Matlab:创建行向量相同的矩阵.使用repmat()或乘以ones() [英] Matlab: create matrix whose rows are identical vector. Use repmat() or multiply by ones()

问题描述

我想通过将向量连接到自身n次来从向量创建矩阵.因此，如果我的向量是mx1，那么我的矩阵将是mxn，并且矩阵的每一列都将等于该向量.

I want to create a matrix from a vector by concatenating the vector onto itself n times. So if my vector is mx1, then my matrix will be mxn and each column of the matrix will be equal to the vector.

以下哪个是最好/正确的方法，或者还有我不知道的更好的方法?

Which of the following is the best/correct way, or maybe there is a better way I do not know?

matrix = repmat(vector, 1, n);
matrix = vector * ones(1, n);

谢谢

推荐答案

以下是使用

Here is some benchmarking using timeit with different vector sizes and repetition factors. The results to be shown are for Matlab R2015b on Windows.

首先为每种考虑的方法定义一个函数:

First define a function for each of the considered approaches:

%// repmat approach
function matrix = f_repmat(vector, n)
matrix = repmat(vector, 1, n);

%// multiply approach
function matrix = f_multiply(vector, n)
matrix = vector * ones(1, n);

%// indexing approach
function matrix = f_indexing(vector,n)
matrix = vector(:,ones(1,n));

然后生成不同大小的向量，并使用不同的重复因子:

Then generate vectors of different size, and use different repetition factors:

M = round(logspace(2,4,15)); %// vector sizes
N = round(logspace(2,3,15)); %// repetition factors
time_repmat   = NaN(numel(M), numel(N)); %// preallocate results
time_multiply = NaN(numel(M), numel(N));
time_indexing = NaN(numel(M), numel(N));
for ind_m = 1:numel(M);
    for ind_n = 1:numel(N);
        vector = (1:M(ind_m)).';
        n = N(ind_n);
        time_repmat(ind_m, ind_n)   = timeit(@() f_repmat(vector, n)); %// measure time
        time_multiply(ind_m, ind_n) = timeit(@() f_multiply(vector, n));
        time_indexing(ind_m, ind_n) = timeit(@() f_indexing(vector, n));
    end
end

结果绘制在以下两个图中，使用repmat作为参考:

The results are plotted in the following two figures, using repmat as reference:

figure
imagesc(time_multiply./time_repmat)
set(gca, 'xtick',1:2:numel(N), 'xticklabels',N(1:2:end))
set(gca, 'ytick',1:2:numel(M), 'yticklabels',M(1:2:end))
title('Time of multiply / time of repmat')
axis image
colorbar

figure
imagesc(time_indexing./time_repmat)
set(gca, 'xtick',1:2:numel(N), 'xticklabels',N(1:2:end))
set(gca, 'ytick',1:2:numel(M), 'yticklabels',M(1:2:end))
title('Time of indexing / time of repmat')
axis image
colorbar

也许更好的比较是针对每个测试的向量大小和重复因子，三种方法中最快的一种:

Perhaps a better comparison is to indicate, for each tested vector size and repetition factor, which of the three approaches is the fastest:

figure
times = cat(3, time_repmat, time_multiply, time_indexing);
[~, fastest] = min(times, [], 3);
imagesc(fastest)
set(gca, 'xtick',1:2:numel(N), 'xticklabels',N(1:2:end))
set(gca, 'ytick',1:2:numel(M), 'yticklabels',M(1:2:end))
title('1: repmat is fastest; 2: multiply is; 3: indexing is')    
axis image
colorbar

一些结论可以从图中得出:

• 基于乘法的方法总是比repmat
慢
• 基于索引的方法类似于repmat.对于向量大小或重复因子较大的值，它往往更快，而对于较小的值，它的速度往往更慢.

• The multiply-based approach is always slower than repmat
• The indexing-based approach is similar to repmat. It tends to be faster for large values of vector size or repetition factor, and slower for small values.

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