映射一个数字范围到另一个 [英] Mapping a numeric range onto another

问题描述

数学从来不是我在学校的强项：（

Math was never my strong suit in school :(

int input_start = 0; // The lowest number of the range input.
int input_end = 254; // The lowest number of the range input.
int output_start = 500; // The lowest number of the range output.
int output_end = 5500; // The largest number of the range ouput.

int input = 127; // Input value.
int output = 0;

我怎样才能输入值转换为一系列相应的产值？

How can I convert the input value to the corresponding output value of that range?

例如，0的输入值应等于500的产值中，254的输入值应等于5500的输出值。我无法弄清楚如何如果输入值是说50或101计算的输出值。

For example, an input value of "0" would equal an output value of "500", an input value of "254" would equal an output value of "5500". I can't figure out how to calculate an output value if an input value is say 50 or 101.

我敢肯定，这很简单，我不认为现在：）

I'm sure it's simple, I can't think right now :)

编辑：我只需要整数，无小数或任何

I just need whole numbers, no fractions or anything.

推荐答案

让我们忘记了数学和尝试直观地解决这个问题。

Let's forget the math and try to solve this intuitively.

首先，如果我们想输入数字范围为[ 0 ， X ]输出范围[图 0 ，是]，我们只需要一个适当的量规模。 0到0， X 进入是，以及一些 T 将前往（Y / X）* T 。

First, if we want to map input numbers in the range [0, x] to output range [0, y], we just need to scale by an appropriate amount. 0 goes to 0, x goes to y, and a number t will go to (y/x)*t.

所以，让我们减少您的问题上面简单的问题。

So, let's reduce your problem to the above simpler problem.

输入电压范围为[ input_start ， input_end ]已 input_end - input_start + 1 号码。因此，它相当于一个范围[ 0 ，研究]，其中 R = input_end - input_start 。

An input range of [input_start, input_end] has input_end - input_start + 1 numbers. So it's equivalent to a range of [0, r], where r = input_end - input_start.

同样，输出范围等效于[ 0 ，研究]，其中 R = OUTPUT_END - OUTPUT_START

Similarly, the output range is equivalent to [0, R], where R = output_end - output_start.

输入的输入等同于 X =输入 - input_start 。这一点，从第一款将转化为 Y =（R / R）* X 。然后，我们可以通过添加 OUTPUT_START 的是值转换回原来的输出范围：输出= OUTPUT_START + Y 。

An input of input is equivalent to x = input - input_start. This, from the first paragraph will translate to y = (R/r)*x. Then, we can translate the y value back to the original output range by adding output_start: output = output_start + y.

这给了我们：

output = output_start + ((output_end - output_start) / (input_end - input_start)) * (input - input_start)

或者另一种方式：

Or, another way:

/* Note, "slope" below is a constant for given numbers, so if you are calculating
   a lot of output values, it makes sense to calculate it once.  It also makes
   understanding the code easier */
slope = (output_end - output_start) / (input_end - input_start)
output = output_start + slope * (input - input_start)

现在，这是C，除用C截断，你应该尝试在浮点计算的东西，以获得更准确的答案：

Now, this being C, and division in C truncates, you should try to get a more accurate answer by calculating things in floating-point:

double slope = 1.0 * (output_end - output_start) / (input_end - input_start)
output = output_start + slope * (input - input_start)

如果希望更加正确，你会做的最后一步，而不是四舍五入截断。您可以通过编写一个简单的圆形功能做到这一点：

If wanted to be even more correct, you would do a rounding instead of truncation in the final step. You can do this by writing a simple round function:

#include <math.h>
double round(double d)
{
    return floor(d + 0.5);
}

然后：

output = output_start + round(slope * (input - input_start))

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