C++ 从 [0.5 - 1] 归一化到 [0 - 1]
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Normalizing from [0.5 - 1] to [0 - 1]
提问by tweetypi
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?
我有点卡在这里，我想这有点脑筋急转弯。如果我有介于 0.5 到 1 之间的数字，我如何将其标准化为介于 0 到 1 之间？
Thanks for any help, maybe I'm just a bit slow since I've been working for the past 24 hours straight O_O
感谢您的帮助，也许我只是有点慢，因为我过去 24 小时一直在工作 O_O
回答by jason
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.
其他人为您提供了公式，但没有提供工作。以下是您处理此类问题的方法。您可能会发现这比仅仅知道答案更有价值。
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.
为了映射[0.5, 1]到，[0, 1]我们将寻找形式为 的线性映射x -> ax + b。我们将要求端点映射到端点并保留该顺序。
Method one:The requirement that endpoints are mapped to endpoints and that order is preserved implies that 0.5is mapped to 0and 1is mapped to 1
方法一：端点映射到端点并保留顺序的要求意味着0.5映射到0和1被映射到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 -2and adding equation (1)to equation (2). Upon doing this we obtain b = -1and substituting this back into equation (2)we obtain that a = 2. Thus the map x -> 2x - 1will do the trick.
这是线性方程的同时系统，并且可以通过乘以等式来解决(1)由-2和添加方程(1)以方程(2)。这样做后，我们得到b = -1并将其代入方程，(2)我们得到a = 2。因此，地图x -> 2x - 1将发挥作用。
Method two:The slope of a line passing through two points (x1, y1)and (x2, y2)is
方法二：一个线通过的通过两个点的斜率(x1, y1)和(x2, y2)是
(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
在这里，我们将使用点(0.5, 0)并(1, 1)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 = 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
我们有那(1, 1)是线上的一个点，因此通过一条线的方程的点斜形式我们有
y - 1 = 2 * (x - 1) = 2x - 2
so that
以便
y = 2x - 1.
Once again we see that x -> 2x - 1is a map that will do the trick.
我们再一次看到这x -> 2x - 1是一张可以解决问题的地图。
回答by Bill the Lizard
Subtract 0.5 (giving you a new range of 0 - 0.5) then multiply by 2.
减去 0.5（给你一个新的范围 0 - 0.5）然后乘以 2。
double normalize( double x )
{
// I'll leave range validation up to you
return (x - 0.5) * 2;
}
回答by Toon Krijthe
To add another generic answer.
添加另一个通用答案。
If you want to map the linear range [A..B] to [C..D], you can apply the following steps:
如果要将线性范围 [A..B] 映射到 [C..D]，可以应用以下步骤：
Shift the range so the lower bound is 0. (subract A from both bounds:
移动范围，使下限为 0。（从两个边界中减去 A：
[A..B] -> [0..B-A]
Scale the range so it is [0..1]. (divide by the upper bound):
缩放范围使其为 [0..1]。（除以上限）：
[0..B-A] -> [0..1]
Scale the range so it has the length of the new range which is D-C. (multiply with D-C):
缩放范围，使其具有新范围的长度，即 DC。（乘以 DC）：
[0..1] -> [0..D-C]
Shift the range so the lower bound is C. (add C to the bounds):
移动范围，使下限为 C。（将 C 添加到边界）：
[0..D-C] -> [C..D]
Combining this to a single formula, we get:
将其组合到一个公式中，我们得到：
(D-C)*(X-A)
X' = ----------- + C
(B-A)
In your case, A=0.5, B=1, C=0, D=1 you get:
在你的情况下，A=0.5, B=1, C=0, D=1 你得到：
(X-0.5)
X' = ------- = 2X-1
(0.5)
Note, if you have to convert a lot of X to X', you can change the formula to:
请注意，如果您必须将很多 X 转换为 X'，则可以将公式更改为：
(D-C) C*B - A*D
X' = ----- * X + ---------
(B-A) (B-A)
It is also interesting to take a look at non linear ranges. You can take the same steps, but you need an extra step to transform the linear range to a nonlinear range.
看看非线性范围也很有趣。您可以采取相同的步骤，但您需要一个额外的步骤来将线性范围转换为非线性范围。
回答by Glenner003
× 2 ? 1
× 2 ? 1
should do the trick
应该做的伎俩
回答by meetar
Lazyweb answer:To convert a value xfrom [minimum..maximum]to [floor..ceil]:
Lazyweb 答案：要将值x从转换[minimum..maximum]为[floor..ceil]：
General case:
一般情况：
normalized_x = ((ceil - floor) * (x - minimum))/(maximum - minimum) + floor
To normalize to [0..255]:
归一化为 [0..255]：
normalized_x = (255 * (x - minimum))/(maximum - minimum)
To normalize to [0..1]:
归一化为 [0..1]：
normalized_x = (x - minimum)/(maximum - minimum)
回答by Whitney Imura
You could always use clamp or saturate within your math to make sure your final value is between 0-1. Some saturate at the end, but I've seen it done during a computation, too.
你总是可以在你的数学中使用clamp或saturate来确保你的最终值在0-1之间。最后有些饱和，但我也看到它在计算过程中完成。
