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

问题描述

我在fortran proramming中有点新鲜感。
任何人都可以请帮我解决方案。
i在使用
random_seed和rand的fortran随机数中产生整数随机数
的范围[0,5]，问题是

解决方案

为了支持Alexander Vogt的答案，我将推广。 / p>

内在 random_number（u）返回一个实数 u （或者这样的一个数组）在区间[0,1）内的均匀分布。为了在整数{n，n + 1，...，m-1上​​有一个离散的均匀分布，它包含0但不是1。]

，m}将连续分布分成m + 1-n个相等大小的块，将每个块映射为一个整数。一种方式可能是：

$ p $ call random_number（u）
j = n + FLOOR（（m + 1-n） *你）！我们想从m-n + 1中选择一个整数


正如您所看到的，对于最初的问题对于{0,1,2,3,4,5}，这减少到

 调用random_number（u）
j = FLOOR（6 * u）！ n = 0和m = 5 
  

另一种情况在您的评论{-1，0， 1}

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

当然，连续的整数，并且应该注意数字问题。

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

解决方案

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

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.]

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

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

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

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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