R 与 Rcpp 中的递归均值 [英] Recursive Mean in R vs. Rcpp

问题描述

我正在尝试使用简单的递归实现来计算变量的平均值:

I am trying to compute the mean of a variable using a simple, recursive implementation:

m <- 0 # initialize mean

for(irep in 0:999){

# new data point
new_data <- rnorm(1,2,1)

# recursive formula for sample mean
m = (irep/(irep+1)) * m + (1/(irep+1)) * new_data

}

在这里，m 将快速收敛到 2，这对应于我们从中生成新数据点的正态分布的均值.在 Rcpp 中实现类似的东西:

Here, m will quickly converge to 2, which corresponds to the mean of the normal distribution where we generate new data points from. Implementing something similar in Rcpp:

#include <RcppArmadillo.h>

// [[Rcpp::depends(RcppArmadillo)]]

using namespace Rcpp;

// [[Rcpp::export]]
double  rec_mean(int sample_size){

  double m = 0; //initialize mean

  for(int irep = 0; irep < sample_size; irep++){

      // new data
      double new_data = R::rnorm(2,1);

      // mean recursive update
      m = ((irep)/(irep+1)) * m + (1/(irep+1)) * new_data;

  }

  return m;

}

此代码未显示预期行为.相反，它返回初始值.有人能告诉我从 R 到 Rcpp 的翻译错误在哪里吗?

This code does not show the expected behavior. Instead, it returns the initial value. Can somebody enlighten me where my error in this translation from R to Rcpp is?

推荐答案

在这一行:

m = ((irep)/(irep+1)) * m + (1/(irep+1)) * new_data;

您将一个 int 除以另一个 int 两次.在 C++ 中，整数除法返回另一个整数，丢弃余数.

You are dividing an int by another int twice. In C++, integer division returns another integer, discarding the remainder.

为了得到你想要的，强制以浮点形式进行除法:

To get what you want, force the divisions to be done in floating point:

m = ((irep)/(irep+1.0)) * m + (1.0/(irep+1.0)) * new_data;

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