一个C ++数组的matlab风格索引

问题描述

我想知道以下matlab惯用法的最有效的c ++实现是什么。

假设我在matlab，x，y和idx中有3个向量。

  x = [13,2,5.5,22,107] 
y = [-3,100,200] 
 idx = [1,2,5] 
  

我要将x的位置1,2和5替换为y的内容。在matlab中，我做

  x [idx] = y 
  

在c ++中最好的做法是什么？

解决方案

图书馆可能是最接近的，因为它的目标之一是使事情容易的人习惯到Matlab。

下面是一个简短示例（和 uvec 是unsigned ints的向量的typdef）

  //将X的四个特定元素设置为1 
 uvec indices; 
 indices<< 2<< 3<< 6<< 8; 
 
 X.elem（indices）= ones< vec>（4）; 
  

显然，右侧可以是与索引维度相同的任何其他向量。 / p>

但是，你不能克服的语言约束限制很少：

• （你可以改变，但很少C / C ++程序员会认为这是一个好主意）


• 某些操作符，包括 [

I would like to know what the most efficient c++ implementation of the following matlab idiom is.

Suppose I have 3 vectors in matlab, x, y and idx.

x = [13,2,5.5,22,107]
y = [-3,100,200]
idx = [1,2,5]

I want to replace positions 1,2 and 5 of x with the contents of y. In matlab I do

x[idx] = y

What is the best way to do this in c++?

解决方案

The Armadillo library probably comes closest as one of its goals is to make things easy for folks used to Matlab.

Here is a short example (and uvec is a typdef for vector of unsigned ints)

// set four specific elements of X to 1
uvec indices;
indices << 2 << 3 << 6 << 8;

X.elem(indices) = ones<vec>(4);

Obviously, the right-hand side could be any other vector of the same dimension as the index.

But there are few language-imposed constraints you cannot overcome:

• zero-based indexing at the C++ level (which you could alter, but few C / C++ programmers will consider it a good idea)
• certain operators, including [

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