从[0.5 - 1]到[0 - 1] [英] Normalizing from [0.5 - 1] to [0 - 1]

问题描述

我被卡在这里，我想这是一个脑筋急转弯。如果我有数字在0.5到1之间的范围，我怎么能规范化它在0到1之间？

感谢任何帮助，也许我只是一点缓慢，因为我一直在过去24小时工作O_O

解决方案

其他人已经提供了你的公式， 。这里是如何解决这样的问题。

要将 [0.5，1] 映射到 [0，1] 我们将寻找形式 x - > ax + b 。我们将要求将端点映射到端点，并保留该顺序。

方法一：端点映射到端点并保留该订单的要求意味着 0.5 映射到 0 和 1 映射到 1

  a *（0.5）+ b = 0（1）
a * 1 + b = 1 2）
  

这是一个线性方程组的同时系统，可以通过乘以等式<$ c $通过 -2 并将等式（1）添加到方程式（2）。这样做，我们得到 b = -1 ，并将它代入方程（2） $ c> a = 2 。因此，映射 x - >方法二：线的斜率通过两点

code>（x1，y1）和（x2，y2）是

 （y2-y1）/（x2-x1）。 
  

这里我们将使用（0.5,0）和（1,1）以满足端点映射到端点并且映射是顺序保留的要求。因此斜率为

  m =（1-0）/（1-0.5）= 1 / 0.5 = $ b  

我们有（1，1）点线上，因此通过线的方程的点斜率形式，我们有

  y  -  1 = 2 *（x  -  1）= 2x  -  2 
  

，以便

  y = 2x  -  1. 
  

我们再次看到 x - > 2x - 1 是一个可以完成这个任务的地图。

I'm kind of stuck here, I guess it's a bit of a brain teaser. If I have numbers in the range between 0.5 to 1 how can I normalize it to be between 0 to 1?

Thanks for any help, maybe I'm just a bit slow since I've been working for the past 24 hours straight O_O

解决方案

Others have provided you the formula, but not the work. Here's how you approach a problem like this. You might find this far more valuable than just knowning the answer.

To map [0.5, 1] to [0, 1] we will seek a linear map of the form x -> ax + b. We will require that endpoints are mapped to endpoints and that order is preserved.

Method one: The requirement that endpoints are mapped to endpoints and that order is preserved implies that 0.5 is mapped to 0 and 1 is mapped to 1

a * (0.5) + b = 0 (1)
a * 1 + b = 1     (2)

This is a simultaneous system of linear equations and can be solved by multiplying equation (1) by -2 and adding equation (1) to equation (2). Upon doing this we obtain b = -1 and substituting this back into equation (2) we obtain that a = 2. Thus the map x -> 2x - 1 will do the trick.

Method two: The slope of a line passing through two points (x1, y1) and (x2, y2) is

(y2 - y1) / (x2 - x1).

Here we will use the points (0.5, 0) and (1, 1) to meet the requirement that endpoints are mapped to endpoints and that the map is order-preserving. Therefore the slope is

m = (1 - 0) / (1 - 0.5) = 1 / 0.5 = 2.

We have that (1, 1) is a point on the line and therefore by the point-slope form of an equation of a line we have that

y - 1 = 2 * (x - 1) = 2x - 2

so that

y = 2x - 1.

Once again we see that x -> 2x - 1 is a map that will do the trick.

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