拉伸数组

Question

我有一个形成曲线的样本矢量.让我们想象其中有1000点.如果我想拉伸它来填充1500点,那么最简单的算法是什么能给出不错的结果呢？我正在寻找只有几行C/C++的东西.

我总是想增加矢量的大小,新矢量可以是当前矢量大小的1.1倍到50倍.

谢谢!

Answer 1

这是用于线性和二次插值的C++.
interp1( 5.3, a, n )是[5] + .3*(a [6] - a [5]),从[5]到[6]的方式的.3;
interp1array( a, 1000, b, 1500 )会延伸a到b.
interp2( 5.3, a, n )通过3个最近点a [4] a [5] a [6]绘制一条抛物线:比interp1更平滑,但仍然很快.
(样条曲线使用4个最近的点,更平滑;如果你阅读python,请参见 基本样条插值 - 几个线条的numpy.

// linear, quadratic interpolation in arrays
// from interpol.py denis 2010-07-23 July

#include <stdio.h>
#include <stdlib.h>

    // linear interpolate x in an array
// inline
float interp1( float x, float a[], int n )
{
    if( x <= 0 )  return a[0];
    if( x >= n - 1 )  return a[n-1];
    int j = int(x);
    return a[j] + (x - j) * (a[j+1] - a[j]);
}

    // linear interpolate array a[] -> array b[]
void inter1parray( float a[], int n, float b[], int m )
{
    float step = float( n - 1 ) / (m - 1);
    for( int j = 0; j < m; j ++ ){
        b[j] = interp1( j*step, a, n );
    }
}

//..............................................................................
    // parabola through 3 points, -1 < x < 1
float parabola( float x, float f_1, float f0, float f1 )
{
    if( x <= -1 )  return f_1; 
    if( x >= 1 )  return f1; 
    float l = f0 - x * (f_1 - f0);
    float r = f0 + x * (f1 - f0);
    return (l + r + x * (r - l)) / 2;
}

    // quadratic interpolate x in an array
float interp2( float x, float a[], int n )
{
    if( x <= .5  ||  x >= n - 1.5 )
        return interp1( x, a, n );
    int j = int( x + .5 );
    float t = 2 * (x - j);  // -1 .. 1
    return parabola( t, (a[j-1] + a[j]) / 2, a[j], (a[j] + a[j+1]) / 2 );
}

    // quadratic interpolate array a[] -> array b[]
void interp2array( float a[], int n, float b[], int m )
{
    float step = float( n - 1 ) / (m - 1);
    for( int j = 0; j < m; j ++ ){
        b[j] = interp2( j*step, a, n );
    }
}

int main( int argc, char* argv[] )
{
        // a.out [n m] --
    int n = 10, m = 100;
    int *ns[] = { &n, &m, 0 },
        **np = ns;
    char* arg;
    for( argv ++;  (arg = *argv) && *np;  argv ++, np ++ )
        **np = atoi( arg );
    printf( "n: %d  m: %d\n", n, m );

    float a[n], b[m];
    for( int j = 0; j < n; j ++ ){
        a[j] = j * j;
    }
    interp2array( a, n, b, m );  // a[] -> b[]

    for( int j = 0; j < m; j ++ ){
        printf( "%.1f ", b[j] );
    }
    printf( "\n" );
}