Question

我想在MATLAB中创建一个矩阵，其中：

第一行由0和1的随机排列组成，均匀分割（即50-50）。

第二行随机分配零到第一行中0和1的50％，以及剩余50％中的0。

第三行随机分配零到50％的0和1在第二行，其余为50％。

非随机化示例：

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

有什么建议吗？

Answer 1

基于检查数字是大于还是小于中位数的解决方案。只要测试的列数是偶数，一组随机双精度的正好一半将大于中位数，一半将更小。这可以保证正好有50％的位被翻转。

nRows = 3;
nCols = 16; %# divisible by 4

%# seed the array
%# assume that the numbers in each row are unique (very, very likely)
array = rand(nRows,nCols); 


out = false(nRows,nCols);

%# first row is special
out(1,:) = array(1,:) > median(array(1,:));

%# for the rest of the row, check median for the zeros/ones in the previous row
for iRow = 2:nRows
    zeroIdx = out(iRow-1,:) == 0;
    %# > or < do not matter, both will replace zeros/ones 
    %# and replace with exactly half zeros and half ones
    out(iRow,zeroIdx) = array(iRow,zeroIdx) > median(array(iRow,zeroIdx));
    out(iRow,~zeroIdx) = array(iRow,~zeroIdx) > median(array(iRow,~zeroIdx));
end

Answer 2

我提供简短的bsxfun解决方案：

%// number of divisions
n = 4;

%// unshuffled matrix like in your example
unshuffled = bsxfun(@(a,b) mod(a,2*b) > b-1, meshgrid(1:n^2,1:n) - 1, (2.^((n-1):-1:0)).') %'

%// shuffle columns
shuffled = unshuffled(:,randperm(n^2))

unshuffled =

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


shuffled =

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

首先，您需要创建非抽动矩阵，这可以通过将meshgrid(1:n^2,1:n)生成的矩阵与行相关模数进行比较来完成。最后你只需要洗牌。

Answer 3

如果您拥有统计工具箱，则可以使用randsample

轻松完成

M = 3; %// number of rows
N = 16; %// number of columns. Should be multiple of 4, according to problem definition 
result = zeros(M,N); %// preallocate and initiallize to zeros
result(1, randsample(1:N,N/2)) = 1; %// first row: half values set to one, half to zero
for m = 2:M
    result(m, :) = result(m-1, :); %// initiallize row m equal to row m-1
    result(m, randsample(find(result(m-1,:)), N/4)) = 0; %// change half of ones
    result(m, randsample(find(~result(m-1,:)), N/4)) = 1; %// change half of zeros
end

示例结果：

result = 
     0 1 0 1 1 0 0 0 0 1 0 1 1 0 1 1
     1 1 0 0 0 1 1 1 0 1 0 1 0 0 0 1
     1 0 0 0 1 0 0 1 0 1 1 0 1 1 0 1

Answer 4

使用Array ( [0] => int ping(string $session_id) [1] => string logout(string $session_id) [2] => string login(string $user, string $password) ) Fatal error: Call to undefined function login() in C:\.... on line 14的解决方案：

randperm

0和1的矩阵，其中后续行中的赋值在前一行中是有效的

4 个答案: