C ++-生成具有可配置平均值"1s到0s"的随机位集的有效方法.比率 [英] C++ - Efficient way to generate random bitset with configurable mean "1s to 0s" ratio
问题描述
我正在寻找一种高效的方法来生成设定长度的随机std::bitset.我还希望能够影响结果中出现1的概率，因此，如果将概率值设置得足够低，则所有结果中只有一小部分甚至会包含1，但是仍然有可能(但极不可能)导致所有1.它将用于大量计算的应用程序，因此欢迎进行所有可能的优化.
I'm looking for a highly efficient way to generate random std::bitset of set length. I'd also like to be able to influence the probability of 1s appearing in the result, so if the probability value is set low enough, only a small percentage of all the results will even contain a 1, but it's still possible (but very unlikely) to result in all 1s. It's going to be used in a very computation-heavy application, so every possible optimization is welcome.
推荐答案
Bernoulli发行版是单个实验中1或0的概率分布.许多这样的分布式变量的总和
Bernoulli distribution is a probability distribution of 1 or 0 in a single experiment. A sum of many such distributed variables
给出一个以均值 n * p 分布的变量(二项式分布).因此，通过取由 p 给出的概率为 1 的 n 伯努利分布位，我们得到了大小为 n 的位集，并且 np 位平均设置为 1 .当然，如果这不能提供足够的效率，这只是下一步优化的起点.
gives a variable distributed with mean n*p (binomial distribution). So by taking n bernoulli distributed bits with probability of 1 given by p we get a bitset of size n and np bits set to 1 on average. Of course this is just a starting point to optimize next if the efficiency this offers is not enough.
#include <iostream>
#include <random>
#include <bitset>
template< size_t size>
typename std::bitset<size> random_bitset( double p = 0.5) {
typename std::bitset<size> bits;
std::random_device rd;
std::mt19937 gen( rd());
std::bernoulli_distribution d( p);
for( int n = 0; n < size; ++n) {
bits[ n] = d( gen);
}
return bits;
}
int main()
{
for( int n = 0; n < 10; ++n) {
std::cout << random_bitset<10>( 0.25) << std::endl;
}
}
结果:
1010101001
1010101001
0001000000
0001000000
1000000000
1000000000
0110010000
0110010000
1000000000
1000000000
0000110100
0000110100
0001000000
0001000000
0000000000
0000000000
1000010000
1000010000
0101010000
0101010000
这篇关于C ++-生成具有可配置平均值"1s到0s"的随机位集的有效方法.比率的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
