Bad linear fit for simple data

Question

I have some data :
data={{1.01074, 0.964488}, {1.08552, 0.993067}, {1.07907, 
  1.01836}, {1.0477, 1.03695}, {1.07717, 1.07973}, {1.10243, 
  1.08195}, {1.12669, 1.09112}, {1.09405, 1.09319}, {1.10857, 
  1.08445}, {1.18604, 1.08802}, {1.13138, 1.08727}, {1.18706, 
  1.08722}, {1.24118, 1.08473}, {1.27214, 1.08528}, {1.22428, 
  1.08384}, {1.30453, 1.08341}, {1.32046, 1.08277}, {1.32045, 
  1.07894}, {1.34901, 1.08288}, {1.35976, 1.08096}, {1.31244, 
  1.08093}, {1.28729, 1.08611}, {1.25115, 1.08975}, {1.18522, 
  1.09474}, {1.11788, 1.09777}, {1.00822, 0.964488}, {1.0938, 
  0.993067}, {1.10913, 1.01836}, {1.01039, 1.03695}, {1.02588, 
  1.07973}, {1.06003, 1.08195}, {1.06165, 1.09112}, {1.03693, 
  1.09319}, {1.01026, 1.08445}, {1.14019, 1.08802}, {1.03334, 
  1.08727}, {1.08583, 1.08722}, {1.17145, 1.08473}, {1.20567, 
  1.08528}, {1.13422, 1.08384}, {1.20849, 1.08341}, {1.27168, 
  1.08277}, {1.24355, 1.07894}, {1.25894, 1.08288}, {1.30205, 
  1.08096}, {1.18572, 1.08093}, {1.14212, 1.08611}, {1.08297, 
  1.08975}, {0.982202, 1.09474}, {0.861208, 1.09777}, {1.01326, 
  0.964488}, {1.07724, 0.993067}, {1.04902, 1.01836}, {1.08501, 
  1.03695}, {1.12847, 1.07973}, {1.14484, 1.08195}, {1.19174, 
  1.09112}, {1.15116, 1.09319}, {1.20687, 1.08445}, {1.23189, 
  1.08802}, {1.22942, 1.08727}, {1.28829, 1.08722}, {1.31091, 
  1.08473}, {1.33861, 1.08528}, {1.31435, 1.08384}, {1.40056, 
  1.08341}, {1.36924, 1.08277}, {1.39734, 1.07894}, {1.43907, 
  1.08288}, {1.41747, 1.08096}, {1.43915, 1.08093}, {1.43246, 
  1.08611}, {1.41933, 1.08975}, {1.38824, 1.09474}, {1.37454, 
  1.09777}}

And I tried to fit them :
ab = Fit[data, {1, x}, x]
Show[{ListPlot[data], Plot[ab, {x, 0, 2}, PlotStyle -> Red]}]

But it gives something very weird :

I don't get what's going on.... Could you help me please ?
Thx

flinty · Accepted Answer

Maybe you could use RANSAC to find inliers by consensus. This implementation isn't exactly right but it finds a pretty decent fit:
samplesize = 30;
inliers[fit_, points_, d_] :=
 Select[points, Abs[#[[2]] - (fit /. x -> #[[1]])] < d &]
votes = Association[# -> 0 & /@ data];
Do[
  sample = RandomSample[data, samplesize];
  fit = Fit[sample, {1, x}, x];
  Scan[votes[#] += 1 &, inliers[fit, data, 0.05]];
  , 2000];
finalfit = Fit[Keys[TakeLargest[votes, samplesize]], {1, x}, x];
Show[{ListPlot[data], Plot[finalfit, {x, 0, 2}, PlotStyle -> Red]}, PlotRange -> All]

Sjoerd Smit · Answer

Use PlotRange -> All. Most plot functions tend to throw away points that aren't nicely clustered with the bulk:
Show[{ListPlot[data, PlotRange -> All], Plot[ab, {x, 0, 2}, PlotStyle -> Red]}]

As you can see, there is a number of points that completely mess up the fit.

Rohit Namjoshi · Answer

You can also try Theil–Sen which is less sensitive to outliers. Using the WL implementation from this answer on your data gives slope, intercept of {0.0037716, 1.07855}. Plot of your data and a line with that slope, intercept.

Anton Antonov · Answer

Use Quantile Regression:
Import["https://raw.githubusercontent.com/antononcube/MathematicaForPrediction/master/MonadicProgramming/MonadicQuantileRegression.m"]

QRMonUnit[data]⟹
  QRMonQuantileRegressionFit[{1, x}]⟹
  QRMonLeastSquaresFit[{1, x}]⟹
  QRMonPlot;

(And, yes, that is a good example of Quantile Regression's robustness.)
Update
Instead of computing with the QRMon package utilized above, the computations can be done with the Wolfram Function Repository function QuantileRegression. That function uses B-splines, but if the fitting is made with one knot and interpolation order one then linear function fits are obtained.
probs = {0.25, 0.5, 0.75};
qFuncs = ResourceFunction["QuantileRegression"][data, 1, probs, InterpolationOrder -> 1];
Simplify[Through[qFuncs[x]]]
Show[{ListPlot[data, PlotStyle -> Gray, PlotRange -> All, ImageSize -> Large]},
 Plot[Evaluate[Through[qFuncs[x]]], {x, Min[data[[All, 1]]], 
   Max[data[[All, 1]]]}, PlotLegends -> probs, PlotTheme -> "Detailed"]]

Bad linear fit for simple data

4 Answers

Update

Add your own answers!

Ask a Question