如何缩小具有已知最小值和最大值的数字范围 [英] How to scale down a range of numbers with a known min and max value

问题描述

所以我想弄清楚如何取一个数字范围并将值缩小以适应一个范围.想要这样做的原因是我试图在 java swing jpanel 中绘制椭圆.我希望每个椭圆的高度和宽度都在 1-30 的范围内.我有一些方法可以从我的数据集中找到最小值和最大值，但直到运行时我才会有最小值和最大值.有没有简单的方法可以做到这一点?

So I am trying to figure out how to take a range of numbers and scale the values down to fit a range. The reason for wanting to do this is that I am trying to draw ellipses in a java swing jpanel. I want the height and width of each ellipse to be in a range of say 1-30. I have methods that find the minimum and maximum values from my data set, but I won't have the min and max until runtime. Is there an easy way to do this?

推荐答案

假设您想将范围 [min,max] 缩放到 [a,b].你正在寻找一个(连续的)函数，它满足

Let's say you want to scale a range [min,max] to [a,b]. You're looking for a (continuous) function that satisfies

f(min) = a
f(max) = b

在您的情况下，a 将是 1 而 b 将是 30，但让我们从更简单的事情开始并尝试映射 [min,max] 进入范围 [0,1].

In your case, a would be 1 and b would be 30, but let's start with something simpler and try to map [min,max] into the range [0,1].

将 min 放入一个函数中，可以用

Putting min into a function and getting out 0 could be accomplished with

f(x) = x - min   ===>   f(min) = min - min = 0

这几乎就是我们想要的.但是当我们真正想要 1 时，输入 max 会给我们 max - min.所以我们必须缩放它:

So that's almost what we want. But putting in max would give us max - min when we actually want 1. So we'll have to scale it:

        x - min                                  max - min
f(x) = ---------   ===>   f(min) = 0;  f(max) =  --------- = 1
       max - min                                 max - min

这是我们想要的.所以我们需要进行平移和缩放.现在，如果我们想要获得 a 和 b 的任意值，我们需要一些更复杂的东西:

which is what we want. So we need to do a translation and a scaling. Now if instead we want to get arbitrary values of a and b, we need something a little more complicated:

       (b-a)(x - min)
f(x) = --------------  + a
          max - min

您可以验证为 x 输入 min 现在给出 a，而输入 max 给出 <代码>b.

You can verify that putting in min for x now gives a, and putting in max gives b.

您可能还注意到 (b-a)/(max-min) 是新范围大小与原始范围大小之间的比例因子.所以实际上我们首先通过 -min 转换 x，将其缩放到正确的因子，然后将其转换回 a.

You might also notice that (b-a)/(max-min) is a scaling factor between the size of the new range and the size of the original range. So really we are first translating x by -min, scaling it to the correct factor, and then translating it back up to the new minimum value of a.

希望这会有所帮助.

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