伸出一个数组 [英] Stretching out an array

问题描述

我有形成一个曲线样品的载体。让我们想象一下有1000点吧。如果我想舒展它来填充1500点，什么是最简单的算法，让体面的结果吗？我在寻找的东西，是C / C ++短短的几行。

I've got a vector of samples that form a curve. Let's imagine there are 1000 points in it. If I want to stretch it to fill 1500 points, what is the simplest algorithm that gives decent results? I'm looking for something that is just a few lines of C/C++.

我会永远想增加向量的大小，而新的载体可以是从1.1倍到50倍的电流矢量的大小的任何地方。

I'll always want to increase the size of the vector, and the new vector can be anywhere from 1.1x to 50x the size of the current vector.

谢谢！

推荐答案

下面是C ++的线性和二次插值。
interp1（5.3，一个，n）的是[5] + 0.3 *（一个[6] - 一个[5]），从0.3的方式〔 5]到[6];
interp1array（一，1000，B，1500）将延伸 A 到 B 。
interp2（5.3，A，N）通过3最近的点绘制一条抛物线A [4] [5] [6]：不是interp1顺畅，但仍快< BR> （花键使用4个最近点，但更流畅;如果你看过蟒蛇，看 <一href="http://advice.mechanicalkern.com/question/22/basic-spline-interpolation-in-a-few-lines-of-numpy">basic-spline-interpolation-in-a-few-lines-of-numpy.

Here's C++ for linear and quadratic interpolation.
interp1( 5.3, a, n ) is a[5] + .3 * (a[6] - a[5]), .3 of the way from a[5] to a[6];
interp1array( a, 1000, b, 1500 ) would stretch a to b.
interp2( 5.3, a, n ) draws a parabola through the 3 nearest points a[4] a[5] a[6]: smoother than interp1 but still fast.
(Splines use 4 nearest points, smoother yet; if you read python, see basic-spline-interpolation-in-a-few-lines-of-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" );
}

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