longbiscuit / EquidistantSamplingPoints

Remove points that are too dense on the curve so that the remaining points are approximately equally spaced

EquidistantSamplingPoints

Remove points that are too dense on the curve so that the remaining points are approximately equally spaced

如何按弧长等分曲线并取出等分点坐标？ 科研中经常需要删除曲线中多余的点，但由于曲线斜率不一致，导致按x或按y等间距取点都不合适，需要按弧长等间距取点，且试验数据是离散点，请问如何做？

原始数据链接：https://pan.baidu.com/s/1vU_qHDDdPDnhrbJCgdyozw?pwd=7bmk

提取码：7bmk

解决：

matlab 2021a 代码如下：

clc;
clear all;
close all;
a=1e-5;
b=1e-2;
x=linspace(0,0.2,1000);% You can replace x here with your own discrete data
y=x./(a+b*x);% You can replace y here with your own discrete data
fig1=figure('Name','Compare');
plot(x,y,'k.-');hold on;


numPoint=40;% self defination of point number you want

% normative approach
% if max(abs(x))>max(abs(y))
%     ti=max(abs(x))/max(abs(y));
%     xo=x;
%     yo=y*ti;
% else
%     ti=max(abs(y))/max(abs(x));
%     xo=x*ti;
%     yo=y;
% end

xo=(x-min(x))/(max(x)-min(x));
yo=(y-min(y))/(max(y)-min(y));

fig2=figure('Name','xo-yo compare');
plot(xo,yo,'k.--');hold on;

% get the curve length
D = pdist([xo',yo']);% solve the distance of points:D(i)=p(i)-p(1)
curLen=D(numel(x)-1);% total length of curve


oneCurLen=curLen/numPoint;

saveIndex(1)=1;
num=1;
if D(1)>oneCurLen % p(2)-p(1)
    saveIndex(2)=1;
    num=2;
else
    saveIndex(2)=0;
end

for ii=2:numel(xo)
    if ((D(ii)-D(num))>oneCurLen)
        saveIndex(ii+1)=1;
        num=ii;
    else
        saveIndex(ii+1)=0;
    end
end

xs=x((saveIndex>0));% xs is the sample point you want
ys=y((saveIndex>0));% ys is the sample point you want

figure(fig1)
plot(xs,ys,'bo');hold on;
xlabel('x')
ylabel('y')
legend({'x-y','xs-ys'});

figure(fig2)
xos=xo(saveIndex>0);
yos=yo(saveIndex>0);
plot(xos,yos,'bo');hold on;
xlabel('x')
ylabel('y')
legend({'xo-yo','xos-yos'});

xs和ys就是抽取的等距样本点，由于没进行曲线插值运算，所以此处为近似等距点。