从小范围生成随机的非重复整数 [英] Generate random non-repeating integers from a small range

问题描述

我要完成的工作如下:

我希望创建一个相对较小范围的整数向量，并确保所有整数都不会跟在同一整数之后.

I wish to create a vector of integers, from a relatively small range, and ensure that none of the integers will be followed by the same integer.

即，这是一个合法"载体: [1 3 4 2 5 3 2 3 5 4]

i.e., This is a "legal" vector: [ 1 3 4 2 5 3 2 3 5 4 ]

，这是一个非法"向量(因为5后接5): [1 3 4 2 5 5 2 3 5 4]

and this is an "illegal" vector (since 5 follows 5): [ 1 3 4 2 5 5 2 3 5 4 ]

我已经尝试了randi，并尝试了randperm的各种变化形式，当我尝试从一个很小的范围(即1到1之间的整数)生成一个大约100个元素的向量时，我总是会卡住5).

I've experimented with randi, and all sorts of variations with randperm, and I always get stuck when i try to generate a vector of around 100 elements, from a small range (i.e., integers between 1 and 5).

该函数运行时间过长.

这是我所做的尝试之一:

Here's one of the attempts that i've made:

function result = nonRepeatingRand(top, count)

    result = randi(top, 1, count);

    while any(diff(result) == 0)
         result = randi(top, 1, count);    
    end

end

任何人和所有帮助将不胜感激.谢谢！

Any and all help will be much appreciated. Thanks !

推荐答案

可以通过生成从1到top - 1的差异，然后计算从一个随机初始值开始的累积总和模量 top:

The kind of sequence you are looking for can be defined by generating differences from 1 to top - 1 and then computing the cumulative sum modulus top, starting from a random initial value:

function result = nonRepeatingRand(top, count)

    diff = randi(top - 1, 1, count);
    result = rem(cumsum(diff) + randi(1, 1, count) - 1, top) + 1;

end

在我的机器上，这会在0.58秒内以1:5的形式生成1000万个数字的非重复序列.

On my machine, this generates a non-repeating sequence of 10 million numbers out of 1:5 in 0.58 seconds.

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