本征：将每一行除以最后一行 [英] Eigen: Divide each row by last row

问题描述

在使用Eigen的 rowwise 操作时，我不太清楚这个语法。

I can't quite figure out the syntax for this when using Eigen's rowwise operations...

我有一个本征矩阵，我想将每一行除以最后一行。因此，如果我们从矩阵开始

I have an Eigen matrix, and I want to divide each row by the last row. So if we started with a matrix

r = [ 0, 1
      2, 3 
      4, 5 ]

然后在进行此转换后，我想拥有

then after this transform, I want to have

r = [  0, .2
      .5, .6
       1,  1 ]

最好执行该操作，覆盖 r 。此外，我不会使用最后一行中的值，因此，最后一行实际上在转换后是否为1都无关紧要。

Preferably the operation would happen in place, overwriting r. Furthermore, I will not be using the values in the last row, so it doesn't matter if the last row is actually 1's after the transform.

以下是我尝试过的一些无法编译的语法：

Here are some syntaxes I have tried that do not compile:

r.rowwise() = (r.array().rowwise() / r.bottomRows(1).array()).eval();
r.rowwise() = (r.rowwise().array() / r.bottomRows(1).array()).eval();
r.rowwise() /= r.bottomRows(1).array();
r = r.rowwise().cwiseQuotient(rrr);

此普通的for循环版本有效

This plain old for-loop version works

int last_row = r.rows() - 1;
for (int row = 0; row < last_row; ++row) {
    r.row(row).array() /= r.row(last_row).array();
}

但是，无论我走到哪里，人们都在倡导使用 rowwise 或 colwise 操作。我无法使用该语法。我想使用 rowwise 运算符做一个简洁的形式吗？

However, everywhere I turn, people are advocating using the rowwise or colwise operations. I can't get this to work with that syntax. Is there a nice concise form of what I want to do using the rowwise operator?

推荐答案

要完成自我回答，如果不需要最后一行，可以使用已标准化：

To complete the self-answer, in case you don't need the last row, you can use hnormalized:

result = r.colwise().hnormalized()

并带有Eigen主干您还可以使用命名空间Eigen :: placeholders :: last;

and with Eigen trunk you can also write:

using namespace Eigen::placeholders::last;
r.array().rowwise() /= r.row(last).array();

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