Multi peaks fit (Voigt, Lorentzian or Gaussian)
4
I have started quite recently to use Mathematica and I don´t have experience in coding.
What I need to do is to fit a peak which probably is a combination of two Voigt or Lorentzian. I have tried with code already present in the forum but without big success. Could you help me in writing this? Maybe with comments in the code so I can understand better what we are doing.
Data
data = {{19.4, 16672.}, {19.41, 16642.}, {19.42, 16778.}, {19.43,
16857.}, {19.44, 16833.}, {19.45, 17086.}, {19.46, 17129.}, {19.47,
17405.}, {19.48, 17483.}, {19.49, 17308.}, {19.5, 17884.}, {19.51,
17950.}, {19.52, 18202.}, {19.53, 18473.}, {19.54, 19021.}, {19.55,
19279.}, {19.56, 20040.}, {19.57, 20399.}, {19.58, 21412.}, {19.59,
22354.}, {19.6, 23334.}, {19.61, 24399.}, {19.62, 25724.}, {19.63,
27133.}, {19.64, 28825.}, {19.65, 30078.}, {19.66, 32224.}, {19.67,
33907.}, {19.68, 36299.}, {19.69, 39152.}, {19.7, 41980.}, {19.71,
45181.}, {19.72, 49547.}, {19.73, 55438.}, {19.74, 62094.}, {19.75,
69884.}, {19.76, 80306.}, {19.77, 92448.}, {19.78, 107115.}, {19.79,
126574.}, {19.8, 148842.}, {19.81, 175298.}, {19.82,
205953.}, {19.83, 240900.}, {19.84, 278834.}, {19.85,
322364.}, {19.86, 365952.}, {19.87, 411105.}, {19.88,
457658.}, {19.89, 500221.}, {19.9, 544824.}, {19.91,
583862.}, {19.92, 619383.}, {19.93, 650362.}, {19.94,
672886.}, {19.95, 690179.}, {19.96, 695603.}, {19.97,
692265.}, {19.98, 677707.}, {19.99, 657226.}, {20.,
630722.}, {20.01, 599184.}, {20.02, 558854.}, {20.03,
514989.}, {20.04, 469037.}, {20.05, 421656.}, {20.06,
370503.}, {20.07, 324609.}, {20.08, 278435.}, {20.09,
233750.}, {20.1, 195167.}, {20.11, 160965.}, {20.12,
131452.}, {20.13, 108026.}, {20.14, 88341.}, {20.15,
71993.}, {20.16, 59909.}, {20.17, 51054.}, {20.18, 44365.}, {20.19,
39526.}, {20.2, 36292.}, {20.21, 34308.}, {20.22, 32666.}, {20.23,
31599.}, {20.24, 30743.}, {20.25, 29621.}, {20.26, 29034.}, {20.27,
28213.}, {20.28, 27597.}, {20.29, 27485.}, {20.3, 26921.}, {20.31,
26588.}, {20.32, 26337.}, {20.33, 25705.}, {20.34, 26199.}, {20.35,
25321.}, {20.36, 25017.}, {20.37, 25011.}, {20.38, 24566.}, {20.39,
24232.}, {20.4, 24005.}}
My starting point is
data1 = Rest@Transpose[Rescale /@ (Transpose@data)];
peakfunc[A_, [Mu]_, [Sigma]_, x_] = A^2 E^(-((x - [Mu])^2/(2 [Sigma]^2)));
Clear[model, modelvalue]
model[data_, n_] :=
Module[{dataconfig, modelfunc, objfunc, fitvar, fitres},
dataconfig = {A[#], [Mu][#], [Sigma][#]} & /@ Range[n];
modelfunc = (peakfunc[##, fitvar] & @@@ dataconfig // Total);
objfunc =
Total[((Sqrt[data[[All, 2]]])/
data[[All,
1]]) (data[[All, 2]] - (modelfunc /. fitvar -> # &) /@
data[[All, 1]])^2];
FindMinimum[objfunc, Join[{}, Flatten@dataconfig]]]
modelvalue[data_, n_] /; NumericQ[n] :=
If[n >= 1, model[data, n][[1]], 0]
fitres = ReleaseHold[
Hold[{Round[n], model[data1, Round[n]]}] /.
FindMinimum[modelvalue[data1, Round[n]], {n, 5},
Method -> "PrincipalAxis"][[2]]] // Quiet
With[{n = 2},
resfunc =
peakfunc[A[#], [Mu][#], [Sigma][#], x] & /@ Range[n] /.
model[data1, n][[2]]]
Show@{Plot[Evaluate[resfunc], {x, 0, 1},
PlotStyle -> ({Directive[Dashed, Thick,
ColorData["Rainbow"][#]]} & /@
Rescale[Range[Length[resfunc]]]), PlotRange -> All,
Frame -> True, Axes -> False],
Plot[Evaluate[Total@resfunc], {x, 0, 1},
PlotStyle -> Directive[Thick, Red], PlotRange -> All,
Frame -> True, Axes -> False],
Graphics[{PointSize[.003], Black, Point@data1}]}
This is already not very clear to me, as you can see the fit is not very good on the right side (I expect two profiles very close as the yellow and green curve in the second picture). I would like also to know how to evaluate if the fit is good enough or not.
Many thanks!
fitting
edited yesterday
Anton Antonov
22.7k164111
asked yesterday
ClarineClarine
212
New contributor
Clarine is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
|
show 3 more comments
4
I have started quite recently to use Mathematica and I don´t have experience in coding.
What I need to do is to fit a peak which probably is a combination of two Voigt or Lorentzian. I have tried with code already present in the forum but without big success. Could you help me in writing this? Maybe with comments in the code so I can understand better what we are doing.
Data
data = {{19.4, 16672.}, {19.41, 16642.}, {19.42, 16778.}, {19.43,
16857.}, {19.44, 16833.}, {19.45, 17086.}, {19.46, 17129.}, {19.47,
17405.}, {19.48, 17483.}, {19.49, 17308.}, {19.5, 17884.}, {19.51,
17950.}, {19.52, 18202.}, {19.53, 18473.}, {19.54, 19021.}, {19.55,
19279.}, {19.56, 20040.}, {19.57, 20399.}, {19.58, 21412.}, {19.59,
22354.}, {19.6, 23334.}, {19.61, 24399.}, {19.62, 25724.}, {19.63,
27133.}, {19.64, 28825.}, {19.65, 30078.}, {19.66, 32224.}, {19.67,
33907.}, {19.68, 36299.}, {19.69, 39152.}, {19.7, 41980.}, {19.71,
45181.}, {19.72, 49547.}, {19.73, 55438.}, {19.74, 62094.}, {19.75,
69884.}, {19.76, 80306.}, {19.77, 92448.}, {19.78, 107115.}, {19.79,
126574.}, {19.8, 148842.}, {19.81, 175298.}, {19.82,
205953.}, {19.83, 240900.}, {19.84, 278834.}, {19.85,
322364.}, {19.86, 365952.}, {19.87, 411105.}, {19.88,
457658.}, {19.89, 500221.}, {19.9, 544824.}, {19.91,
583862.}, {19.92, 619383.}, {19.93, 650362.}, {19.94,
672886.}, {19.95, 690179.}, {19.96, 695603.}, {19.97,
692265.}, {19.98, 677707.}, {19.99, 657226.}, {20.,
630722.}, {20.01, 599184.}, {20.02, 558854.}, {20.03,
514989.}, {20.04, 469037.}, {20.05, 421656.}, {20.06,
370503.}, {20.07, 324609.}, {20.08, 278435.}, {20.09,
233750.}, {20.1, 195167.}, {20.11, 160965.}, {20.12,
131452.}, {20.13, 108026.}, {20.14, 88341.}, {20.15,
71993.}, {20.16, 59909.}, {20.17, 51054.}, {20.18, 44365.}, {20.19,
39526.}, {20.2, 36292.}, {20.21, 34308.}, {20.22, 32666.}, {20.23,
31599.}, {20.24, 30743.}, {20.25, 29621.}, {20.26, 29034.}, {20.27,
28213.}, {20.28, 27597.}, {20.29, 27485.}, {20.3, 26921.}, {20.31,
26588.}, {20.32, 26337.}, {20.33, 25705.}, {20.34, 26199.}, {20.35,
25321.}, {20.36, 25017.}, {20.37, 25011.}, {20.38, 24566.}, {20.39,
24232.}, {20.4, 24005.}}
My starting point is
data1 = Rest@Transpose[Rescale /@ (Transpose@data)];
peakfunc[A_, [Mu]_, [Sigma]_, x_] = A^2 E^(-((x - [Mu])^2/(2 [Sigma]^2)));
Clear[model, modelvalue]
model[data_, n_] :=
Module[{dataconfig, modelfunc, objfunc, fitvar, fitres},
dataconfig = {A[#], [Mu][#], [Sigma][#]} & /@ Range[n];
modelfunc = (peakfunc[##, fitvar] & @@@ dataconfig // Total);
objfunc =
Total[((Sqrt[data[[All, 2]]])/
data[[All,
1]]) (data[[All, 2]] - (modelfunc /. fitvar -> # &) /@
data[[All, 1]])^2];
FindMinimum[objfunc, Join[{}, Flatten@dataconfig]]]
modelvalue[data_, n_] /; NumericQ[n] :=
If[n >= 1, model[data, n][[1]], 0]
fitres = ReleaseHold[
Hold[{Round[n], model[data1, Round[n]]}] /.
FindMinimum[modelvalue[data1, Round[n]], {n, 5},
Method -> "PrincipalAxis"][[2]]] // Quiet
With[{n = 2},
resfunc =
peakfunc[A[#], [Mu][#], [Sigma][#], x] & /@ Range[n] /.
model[data1, n][[2]]]
Show@{Plot[Evaluate[resfunc], {x, 0, 1},
PlotStyle -> ({Directive[Dashed, Thick,
ColorData["Rainbow"][#]]} & /@
Rescale[Range[Length[resfunc]]]), PlotRange -> All,
Frame -> True, Axes -> False],
Plot[Evaluate[Total@resfunc], {x, 0, 1},
PlotStyle -> Directive[Thick, Red], PlotRange -> All,
Frame -> True, Axes -> False],
Graphics[{PointSize[.003], Black, Point@data1}]}
This is already not very clear to me, as you can see the fit is not very good on the right side (I expect two profiles very close as the yellow and green curve in the second picture). I would like also to know how to evaluate if the fit is good enough or not.
Many thanks!
fitting
edited yesterday
Anton Antonov
22.7k164111
asked yesterday
ClarineClarine
212
New contributor
Clarine is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
Welcome on Mathematic.StackExchange. MaybeNonlinearModelFitmight be helpful to you? At least, make sure to have a look at its documentation
– Henrik Schumacher
yesterday
3
As a reminder, make sure you have good initial guesses for the parameters of your fitting function.
– J. M. is computer-less♦
yesterday
1
Are the integer values counts or measurements that just recorded as integers? I ask because it is important to know if what you have is a random sample from a maybe not-so-simple probability distribution or you're performing a regression that just might happen to have the shape of a mixture of probability distributions? In other words, are there 16,928,270 observations given as a histogram? (I'm also curious as to how you might have a "feeling" about the shape of the distributions - but that's not a Mathematica issue.)
– JimB
yesterday
This question seems to be a duplicate of 98226.
– Anton Antonov
yesterday
If you do have a histogram of counts, I would have to assume that you've got a truncated sample (truncated on the left at 19.395 and truncated on the right at 20.405). If so, please add that information to the body of your question.
– JimB
yesterday
|
show 3 more comments
4
4
4
I have started quite recently to use Mathematica and I don´t have experience in coding.
What I need to do is to fit a peak which probably is a combination of two Voigt or Lorentzian. I have tried with code already present in the forum but without big success. Could you help me in writing this? Maybe with comments in the code so I can understand better what we are doing.
Data
data = {{19.4, 16672.}, {19.41, 16642.}, {19.42, 16778.}, {19.43,
16857.}, {19.44, 16833.}, {19.45, 17086.}, {19.46, 17129.}, {19.47,
17405.}, {19.48, 17483.}, {19.49, 17308.}, {19.5, 17884.}, {19.51,
17950.}, {19.52, 18202.}, {19.53, 18473.}, {19.54, 19021.}, {19.55,
19279.}, {19.56, 20040.}, {19.57, 20399.}, {19.58, 21412.}, {19.59,
22354.}, {19.6, 23334.}, {19.61, 24399.}, {19.62, 25724.}, {19.63,
27133.}, {19.64, 28825.}, {19.65, 30078.}, {19.66, 32224.}, {19.67,
33907.}, {19.68, 36299.}, {19.69, 39152.}, {19.7, 41980.}, {19.71,
45181.}, {19.72, 49547.}, {19.73, 55438.}, {19.74, 62094.}, {19.75,
69884.}, {19.76, 80306.}, {19.77, 92448.}, {19.78, 107115.}, {19.79,
126574.}, {19.8, 148842.}, {19.81, 175298.}, {19.82,
205953.}, {19.83, 240900.}, {19.84, 278834.}, {19.85,
322364.}, {19.86, 365952.}, {19.87, 411105.}, {19.88,
457658.}, {19.89, 500221.}, {19.9, 544824.}, {19.91,
583862.}, {19.92, 619383.}, {19.93, 650362.}, {19.94,
672886.}, {19.95, 690179.}, {19.96, 695603.}, {19.97,
692265.}, {19.98, 677707.}, {19.99, 657226.}, {20.,
630722.}, {20.01, 599184.}, {20.02, 558854.}, {20.03,
514989.}, {20.04, 469037.}, {20.05, 421656.}, {20.06,
370503.}, {20.07, 324609.}, {20.08, 278435.}, {20.09,
233750.}, {20.1, 195167.}, {20.11, 160965.}, {20.12,
131452.}, {20.13, 108026.}, {20.14, 88341.}, {20.15,
71993.}, {20.16, 59909.}, {20.17, 51054.}, {20.18, 44365.}, {20.19,
39526.}, {20.2, 36292.}, {20.21, 34308.}, {20.22, 32666.}, {20.23,
31599.}, {20.24, 30743.}, {20.25, 29621.}, {20.26, 29034.}, {20.27,
28213.}, {20.28, 27597.}, {20.29, 27485.}, {20.3, 26921.}, {20.31,
26588.}, {20.32, 26337.}, {20.33, 25705.}, {20.34, 26199.}, {20.35,
25321.}, {20.36, 25017.}, {20.37, 25011.}, {20.38, 24566.}, {20.39,
24232.}, {20.4, 24005.}}
My starting point is
data1 = Rest@Transpose[Rescale /@ (Transpose@data)];
peakfunc[A_, [Mu]_, [Sigma]_, x_] = A^2 E^(-((x - [Mu])^2/(2 [Sigma]^2)));
Clear[model, modelvalue]
model[data_, n_] :=
Module[{dataconfig, modelfunc, objfunc, fitvar, fitres},
dataconfig = {A[#], [Mu][#], [Sigma][#]} & /@ Range[n];
modelfunc = (peakfunc[##, fitvar] & @@@ dataconfig // Total);
objfunc =
Total[((Sqrt[data[[All, 2]]])/
data[[All,
1]]) (data[[All, 2]] - (modelfunc /. fitvar -> # &) /@
data[[All, 1]])^2];
FindMinimum[objfunc, Join[{}, Flatten@dataconfig]]]
modelvalue[data_, n_] /; NumericQ[n] :=
If[n >= 1, model[data, n][[1]], 0]
fitres = ReleaseHold[
Hold[{Round[n], model[data1, Round[n]]}] /.
FindMinimum[modelvalue[data1, Round[n]], {n, 5},
Method -> "PrincipalAxis"][[2]]] // Quiet
With[{n = 2},
resfunc =
peakfunc[A[#], [Mu][#], [Sigma][#], x] & /@ Range[n] /.
model[data1, n][[2]]]
Show@{Plot[Evaluate[resfunc], {x, 0, 1},
PlotStyle -> ({Directive[Dashed, Thick,
ColorData["Rainbow"][#]]} & /@
Rescale[Range[Length[resfunc]]]), PlotRange -> All,
Frame -> True, Axes -> False],
Plot[Evaluate[Total@resfunc], {x, 0, 1},
PlotStyle -> Directive[Thick, Red], PlotRange -> All,
Frame -> True, Axes -> False],
Graphics[{PointSize[.003], Black, Point@data1}]}
This is already not very clear to me, as you can see the fit is not very good on the right side (I expect two profiles very close as the yellow and green curve in the second picture). I would like also to know how to evaluate if the fit is good enough or not.
Many thanks!
fitting
edited yesterday
Anton Antonov
22.7k164111
asked yesterday
ClarineClarine
212
New contributor
Clarine is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have started quite recently to use Mathematica and I don´t have experience in coding.
What I need to do is to fit a peak which probably is a combination of two Voigt or Lorentzian. I have tried with code already present in the forum but without big success. Could you help me in writing this? Maybe with comments in the code so I can understand better what we are doing.
Data
data = {{19.4, 16672.}, {19.41, 16642.}, {19.42, 16778.}, {19.43,
16857.}, {19.44, 16833.}, {19.45, 17086.}, {19.46, 17129.}, {19.47,
17405.}, {19.48, 17483.}, {19.49, 17308.}, {19.5, 17884.}, {19.51,
17950.}, {19.52, 18202.}, {19.53, 18473.}, {19.54, 19021.}, {19.55,
19279.}, {19.56, 20040.}, {19.57, 20399.}, {19.58, 21412.}, {19.59,
22354.}, {19.6, 23334.}, {19.61, 24399.}, {19.62, 25724.}, {19.63,
27133.}, {19.64, 28825.}, {19.65, 30078.}, {19.66, 32224.}, {19.67,
33907.}, {19.68, 36299.}, {19.69, 39152.}, {19.7, 41980.}, {19.71,
45181.}, {19.72, 49547.}, {19.73, 55438.}, {19.74, 62094.}, {19.75,
69884.}, {19.76, 80306.}, {19.77, 92448.}, {19.78, 107115.}, {19.79,
126574.}, {19.8, 148842.}, {19.81, 175298.}, {19.82,
205953.}, {19.83, 240900.}, {19.84, 278834.}, {19.85,
322364.}, {19.86, 365952.}, {19.87, 411105.}, {19.88,
457658.}, {19.89, 500221.}, {19.9, 544824.}, {19.91,
583862.}, {19.92, 619383.}, {19.93, 650362.}, {19.94,
672886.}, {19.95, 690179.}, {19.96, 695603.}, {19.97,
692265.}, {19.98, 677707.}, {19.99, 657226.}, {20.,
630722.}, {20.01, 599184.}, {20.02, 558854.}, {20.03,
514989.}, {20.04, 469037.}, {20.05, 421656.}, {20.06,
370503.}, {20.07, 324609.}, {20.08, 278435.}, {20.09,
233750.}, {20.1, 195167.}, {20.11, 160965.}, {20.12,
131452.}, {20.13, 108026.}, {20.14, 88341.}, {20.15,
71993.}, {20.16, 59909.}, {20.17, 51054.}, {20.18, 44365.}, {20.19,
39526.}, {20.2, 36292.}, {20.21, 34308.}, {20.22, 32666.}, {20.23,
31599.}, {20.24, 30743.}, {20.25, 29621.}, {20.26, 29034.}, {20.27,
28213.}, {20.28, 27597.}, {20.29, 27485.}, {20.3, 26921.}, {20.31,
26588.}, {20.32, 26337.}, {20.33, 25705.}, {20.34, 26199.}, {20.35,
25321.}, {20.36, 25017.}, {20.37, 25011.}, {20.38, 24566.}, {20.39,
24232.}, {20.4, 24005.}}
My starting point is
data1 = Rest@Transpose[Rescale /@ (Transpose@data)];
peakfunc[A_, [Mu]_, [Sigma]_, x_] = A^2 E^(-((x - [Mu])^2/(2 [Sigma]^2)));
Clear[model, modelvalue]
model[data_, n_] :=
Module[{dataconfig, modelfunc, objfunc, fitvar, fitres},
dataconfig = {A[#], [Mu][#], [Sigma][#]} & /@ Range[n];
modelfunc = (peakfunc[##, fitvar] & @@@ dataconfig // Total);
objfunc =
Total[((Sqrt[data[[All, 2]]])/
data[[All,
1]]) (data[[All, 2]] - (modelfunc /. fitvar -> # &) /@
data[[All, 1]])^2];
FindMinimum[objfunc, Join[{}, Flatten@dataconfig]]]
modelvalue[data_, n_] /; NumericQ[n] :=
If[n >= 1, model[data, n][[1]], 0]
fitres = ReleaseHold[
Hold[{Round[n], model[data1, Round[n]]}] /.
FindMinimum[modelvalue[data1, Round[n]], {n, 5},
Method -> "PrincipalAxis"][[2]]] // Quiet
With[{n = 2},
resfunc =
peakfunc[A[#], [Mu][#], [Sigma][#], x] & /@ Range[n] /.
model[data1, n][[2]]]
Show@{Plot[Evaluate[resfunc], {x, 0, 1},
PlotStyle -> ({Directive[Dashed, Thick,
ColorData["Rainbow"][#]]} & /@
Rescale[Range[Length[resfunc]]]), PlotRange -> All,
Frame -> True, Axes -> False],
Plot[Evaluate[Total@resfunc], {x, 0, 1},
PlotStyle -> Directive[Thick, Red], PlotRange -> All,
Frame -> True, Axes -> False],
Graphics[{PointSize[.003], Black, Point@data1}]}
This is already not very clear to me, as you can see the fit is not very good on the right side (I expect two profiles very close as the yellow and green curve in the second picture). I would like also to know how to evaluate if the fit is good enough or not.
Many thanks!
fitting
fitting
edited yesterday
Anton Antonov
22.7k164111
asked yesterday
ClarineClarine
212
New contributor
Clarine is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Anton Antonov
22.7k164111
asked yesterday
ClarineClarine
212
New contributor
Clarine is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Anton Antonov
22.7k164111
edited yesterday
Anton Antonov
22.7k164111
edited yesterday
Anton Antonov
22.7k164111
22.7k164111
asked yesterday
ClarineClarine
212
New contributor
Clarine is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
ClarineClarine
212
asked yesterday
ClarineClarine
212
212
New contributor
Clarine is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Clarine is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Clarine is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
Welcome on Mathematic.StackExchange. MaybeNonlinearModelFitmight be helpful to you? At least, make sure to have a look at its documentation
– Henrik Schumacher
yesterday
3
As a reminder, make sure you have good initial guesses for the parameters of your fitting function.
– J. M. is computer-less♦
yesterday
1
Are the integer values counts or measurements that just recorded as integers? I ask because it is important to know if what you have is a random sample from a maybe not-so-simple probability distribution or you're performing a regression that just might happen to have the shape of a mixture of probability distributions? In other words, are there 16,928,270 observations given as a histogram? (I'm also curious as to how you might have a "feeling" about the shape of the distributions - but that's not a Mathematica issue.)
– JimB
yesterday
This question seems to be a duplicate of 98226.
– Anton Antonov
yesterday
If you do have a histogram of counts, I would have to assume that you've got a truncated sample (truncated on the left at 19.395 and truncated on the right at 20.405). If so, please add that information to the body of your question.
– JimB
yesterday
|
show 3 more comments
2
Welcome on Mathematic.StackExchange. MaybeNonlinearModelFitmight be helpful to you? At least, make sure to have a look at its documentation
– Henrik Schumacher
yesterday
3
As a reminder, make sure you have good initial guesses for the parameters of your fitting function.
– J. M. is computer-less♦
yesterday
1
Are the integer values counts or measurements that just recorded as integers? I ask because it is important to know if what you have is a random sample from a maybe not-so-simple probability distribution or you're performing a regression that just might happen to have the shape of a mixture of probability distributions? In other words, are there 16,928,270 observations given as a histogram? (I'm also curious as to how you might have a "feeling" about the shape of the distributions - but that's not a Mathematica issue.)
– JimB
yesterday
This question seems to be a duplicate of 98226.
– Anton Antonov
yesterday
If you do have a histogram of counts, I would have to assume that you've got a truncated sample (truncated on the left at 19.395 and truncated on the right at 20.405). If so, please add that information to the body of your question.
– JimB
yesterday
2
2
Welcome on Mathematic.StackExchange. Maybe
NonlinearModelFit might be helpful to you? At least, make sure to have a look at its documentation– Henrik Schumacher
yesterday
Welcome on Mathematic.StackExchange. Maybe
NonlinearModelFit might be helpful to you? At least, make sure to have a look at its documentation– Henrik Schumacher
yesterday
3
3
As a reminder, make sure you have good initial guesses for the parameters of your fitting function.
– J. M. is computer-less♦
yesterday
As a reminder, make sure you have good initial guesses for the parameters of your fitting function.
– J. M. is computer-less♦
yesterday
1
1
Are the integer values counts or measurements that just recorded as integers? I ask because it is important to know if what you have is a random sample from a maybe not-so-simple probability distribution or you're performing a regression that just might happen to have the shape of a mixture of probability distributions? In other words, are there 16,928,270 observations given as a histogram? (I'm also curious as to how you might have a "feeling" about the shape of the distributions - but that's not a Mathematica issue.)
– JimB
yesterday
Are the integer values counts or measurements that just recorded as integers? I ask because it is important to know if what you have is a random sample from a maybe not-so-simple probability distribution or you're performing a regression that just might happen to have the shape of a mixture of probability distributions? In other words, are there 16,928,270 observations given as a histogram? (I'm also curious as to how you might have a "feeling" about the shape of the distributions - but that's not a Mathematica issue.)
– JimB
yesterday
This question seems to be a duplicate of 98226.
– Anton Antonov
yesterday
This question seems to be a duplicate of 98226.
– Anton Antonov
yesterday
If you do have a histogram of counts, I would have to assume that you've got a truncated sample (truncated on the left at 19.395 and truncated on the right at 20.405). If so, please add that information to the body of your question.
– JimB
yesterday
If you do have a histogram of counts, I would have to assume that you've got a truncated sample (truncated on the left at 19.395 and truncated on the right at 20.405). If so, please add that information to the body of your question.
– JimB
yesterday
|
show 3 more comments
1 Answer
1
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6
Here is a fit using Fit and a basis of functions. I am not sure is this a solution OP wants -- may be only one Exp term is desired.
(The data variable data is taken from the question.)
(* Taken from the question. *)
Clear[peakfunc]
peakfunc[A_, [Mu]_, [Sigma]_, x_] = A^2 E^(-((x - [Mu])^2/(2 [Sigma]^2)));
(* Generate a basis of functions. *)
bFuncs = Flatten@
Table[peakfunc[1, m, s, x], {m, 19.4, 20.4, 0.2}, {s, 0.1, 0.4, 0.1}];
Length[bFuncs]
(* 24 *)
(* Load the QRMon package. *)
Import["https://raw.githubusercontent.com/antononcube/MathematicaForPrediction/master/MonadicProgramming/MonadicQuantileRegression.m"]
(* Do a fit with the basis, plot it, and plot the relative errors. *)
qmon =
QRMonUnit[data]⟹
QRMonEchoDataSummary⟹
QRMonFit[bFuncs]⟹
QRMonPlot⟹
QRMonErrorPlots;
(* Take the regression function from the monad object and show it. *)
qFunc = (qmon⟹QRMonTakeRegressionFunctions)["mean"];
qFunc[x]
(* -2.445*10^7 E^(-50. (-20.4 + x)^2) +
3.41622*10^10 E^(-12.5 (-20.4 + x)^2) -
1.63757*10^11 E^(-5.55556 (-20.4 + x)^2) -
2.91489*10^11 E^(-3.125 (-20.4 + x)^2) +
1.83408*10^7 E^(-50. (-20.2 + x)^2) +
6.05021*10^10 E^(-12.5 (-20.2 + x)^2) -
4.04765*10^11 E^(-5.55556 (-20.2 + x)^2) +
8.90334*10^11 E^(-3.125 (-20.2 + x)^2) -
1.47152*10^7 E^(-50. (-20. + x)^2) +
5.14204*10^10 E^(-12.5 (-20. + x)^2) -
1.34254*10^11 E^(-5.55556 (-20. + x)^2) +
1.54544*10^11 E^(-3.125 (-20. + x)^2) +
1.35117*10^7 E^(-50. (-19.8 + x)^2) +
1.77025*10^10 E^(-12.5 (-19.8 + x)^2) -
2.38068*10^10 E^(-5.55556 (-19.8 + x)^2) -
1.63691*10^11 E^(-3.125 (-19.8 + x)^2) -
1.75233*10^7 E^(-50. (-19.6 + x)^2) -
1.74451*10^10 E^(-12.5 (-19.6 + x)^2) +
2.91813*10^11 E^(-5.55556 (-19.6 + x)^2) -
6.2293*10^11 E^(-3.125 (-19.6 + x)^2) +
3.084*10^7 E^(-50. (-19.4 + x)^2) -
2.02534*10^10 E^(-12.5 (-19.4 + x)^2) +
1.21184*10^11 E^(-5.55556 (-19.4 + x)^2) +
2.04278*10^11 E^(-3.125 (-19.4 + x)^2) *)
answered yesterday
Anton AntonovAnton Antonov
22.7k164111
First, thank you for the answer. Is it for a multi peak fitting? Where you define the number of peak that you think there are in your set of data?
– Clarine
5 hours ago
@Clarine Your comment indicates that you have a different question in mind that the one you posted. Is this discussion (175989) more in line if with what you are looking for?
– Anton Antonov
4 hours ago
Yes and no, so the thing is that I believe this dataset could be fit by two peaks (voigt, lorentzian, Gaussian or maybe a combination) but the problem is that I don't know how to write where I believe the centre of the peak is. What I want to do is basically sat "ok we have this peak, I want to try and fit these like we have two peaks and not only one" and check whether my guess makes sense. In the case that you have linked the peak positions were clear
– Clarine
16 mins ago
add a comment |
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1 Answer
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1 Answer
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6
Here is a fit using Fit and a basis of functions. I am not sure is this a solution OP wants -- may be only one Exp term is desired.
(The data variable data is taken from the question.)
(* Taken from the question. *)
Clear[peakfunc]
peakfunc[A_, [Mu]_, [Sigma]_, x_] = A^2 E^(-((x - [Mu])^2/(2 [Sigma]^2)));
(* Generate a basis of functions. *)
bFuncs = Flatten@
Table[peakfunc[1, m, s, x], {m, 19.4, 20.4, 0.2}, {s, 0.1, 0.4, 0.1}];
Length[bFuncs]
(* 24 *)
(* Load the QRMon package. *)
Import["https://raw.githubusercontent.com/antononcube/MathematicaForPrediction/master/MonadicProgramming/MonadicQuantileRegression.m"]
(* Do a fit with the basis, plot it, and plot the relative errors. *)
qmon =
QRMonUnit[data]⟹
QRMonEchoDataSummary⟹
QRMonFit[bFuncs]⟹
QRMonPlot⟹
QRMonErrorPlots;
(* Take the regression function from the monad object and show it. *)
qFunc = (qmon⟹QRMonTakeRegressionFunctions)["mean"];
qFunc[x]
(* -2.445*10^7 E^(-50. (-20.4 + x)^2) +
3.41622*10^10 E^(-12.5 (-20.4 + x)^2) -
1.63757*10^11 E^(-5.55556 (-20.4 + x)^2) -
2.91489*10^11 E^(-3.125 (-20.4 + x)^2) +
1.83408*10^7 E^(-50. (-20.2 + x)^2) +
6.05021*10^10 E^(-12.5 (-20.2 + x)^2) -
4.04765*10^11 E^(-5.55556 (-20.2 + x)^2) +
8.90334*10^11 E^(-3.125 (-20.2 + x)^2) -
1.47152*10^7 E^(-50. (-20. + x)^2) +
5.14204*10^10 E^(-12.5 (-20. + x)^2) -
1.34254*10^11 E^(-5.55556 (-20. + x)^2) +
1.54544*10^11 E^(-3.125 (-20. + x)^2) +
1.35117*10^7 E^(-50. (-19.8 + x)^2) +
1.77025*10^10 E^(-12.5 (-19.8 + x)^2) -
2.38068*10^10 E^(-5.55556 (-19.8 + x)^2) -
1.63691*10^11 E^(-3.125 (-19.8 + x)^2) -
1.75233*10^7 E^(-50. (-19.6 + x)^2) -
1.74451*10^10 E^(-12.5 (-19.6 + x)^2) +
2.91813*10^11 E^(-5.55556 (-19.6 + x)^2) -
6.2293*10^11 E^(-3.125 (-19.6 + x)^2) +
3.084*10^7 E^(-50. (-19.4 + x)^2) -
2.02534*10^10 E^(-12.5 (-19.4 + x)^2) +
1.21184*10^11 E^(-5.55556 (-19.4 + x)^2) +
2.04278*10^11 E^(-3.125 (-19.4 + x)^2) *)
answered yesterday
Anton AntonovAnton Antonov
22.7k164111
First, thank you for the answer. Is it for a multi peak fitting? Where you define the number of peak that you think there are in your set of data?
– Clarine
5 hours ago
@Clarine Your comment indicates that you have a different question in mind that the one you posted. Is this discussion (175989) more in line if with what you are looking for?
– Anton Antonov
4 hours ago
Yes and no, so the thing is that I believe this dataset could be fit by two peaks (voigt, lorentzian, Gaussian or maybe a combination) but the problem is that I don't know how to write where I believe the centre of the peak is. What I want to do is basically sat "ok we have this peak, I want to try and fit these like we have two peaks and not only one" and check whether my guess makes sense. In the case that you have linked the peak positions were clear
– Clarine
16 mins ago
add a comment |
6
Here is a fit using Fit and a basis of functions. I am not sure is this a solution OP wants -- may be only one Exp term is desired.
(The data variable data is taken from the question.)
(* Taken from the question. *)
Clear[peakfunc]
peakfunc[A_, [Mu]_, [Sigma]_, x_] = A^2 E^(-((x - [Mu])^2/(2 [Sigma]^2)));
(* Generate a basis of functions. *)
bFuncs = Flatten@
Table[peakfunc[1, m, s, x], {m, 19.4, 20.4, 0.2}, {s, 0.1, 0.4, 0.1}];
Length[bFuncs]
(* 24 *)
(* Load the QRMon package. *)
Import["https://raw.githubusercontent.com/antononcube/MathematicaForPrediction/master/MonadicProgramming/MonadicQuantileRegression.m"]
(* Do a fit with the basis, plot it, and plot the relative errors. *)
qmon =
QRMonUnit[data]⟹
QRMonEchoDataSummary⟹
QRMonFit[bFuncs]⟹
QRMonPlot⟹
QRMonErrorPlots;
(* Take the regression function from the monad object and show it. *)
qFunc = (qmon⟹QRMonTakeRegressionFunctions)["mean"];
qFunc[x]
(* -2.445*10^7 E^(-50. (-20.4 + x)^2) +
3.41622*10^10 E^(-12.5 (-20.4 + x)^2) -
1.63757*10^11 E^(-5.55556 (-20.4 + x)^2) -
2.91489*10^11 E^(-3.125 (-20.4 + x)^2) +
1.83408*10^7 E^(-50. (-20.2 + x)^2) +
6.05021*10^10 E^(-12.5 (-20.2 + x)^2) -
4.04765*10^11 E^(-5.55556 (-20.2 + x)^2) +
8.90334*10^11 E^(-3.125 (-20.2 + x)^2) -
1.47152*10^7 E^(-50. (-20. + x)^2) +
5.14204*10^10 E^(-12.5 (-20. + x)^2) -
1.34254*10^11 E^(-5.55556 (-20. + x)^2) +
1.54544*10^11 E^(-3.125 (-20. + x)^2) +
1.35117*10^7 E^(-50. (-19.8 + x)^2) +
1.77025*10^10 E^(-12.5 (-19.8 + x)^2) -
2.38068*10^10 E^(-5.55556 (-19.8 + x)^2) -
1.63691*10^11 E^(-3.125 (-19.8 + x)^2) -
1.75233*10^7 E^(-50. (-19.6 + x)^2) -
1.74451*10^10 E^(-12.5 (-19.6 + x)^2) +
2.91813*10^11 E^(-5.55556 (-19.6 + x)^2) -
6.2293*10^11 E^(-3.125 (-19.6 + x)^2) +
3.084*10^7 E^(-50. (-19.4 + x)^2) -
2.02534*10^10 E^(-12.5 (-19.4 + x)^2) +
1.21184*10^11 E^(-5.55556 (-19.4 + x)^2) +
2.04278*10^11 E^(-3.125 (-19.4 + x)^2) *)
answered yesterday
Anton AntonovAnton Antonov
22.7k164111
First, thank you for the answer. Is it for a multi peak fitting? Where you define the number of peak that you think there are in your set of data?
– Clarine
5 hours ago
@Clarine Your comment indicates that you have a different question in mind that the one you posted. Is this discussion (175989) more in line if with what you are looking for?
– Anton Antonov
4 hours ago
Yes and no, so the thing is that I believe this dataset could be fit by two peaks (voigt, lorentzian, Gaussian or maybe a combination) but the problem is that I don't know how to write where I believe the centre of the peak is. What I want to do is basically sat "ok we have this peak, I want to try and fit these like we have two peaks and not only one" and check whether my guess makes sense. In the case that you have linked the peak positions were clear
– Clarine
16 mins ago
add a comment |
6
6
6
Here is a fit using Fit and a basis of functions. I am not sure is this a solution OP wants -- may be only one Exp term is desired.
(The data variable data is taken from the question.)
(* Taken from the question. *)
Clear[peakfunc]
peakfunc[A_, [Mu]_, [Sigma]_, x_] = A^2 E^(-((x - [Mu])^2/(2 [Sigma]^2)));
(* Generate a basis of functions. *)
bFuncs = Flatten@
Table[peakfunc[1, m, s, x], {m, 19.4, 20.4, 0.2}, {s, 0.1, 0.4, 0.1}];
Length[bFuncs]
(* 24 *)
(* Load the QRMon package. *)
Import["https://raw.githubusercontent.com/antononcube/MathematicaForPrediction/master/MonadicProgramming/MonadicQuantileRegression.m"]
(* Do a fit with the basis, plot it, and plot the relative errors. *)
qmon =
QRMonUnit[data]⟹
QRMonEchoDataSummary⟹
QRMonFit[bFuncs]⟹
QRMonPlot⟹
QRMonErrorPlots;
(* Take the regression function from the monad object and show it. *)
qFunc = (qmon⟹QRMonTakeRegressionFunctions)["mean"];
qFunc[x]
(* -2.445*10^7 E^(-50. (-20.4 + x)^2) +
3.41622*10^10 E^(-12.5 (-20.4 + x)^2) -
1.63757*10^11 E^(-5.55556 (-20.4 + x)^2) -
2.91489*10^11 E^(-3.125 (-20.4 + x)^2) +
1.83408*10^7 E^(-50. (-20.2 + x)^2) +
6.05021*10^10 E^(-12.5 (-20.2 + x)^2) -
4.04765*10^11 E^(-5.55556 (-20.2 + x)^2) +
8.90334*10^11 E^(-3.125 (-20.2 + x)^2) -
1.47152*10^7 E^(-50. (-20. + x)^2) +
5.14204*10^10 E^(-12.5 (-20. + x)^2) -
1.34254*10^11 E^(-5.55556 (-20. + x)^2) +
1.54544*10^11 E^(-3.125 (-20. + x)^2) +
1.35117*10^7 E^(-50. (-19.8 + x)^2) +
1.77025*10^10 E^(-12.5 (-19.8 + x)^2) -
2.38068*10^10 E^(-5.55556 (-19.8 + x)^2) -
1.63691*10^11 E^(-3.125 (-19.8 + x)^2) -
1.75233*10^7 E^(-50. (-19.6 + x)^2) -
1.74451*10^10 E^(-12.5 (-19.6 + x)^2) +
2.91813*10^11 E^(-5.55556 (-19.6 + x)^2) -
6.2293*10^11 E^(-3.125 (-19.6 + x)^2) +
3.084*10^7 E^(-50. (-19.4 + x)^2) -
2.02534*10^10 E^(-12.5 (-19.4 + x)^2) +
1.21184*10^11 E^(-5.55556 (-19.4 + x)^2) +
2.04278*10^11 E^(-3.125 (-19.4 + x)^2) *)
answered yesterday
Anton AntonovAnton Antonov
22.7k164111
Here is a fit using Fit and a basis of functions. I am not sure is this a solution OP wants -- may be only one Exp term is desired.
(The data variable data is taken from the question.)
(* Taken from the question. *)
Clear[peakfunc]
peakfunc[A_, [Mu]_, [Sigma]_, x_] = A^2 E^(-((x - [Mu])^2/(2 [Sigma]^2)));
(* Generate a basis of functions. *)
bFuncs = Flatten@
Table[peakfunc[1, m, s, x], {m, 19.4, 20.4, 0.2}, {s, 0.1, 0.4, 0.1}];
Length[bFuncs]
(* 24 *)
(* Load the QRMon package. *)
Import["https://raw.githubusercontent.com/antononcube/MathematicaForPrediction/master/MonadicProgramming/MonadicQuantileRegression.m"]
(* Do a fit with the basis, plot it, and plot the relative errors. *)
qmon =
QRMonUnit[data]⟹
QRMonEchoDataSummary⟹
QRMonFit[bFuncs]⟹
QRMonPlot⟹
QRMonErrorPlots;
(* Take the regression function from the monad object and show it. *)
qFunc = (qmon⟹QRMonTakeRegressionFunctions)["mean"];
qFunc[x]
(* -2.445*10^7 E^(-50. (-20.4 + x)^2) +
3.41622*10^10 E^(-12.5 (-20.4 + x)^2) -
1.63757*10^11 E^(-5.55556 (-20.4 + x)^2) -
2.91489*10^11 E^(-3.125 (-20.4 + x)^2) +
1.83408*10^7 E^(-50. (-20.2 + x)^2) +
6.05021*10^10 E^(-12.5 (-20.2 + x)^2) -
4.04765*10^11 E^(-5.55556 (-20.2 + x)^2) +
8.90334*10^11 E^(-3.125 (-20.2 + x)^2) -
1.47152*10^7 E^(-50. (-20. + x)^2) +
5.14204*10^10 E^(-12.5 (-20. + x)^2) -
1.34254*10^11 E^(-5.55556 (-20. + x)^2) +
1.54544*10^11 E^(-3.125 (-20. + x)^2) +
1.35117*10^7 E^(-50. (-19.8 + x)^2) +
1.77025*10^10 E^(-12.5 (-19.8 + x)^2) -
2.38068*10^10 E^(-5.55556 (-19.8 + x)^2) -
1.63691*10^11 E^(-3.125 (-19.8 + x)^2) -
1.75233*10^7 E^(-50. (-19.6 + x)^2) -
1.74451*10^10 E^(-12.5 (-19.6 + x)^2) +
2.91813*10^11 E^(-5.55556 (-19.6 + x)^2) -
6.2293*10^11 E^(-3.125 (-19.6 + x)^2) +
3.084*10^7 E^(-50. (-19.4 + x)^2) -
2.02534*10^10 E^(-12.5 (-19.4 + x)^2) +
1.21184*10^11 E^(-5.55556 (-19.4 + x)^2) +
2.04278*10^11 E^(-3.125 (-19.4 + x)^2) *)
answered yesterday
Anton AntonovAnton Antonov
22.7k164111
edited yesterday
answered yesterday
Anton AntonovAnton Antonov
22.7k164111
answered yesterday
Anton AntonovAnton Antonov
22.7k164111
answered yesterday
Anton AntonovAnton Antonov
22.7k164111
22.7k164111
First, thank you for the answer. Is it for a multi peak fitting? Where you define the number of peak that you think there are in your set of data?
– Clarine
5 hours ago
@Clarine Your comment indicates that you have a different question in mind that the one you posted. Is this discussion (175989) more in line if with what you are looking for?
– Anton Antonov
4 hours ago
Yes and no, so the thing is that I believe this dataset could be fit by two peaks (voigt, lorentzian, Gaussian or maybe a combination) but the problem is that I don't know how to write where I believe the centre of the peak is. What I want to do is basically sat "ok we have this peak, I want to try and fit these like we have two peaks and not only one" and check whether my guess makes sense. In the case that you have linked the peak positions were clear
– Clarine
16 mins ago
add a comment |
First, thank you for the answer. Is it for a multi peak fitting? Where you define the number of peak that you think there are in your set of data?
– Clarine
5 hours ago
@Clarine Your comment indicates that you have a different question in mind that the one you posted. Is this discussion (175989) more in line if with what you are looking for?
– Anton Antonov
4 hours ago
Yes and no, so the thing is that I believe this dataset could be fit by two peaks (voigt, lorentzian, Gaussian or maybe a combination) but the problem is that I don't know how to write where I believe the centre of the peak is. What I want to do is basically sat "ok we have this peak, I want to try and fit these like we have two peaks and not only one" and check whether my guess makes sense. In the case that you have linked the peak positions were clear
– Clarine
16 mins ago
First, thank you for the answer. Is it for a multi peak fitting? Where you define the number of peak that you think there are in your set of data?
– Clarine
5 hours ago
First, thank you for the answer. Is it for a multi peak fitting? Where you define the number of peak that you think there are in your set of data?
– Clarine
5 hours ago
@Clarine Your comment indicates that you have a different question in mind that the one you posted. Is this discussion (175989) more in line if with what you are looking for?
– Anton Antonov
4 hours ago
@Clarine Your comment indicates that you have a different question in mind that the one you posted. Is this discussion (175989) more in line if with what you are looking for?
– Anton Antonov
4 hours ago
Yes and no, so the thing is that I believe this dataset could be fit by two peaks (voigt, lorentzian, Gaussian or maybe a combination) but the problem is that I don't know how to write where I believe the centre of the peak is. What I want to do is basically sat "ok we have this peak, I want to try and fit these like we have two peaks and not only one" and check whether my guess makes sense. In the case that you have linked the peak positions were clear
– Clarine
16 mins ago
Yes and no, so the thing is that I believe this dataset could be fit by two peaks (voigt, lorentzian, Gaussian or maybe a combination) but the problem is that I don't know how to write where I believe the centre of the peak is. What I want to do is basically sat "ok we have this peak, I want to try and fit these like we have two peaks and not only one" and check whether my guess makes sense. In the case that you have linked the peak positions were clear
– Clarine
16 mins ago
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2
Welcome on Mathematic.StackExchange. Maybe
NonlinearModelFitmight be helpful to you? At least, make sure to have a look at its documentation– Henrik Schumacher
yesterday
3
As a reminder, make sure you have good initial guesses for the parameters of your fitting function.
– J. M. is computer-less♦
yesterday
1
Are the integer values counts or measurements that just recorded as integers? I ask because it is important to know if what you have is a random sample from a maybe not-so-simple probability distribution or you're performing a regression that just might happen to have the shape of a mixture of probability distributions? In other words, are there 16,928,270 observations given as a histogram? (I'm also curious as to how you might have a "feeling" about the shape of the distributions - but that's not a Mathematica issue.)
– JimB
yesterday
This question seems to be a duplicate of 98226.
– Anton Antonov
yesterday
If you do have a histogram of counts, I would have to assume that you've got a truncated sample (truncated on the left at 19.395 and truncated on the right at 20.405). If so, please add that information to the body of your question.
– JimB
yesterday