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Fitting two-dimensional data
How to fit polynomial curve using Mathematica for multiple inputs?How to perform a multi-peak fitting?Obtaining the envelope of an oscillating function and perform fittingStill more problems with Fitting, please helpHow to find the best fit for given input data?Smoothing out data points in a ListDensityPlotHow to do deconvolution in spectrum with some peaks?Obtaining the interpolation function of a surfaceUsing tables to find values of functions from given dataFind elementary function that fits the dataMulti-peak fitting for peak position
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
2
$begingroup$
For this question, I cannot use random sample data. So the actual data can be found here. The data file contains three columns, where the first two are the coordinates $(x,y)$, while the third is the value of a function $f$. Now we plot them, thus obtaining the shape of $f$
data = Import["L1.dat", "Table"];
or
data = Import["https://pastebin.com/raw/YMCFB4mK", "TSV"]
Plot
L0 = ListPlot3D[data]
My question is the following: is there a way to interpolate the data and obtain an analytical fitting function $f(x,y)$? Taking into account that the distribution of $f$ is rather smooth, without peaks and holes, I suppose it should be rather easy to obtain its fitting function. Any ideas?
fitting interpolation data
edited Jul 21 at 10:53
rhermans
23.5k4 gold badges42 silver badges110 bronze badges
asked Jul 21 at 10:25
Vaggelis_ZVaggelis_Z
3,7662 gold badges24 silver badges58 bronze badges
$endgroup$
add a comment |
2
$begingroup$
For this question, I cannot use random sample data. So the actual data can be found here. The data file contains three columns, where the first two are the coordinates $(x,y)$, while the third is the value of a function $f$. Now we plot them, thus obtaining the shape of $f$
data = Import["L1.dat", "Table"];
or
data = Import["https://pastebin.com/raw/YMCFB4mK", "TSV"]
Plot
L0 = ListPlot3D[data]
My question is the following: is there a way to interpolate the data and obtain an analytical fitting function $f(x,y)$? Taking into account that the distribution of $f$ is rather smooth, without peaks and holes, I suppose it should be rather easy to obtain its fitting function. Any ideas?
fitting interpolation data
edited Jul 21 at 10:53
rhermans
23.5k4 gold badges42 silver badges110 bronze badges
asked Jul 21 at 10:25
Vaggelis_ZVaggelis_Z
3,7662 gold badges24 silver badges58 bronze badges
$endgroup$
$begingroup$
have you triedInterpolation? E.g.,iF = Interpolation[data];Plot3D[iF[x, y], x, 0, .5, y, 0, 1]?
$endgroup$
– kglr
Jul 21 at 10:39
$begingroup$
@kglr I want to obtain an analytical equation of the form f = ax^2 + b*y^2 +....
$endgroup$
– Vaggelis_Z
Jul 21 at 10:42
$begingroup$
It's eitherInterpolationor fit. What is it?
$endgroup$
– rhermans
Jul 21 at 10:54
$begingroup$
Thanks for accepting my answer, but I think you were too hasty doing that. While accepting is one of the things to do after your question is answered, we recommend that users should test answers before voting and wait 24 hours before accepting the best one. That allows people in all timezones to answer your question and an opportunity for other users to point alternatives, caveats or limitations of the available answers. I have edited my answer so you can accept the best one later.
$endgroup$
– rhermans
Jul 21 at 11:28
add a comment |
2
2
2
3
$begingroup$
For this question, I cannot use random sample data. So the actual data can be found here. The data file contains three columns, where the first two are the coordinates $(x,y)$, while the third is the value of a function $f$. Now we plot them, thus obtaining the shape of $f$
data = Import["L1.dat", "Table"];
or
data = Import["https://pastebin.com/raw/YMCFB4mK", "TSV"]
Plot
L0 = ListPlot3D[data]
My question is the following: is there a way to interpolate the data and obtain an analytical fitting function $f(x,y)$? Taking into account that the distribution of $f$ is rather smooth, without peaks and holes, I suppose it should be rather easy to obtain its fitting function. Any ideas?
fitting interpolation data
edited Jul 21 at 10:53
rhermans
23.5k4 gold badges42 silver badges110 bronze badges
asked Jul 21 at 10:25
Vaggelis_ZVaggelis_Z
3,7662 gold badges24 silver badges58 bronze badges
$endgroup$
For this question, I cannot use random sample data. So the actual data can be found here. The data file contains three columns, where the first two are the coordinates $(x,y)$, while the third is the value of a function $f$. Now we plot them, thus obtaining the shape of $f$
data = Import["L1.dat", "Table"];
or
data = Import["https://pastebin.com/raw/YMCFB4mK", "TSV"]
Plot
L0 = ListPlot3D[data]
My question is the following: is there a way to interpolate the data and obtain an analytical fitting function $f(x,y)$? Taking into account that the distribution of $f$ is rather smooth, without peaks and holes, I suppose it should be rather easy to obtain its fitting function. Any ideas?
fitting interpolation data
fitting interpolation data
edited Jul 21 at 10:53
rhermans
23.5k4 gold badges42 silver badges110 bronze badges
asked Jul 21 at 10:25
Vaggelis_ZVaggelis_Z
3,7662 gold badges24 silver badges58 bronze badges
edited Jul 21 at 10:53
rhermans
23.5k4 gold badges42 silver badges110 bronze badges
asked Jul 21 at 10:25
Vaggelis_ZVaggelis_Z
3,7662 gold badges24 silver badges58 bronze badges
edited Jul 21 at 10:53
rhermans
23.5k4 gold badges42 silver badges110 bronze badges
edited Jul 21 at 10:53
rhermans
23.5k4 gold badges42 silver badges110 bronze badges
edited Jul 21 at 10:53
rhermans
23.5k4 gold badges42 silver badges110 bronze badges
23.5k4 gold badges42 silver badges110 bronze badges
asked Jul 21 at 10:25
Vaggelis_ZVaggelis_Z
3,7662 gold badges24 silver badges58 bronze badges
asked Jul 21 at 10:25
Vaggelis_ZVaggelis_Z
3,7662 gold badges24 silver badges58 bronze badges
asked Jul 21 at 10:25
Vaggelis_ZVaggelis_Z
3,7662 gold badges24 silver badges58 bronze badges
3,7662 gold badges24 silver badges58 bronze badges
$begingroup$
have you triedInterpolation? E.g.,iF = Interpolation[data];Plot3D[iF[x, y], x, 0, .5, y, 0, 1]?
$endgroup$
– kglr
Jul 21 at 10:39
$begingroup$
@kglr I want to obtain an analytical equation of the form f = ax^2 + b*y^2 +....
$endgroup$
– Vaggelis_Z
Jul 21 at 10:42
$begingroup$
It's eitherInterpolationor fit. What is it?
$endgroup$
– rhermans
Jul 21 at 10:54
$begingroup$
Thanks for accepting my answer, but I think you were too hasty doing that. While accepting is one of the things to do after your question is answered, we recommend that users should test answers before voting and wait 24 hours before accepting the best one. That allows people in all timezones to answer your question and an opportunity for other users to point alternatives, caveats or limitations of the available answers. I have edited my answer so you can accept the best one later.
$endgroup$
– rhermans
Jul 21 at 11:28
add a comment |
$begingroup$
have you triedInterpolation? E.g.,iF = Interpolation[data];Plot3D[iF[x, y], x, 0, .5, y, 0, 1]?
$endgroup$
– kglr
Jul 21 at 10:39
$begingroup$
@kglr I want to obtain an analytical equation of the form f = ax^2 + b*y^2 +....
$endgroup$
– Vaggelis_Z
Jul 21 at 10:42
$begingroup$
It's eitherInterpolationor fit. What is it?
$endgroup$
– rhermans
Jul 21 at 10:54
$begingroup$
Thanks for accepting my answer, but I think you were too hasty doing that. While accepting is one of the things to do after your question is answered, we recommend that users should test answers before voting and wait 24 hours before accepting the best one. That allows people in all timezones to answer your question and an opportunity for other users to point alternatives, caveats or limitations of the available answers. I have edited my answer so you can accept the best one later.
$endgroup$
– rhermans
Jul 21 at 11:28
$begingroup$
have you tried
Interpolation? E.g., iF = Interpolation[data];Plot3D[iF[x, y], x, 0, .5, y, 0, 1]?$endgroup$
– kglr
Jul 21 at 10:39
$begingroup$
have you tried
Interpolation? E.g., iF = Interpolation[data];Plot3D[iF[x, y], x, 0, .5, y, 0, 1]?$endgroup$
– kglr
Jul 21 at 10:39
$begingroup$
@kglr I want to obtain an analytical equation of the form f = ax^2 + b*y^2 +....
$endgroup$
– Vaggelis_Z
Jul 21 at 10:42
$begingroup$
@kglr I want to obtain an analytical equation of the form f = ax^2 + b*y^2 +....
$endgroup$
– Vaggelis_Z
Jul 21 at 10:42
$begingroup$
It's either
Interpolation or fit. What is it?$endgroup$
– rhermans
Jul 21 at 10:54
$begingroup$
It's either
Interpolation or fit. What is it?$endgroup$
– rhermans
Jul 21 at 10:54
$begingroup$
Thanks for accepting my answer, but I think you were too hasty doing that. While accepting is one of the things to do after your question is answered, we recommend that users should test answers before voting and wait 24 hours before accepting the best one. That allows people in all timezones to answer your question and an opportunity for other users to point alternatives, caveats or limitations of the available answers. I have edited my answer so you can accept the best one later.
$endgroup$
– rhermans
Jul 21 at 11:28
$begingroup$
Thanks for accepting my answer, but I think you were too hasty doing that. While accepting is one of the things to do after your question is answered, we recommend that users should test answers before voting and wait 24 hours before accepting the best one. That allows people in all timezones to answer your question and an opportunity for other users to point alternatives, caveats or limitations of the available answers. I have edited my answer so you can accept the best one later.
$endgroup$
– rhermans
Jul 21 at 11:28
add a comment |
2 Answers
2
active
oldest
votes
2
$begingroup$
Interpolation
intFunc = With[
order = 1,
dat = Import["https://pastebin.com/raw/YMCFB4mK", "TSV"]
,
Interpolation[#1, #2, #3 & @@@ dat, InterpolationOrder -> order]
]
Plot3D[
intFunc[x, y]
, x, 0.001, 1/2
, y, 0, 1
, PlotStyle -> Green
]
Fit
fitModel = With[
order = 4,
dat = Import["https://pastebin.com/raw/YMCFB4mK", "TSV"]
,
LinearModelFit[
dat
, Flatten@Outer[Times, Sequence @@ Transpose@Array[Power[x, y, # - 1] &, order + 1]]
, x, y
]
]
fitModel[x,y]
(* 0.839678 - 3.39587 x + 10.2762 x^2 - 23.2243 x^3 + 18.9954 x^4 -
0.870772 y - 1.85421 x y + 14.7251 x^2 y - 36.8365 x^3 y +
31.9079 x^4 y + 1.78944 y^2 + 5.55233 x y^2 - 43.8506 x^2 y^2 +
111.182 x^3 y^2 - 96.807 x^4 y^2 - 1.99269 y^3 - 5.49884 x y^3 +
46.2492 x^2 y^3 - 118.606 x^3 y^3 + 103.764 x^4 y^3 + 0.819593 y^4 +
1.98819 x y^4 - 17.6815 x^2 y^4 + 45.7184 x^3 y^4 - 40.1317 x^4 y^4 *)
Show[
Plot3D[
fitModel[x, y]
, x, 0, 1/2
, y, 0, 1
, PlotStyle -> Blue
]
, ListPlot3D[
dat
, PlotStyle -> Directive[Red, Opacity[0.5]]
]
]
answered Jul 21 at 10:52
rhermansrhermans
23.5k4 gold badges42 silver badges110 bronze badges
$endgroup$
$begingroup$
@Vaggelis_Z I think you misunderstand the output ofLinearModelFit. The fit is linear only on the coefficients. See my edit.
$endgroup$
– rhermans
Jul 21 at 10:59
add a comment |
4
$begingroup$
Adapting the approach from this answer:
vars = x, y;
maxdegree = 3;
cols = Join @@ (MonomialList[(Plus @@ vars)^#] /. _Integer x_ :> x & /@ Range[maxdegree]);
nparams = 5;
models = Prepend[#, 1] & /@ Subsets[cols, 1, nparams];
Length@models
381
fits = Table[Join[j, Length@j, LinearModelFit[l1dat, j, vars]["BestFit", "AICc", "BIC",
"AdjustedRSquared", "RSquared"]], j, models]
topTenByAICc = SortBy[fits, #[[4]] &][[;; 10]];
Style[# /. x_Real :> Round[x, .00001]] &@
Grid["Model", "Length", "BestFit", "AICc", "BIC",
"AdjustedRSquared", "RSquared", ## & @@ topTenByAICc, Dividers -> All]
bestmodel = topTenByAICc[[1, 3]];
Show[Plot3D[bestmodel, x, 0, .6, y, 0, .1, Mesh -> None],
ListPointPlot3D[l1dat, PlotStyle -> Opacity[.5, Red]]]
answered Jul 21 at 11:10
kglrkglr
208k10 gold badges239 silver badges473 bronze badges
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
$begingroup$
Interpolation
intFunc = With[
order = 1,
dat = Import["https://pastebin.com/raw/YMCFB4mK", "TSV"]
,
Interpolation[#1, #2, #3 & @@@ dat, InterpolationOrder -> order]
]
Plot3D[
intFunc[x, y]
, x, 0.001, 1/2
, y, 0, 1
, PlotStyle -> Green
]
Fit
fitModel = With[
order = 4,
dat = Import["https://pastebin.com/raw/YMCFB4mK", "TSV"]
,
LinearModelFit[
dat
, Flatten@Outer[Times, Sequence @@ Transpose@Array[Power[x, y, # - 1] &, order + 1]]
, x, y
]
]
fitModel[x,y]
(* 0.839678 - 3.39587 x + 10.2762 x^2 - 23.2243 x^3 + 18.9954 x^4 -
0.870772 y - 1.85421 x y + 14.7251 x^2 y - 36.8365 x^3 y +
31.9079 x^4 y + 1.78944 y^2 + 5.55233 x y^2 - 43.8506 x^2 y^2 +
111.182 x^3 y^2 - 96.807 x^4 y^2 - 1.99269 y^3 - 5.49884 x y^3 +
46.2492 x^2 y^3 - 118.606 x^3 y^3 + 103.764 x^4 y^3 + 0.819593 y^4 +
1.98819 x y^4 - 17.6815 x^2 y^4 + 45.7184 x^3 y^4 - 40.1317 x^4 y^4 *)
Show[
Plot3D[
fitModel[x, y]
, x, 0, 1/2
, y, 0, 1
, PlotStyle -> Blue
]
, ListPlot3D[
dat
, PlotStyle -> Directive[Red, Opacity[0.5]]
]
]
answered Jul 21 at 10:52
rhermansrhermans
23.5k4 gold badges42 silver badges110 bronze badges
$endgroup$
$begingroup$
@Vaggelis_Z I think you misunderstand the output ofLinearModelFit. The fit is linear only on the coefficients. See my edit.
$endgroup$
– rhermans
Jul 21 at 10:59
add a comment |
2
$begingroup$
Interpolation
intFunc = With[
order = 1,
dat = Import["https://pastebin.com/raw/YMCFB4mK", "TSV"]
,
Interpolation[#1, #2, #3 & @@@ dat, InterpolationOrder -> order]
]
Plot3D[
intFunc[x, y]
, x, 0.001, 1/2
, y, 0, 1
, PlotStyle -> Green
]
Fit
fitModel = With[
order = 4,
dat = Import["https://pastebin.com/raw/YMCFB4mK", "TSV"]
,
LinearModelFit[
dat
, Flatten@Outer[Times, Sequence @@ Transpose@Array[Power[x, y, # - 1] &, order + 1]]
, x, y
]
]
fitModel[x,y]
(* 0.839678 - 3.39587 x + 10.2762 x^2 - 23.2243 x^3 + 18.9954 x^4 -
0.870772 y - 1.85421 x y + 14.7251 x^2 y - 36.8365 x^3 y +
31.9079 x^4 y + 1.78944 y^2 + 5.55233 x y^2 - 43.8506 x^2 y^2 +
111.182 x^3 y^2 - 96.807 x^4 y^2 - 1.99269 y^3 - 5.49884 x y^3 +
46.2492 x^2 y^3 - 118.606 x^3 y^3 + 103.764 x^4 y^3 + 0.819593 y^4 +
1.98819 x y^4 - 17.6815 x^2 y^4 + 45.7184 x^3 y^4 - 40.1317 x^4 y^4 *)
Show[
Plot3D[
fitModel[x, y]
, x, 0, 1/2
, y, 0, 1
, PlotStyle -> Blue
]
, ListPlot3D[
dat
, PlotStyle -> Directive[Red, Opacity[0.5]]
]
]
answered Jul 21 at 10:52
rhermansrhermans
23.5k4 gold badges42 silver badges110 bronze badges
$endgroup$
$begingroup$
@Vaggelis_Z I think you misunderstand the output ofLinearModelFit. The fit is linear only on the coefficients. See my edit.
$endgroup$
– rhermans
Jul 21 at 10:59
add a comment |
2
2
2
$begingroup$
Interpolation
intFunc = With[
order = 1,
dat = Import["https://pastebin.com/raw/YMCFB4mK", "TSV"]
,
Interpolation[#1, #2, #3 & @@@ dat, InterpolationOrder -> order]
]
Plot3D[
intFunc[x, y]
, x, 0.001, 1/2
, y, 0, 1
, PlotStyle -> Green
]
Fit
fitModel = With[
order = 4,
dat = Import["https://pastebin.com/raw/YMCFB4mK", "TSV"]
,
LinearModelFit[
dat
, Flatten@Outer[Times, Sequence @@ Transpose@Array[Power[x, y, # - 1] &, order + 1]]
, x, y
]
]
fitModel[x,y]
(* 0.839678 - 3.39587 x + 10.2762 x^2 - 23.2243 x^3 + 18.9954 x^4 -
0.870772 y - 1.85421 x y + 14.7251 x^2 y - 36.8365 x^3 y +
31.9079 x^4 y + 1.78944 y^2 + 5.55233 x y^2 - 43.8506 x^2 y^2 +
111.182 x^3 y^2 - 96.807 x^4 y^2 - 1.99269 y^3 - 5.49884 x y^3 +
46.2492 x^2 y^3 - 118.606 x^3 y^3 + 103.764 x^4 y^3 + 0.819593 y^4 +
1.98819 x y^4 - 17.6815 x^2 y^4 + 45.7184 x^3 y^4 - 40.1317 x^4 y^4 *)
Show[
Plot3D[
fitModel[x, y]
, x, 0, 1/2
, y, 0, 1
, PlotStyle -> Blue
]
, ListPlot3D[
dat
, PlotStyle -> Directive[Red, Opacity[0.5]]
]
]
answered Jul 21 at 10:52
rhermansrhermans
23.5k4 gold badges42 silver badges110 bronze badges
$endgroup$
Interpolation
intFunc = With[
order = 1,
dat = Import["https://pastebin.com/raw/YMCFB4mK", "TSV"]
,
Interpolation[#1, #2, #3 & @@@ dat, InterpolationOrder -> order]
]
Plot3D[
intFunc[x, y]
, x, 0.001, 1/2
, y, 0, 1
, PlotStyle -> Green
]
Fit
fitModel = With[
order = 4,
dat = Import["https://pastebin.com/raw/YMCFB4mK", "TSV"]
,
LinearModelFit[
dat
, Flatten@Outer[Times, Sequence @@ Transpose@Array[Power[x, y, # - 1] &, order + 1]]
, x, y
]
]
fitModel[x,y]
(* 0.839678 - 3.39587 x + 10.2762 x^2 - 23.2243 x^3 + 18.9954 x^4 -
0.870772 y - 1.85421 x y + 14.7251 x^2 y - 36.8365 x^3 y +
31.9079 x^4 y + 1.78944 y^2 + 5.55233 x y^2 - 43.8506 x^2 y^2 +
111.182 x^3 y^2 - 96.807 x^4 y^2 - 1.99269 y^3 - 5.49884 x y^3 +
46.2492 x^2 y^3 - 118.606 x^3 y^3 + 103.764 x^4 y^3 + 0.819593 y^4 +
1.98819 x y^4 - 17.6815 x^2 y^4 + 45.7184 x^3 y^4 - 40.1317 x^4 y^4 *)
Show[
Plot3D[
fitModel[x, y]
, x, 0, 1/2
, y, 0, 1
, PlotStyle -> Blue
]
, ListPlot3D[
dat
, PlotStyle -> Directive[Red, Opacity[0.5]]
]
]
answered Jul 21 at 10:52
rhermansrhermans
23.5k4 gold badges42 silver badges110 bronze badges
edited Jul 21 at 11:25
answered Jul 21 at 10:52
rhermansrhermans
23.5k4 gold badges42 silver badges110 bronze badges
answered Jul 21 at 10:52
rhermansrhermans
23.5k4 gold badges42 silver badges110 bronze badges
answered Jul 21 at 10:52
rhermansrhermans
23.5k4 gold badges42 silver badges110 bronze badges
23.5k4 gold badges42 silver badges110 bronze badges
$begingroup$
@Vaggelis_Z I think you misunderstand the output ofLinearModelFit. The fit is linear only on the coefficients. See my edit.
$endgroup$
– rhermans
Jul 21 at 10:59
add a comment |
$begingroup$
@Vaggelis_Z I think you misunderstand the output ofLinearModelFit. The fit is linear only on the coefficients. See my edit.
$endgroup$
– rhermans
Jul 21 at 10:59
$begingroup$
@Vaggelis_Z I think you misunderstand the output of
LinearModelFit. The fit is linear only on the coefficients. See my edit.$endgroup$
– rhermans
Jul 21 at 10:59
$begingroup$
@Vaggelis_Z I think you misunderstand the output of
LinearModelFit. The fit is linear only on the coefficients. See my edit.$endgroup$
– rhermans
Jul 21 at 10:59
add a comment |
4
$begingroup$
Adapting the approach from this answer:
vars = x, y;
maxdegree = 3;
cols = Join @@ (MonomialList[(Plus @@ vars)^#] /. _Integer x_ :> x & /@ Range[maxdegree]);
nparams = 5;
models = Prepend[#, 1] & /@ Subsets[cols, 1, nparams];
Length@models
381
fits = Table[Join[j, Length@j, LinearModelFit[l1dat, j, vars]["BestFit", "AICc", "BIC",
"AdjustedRSquared", "RSquared"]], j, models]
topTenByAICc = SortBy[fits, #[[4]] &][[;; 10]];
Style[# /. x_Real :> Round[x, .00001]] &@
Grid["Model", "Length", "BestFit", "AICc", "BIC",
"AdjustedRSquared", "RSquared", ## & @@ topTenByAICc, Dividers -> All]
bestmodel = topTenByAICc[[1, 3]];
Show[Plot3D[bestmodel, x, 0, .6, y, 0, .1, Mesh -> None],
ListPointPlot3D[l1dat, PlotStyle -> Opacity[.5, Red]]]
answered Jul 21 at 11:10
kglrkglr
208k10 gold badges239 silver badges473 bronze badges
$endgroup$
add a comment |
4
$begingroup$
Adapting the approach from this answer:
vars = x, y;
maxdegree = 3;
cols = Join @@ (MonomialList[(Plus @@ vars)^#] /. _Integer x_ :> x & /@ Range[maxdegree]);
nparams = 5;
models = Prepend[#, 1] & /@ Subsets[cols, 1, nparams];
Length@models
381
fits = Table[Join[j, Length@j, LinearModelFit[l1dat, j, vars]["BestFit", "AICc", "BIC",
"AdjustedRSquared", "RSquared"]], j, models]
topTenByAICc = SortBy[fits, #[[4]] &][[;; 10]];
Style[# /. x_Real :> Round[x, .00001]] &@
Grid["Model", "Length", "BestFit", "AICc", "BIC",
"AdjustedRSquared", "RSquared", ## & @@ topTenByAICc, Dividers -> All]
bestmodel = topTenByAICc[[1, 3]];
Show[Plot3D[bestmodel, x, 0, .6, y, 0, .1, Mesh -> None],
ListPointPlot3D[l1dat, PlotStyle -> Opacity[.5, Red]]]
answered Jul 21 at 11:10
kglrkglr
208k10 gold badges239 silver badges473 bronze badges
$endgroup$
add a comment |
4
4
4
$begingroup$
Adapting the approach from this answer:
vars = x, y;
maxdegree = 3;
cols = Join @@ (MonomialList[(Plus @@ vars)^#] /. _Integer x_ :> x & /@ Range[maxdegree]);
nparams = 5;
models = Prepend[#, 1] & /@ Subsets[cols, 1, nparams];
Length@models
381
fits = Table[Join[j, Length@j, LinearModelFit[l1dat, j, vars]["BestFit", "AICc", "BIC",
"AdjustedRSquared", "RSquared"]], j, models]
topTenByAICc = SortBy[fits, #[[4]] &][[;; 10]];
Style[# /. x_Real :> Round[x, .00001]] &@
Grid["Model", "Length", "BestFit", "AICc", "BIC",
"AdjustedRSquared", "RSquared", ## & @@ topTenByAICc, Dividers -> All]
bestmodel = topTenByAICc[[1, 3]];
Show[Plot3D[bestmodel, x, 0, .6, y, 0, .1, Mesh -> None],
ListPointPlot3D[l1dat, PlotStyle -> Opacity[.5, Red]]]
answered Jul 21 at 11:10
kglrkglr
208k10 gold badges239 silver badges473 bronze badges
$endgroup$
Adapting the approach from this answer:
vars = x, y;
maxdegree = 3;
cols = Join @@ (MonomialList[(Plus @@ vars)^#] /. _Integer x_ :> x & /@ Range[maxdegree]);
nparams = 5;
models = Prepend[#, 1] & /@ Subsets[cols, 1, nparams];
Length@models
381
fits = Table[Join[j, Length@j, LinearModelFit[l1dat, j, vars]["BestFit", "AICc", "BIC",
"AdjustedRSquared", "RSquared"]], j, models]
topTenByAICc = SortBy[fits, #[[4]] &][[;; 10]];
Style[# /. x_Real :> Round[x, .00001]] &@
Grid["Model", "Length", "BestFit", "AICc", "BIC",
"AdjustedRSquared", "RSquared", ## & @@ topTenByAICc, Dividers -> All]
bestmodel = topTenByAICc[[1, 3]];
Show[Plot3D[bestmodel, x, 0, .6, y, 0, .1, Mesh -> None],
ListPointPlot3D[l1dat, PlotStyle -> Opacity[.5, Red]]]
answered Jul 21 at 11:10
kglrkglr
208k10 gold badges239 silver badges473 bronze badges
edited Jul 21 at 11:16
answered Jul 21 at 11:10
kglrkglr
208k10 gold badges239 silver badges473 bronze badges
answered Jul 21 at 11:10
kglrkglr
208k10 gold badges239 silver badges473 bronze badges
answered Jul 21 at 11:10
kglrkglr
208k10 gold badges239 silver badges473 bronze badges
208k10 gold badges239 silver badges473 bronze badges
add a comment |
add a comment |
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$begingroup$
have you tried
Interpolation? E.g.,iF = Interpolation[data];Plot3D[iF[x, y], x, 0, .5, y, 0, 1]?$endgroup$
– kglr
Jul 21 at 10:39
$begingroup$
@kglr I want to obtain an analytical equation of the form f = ax^2 + b*y^2 +....
$endgroup$
– Vaggelis_Z
Jul 21 at 10:42
$begingroup$
It's either
Interpolationor fit. What is it?$endgroup$
– rhermans
Jul 21 at 10:54
$begingroup$
Thanks for accepting my answer, but I think you were too hasty doing that. While accepting is one of the things to do after your question is answered, we recommend that users should test answers before voting and wait 24 hours before accepting the best one. That allows people in all timezones to answer your question and an opportunity for other users to point alternatives, caveats or limitations of the available answers. I have edited my answer so you can accept the best one later.
$endgroup$
– rhermans
Jul 21 at 11:28