我怎样才能让一个随机选择的逆加权列表？ [英] How can I make a random selection from an inversely-weighted list?

问题描述

鉴于整数，例如一个列表 1，2，3，4 ，我知道如何选择基于其权重的物品。的示例项目将具有10％，20％，30％和40％的概率，分别

Given a list of integers, e.g. 1, 2, 3, 4, I know how to select items based on their weight. The example items would have probabilities of 10%, 20%, 30%, and 40%, respectively.

有没有基于其权重的逆选择项目的一个同样简单的方法？用这种方法，例如列表将等于的加权列表 1，1/2，1/3，1/4 （48％，24％，16％ ，12％），但我想，以避免浮点运算的转化和利用。 （假设所有整数的是积极的，非零。）

Is there an equally simple method of selecting items based on the inverse of their weight? With this method, the example list would be equal to a weighted list of 1, 1/2, 1/3, 1/4 (48%, 24%, 16%, 12%), but I want to avoid the conversion and use of floating-point arithmetic. (Assume all of the integers are positive and non-zero.)

推荐答案

您可以划分数最不常见多通过每个号码并获得积分的地步。

You could divide the numbers' least common multiple by each number and get integral proportions.

有关 [1，2，3，4] ，这是12。你的权重为12/1 = 12，12/2 = 6，12/3 = 4，12/4 = 3。

For [1, 2, 3, 4], this is 12. Your weights are 12/1=12, 12/2=6, 12/3=4, 12/4=3.

您也可以乘它们放在一起，而不是与LCM打扰为好。的数字将更高，但比例将是相同的：24/1 = 24，24/2 = 12，24/3 = 8，24/4 = 6

You could also multiply them all together and not bother with the LCM as well. The numbers will be higher but the proportions will be the same: 24/1=24, 24/2=12, 24/3=8, 24/4=6.

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