Question

对于学校练习我需要使用逐元素函数构建乘法表10x10，并使其尽可能短。这是我写的代码（工作但太长），请为此代码提出一些建议。提前谢谢（：

base=zeros(10);
oneten=[1:1:10];
base(1,:)=1.*oneten;
base(2,:)=2.*oneten;
base(3,:)=3.*oneten;
base(4,:)=4.*oneten;
base(5,:)=5.*oneten;
base(6,:)=6.*oneten;
base(7,:)=7.*oneten;
base(8,:)=8.*oneten;
base(9,:)=9.*oneten;
base(10,:)=10.*oneten

Answer 1

做这样的事情：

(1:10)' * (1:10)

编辑---＆gt;

当N很大时，我测试了Daniel，我，Luis Mendo和David建议的解决方案的速度：

N = 100;    % number of iterations
runtime_a = zeros(N, 1);    % runtime of Daniel's solution
runtime_b = zeros(N, 1);    % runtime of the obvious solution
runtime_c = zeros(N, 1);    % runtime of Luis Mendo's solution
runtime_d = zeros(N, 1);    % runtime of Luis Mendo's solution
runtime_e = zeros(N, 1);    % runtime of David's solution
n = 5000;    % number of elements
one_to_n = 1:n;
for hh = 1:N
  % Solution by Daniel R.
  tic, a = bsxfun(@times, one_to_n, one_to_n'); runtime_a(hh) = toc;
  clear a
  tic, b = one_to_n' * one_to_n; runtime_b(hh) = toc;
  clear b
  % Solution by Luis Mendo
  tic, c = cell2mat(arrayfun(@(x) (x:x:n*x).', one_to_n, 'uni', false)); runtime_c(hh) = toc;
  clear c
  % Solution by Luis Mendo.
  tic, d = cumsum(repmat(one_to_n, [n 1])); runtime_d(hh) = toc;
  clear d
  % Solution by David
  tic, [A, B] = meshgrid(one_to_n); e = A.*B; runtime_e(hh) = toc;
  clear e
end

% Check mean and standard deviation:
mean([runtime_a, runtime_b, runtime_c, runtime_d, runtime_e])
std([runtime_a, runtime_b, runtime_c, runtime_d, runtime_e])

结果是：

% mean:
0.105048900691251   0.188570704057917   0.491929288458701   0.787045026437718   0.979624197407329
% standard deviation:
0.034274873626281   0.077388368324427   0.163983982925543   0.285395301735065   0.372693172505310

所以，显然，当N很大时，Daniel的解决方案是最快的。

Answer 2

只是为了它的乐趣：其他可能性

```
cumsum(repmat(1:10,[10 1]))
```

cell2mat(arrayfun(@(n) (n:n:10*n).',1:10,'uni',false))

Answer 3

oneten=[(1:10)]
base = bsxfun(@times,oneten,oneten')

在这种情况下，

预分配（base=zeros(10);）是无效的。

另一个更容易理解的解决方案：

base=zeros(10);
oneten=[(1:10)];
for k=oneten
    base(k,:)=k.*oneten;
end

Answer 4

[A,B]=meshgrid(1:10);
A.*B

使用逐元素乘法

Answer 5

我的回答是使用嵌套for循环：

for i = (1:10)
    for j = (1:10) 
        fprintf('%d\t',i*j); 
    end 
    fprintf('\n'); 
end

在matlab中构建乘法表

5 个答案: