fitting a curve on 3d data

i have 3d data and I have plotted it using

scatter3(data(:,3),data(:,1) ,data(:,2));

I want to plot a curve along this 3d data, something like

Also how once i have plotted i want to extend the curve and get the x and y points (data(:,1) and data(:,2)) till my z points (data(:,3)) becomes 0. z usually starts from 35 and goes till 128. once the curve is fit i want to know the x and y points corresponding to z(0:34). Uploading the data.mat.

[Matt J]

The extrapolation that you get will depend very much on the model you select for the curve. I assume you have a model in mind? Why else would you have you departed from the initial model of the points as a line?

[Akira Agata]

Looking at you plot, several points are not on the fitting curve. Is it possible to ignore them when calculating the fitting curve ?

[prashant singh]

@Matt the line was ok but it only covers like 10% of the data. I have a lot data like this the line wasn't as generalize solution as i thought it would be.

[prashant singh]

@Akria Agata Yes its fine if we ignore the noisy part.

Answer 1

[John D'Errico]

UGH. If I understand you, this curve has a set of points that follow no known model, but you want to fit them, in 3-D. So this becomes an errors in variables kind of thing, but with no known model, what must be effectively a spline is the only thing that makes sense here.

I think, if I understand what you said, that those points that are not on the curve are identifiable (and can thus be segregated from the fit) have z==0. So at least you know which points do not belong on the curve.

That first part of the curve seems to be pretty linear to me. Then there is a fairly sharp break into a curved segment. And you are telling us that you now want to extrapolate that curved segment over a moderately significant distance? Sigh. Why do I want to insert Mark Twain's comments on the dangers of extrapolation here? Worse, this is also moderately noisy data?

That is, when you use a model that can fit a curved segment there, the fit at the end (and any extrapolation that you do) will be primarily based on the last few points on the curve. So if you were able to put confidence limits around the extrapolated curve near the end, they would get very wide very quickly.

Do you know where/when the break occurs?

I might also note that it LOOKS like there are linear segments on this curve. Thus the curved region even seems to be composed of approximately linear segments. I might guess that curved region is three linear segments. But finding break points, even in a piecewise linear, but continuous curve, in three dimensions, with errors in variables is not a trivial thing.

Do you know some reason why the curve appears to have vaguely linear segments? That seems to be suggested, given that is what you drew.

Did I say UGH? Is this something you need to do ONCE? Or is it something you need to do automatically for multiple sets of data? (Please don't tell me the latter is the case.)

You have not provided the data itself. That makes me worry/wonder if this is something you will be doing multiple times. (Edit: oops, I see you did attach the data!)

For example, if I were needing to do this ONCE, I might artificially pick the breaks by eye. Once I had those points, I could at least use the last segment to fit a straight line using an SVD to do the errors in variables fit, then extrapolating that line.

[John D'Errico]

Here is the data you have given us:

data =
 252   694    36
 249   686    37
 253   669    38
 252   671    39
 250   663    40
 250   655    41
 247   654    42
 248   644    43
 250   633    44
 246   632    45
 243   623    46
 249   615    47
 247   610    48
 244   602    49
 247   591    50
 242   586    51
 242   584    52
 248   570    53
 244   568    54
 244   560    55
 240   557    56
 242   550    57
 242   541    58
 245   532    59
 244   525    60
 242   518    61
 246   514    62
 237   504    63
 191     1    64
 184     2    65
 173     1    66
 169     2    67
 168     2    68
 174     1    69
 182     2    70
 245   447    71
 238   444    72
 239   435    73
 241   428    74
 238   423    75
 239   416    76
 238   403    77
 238   400    78
 235   387    79
 243   385    80
 238   381    81
 236   370    82
 231   364    83
 232   356    84
 233   342    85
 231   344    86
 230   335    87
 233   325    88
 235   313    89
 233   306    90
 229   299    91
 234   283    92
 225   252    93
 231   230    94
 225   200    95
 228   185    96
 233   174    97
 231   161    98
 224   151    99
 225   139   100
 229   130   101
 241   120   102
 230   115   103
 234   108   104
 235    98   105
 228    91   106
 230    82   107
 231    79   108
 224    70   109
 228    59   110
 227    51   111
 229    45   112
 226    39   113
 227    24   114
 237    19   115
 226    17   116
 222     5   117
 218     2   118
 211     3   119
 209     2   120
 205     1   121
 207     1   122
 202     1   123
 199     3   124
 196     1   125
 195     3   126
 262     1   127

So, not only is this an errors in variables thing, but it is a curve defined by points that are not even apparently connected in time. Only a loose connection exists, because there appear to be groups of points that ARE connected.

My headache is starting to get worse. Yes, this is early New Years day, but I don't drink, so that is not the issue. ;-)

Worse yet.

So the data you have attached is not the same curve as the one you show in the picture, and the noise appears to be significantly larger than it did for the curve you drew for us.

Multiple sets of data tells me this is a problem you will face multiple times, so you are looking for some sort of automatic solution.

SIGH.

[prashant singh]

To view the data, please switch the dimension to

scatter3(data(:,3),data(:,1) ,data(:,2));

Also you can ignore the noisy part .. i just want to fit some curve even if it covers like 50% to 60% of data. I have a lot other data like this and i want to find some generalised solution for that

[John D'Errico]

It does not matter what variable is plotted on which axis. That does not improve the data or make the problem simpler.

Answer 2

[Image Analyst]

I thought I already answered this in your prior post. Basically you parameterize the 3 coordinates and do a fit, cleaning the data beforehand if desired. What was wrong with that solution, except that you didn't want the fitted red line to actually be plotted, which is an easy thing to get rid of?