Should a new user just default to LinearModelFit (vs Fit)Difference between Fitting AlgorithmsFunctionalize...
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Should a new user just default to LinearModelFit (vs Fit)
Difference between Fitting AlgorithmsFunctionalize fitTrying to fit an unknown extreme distributionModule in numerical model for NonlinearModelFit is slow and leaks memoryFit linear data with weights for y and x in LinearModelFitProblem fitting when some data is missingLinearModelFit gives bad fit for simple data setcalculate new table from values of old table and find fit model curveUnusual memory usage in LinearModelFit in version 11.1Increasing the accuracy of the fit when using LinearModelFitNew issue or new setting(s) of Goodness-of-Fit Tests in 11.3@student editionCreating a new list made up by just a particular coordinate of each point in a given set
$begingroup$
I tried searching but the noise in the responses is high due to 'fit' being both a function and a valid term.
If I'm new to Mathematica should I just default to LinearModelFit
(v7) and recognise Fit
as a v1 version that got superceded?
My instinct is 'yes' but wondered if there were valid use cases where Fit would be preferred over LinearModelFit.
fitting
$endgroup$
add a comment |
$begingroup$
I tried searching but the noise in the responses is high due to 'fit' being both a function and a valid term.
If I'm new to Mathematica should I just default to LinearModelFit
(v7) and recognise Fit
as a v1 version that got superceded?
My instinct is 'yes' but wondered if there were valid use cases where Fit would be preferred over LinearModelFit.
fitting
$endgroup$
3
$begingroup$
You may want to take a look at mathematica.stackexchange.com/questions/182053/… for an example of some of the usage differences.LinearModelFit
generates a model object which has some useful properties (e.g. "RSquared"), in addition to the fit itself. In short, I'd only recommendFit
if you just want the shortest code for getting the linear fit expression without any other baggage.
$endgroup$
– eyorble
15 hours ago
$begingroup$
thx, if i'd found that i wouldn't have asked.
$endgroup$
– Joe
14 hours ago
3
$begingroup$
See also Difference between fitting algorithms
$endgroup$
– MarcoB
10 hours ago
1
$begingroup$
Possible duplicate of Difference between Fitting Algorithms
$endgroup$
– Henrik Schumacher
2 hours ago
add a comment |
$begingroup$
I tried searching but the noise in the responses is high due to 'fit' being both a function and a valid term.
If I'm new to Mathematica should I just default to LinearModelFit
(v7) and recognise Fit
as a v1 version that got superceded?
My instinct is 'yes' but wondered if there were valid use cases where Fit would be preferred over LinearModelFit.
fitting
$endgroup$
I tried searching but the noise in the responses is high due to 'fit' being both a function and a valid term.
If I'm new to Mathematica should I just default to LinearModelFit
(v7) and recognise Fit
as a v1 version that got superceded?
My instinct is 'yes' but wondered if there were valid use cases where Fit would be preferred over LinearModelFit.
fitting
fitting
asked 15 hours ago
JoeJoe
687213
687213
3
$begingroup$
You may want to take a look at mathematica.stackexchange.com/questions/182053/… for an example of some of the usage differences.LinearModelFit
generates a model object which has some useful properties (e.g. "RSquared"), in addition to the fit itself. In short, I'd only recommendFit
if you just want the shortest code for getting the linear fit expression without any other baggage.
$endgroup$
– eyorble
15 hours ago
$begingroup$
thx, if i'd found that i wouldn't have asked.
$endgroup$
– Joe
14 hours ago
3
$begingroup$
See also Difference between fitting algorithms
$endgroup$
– MarcoB
10 hours ago
1
$begingroup$
Possible duplicate of Difference between Fitting Algorithms
$endgroup$
– Henrik Schumacher
2 hours ago
add a comment |
3
$begingroup$
You may want to take a look at mathematica.stackexchange.com/questions/182053/… for an example of some of the usage differences.LinearModelFit
generates a model object which has some useful properties (e.g. "RSquared"), in addition to the fit itself. In short, I'd only recommendFit
if you just want the shortest code for getting the linear fit expression without any other baggage.
$endgroup$
– eyorble
15 hours ago
$begingroup$
thx, if i'd found that i wouldn't have asked.
$endgroup$
– Joe
14 hours ago
3
$begingroup$
See also Difference between fitting algorithms
$endgroup$
– MarcoB
10 hours ago
1
$begingroup$
Possible duplicate of Difference between Fitting Algorithms
$endgroup$
– Henrik Schumacher
2 hours ago
3
3
$begingroup$
You may want to take a look at mathematica.stackexchange.com/questions/182053/… for an example of some of the usage differences.
LinearModelFit
generates a model object which has some useful properties (e.g. "RSquared"), in addition to the fit itself. In short, I'd only recommend Fit
if you just want the shortest code for getting the linear fit expression without any other baggage.$endgroup$
– eyorble
15 hours ago
$begingroup$
You may want to take a look at mathematica.stackexchange.com/questions/182053/… for an example of some of the usage differences.
LinearModelFit
generates a model object which has some useful properties (e.g. "RSquared"), in addition to the fit itself. In short, I'd only recommend Fit
if you just want the shortest code for getting the linear fit expression without any other baggage.$endgroup$
– eyorble
15 hours ago
$begingroup$
thx, if i'd found that i wouldn't have asked.
$endgroup$
– Joe
14 hours ago
$begingroup$
thx, if i'd found that i wouldn't have asked.
$endgroup$
– Joe
14 hours ago
3
3
$begingroup$
See also Difference between fitting algorithms
$endgroup$
– MarcoB
10 hours ago
$begingroup$
See also Difference between fitting algorithms
$endgroup$
– MarcoB
10 hours ago
1
1
$begingroup$
Possible duplicate of Difference between Fitting Algorithms
$endgroup$
– Henrik Schumacher
2 hours ago
$begingroup$
Possible duplicate of Difference between Fitting Algorithms
$endgroup$
– Henrik Schumacher
2 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There are a few major differences between Fit
and LinearModelFit
.
LinearModelFit
generates a full model object as its output, which maintains a large set of interesting properties (such as RSquared
or ANOVATable
). These can be accessed by:
lmf = LinearModelFit[{2, 3, 4, 5}, {x}, x];
lmf["Properties"]
lmf["RSquared"]
By default, LinearModelFit
assumes that the zero-intercept may be non-zero. This property is controlled by the IncludeConstantBasis
option. LinearModelFit
has several interesting but advanced options available as well, such as the ability to define weights and perform error propagation through the model.
The final fit expression from LinearModelFit
can be extracted with Normal
.
Fit
does not assume a constant basis, and returns the fit expression directly. It does not calculate additional properties, will automatically convert exact numbers to machine precision, and only finds the least-squares fit. LinearModelFit
and Fit
can work at arbitrary precision (as unlikely as that may be to be useful), but only LinearModelFit
allows the specification of WorkingPrecision
as an option.
The primary advantage of Fit
would appear to be its simplicity and brevity. If you only need the least-squares fit expression or are doing code-golf, Fit
should suffice. In most other cases, I would recommend using LinearModelFit
, if only to be able to access the various quality measures it can calculate on the side.
Note, however, that I wouldn't expect either function to have significant errors -- they ultimately should find the same model, assuming that the constant basis is either included or excluded in both, and that no special options are specified in LinearModelFit
to adapt the model to a specific use case. It's just a question of if you want/need the additional data of the model object.
$endgroup$
$begingroup$
I believeLinearModelFit
usesFit
behind the scenes (not sure).
$endgroup$
– Szabolcs
5 hours ago
1
$begingroup$
@Szabolcs I just spent a few minutesPrintDefinitions
ingLinearModelFit
and didn't see any usage ofFit
specifically.
$endgroup$
– Carl Lange
5 hours ago
add a comment |
Your Answer
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There are a few major differences between Fit
and LinearModelFit
.
LinearModelFit
generates a full model object as its output, which maintains a large set of interesting properties (such as RSquared
or ANOVATable
). These can be accessed by:
lmf = LinearModelFit[{2, 3, 4, 5}, {x}, x];
lmf["Properties"]
lmf["RSquared"]
By default, LinearModelFit
assumes that the zero-intercept may be non-zero. This property is controlled by the IncludeConstantBasis
option. LinearModelFit
has several interesting but advanced options available as well, such as the ability to define weights and perform error propagation through the model.
The final fit expression from LinearModelFit
can be extracted with Normal
.
Fit
does not assume a constant basis, and returns the fit expression directly. It does not calculate additional properties, will automatically convert exact numbers to machine precision, and only finds the least-squares fit. LinearModelFit
and Fit
can work at arbitrary precision (as unlikely as that may be to be useful), but only LinearModelFit
allows the specification of WorkingPrecision
as an option.
The primary advantage of Fit
would appear to be its simplicity and brevity. If you only need the least-squares fit expression or are doing code-golf, Fit
should suffice. In most other cases, I would recommend using LinearModelFit
, if only to be able to access the various quality measures it can calculate on the side.
Note, however, that I wouldn't expect either function to have significant errors -- they ultimately should find the same model, assuming that the constant basis is either included or excluded in both, and that no special options are specified in LinearModelFit
to adapt the model to a specific use case. It's just a question of if you want/need the additional data of the model object.
$endgroup$
$begingroup$
I believeLinearModelFit
usesFit
behind the scenes (not sure).
$endgroup$
– Szabolcs
5 hours ago
1
$begingroup$
@Szabolcs I just spent a few minutesPrintDefinitions
ingLinearModelFit
and didn't see any usage ofFit
specifically.
$endgroup$
– Carl Lange
5 hours ago
add a comment |
$begingroup$
There are a few major differences between Fit
and LinearModelFit
.
LinearModelFit
generates a full model object as its output, which maintains a large set of interesting properties (such as RSquared
or ANOVATable
). These can be accessed by:
lmf = LinearModelFit[{2, 3, 4, 5}, {x}, x];
lmf["Properties"]
lmf["RSquared"]
By default, LinearModelFit
assumes that the zero-intercept may be non-zero. This property is controlled by the IncludeConstantBasis
option. LinearModelFit
has several interesting but advanced options available as well, such as the ability to define weights and perform error propagation through the model.
The final fit expression from LinearModelFit
can be extracted with Normal
.
Fit
does not assume a constant basis, and returns the fit expression directly. It does not calculate additional properties, will automatically convert exact numbers to machine precision, and only finds the least-squares fit. LinearModelFit
and Fit
can work at arbitrary precision (as unlikely as that may be to be useful), but only LinearModelFit
allows the specification of WorkingPrecision
as an option.
The primary advantage of Fit
would appear to be its simplicity and brevity. If you only need the least-squares fit expression or are doing code-golf, Fit
should suffice. In most other cases, I would recommend using LinearModelFit
, if only to be able to access the various quality measures it can calculate on the side.
Note, however, that I wouldn't expect either function to have significant errors -- they ultimately should find the same model, assuming that the constant basis is either included or excluded in both, and that no special options are specified in LinearModelFit
to adapt the model to a specific use case. It's just a question of if you want/need the additional data of the model object.
$endgroup$
$begingroup$
I believeLinearModelFit
usesFit
behind the scenes (not sure).
$endgroup$
– Szabolcs
5 hours ago
1
$begingroup$
@Szabolcs I just spent a few minutesPrintDefinitions
ingLinearModelFit
and didn't see any usage ofFit
specifically.
$endgroup$
– Carl Lange
5 hours ago
add a comment |
$begingroup$
There are a few major differences between Fit
and LinearModelFit
.
LinearModelFit
generates a full model object as its output, which maintains a large set of interesting properties (such as RSquared
or ANOVATable
). These can be accessed by:
lmf = LinearModelFit[{2, 3, 4, 5}, {x}, x];
lmf["Properties"]
lmf["RSquared"]
By default, LinearModelFit
assumes that the zero-intercept may be non-zero. This property is controlled by the IncludeConstantBasis
option. LinearModelFit
has several interesting but advanced options available as well, such as the ability to define weights and perform error propagation through the model.
The final fit expression from LinearModelFit
can be extracted with Normal
.
Fit
does not assume a constant basis, and returns the fit expression directly. It does not calculate additional properties, will automatically convert exact numbers to machine precision, and only finds the least-squares fit. LinearModelFit
and Fit
can work at arbitrary precision (as unlikely as that may be to be useful), but only LinearModelFit
allows the specification of WorkingPrecision
as an option.
The primary advantage of Fit
would appear to be its simplicity and brevity. If you only need the least-squares fit expression or are doing code-golf, Fit
should suffice. In most other cases, I would recommend using LinearModelFit
, if only to be able to access the various quality measures it can calculate on the side.
Note, however, that I wouldn't expect either function to have significant errors -- they ultimately should find the same model, assuming that the constant basis is either included or excluded in both, and that no special options are specified in LinearModelFit
to adapt the model to a specific use case. It's just a question of if you want/need the additional data of the model object.
$endgroup$
There are a few major differences between Fit
and LinearModelFit
.
LinearModelFit
generates a full model object as its output, which maintains a large set of interesting properties (such as RSquared
or ANOVATable
). These can be accessed by:
lmf = LinearModelFit[{2, 3, 4, 5}, {x}, x];
lmf["Properties"]
lmf["RSquared"]
By default, LinearModelFit
assumes that the zero-intercept may be non-zero. This property is controlled by the IncludeConstantBasis
option. LinearModelFit
has several interesting but advanced options available as well, such as the ability to define weights and perform error propagation through the model.
The final fit expression from LinearModelFit
can be extracted with Normal
.
Fit
does not assume a constant basis, and returns the fit expression directly. It does not calculate additional properties, will automatically convert exact numbers to machine precision, and only finds the least-squares fit. LinearModelFit
and Fit
can work at arbitrary precision (as unlikely as that may be to be useful), but only LinearModelFit
allows the specification of WorkingPrecision
as an option.
The primary advantage of Fit
would appear to be its simplicity and brevity. If you only need the least-squares fit expression or are doing code-golf, Fit
should suffice. In most other cases, I would recommend using LinearModelFit
, if only to be able to access the various quality measures it can calculate on the side.
Note, however, that I wouldn't expect either function to have significant errors -- they ultimately should find the same model, assuming that the constant basis is either included or excluded in both, and that no special options are specified in LinearModelFit
to adapt the model to a specific use case. It's just a question of if you want/need the additional data of the model object.
answered 15 hours ago
eyorbleeyorble
5,61311028
5,61311028
$begingroup$
I believeLinearModelFit
usesFit
behind the scenes (not sure).
$endgroup$
– Szabolcs
5 hours ago
1
$begingroup$
@Szabolcs I just spent a few minutesPrintDefinitions
ingLinearModelFit
and didn't see any usage ofFit
specifically.
$endgroup$
– Carl Lange
5 hours ago
add a comment |
$begingroup$
I believeLinearModelFit
usesFit
behind the scenes (not sure).
$endgroup$
– Szabolcs
5 hours ago
1
$begingroup$
@Szabolcs I just spent a few minutesPrintDefinitions
ingLinearModelFit
and didn't see any usage ofFit
specifically.
$endgroup$
– Carl Lange
5 hours ago
$begingroup$
I believe
LinearModelFit
uses Fit
behind the scenes (not sure).$endgroup$
– Szabolcs
5 hours ago
$begingroup$
I believe
LinearModelFit
uses Fit
behind the scenes (not sure).$endgroup$
– Szabolcs
5 hours ago
1
1
$begingroup$
@Szabolcs I just spent a few minutes
PrintDefinitions
ing LinearModelFit
and didn't see any usage of Fit
specifically.$endgroup$
– Carl Lange
5 hours ago
$begingroup$
@Szabolcs I just spent a few minutes
PrintDefinitions
ing LinearModelFit
and didn't see any usage of Fit
specifically.$endgroup$
– Carl Lange
5 hours ago
add a comment |
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3
$begingroup$
You may want to take a look at mathematica.stackexchange.com/questions/182053/… for an example of some of the usage differences.
LinearModelFit
generates a model object which has some useful properties (e.g. "RSquared"), in addition to the fit itself. In short, I'd only recommendFit
if you just want the shortest code for getting the linear fit expression without any other baggage.$endgroup$
– eyorble
15 hours ago
$begingroup$
thx, if i'd found that i wouldn't have asked.
$endgroup$
– Joe
14 hours ago
3
$begingroup$
See also Difference between fitting algorithms
$endgroup$
– MarcoB
10 hours ago
1
$begingroup$
Possible duplicate of Difference between Fitting Algorithms
$endgroup$
– Henrik Schumacher
2 hours ago