在一定范围内并以最小的距离创建整数的随机序列 [英] Create random sequence of integers within a range and with a minimum distance between them

问题描述

生成特定nr个随机值整数的最快方法是什么?该整数均匀分布在特定范围内，并且每个元素之间的距离最短?

What is the fastest way to generate a specific nr of integers of random value uniformly distributed within a specific range and with a minimum distance between each element?

例如，给定一个介于0到20之间的序列范围，我们想要创建5个元素，每个元素之间的距离至少为3点，结果可能是这样的[0,5,11,14,19]或[2,5,9,13,18]

For example, given a sequence range between 0 and 20, we want to create 5 elements with at least 3 points distance between each element, the result could be something like this [0,5,11,14,19] or [2,5,9,13,18]

我创建了一个可以达到此目的的循环，但是当我要创建数百万个范围时，速度非常慢.

I created a loop that achieves this but it is very slow when i want to create ranges in the order of millions.

推荐答案

如何处理以下食谱:如果您希望相邻的5个元素之间的间距为3，但总范围为20，则您实际上拥有松弛"步骤，您可以在元素之间的间隙中随机分布.假设我们生成一个在该范围内的数字，并将其分散在元素之间，然后最终得到类似以下的代码:

How about the following recipe: If you want a gap of 3 between your 5 adjacent elements, but want a total range of 20, then you effectively have 20 - (5-1)*3 steps of "slack" that you can randomly distribute in the gaps between your elements. Suppose we generate a number in that range, and scatter it between the elements, then we end up with code something like the following:

import numpy, random

n = 5
limit = 20
mingap = 3

slack = 20 - mingap * (n - 1)

def generate():
    steps = random.randint(0, slack)

    increments = numpy.hstack([numpy.ones((steps,)), numpy.zeros((n,))])
    numpy.random.shuffle(increments)

    locs = numpy.argwhere(increments == 0).flatten()
    return numpy.cumsum(increments)[locs] + mingap * numpy.arange(0, n)

如果您随后调用此generate()函数十次，您将获得类似于以下内容的向量集合:

If you then invoke this generate() function ten times, you get a collection of vectors something like the following:

[  0.   3.   6.   9.  12.]
[  0.   3.   6.  10.  13.]
[  2.   5.   8.  12.  15.]
[  1.   4.   7.  12.  16.]
[  0.   4.   7.  10.  13.]
[  0.   3.   6.   9.  12.]
[  1.   4.   9.  12.  16.]
[  0.   7.  10.  13.  16.]
[  0.   5.   8.  11.  14.]
[  1.   4.   8.  11.  17.]

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