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c++ - 使用 C++ 将曲线重新采样为偶数长度的段

转载 作者:塔克拉玛干 更新时间:2023-11-02 23:36:44 27 4
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使用 C++ 将曲线重采样为等长段的最佳方法是什么?我所拥有的是一组代表二维曲线的点。在下面的示例中,我有一个带有 x 和 y 分量的点结构和一个带有测试位置的点 vector 。每对点代表曲线上的一段。示例重采样曲线如下图所示。红色圆圈是原始位置,绿色圆圈是重采样后的目标位置。

Resample 1 enter image description here

struct Point
{
float x, y;
};


std::vector<Point> Points;
int numPoints = 5;

float positions[] = {
0.0350462, -0.0589667,
0.0688311, 0.240896,
0.067369, 0.557199,
-0.024258, 0.715255,
0.0533231, 0.948694,
};

// Add points
int offset = 2;
for (int i =0; i < numPoints; i++)
{
offset = i * 2;
Point pt;
pt.x = positions[offset];
pt.y = positions[offset+1];
Points.push_back(pt);

}

最佳答案

看看这是否适合您。重采样点在源 vector 点的线性插值上彼此等距。

#include <iostream>
#include <iomanip>
#include <vector>
#include <cmath>

struct Point {
double x, y;
};

// Distance gives the Euclidean distance between two Points
double Distance(const Point& a, const Point& b) {
const double dx = b.x - a.x;
const double dy = b.y - a.y;
const double lsq = dx*dx + dy*dy;
return std::sqrt(lsq);
}

// LinearCurveLength calculates the total length of the linear
// interpolation through a vector of Points. It is the sum of
// the Euclidean distances between all consecutive points in
// the vector.
double LinearCurveLength(std::vector<Point> const &points) {
auto start = points.begin();
if(start == points.end()) return 0;
auto finish = start + 1;
double sum = 0;
while(finish != points.end()) {
sum += Distance(*start, *finish);
start = finish++;
}
return sum;
}

// Gives a vector of Points which are sampled as equally-spaced segments
// taken along the linear interpolation between points in the source.
// In general, consecutive points in the result will not be equidistant,
// because of a corner-cutting effect.
std::vector<Point> UniformLinearInterpolation(std::vector<Point> const &source, std::size_t target_count) {
std::vector<Point> result;
if(source.size() < 2 || target_count < 2) {
// degenerate source vector or target_count value
// for simplicity, this returns an empty result
// but special cases may be handled when appropriate for the application
return result;
}
// total_length is the total length along a linear interpolation
// of the source points.
const double total_length = LinearCurveLength(source);

// segment_length is the length between result points, taken as
// distance traveled between these points on a linear interpolation
// of the source points. The actual Euclidean distance between
// points in the result vector can vary, and is always less than
// or equal to segment_length.
const double segment_length = total_length / (target_count - 1);

// start and finish are the current source segment's endpoints
auto start = source.begin();
auto finish = start + 1;

// src_segment_offset is the distance along a linear interpolation
// of the source curve from its first point to the start of the current
// source segment.
double src_segment_offset = 0;

// src_segment_length is the length of a line connecting the current
// source segment's start and finish points.
double src_segment_length = Distance(*start, *finish);

// The first point in the result is the same as the first point
// in the source.
result.push_back(*start);

for(std::size_t i=1; i<target_count-1; ++i) {
// next_offset is the distance along a linear interpolation
// of the source curve from its beginning to the location
// of the i'th point in the result.
// segment_length is multiplied by i here because iteratively
// adding segment_length could accumulate error.
const double next_offset = segment_length * i;

// Check if next_offset lies inside the current source segment.
// If not, move to the next source segment and update the
// source segment offset and length variables.
while(src_segment_offset + src_segment_length < next_offset) {
src_segment_offset += src_segment_length;
start = finish++;
src_segment_length = Distance(*start, *finish);
}
// part_offset is the distance into the current source segment
// associated with the i'th point's offset.
const double part_offset = next_offset - src_segment_offset;

// part_ratio is part_offset's normalized distance into the
// source segment. Its value is between 0 and 1,
// where 0 locates the next point at "start" and 1
// locates it at "finish". In-between values represent a
// weighted location between these two extremes.
const double part_ratio = part_offset / src_segment_length;

// Use part_ratio to calculate the next point's components
// as weighted averages of components of the current
// source segment's points.
result.push_back({
start->x + part_ratio * (finish->x - start->x),
start->y + part_ratio * (finish->y - start->y)
});
}

// The first and last points of the result are exactly
// the same as the first and last points from the input,
// so the iterated calculation above skips calculating
// the last point in the result, which is instead copied
// directly from the source vector here.
result.push_back(source.back());
return result;
}

int main() {
std::vector<Point> points = {
{ 0.0350462, -0.0589667},
{ 0.0688311, 0.240896 },
{ 0.067369, 0.557199 },
{-0.024258, 0.715255 },
{ 0.0533231, 0.948694 }
};

std::cout << "Source Points:\n";
for(const auto& point : points) {
std::cout << std::setw(14) << point.x << " " << std::setw(14) << point.y << '\n';
}
std::cout << '\n';
auto interpolated = UniformLinearInterpolation(points, 7);
std::cout << "Interpolated Points:\n";
for(const auto& point : interpolated) {
std::cout << std::setw(14) << point.x << " " << std::setw(14) << point.y << '\n';
}
std::cout << '\n';
std::cout << "Source linear interpolated length: " << LinearCurveLength(points) << '\n';
std::cout << "Interpolation's linear interpolated length: " << LinearCurveLength(interpolated) << '\n';
}

关于c++ - 使用 C++ 将曲线重新采样为偶数长度的段,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/43726836/

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