具有行均值的列向量-具有std :: accumulate? [英] column vector with row means -- with std::accumulate?

问题描述

为了尽可能地懒惰，我在矩阵中读为

In an effort to be as lazy as possible I read in a matrix as

vector< vector<double> > data ( rows, vector<double> ( columns ) );

并尝试使用尽可能多的STL好东西.

and try to use as many STL goodies as I can.

接下来我要做的一件事是计算行均值.在C风格的编程中，

One thing I need to do next is to compute the row means. In C-style programming that would be

vector<double> rowmeans( data.size() );
for ( int i=0; i<data.size(); i++ )
    for ( int j=0; j<data[i].size(); j++ )
        rowmeans[i] += data[i][j]/data[i].size();

在在C ++中，如何使用向量视图和gsl_stats_mean计算整数向量的均值?解释说:对于数字向量，您可以在一行中计算向量均值，而无需在每一步都调用size()运算符:

In In C++, how to compute the mean of a vector of integers using a vector view and gsl_stats_mean? it is explained that for a vector of numbers you can compute a vector mean in one line without calling the size() operator at every step:

double mean = std::accumulate(stl_v.begin(), stl_v.end(), 0.0) / stl_v.size();

是否可以在向量的向量上使用这些迭代器?中间形式是

Is it possible to use these iterators over a vector of vectors? An intermediate form is

vector<double> rowmeans( rows );
    for ( int i=0; i<data.size(); i++ )
        rowmeans[i] = std::accumulate(data[i].begin(), data[i].end(), 0.0) / data[i].size();

已经有1条线消失了！但是使用STL函数是否也可以摆脱[i]索引? (在顶层，只需收集行平均值即可.)

already 1 line gone! but using STL functions is it possible to get rid of the [i] index as well? (on the top level it's just a matter of collecting the row means).

推荐答案

std::transform(data.begin(), data.end(), rowmeans.begin(),
    [](std::vector<double> const& d) {
        return std::accumulate(d.begin(), d.end(), 0.0) / d.size();
    });

尽管，我的个人风格将涉及一个命名的lambda或函数，因为我会发现更多的自我证明:

Although, my personal style would involve a named lambda or function, because I would find that more self-documenting:

auto Mean = [](std::vector<double> const& d) { return std::accumulate(d.begin(), d.end(), 0.0) / d.size(); };
std::transform(data.begin(), data.end(), rowmeans.begin(), Mean);

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