Fit scatter plot with a curve

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,'-')

Answer 1

[Alan Stevens]

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,'-')

[giannit]

Thanks for the kind reply, this is very good.

What if we want to obtain a curve looking more sensible to the variation in the data? If you look closely at the second image, you see that there is a small jump at about x=0.9. You can see this effect more clearly in the following two plots.

Should we use something like smoothdata? Many thanks

[Alan Stevens]

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!

[giannit]

Thank you very much!

Answer 2

[Image Analyst]

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.

https://en.wikipedia.org/wiki/Random_sample_consensus

[giannit]

Thank you very much!