Vjylku

Is there a way to pause a running process on Linux systems and resume later?

If I tried and failed to start my own business, how do I apply for a job without job experience?

Why do objects rebound after hitting the ground?

Are all power cords made equal?

"Starve to death" Vs. "Starve to the point of death"

Why is Shelob considered evil?

What does an unprocessed RAW file look like?

How unreachable are Jupiter's moons from Mars with the technology developed for going to Mars?

How can I handle players killing my NPC outside of combat?

What is the draw frequency for 3 consecutive games (same players; amateur level)?

What is a good way to explain how a character can produce flames from their body?

Is .NET Framework 3.5 still needed with a SQL Server 2017 installation to utilize Database Mail?

"I showed the monkey himself in the mirror". Why is this sentence grammatical?

How can I give a Ranger advantage on a check due to Favored Enemy without spoiling the story for the player?

Crack the bank account's password!

How do you get out of your own psychology to write characters?

How do dictionaries source attestation?

Critique vs nitpicking

How to put text above column in minipage?

What's the reason that we have a different number of days each month?

How many proficiencies and languages does a noble half-elf Knowledge Domain cleric start with?

XOR-free sets: Maximum density?

Why do single electrical receptacles exist?

How bad is a Computer Science course that doesn't teach Design Patterns?

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

1

$begingroup$

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

share|improve this question

asked 6 hours ago

JoeJoe

667213

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  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
  6 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thx, if i'd found that i wouldn't have asked.
  $endgroup$
  – Joe
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See also Difference between fitting algorithms
  $endgroup$
  – MarcoB
  37 mins ago
  

  
  
  
  

add a comment | 

1

$begingroup$

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

share|improve this question

asked 6 hours ago

JoeJoe

667213

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  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
  6 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thx, if i'd found that i wouldn't have asked.
  $endgroup$
  – Joe
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See also Difference between fitting algorithms
  $endgroup$
  – MarcoB
  37 mins 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

share|improve this question

asked 6 hours ago

JoeJoe

667213

$endgroup$

share|improve this question

asked 6 hours ago

JoeJoe

667213

share|improve this question

asked 6 hours ago

JoeJoe

667213

asked 6 hours ago

JoeJoe

667213

asked 6 hours ago

JoeJoe

667213

1 Answer
1

active

oldest

votes

6

$begingroup$

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.

share|improve this answer

answered 5 hours ago

eyorbleeyorble

5,5531927

$endgroup$

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "387"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f192147%2fshould-a-new-user-just-default-to-linearmodelfit-vs-fit%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

6

$begingroup$

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.

share|improve this answer

answered 5 hours ago

eyorbleeyorble

5,5531927

$endgroup$

add a comment | 

6

$begingroup$

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.

share|improve this answer

answered 5 hours ago

eyorbleeyorble

5,5531927

$endgroup$

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.

share|improve this answer

answered 5 hours ago

eyorbleeyorble

5,5531927

$endgroup$

lmf = LinearModelFit[{2, 3, 4, 5}, {x}, x];

lmf["Properties"]

lmf["RSquared"]

share|improve this answer

answered 5 hours ago

eyorbleeyorble

5,5531927

answered 5 hours ago

eyorbleeyorble

5,5531927

answered 5 hours ago

eyorbleeyorble

5,5531927