Question

我有N个时间步，想要创建与每个时间步对应的N个实数，如下所示：数字应该在[a，b]范围内，切换概率为p。即，如果第k个数字（对应于第k个时间步长）是n，那么第k + 1个数字（对应于第k + 1个时间步长）是不同于n的任何其他数字的概率是p。所有创建的数字应该在[a，b]范围内。

如何在matlab中完成？感谢。

Answer 1

我不确定我是否满足您的所有要求，但这可能是您要搜索的脚本：

N = 10;     % count of numbers
p = 0.2;    % switching probability
a = 5;
b = 20; 

% init empty numbers and get the first random number
numbers = zeros(N,1);
numbers(1) = rand(1) * (b-a) + a;

for iNumber = 2:N
    % test if number must change
    if rand(1) < (1-p)
        % the number must be the same than the previous
        % copy the value and go to next number
        numbers(iNumber) = numbers(iNumber-1);
        continue;

    else 
        % a new number must be found
        while 1
            % get a new random number 
            numbers(iNumber) = rand(1) * (b-a) + a;

            % check if the new random number is different from the previous
            if numbers(iNumber) ~= numbers(iNumber-1)
                % in case they are different, the while 1 can be stopped
                break;
            end

        end % while 1

    end % if rand(1) < (1-p)
end % for iNumber = 2:N

Answer 2

保证超级随机：

N = randi(100);
p = rand;
l1 = rand*randi(100);
l2 = rand*randi(100);

if l2 < l1
  [l2 l1] = deal(l1,l2);
end

out = []

while isempty(out)
    if rand>rand
        n = 2:N
        a = rand([N,1])*(l2-l1)+l1;
        x = rand([N-1,1])<p;
        n(x==0)=n(x==0)-1;
        n = [1 n];
        out = a(n);
    end
end

在matlab中切换数字

2 个答案: