将浮点数转换为负数或正数

Question

I have a float from 0 to 100 and I want to translate that into a number from -20 to 20 an example:

float is 100 then translated it would be -20

float is 50 then translated it would be 0

float is 0 then translated it would be 20

What is the best method of doing this?

Answer 1

[I'm going to give you the approach to figuring this out, rather than just the answer, because it'll be more useful in the long-run.]

You need a linear transform, of the form y = mx + c , where x is your input number, y is your output number, and m and c are constants that you need to determine ahead of time.

To do so, you need to solve the following simultaneous equations:

-20 = m * 100 + c
+20 = m * 0   + c

Note that I've picked two of your required transformation examples, and plugged them into the equation. I could have picked any two.

Answer 2

怎么样： (50. - x) * 0.4

Answer 3

这样的事情应该做：

20 - val * 0.4

Answer 4

 float procent = (myval - 50)/2.5f;

如果需要整数，请使用(int) floor(procent) ...

Answer 5

float translate(float f)
{
  return 20.0f - ((20.0f * f) / 50.0f);
}

Answer 6

What you want to achieve is called "Linear interpolation" and can be done in a general function like this:

float linear_interpolate(float x, float x0, float x1, float y0, float y1)
{
  return y0 + (x - x0)*((y1-y0)/(x1-x0));
}

In your case you would call it like (replace x with your in value):

float value = linear_interpolate(x, 0.0f, 100.0f, -20.0f, 20.0f);

See http://en.wikipedia.org/wiki/Linear_interpolation for a reference article.

Answer 7

((x / 2.5)- 20) * -1

这应该做

Answer 8

You need to calculate the slope. Since you have already 3 points (0, -20) (50, 0) (100, 20), you can do dx = 40/100 = 2/5 (change in y / change in x) and b = -20. Then you define a function (f(x) = mx + b) f(x) = (2/5)*x - 20, 0 <= x <= 100.