Fit scatter plot with a curve
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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 Comments
Accepted Answer
Alan Stevens
on 2 Dec 2021
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 Comments
Alan Stevens
on 2 Dec 2021
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!
More Answers (1)
Image Analyst
on 2 Dec 2021
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.
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