如何在[0,5]范围内在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.任何人都可以帮我解决这个问题.我在生成整数随机数时遇到问题在 [0,5] 范围内 fortran 随机数使用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 answer,我将概括.
To support the answer by Alexander Vogt, I'll generalize.
内在的 random_number(u)
从区间 [0,1) 上的均匀分布返回一个实数 u
(或这样的数组).[也就是说,它包括0但不包括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.]
为了在整数 {n, n+1, ..., m-1, m} 上有一个离散的均匀分布,将连续分布分割成 m+1-n 个相等大小的块,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,还需要其他变换,需要注意数值问题.
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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