如何在matlab中生成一个具有特定条件的随机二进制矩阵？ [英] how to generate a random binary matrix with a specific condition in matlab?
问题描述
如果矩阵A显示G（n，m）矩阵中的一组数目，那么
G = [ 1 1 0 0 1
0 1 1 1 0
1 0 1 1 1]
所以A矩阵将会是
A = [2 1
3 0
1 3 ]
然后我想生成（n，m）随机矩阵，矩阵中的矩阵取决于A按照同样的顺序出现
一个解决方案将是
x = [0 1 1 0 1
0 0 1 1 1
1 0 1 1 1]
另一个解决方案
x = [1 1 0 1 0
1 1 1 0 0
1 0 1 1 1]
在对OP的评论中，StackOverflow不是一个代码写入服务。话虽如此，这是一个有趣的问题，我决定作出例外并回答它。
手续费不计入...
我想我有一个通用的解决方案，它也可以处理一些（全部）边缘案例，如 G ，其中零行等。下面的代码生成一个 x 矩阵。
创建一个 n - by - m 这些 x s留给作为读者的练习（主要是因为它不确定OP是否希望它作为矩阵的单元阵列或4-D逻辑/双数组）。
要理解它的作用，请参阅代码+变量名称中的注释。
function x = q37055681（G）
%%输入
如果nargin < 1
G = [1 1 0 0 1
0 1 1 1 0
1 0 1 1 1];
end
%%输入分析：
[A_ROWS，OUT_COLS] = size（G）;
transitions = find（diff（padarray（G。'，1，false，'both'）。'，1,2）。'）;
tr_per_ln = hist（ceil（transitions /（size（G，2）+1）），size（G，1））/ 2;
A_COLS = max（tr_per_ln）;
missing_trans_per_line = A_COLS - tr_per_ln;
groups_of_ones = diff（reshape（transitions，2，[]），1,1）; ％<忽略0的
％的RLE结果基于外部定义，每个
％行只有1个组的外部定义（在这种情况下，将其计算在A的第一个元素中）：
insrt = @（what，into，where）cat（2，into（1：where-1），what，into（where：end））;
for indZ = 1：sum（missing_trans_per_line（:)）
next_incomplete = find（missing_trans_per_line，1）;
groups_of_ones = insrt（0，groups_of_ones，A_COLS * next_incomplete -...
（missing_trans_per_line（next_incomplete）-1））;
missing_trans_per_line（next_incomplete）= missing_trans_per_line（next_incomplete）-1;
end
A = reshape（groups_of_ones，A_COLS，[]）。';
%%产出的产生：
x =零（大小（G））;
for indR = 1：A_ROWS
tokens = cell（sum（A（indR，:)〜= 0），1）;
开关numel（令牌）
情况0
％我们不需要做任何事情，整个行都是0.
continue;
case 1
tokens {1} = repelem（true，A（indR，1））;
否则
对于indT = 1：numel（令牌）-1
令牌{indT} = [repelem（true，A（indR，indT））0];
end
tokens {end} = repelem（true，A（indR，find（A（indR，:)，1，'last'）））;
end
zero_tokens = repmat（{0}，[OUT_COLS-sum（A（indR，：）） - （numel（tokens）-1），1]）;
％现在我们需要构建一个随机选择的矢量，但是从
％或者标记或zero_tokens中选择一个。
cell_to_pick_from = [ones（1，numel（tokens））zeros（1，numel（zero_tokens））];
choices_order = cell_to_pick_from（randperm（numel（cell_to_pick_from）））;
％^这就是随机性出现的地方：
x_line = [];
for indC = 1：numel（choices_order）
如果选择订单（indC）
％如果它是1，从标记中选择
标记=标记{sum（choices_order（1： INDC）== 1）};
else
％如果它是0，则从零中选择
token = zero_tokens {sum（chosen_order（1：indC）== 0）};
结束
x_line = [x_line令牌]; ％＃ok
end
x（indR，:) = x_line;
结束
结束
If matrix A shows the number of group of ones in G (n,m) matrix like that
G = [ 1 1 0 0 1
0 1 1 1 0
1 0 1 1 1 ]
so the A matrix will be
A = [ 2 1
3 0
1 3 ]
Then i want to generate (n,m) random matrix which the ones in this Matrix depend on A in the same order they appear
one solution will be
x = [ 0 1 1 0 1
0 0 1 1 1
1 0 1 1 1 ]
another solution
x = [ 1 1 0 1 0
1 1 1 0 0
1 0 1 1 1 ]
As mentions in comments to the OP, StackOverflow is not a code-writing service. Having said that, this was an interesting problem and I decided to make an exception and answer it anyway.
Formalities aside...
I think I have a general solution that can also handle some (all?) edge cases like G with zero rows etc. The code below generates one instance of the x matrix.
Creating an n-by-m array of these xs is left "as an exercise to the reader" (mainly because it's undefined whether OP wants it as a cell array of matrices or a 4-D logical/double array).
To understand what it does, see comments in the code + variable names. I hope it is clear enough (and that I didn't miss any edge cases).
function x = q37055681(G)
%% Input
if nargin < 1
G = [ 1 1 0 0 1
0 1 1 1 0
1 0 1 1 1 ];
end
%% Input analysis:
[A_ROWS,OUT_COLS] = size(G);
transitions = find(diff(padarray(G.',1,false,'both').',1,2).');
tr_per_ln = hist(ceil(transitions/(size(G,2)+1)),size(G,1))/2;
A_COLS = max(tr_per_ln);
missing_trans_per_line = A_COLS - tr_per_ln;
groups_of_ones = diff(reshape(transitions,2,[]),1,1); % < result of RLE which ignores 0's
% Count "fixing" based on the external definition of what to do with only 1 group per
% line (in this case, count it in the first element of A):
insrt = @(what, into, where) cat(2, into(1:where-1), what, into(where:end));
for indZ = 1:sum(missing_trans_per_line(:))
next_incomplete = find(missing_trans_per_line,1);
groups_of_ones = insrt(0, groups_of_ones, A_COLS*next_incomplete-...
(missing_trans_per_line(next_incomplete)-1));
missing_trans_per_line(next_incomplete) = missing_trans_per_line(next_incomplete)-1;
end
A = reshape(groups_of_ones,A_COLS,[]).';
%% Generation of output:
x = zeros(size(G));
for indR = 1:A_ROWS
tokens = cell(sum(A(indR,:)~=0),1);
switch numel(tokens)
case 0
% we don't need to do anything, the entire line is 0.
continue;
case 1
tokens{1} = repelem(true,A(indR,1));
otherwise
for indT = 1:numel(tokens)-1
tokens{indT} = [repelem(true,A(indR,indT)) 0];
end
tokens{end} = repelem(true,A(indR,find(A(indR,:),1,'last')));
end
zero_tokens = repmat({0},[OUT_COLS-sum(A(indR,:))-(numel(tokens)-1),1]);
% Now we need to build a vector that selects randomly but SEQUENTIALLY from
% EITHER tokens or zero_tokens.
cell_to_pick_from = [ones(1,numel(tokens)) zeros(1,numel(zero_tokens))];
choosing_order = cell_to_pick_from(randperm(numel(cell_to_pick_from)));
% ^ Here's where the randomness comes in:
x_line = [];
for indC = 1:numel(choosing_order)
if choosing_order(indC)
% if it's 1, choose from "tokens"
token = tokens{sum(choosing_order(1:indC)==1)};
else
% if it's 0, choose from "zeros"
token = zero_tokens{sum(choosing_order(1:indC)==0)};
end
x_line = [x_line token]; %#ok
end
x(indR,:) = x_line;
end
end
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