of probabilities, use
// std::discrete_distribution distribution(cbegin(probabilities), cend(probabilities));
int const outputSize = 10;
std::vector<decltype(distribution)::result_type> indices;
indices.reserve(outputSize); // reserve to prevent reallocation
// use a generator lambda to draw random indices based on distribution
std::generate_n(back_inserter(indices), outputSize,
[distribution = std::move(distribution), // could also capture by reference (&) or construct in the capture list
generator = std::default_random_engine{} //pseudo random. Fixed seed! Always same output.
]() mutable { // mutable required for generator
return distribution(generator);
});
std::cout << "Indices: ";
for(auto const index : indices) std::cout << index << " ";
std::cout << '\n';
// just a trick to get the underlying type of samples. Works for std::initializer list, std::vector and std::array
std::vector<decltype(samples)::value_type> output;
output.reserve(outputSize); // reserve to prevent reallocation
std::transform(cbegin(indices), cend(indices),
back_inserter(output),
[&samples](auto const index) {
return *std::next(cbegin(samples), index);
// note, for std::vector or std::array of samples, you can use
// return samples[index];
});
std::cout << "Output samples: ";
for(auto const sample : output) std::cout << sample << " ";
std::cout << '\n';
}
编辑:链接似乎建议执行带替换的std::default_random_engine
采样。
似乎您正在寻找从离散随机分布中采样
该页面上的示例相当具有示范性:
// discrete_distribution
#include <iostream>
#include <random>
int main()
{
const int nrolls = 10000; // number of experiments
const int nstars = 100; // maximum number of stars to distribute
std::default_random_engine generator;
std::discrete_distribution<int> distribution {2,2,1,1,2,2,1,1,2,2};
int p[10]={};
for (int i=0; i<nrolls; ++i) {
int number = distribution(generator);
++p[number];
}
std::cout << "a discrete_distribution:" << std::endl;
for (int i=0; i<10; ++i)
std::cout << i << ": " << std::string(p[i]*nstars/nrolls,'*') << std::endl;
return 0;
}
我不认为有一个功能可以免费为您提供。你可能需要自己写。
关于如何编写这样一个函数的一些提示:
vector<float>
存储你的概率。首先使用std::partial_sum
这个向量来获得元素的累积概率。random_value
。迭代累积概率向量,直到找到大于 的值random_value
。此时的索引是您的示例索引。获取向量中此索引处的值samples
,将其存储在某处并重复。