如何在[0,5]范围内的fortran 90中生成整数随机数? [英] how to generate integer random number in fortran 90 in the range [0,5]?

问题描述

我是 Fortran 编程的新手.谁能帮我解决一下.我遇到了生成整数随机数的问题在 fortran 随机数的 [0,5] 范围内使用random_seed 和 rand

I am kind of new in the fortran proramming. Can anyone please help me out with the solution. i am having a problem of generating integer random number in the range [0,5] in fortran random number using random_seed and rand

推荐答案

为了支持 Alexander Vogt 的答案，我'将概括.

To support the answer by Alexander Vogt, I'll generalize.

内在 random_number(u) 从区间 [0,1) 上的均匀分布返回实数 u(或此类数组).【即包含0但不包含1】

The intrinsic random_number(u) returns a real number u (or an array of such) from the uniform distribution over the interval [0,1). [That is, it includes 0 but not 1.]

为了对整数 {n, n+1, ..., m-1, m} 进行离散均匀分布，将连续分布划分为 m+1-n 个大小相等的块，将每个块映射到一个整数.一种方法可能是:

To have a discrete uniform distribution on the integers {n, n+1, ..., m-1, m} carve the continuous distribution up into m+1-n equal sized chunks, mapping each chunk to an integer. One way could be:

call random_number(u)
j = n + FLOOR((m+1-n)*u)  ! We want to choose one from m-n+1 integers

如您所见，对于 {0, 1, 2, 3, 4, 5} 的初始问题，这简化为

As you can see, for the initial question for {0, 1, 2, 3, 4, 5} this reduces to

call random_number(u)
j = FLOOR(6*u)            ! n=0 and m=5

对于您评论中的另一种情况 {-1, 0, 1}

and for the other case in your comment {-1, 0, 1}

call random_number(u)
j = -1 + FLOOR(3*u)       ! n=-1 and m=1

当然，对于不连续的整数集合，需要进行其他的转换，并且应该注意数值问题.

Of course, other transformations will be required for sets of non-contiguous integers, and one should pay attention to numerical issues.

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