Question

对于LPC922微控制器（带SDCC），我想创建一个带线性插值的查找表。让我们假设我得到x和y值，如

x=300 y=10,0201 
x=700 y=89,542 
x=800 y=126,452 
x=900 y=171,453 
x=1500 y=225,123

具有线性插值的查找表的代码怎么样，所以我得到x = 850的正确值y（（171,453 + 126,452）/ 2）？

Answer 1

typedef struct { double x; double y; } coord_t;

coord_t c[5] = 
{
    {300, 10.02},
    {700, 89.542},
    {800, 126.452}, 
    {900, 171.453}, 
    {1500,225.123}
};    

double interp( coord_t* c, double x, int n )
{
    int i;

    for( i = 0; i < n-1; i++ )
    {
        if ( c[i].x <= x && c[i+1].x >= x )
        {
            double diffx = x - c[i].x;
            double diffn = c[i+1].x - c[i].x;

            return c[i].y + ( c[i+1].y - c[i].y ) * diffx / diffn; 
        }
    }

    return 0; // Not in Range
}

int main(int argc, char** argv) 
{
    double y = interp( c, 850, 5 );
}

Answer 2

double get_value(double x)
{
    /* NOTE: xs MUST be sorted */
    static const double xs[] = { 300, 700, 800, 900, 1500 };
    static const double ys[] = { 10.0201, 89.542, 126.452, 171.453, 225.123 };

    /* number of elements in the array */
    static const int count = sizeof(xs)/sizeof(xs[0]);

    int i;
    double dx, dy;

    if (x < xs[0]) {
        /* x is less than the minimum element
         * handle error here if you want */
        return ys[0]; /* return minimum element */
    }

    if (x > xs[count-1]) {
        return ys[count-1]; /* return maximum */
    }

    /* find i, such that xs[i] <= x < xs[i+1] */
    for (i = 0; i < count-1; i++) {
        if (xs[i+1] > x) {
            break;
        }
    }

    /* interpolate */
    dx = xs[i+1] - xs[i];
    dy = ys[i+1] - ys[i];
    return ys[i] + (x - xs[i]) * dy / dx;
}

如果您愿意，可以相当容易地将其扩展到其他interpolation methods。请注意，您可能必须扩展边界区域的特殊情况，但是您希望处理它。一种常见的方法是在没有足够的相邻值可用于首选方法时进行线性插值。

当值的数量开始增长时，我建议使用binary search方法来计算起点。但这对于这几个值来说应该不是问题。

更新：由于OP正在有限的平台上运行，以下是使用libfixmath的上述版本：

/* NOTE: xs MUST be sorted */
static const fix16_t xs[] = { 300<<16, 700<<16, 800<<16, 900<<16, 1500<<16 };
static const fix16_t ys[] = { (fix16_t)(65536.0*10.0201+0.5), (fix16_t)(65536.0*89.542+0.5), (fix16_t)(65536.0*126.452+0.5), (fix16_t)(65536.0*171.453+0.5), (fix16_t)(65536.0*225.123+0.5) };

fix16_t get_value_fix(fix16_t x)
{    
    /* number of elements in the array */
    static const int count = sizeof(xs)/sizeof(xs[0]);
    int i;
    fix16_t dx, dy;

    if (x < xs[0]) {
        /* x is less than the minimum element
         * handle error here if you want */
        return ys[0]; /* return minimum element */
    }

    if (x > xs[count-1]) {
        return ys[count-1]; /* return maximum */
    }

    /* find i, such that xs[i] <= x < xs[i+1] */
    for (i = 0; i < count-1; i++) {
        if (xs[i+1] > x) {
            break;
        }
    }

    /* interpolate */
    dx = fix16_sub(xs[i+1], xs[i]);
    dy = fix16_sub(ys[i+1], ys[i]);
    return fix16_add(ys[i],  fix16_div(fix16_mul(fix16_sub(x, xs[i]), dy), dx));
}

Answer 3

将所有值放在数组中。然后搜索x ..如果你没有找到正确的值，你有一个索引和一个邻居索引，然后使用你的公式计算值..

如何使用线性插值在C（SDCC编译器）中构建查找表

3 个答案: