Fit scatter plot with a curve
28 次查看(过去 30 天)
显示 更早的评论
I'm trying to fit the following data (here plotted using scatter)
with a curve so that the result will be something like this
I tried with polyfit and polyval but failed, so probably I used them in the wrong way, any help?
x = [0.2337;0.296;0.3071;0.4208;0.2055;0.9597;0.8683;0.243;0.3363;0.2793;0.5292;0.2471;0.2282;0.4774;1.0392;0.4361;0.1367;0.2952;0.1983;1.0468;0.906;0.9578;0.5368;0.5956;0.8616;0.1641;0.1312;1.0381;0.2361;0.4668;0.7477;0.5303;1.367;1.0894;1.2836;0.2487;0.5869;0.8664;0.3446;0.5062;0.7245;1.3289;0.4958;1.6644;0.2826;0.6825;0.103;0.3205;0.4456;0.1835;0.2622;0.0673;0.4219;0.639;0.7599;0.2172;0.5491;0.6694;0.3774;1.1869;0.7206;0.9669;0.0672;0.6705;0.1681;1.5364;0.3779;0.3483;0.5097;1.7493;0.5388;0.4481;0.2657;1.2815;0.9019;0.9402;0.12;0.4465;1.0316;0.5493;1.0942;0.2359;0.1906;2.1019;0.9408;0.8557;0.1598;0.9746;0.3083;1.0001;0.9645;0.498;0.0614;0.1956;0.7869;1.2872;0.4342;0.0462];
y = [0.0136;0.0075;0.0089;0.0088;0.0104;0.0153;0.0024;0.01;0.0047;0.0137;0.0026;0.0094;0.0093;0.0044;0.013;0.0018;0.0154;0.0058;0.0107;0.011;0.0019;0.013;0.0078;0.0071;0.0018;0.0204;0.0179;0.007;0.0119;0.0013;0.0142;0.022;0.0182;0.0054;0.0434;0.0079;0.0008;0.0066;0.0039;0.0009;0.0018;0.0199;0.0107;0.0326;0.0092;0.0013;0.0194;0.0057;0.0471;0.0133;0.0139;0.0255;0.016;0.0016;0.0013;0.0093;0.0011;0.0014;0.0233;0.0217;0.0003;0.004;0.0251;0.0049;0.0133;0.0316;0.0029;0.0082;0.0156;0.0476;0.0013;0.0016;0.0079;0.0307;0.013;0.012;0.0239;0.0134;0.0203;0.0007;0.013;0.009;0.0146;0.0556;0.0221;0.0027;0.0159;0.0037;0.0053;0.0035;0.0076;0.0008;0.0291;0.0162;0.017;0.0418;0.0146;0.033];
f = polyfit(x, y, 5);
v = polyval(f, x);
plot(x,y,'o', x,v,'-')
0 个评论
采纳的回答
Alan Stevens
2021-12-2
More like ths?
x = [0.2337;0.296;0.3071;0.4208;0.2055;0.9597;0.8683;0.243;0.3363;0.2793;0.5292;0.2471;0.2282;0.4774;1.0392;0.4361;0.1367;0.2952;0.1983;1.0468;0.906;0.9578;0.5368;0.5956;0.8616;0.1641;0.1312;1.0381;0.2361;0.4668;0.7477;0.5303;1.367;1.0894;1.2836;0.2487;0.5869;0.8664;0.3446;0.5062;0.7245;1.3289;0.4958;1.6644;0.2826;0.6825;0.103;0.3205;0.4456;0.1835;0.2622;0.0673;0.4219;0.639;0.7599;0.2172;0.5491;0.6694;0.3774;1.1869;0.7206;0.9669;0.0672;0.6705;0.1681;1.5364;0.3779;0.3483;0.5097;1.7493;0.5388;0.4481;0.2657;1.2815;0.9019;0.9402;0.12;0.4465;1.0316;0.5493;1.0942;0.2359;0.1906;2.1019;0.9408;0.8557;0.1598;0.9746;0.3083;1.0001;0.9645;0.498;0.0614;0.1956;0.7869;1.2872;0.4342;0.0462];
y = [0.0136;0.0075;0.0089;0.0088;0.0104;0.0153;0.0024;0.01;0.0047;0.0137;0.0026;0.0094;0.0093;0.0044;0.013;0.0018;0.0154;0.0058;0.0107;0.011;0.0019;0.013;0.0078;0.0071;0.0018;0.0204;0.0179;0.007;0.0119;0.0013;0.0142;0.022;0.0182;0.0054;0.0434;0.0079;0.0008;0.0066;0.0039;0.0009;0.0018;0.0199;0.0107;0.0326;0.0092;0.0013;0.0194;0.0057;0.0471;0.0133;0.0139;0.0255;0.016;0.0016;0.0013;0.0093;0.0011;0.0014;0.0233;0.0217;0.0003;0.004;0.0251;0.0049;0.0133;0.0316;0.0029;0.0082;0.0156;0.0476;0.0013;0.0016;0.0079;0.0307;0.013;0.012;0.0239;0.0134;0.0203;0.0007;0.013;0.009;0.0146;0.0556;0.0221;0.0027;0.0159;0.0037;0.0053;0.0035;0.0076;0.0008;0.0291;0.0162;0.017;0.0418;0.0146;0.033];
f = polyfit(x, y, 5);
xx = 0:0.1:2.5; %%%%%%%%%%%%%%%%%%%%%%
v = polyval(f, xx); %%%%%%%%%%%%%%%%%%%
plot(x,y,'o', xx,v,'-')
3 个评论
Alan Stevens
2021-12-2
You could split it into regions. For example:
x = [0.2337;0.296;0.3071;0.4208;0.2055;0.9597;0.8683;0.243;0.3363;0.2793;0.5292;0.2471;0.2282;0.4774;1.0392;0.4361;0.1367;0.2952;0.1983;1.0468;0.906;0.9578;0.5368;0.5956;0.8616;0.1641;0.1312;1.0381;0.2361;0.4668;0.7477;0.5303;1.367;1.0894;1.2836;0.2487;0.5869;0.8664;0.3446;0.5062;0.7245;1.3289;0.4958;1.6644;0.2826;0.6825;0.103;0.3205;0.4456;0.1835;0.2622;0.0673;0.4219;0.639;0.7599;0.2172;0.5491;0.6694;0.3774;1.1869;0.7206;0.9669;0.0672;0.6705;0.1681;1.5364;0.3779;0.3483;0.5097;1.7493;0.5388;0.4481;0.2657;1.2815;0.9019;0.9402;0.12;0.4465;1.0316;0.5493;1.0942;0.2359;0.1906;2.1019;0.9408;0.8557;0.1598;0.9746;0.3083;1.0001;0.9645;0.498;0.0614;0.1956;0.7869;1.2872;0.4342;0.0462];
y = [0.0136;0.0075;0.0089;0.0088;0.0104;0.0153;0.0024;0.01;0.0047;0.0137;0.0026;0.0094;0.0093;0.0044;0.013;0.0018;0.0154;0.0058;0.0107;0.011;0.0019;0.013;0.0078;0.0071;0.0018;0.0204;0.0179;0.007;0.0119;0.0013;0.0142;0.022;0.0182;0.0054;0.0434;0.0079;0.0008;0.0066;0.0039;0.0009;0.0018;0.0199;0.0107;0.0326;0.0092;0.0013;0.0194;0.0057;0.0471;0.0133;0.0139;0.0255;0.016;0.0016;0.0013;0.0093;0.0011;0.0014;0.0233;0.0217;0.0003;0.004;0.0251;0.0049;0.0133;0.0316;0.0029;0.0082;0.0156;0.0476;0.0013;0.0016;0.0079;0.0307;0.013;0.012;0.0239;0.0134;0.0203;0.0007;0.013;0.009;0.0146;0.0556;0.0221;0.0027;0.0159;0.0037;0.0053;0.0035;0.0076;0.0008;0.0291;0.0162;0.017;0.0418;0.0146;0.033];
xbreak = 1.02; % change value as desired
n = size(find(x<=xbreak),1); % Number of values <= xbreak
z = [x y]; % Combine for sorting
z = sortrows(z); % Sort rows based on first column of z, namely the x values
% Separate data into two sets
xlo = z(1:n,1); ylo = z(1:n,2);
xhi = z(n+1:end,1); yhi = z(n+1:end,2);
plot(xlo,ylo,'o',xhi,yhi,'s'),grid % plot data
hold on
% Fit separate curves to each set
flo = polyfit(xlo,ylo,2);
fhi = polyfit(xhi,yhi,2);
% Construct fitted curves
xxlo = 0:0.1:xbreak;
xxhi = xbreak:0.1:2.5;
vlo = polyval(flo, xxlo);
vhi = polyval(fhi, xxhi);
% plot fitted curves on top of data
plot(xxlo,vlo,xxhi,vhi)
Only you can decide if the results make sense!
更多回答(1 个)
Image Analyst
2021-12-2
Not sure if the scattered data is legitimate or noise. The bottom of the data looks like a nice polynomial. So if you want to fit just the highly clustered points along the bottom and ignore some of the outliers in the middle, you could try fitPolynomialRANSAC if you have the Computer Vision Toolbox.
另请参阅
类别
在 Help Center 和 File Exchange 中查找有关 Fit Postprocessing 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!