将数字范围映射到另一个 [英] Mapping a numeric range onto another

问题描述

数学从来都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 largest 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 output.

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

如何将输入值转换为该范围对应的输出值?

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

例如，输入值0"将等于输出值500"，输入值254"将等于输出值5500".如果输入值为 50 或 101，我无法弄清楚如何计算输出值.

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.

我确定这很简单，我现在想不出来:)

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

我只需要整数，没有分数或任何东西.

I just need whole numbers, no fractions or anything.

推荐答案

让我们忘记数学，试着凭直觉解决这个问题.

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

首先，如果我们想将范围[0,x]内的输入数字映射到输出范围[0,y]，我们只需要适当的缩放.0到0，x到y，数字t到(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.

那么，让我们将您的问题简化为上述更简单的问题.

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

[input_start, input_end] 的输入范围有 input_end - input_start + 1 个数字.所以它等价于一个范围[0, r]，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.

同理，输出范围相当于[0,R]，其中R = output_end - output_start.

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

input 的输入等价于 x = input - input_start.这，从第一段将转换为 y = (R/r)*x.然后，我们可以通过添加 output_start 将 y 值转换回原始输出范围:output = 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:

/* 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，C 中的除法会截断，您应该尝试通过计算浮点数来获得更准确的答案:

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)

如果想要更正确，您可以在最后一步进行四舍五入而不是截断.您可以通过编写一个简单的 round 函数来实现:

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);
}

那么:

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

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