用常量列向量替换矩阵中的特定列 [英] Replace specific columns in a matrix with a constant column vector

问题描述

对于神经网络，我想在矩阵中表示列向量 y = [1;2;3]，如下所示:

For neural networking, I would like to represent a column vector y = [1;2;3] in a matrix like so:

y = [1 0 0;
     0 1 0;
     0 0 1]

我的向量 y 非常大，所以硬编码不是一种选择.另外，我想避免使用 for-loops.

My vector y is very large, and so hardcoding is not an option. Also, I would like to avoid using for-loops.

到目前为止我做了什么:

What I did so far:

y1 =[y; zeros(1,length(y)) ;zeros(1,length(y))] % add two rows with zeros in orde to give y the right format

idx = find(y1(1,:) == 2); % find all the columns containing a 2
y1(:,idx(1):idx(end)) = y1(:,[0;1;0]); % this does not work because now I am comparing a matrix with a vector

我也试过这个:

y1( y1 == [2;0;0] )=[0;1;0]; % This of course does not work 

有没有办法指定我要比较 y1 == [2;0;0] 中的列，或者有其他方法可以解决这个问题?

Is there a way to specify I want to compare columns in y1 == [2;0;0], or is there another way to solve this?

推荐答案

根据您问题的上下文，您希望找到一个矩阵，其中每一列都是一个身份向量.对于单位向量，该矩阵中的每一列都是一个非零向量，其中在由 y 的每个位置表示的向量位置处设置为 1，否则为 0.因此，假设我们有以下示例:

From the context of your question, you wish to find a matrix where each column is an identity vector. For an identity vector, each column in this matrix is a non-zero vector where 1 is set in the position of the vector denoted by each position of y and 0 otherwise. Therefore, let's say we had the following example:

y = [1 5 4 3]

您将 y_out 作为最终矩阵，即:

You would have y_out as the final matrix, which is:

y_out =

     1     0     0     0
     0     0     0     0
     0     0     0     1
     0     0     1     0
     0     1     0     0

<小时>

有几种方法可以做到这一点.最简单的方法是使用 eye 声明单位矩阵，然后让 y 从这个矩阵中挑选出你想要的那些列，并将它们作为列放入你的最终矩阵中.如果 y 具有所有唯一值，那么我们将简单地根据 y 重新排列此单位矩阵的列.因此:

---

There are several ways to do this. The easiest one would be to declare the identity matrix with eye, then let y pick out those columns that you want from this matrix and place them as columns into your final matrix. If y had all unique values, then we would simply be rearranging the columns of this identity matrix based on y. As such:

y_out = eye(max(y));
y_out = y_out(:,y)

y_out =

     1     0     0     0
     0     0     0     0
     0     0     0     1
     0     0     1     0
     0     1     0     0

<小时>

另一种方法是声明一个 sparse 矩阵，其中每个行索引只是来自 y 的那些元素，每个列索引从 1 增加到与 y 一样多的元素:

---

Another way would be to declare a sparse matrix, where each row index is simply those elements from y and each column index is increasing from 1 up to as many elements as we have y:

y_out = sparse(y, 1:numel(y), 1, max(y), numel(y));
y_out = full(y_out)

y_out =

     1     0     0     0
     0     0     0     0
     0     0     0     1
     0     0     1     0
     0     1     0     0

<小时>

另一种方法是使用 sub2ind 找到矩阵的线性索引，然后访问这些元素并将它们设置为 1.因此:

---

One more way would be to use sub2ind to find linear indices into your matrix, then access those elements and set them to 1. Therefore:

ind = sub2ind([max(y) numel(y)], y, 1:numel(y));
y_out = zeros(max(y), numel(y));
y_out(ind) = 1

y_out =

     1     0     0     0
     0     0     0     0
     0     0     0     1
     0     0     1     0
     0     1     0     0

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