How can I perform a piecewise linear fit to data?

I have y-axis data that varies linearly in three regions over 0-1 on the x-axis. I would like to obtain a simple piecewise linear fit to get the 2-break points and the three linear fits. Is there a function that I could use? I have found nlhw that might do it, but I cannot figure out how to use it from the documentation. Would appreciate any help in using it or any other alternative.

Answer 1

[Image Analyst]

Do you have the location of the breakpoints? If so, simply use polyfit(). If not, then you should fit a line through some number of points, like 3 or 5, and scan that across your data. You can get the estimates slope at every location. Then plot the slopes and look for where it changes a lot. You can use diff() to find out where the slope changes substantially. Attach some sample data if you want people to kindly help you by writing code for you.

[Priyank Pathak]

Please can you solve my issue?

I have same issue as @Ranjan Sonalkar

I have y-axis data that varies nearly linearly in three regions over 0-0.6 on the x-axis. I would like to obtain a simple piecewise linear fit to get the 2-break points and the three linear fits. Is there a function that I could use? Or can you develop some code?

From left(x=0) the first slope should be around 95, and the second clope around 33.

x=linspace(0,0.06,129);
y=[5.07742010175952
5.31185099410990
5.10691673392019
4.74615958521947
4.19177682576652
3.37068344802045
2.37609787828215
2.12259276016394
1.85196259843330
1.56974576757917
1.48983435635157
1.50485054453710
1.20735946054088
0.661344454963583
1.13352524509828
1.06028957687000
0.599328807494000
1.03613472954618
0.987301049044117
0.217298697691901
0.920810108362981
1.05394744136997
0.281153643572616
0.469681534710498
0.897003388264222
0.327421850031165
0.183590736373175
0.857998112549593
0.581121685434373
-0.125993250940098
0.683500160910840
0.534539058937860
-0.396726323212621
0.634482480611096
0.658943411183352
-0.438952316218105
0.345284196220544
0.600667158720749
-0.115141464323266
0.120829134737583
0.548098646887191
0.0463178869874998
-0.113048252136384
0.512689921996228
0.231986574873253
-0.207338103120708
0.504643103976643
0.434781176498638
-0.0110627611202763
0.489236905438070
0.571542855334634
0.195839245448528
0.318850093243477
0.454160515353964
0.142640929539695
0.310785155059393
0.512602213738191
0.162014480668000
-0.0607597283337748
0.358834860427813
0.138829100632712
-0.384114894557834
0.196010083174632
0.0877202333255821
-0.439559931012924
0.240175548664580
0.281973380388760
-0.417220964330667
-0.0796052840983549
0.154742142117545
-0.393344891329877
-0.0605607647867973
0.277797879414859
-0.0511663599311094
-0.188885994780561
0.236644526326914
0.0320718007648540
-0.420498358462674
0.116044715556200
0.0456658813295294
-0.481396963008032
0.0686941440278348
0.134350947055467
-0.392697383096293
-0.0822493346453244
0.0840913913909094
-0.560007006542937
-0.309347358363015
0.100439695825144
-0.305329002501373
-0.634393763791120
0.00828091935610617
-0.199591976608657
-0.527071260057054
0.0538501718509587
-0.0536584297550401
-1.06586660939429
-0.192609804740431
-0.117750795306160
-1.23944999109543
-0.238042348232044
0.0205644121511571
-0.516135755182763
-0.362229685328809
0.00351231197974391
-0.465633808298506
-0.632540256198356
0.0215932583608414
-0.187152394729513
-1.07708193704972
-0.198132189949225
-0.270724057886369
-0.967609157331435
-0.109900894024127
-0.0638341059632911
-1.06220258506714
-0.387954592612164
-0.115951129740455
-0.927157022631441
-0.440143390491659
0.0171476050596097
-0.365295340437865
-0.922551001315555
-0.179188776250813
-0.469657233191098
-0.965827555014108
-0.0694730732513197
-0.154504170420157
-1.25531875940010]

Thanks in advance.

[Image Analyst]

@Priyank Pathak, this does not look like a piecewise linear plot:

It looks more like an exponential decay. I'm attaching a demo for that. Or if piecewise linear, just two regions, not three, and so my existing code should work. You may have to adjust the range over where the split occurs.

[Priyank Pathak]

okay @Image Analyst ......

i will try

thank you

Answer 2

[Greg Dionne]

If you have the R2016a Signal Processing Toolbox, you can try FINDCHANGEPTS.