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

问题描述

所以我试图弄清楚如何取一个数字范围并将值缩小以适合一个范围.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在1到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，但是直到运行时我都没有最小值和最大值.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

假设您要将范围从[min,max]缩放到[a,b].您正在寻找一个满足要求的(连续)功能

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为1，b为30，但让我们从一个简单的例子开始，尝试将[min,max]映射到[0,1]范围内.

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放入函数中并得出0可以通过

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，当我们实际需要1时，放入max会给我们max - min.因此，我们必须对其进行缩放:

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

这就是我们想要的.因此，我们需要进行平移和缩放.现在，如果要获取a和b的任意值，则需要更复杂一些的东西:

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

您可以验证将min放入x现在可以得到a，而放入max可以得到b.

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

您可能还会注意到，(b-a)/(max-min)是新范围的大小和原始范围的大小之间的比例因子.因此，实际上我们首先是将x转换为-min，将其缩放为正确的因子，然后再将其转换回新的最小值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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