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Re: Model object as output for linear regression in Octave 3.8
From: |
Juan Pablo Carbajal |
Subject: |
Re: Model object as output for linear regression in Octave 3.8 |
Date: |
Mon, 22 Sep 2014 14:51:40 +0200 |
On Mon, Sep 22, 2014 at 2:50 PM, Juan Pablo Carbajal
<address@hidden> wrote:
> On Mon, Sep 22, 2014 at 1:40 PM, Narayanan, Krishnaprasad
> <address@hidden> wrote:
>> Hallo all,
>>
>>
>>
>> I am using Octave 3.8.0 to build a linear model that has several input
>> features and one output feature. I will be using this model in order to
>> predict the output for the given set of input features. When I searched on
>> the web for statistical packages, I found the following functions: polyfit
>> and regress. But none of these functions returns me a model object which I
>> can use it for prediction.
>>
>>
>>
>> Can I kindly know from the forum is there a function in octave that returns
>> me a model object for linear regression?
>>
>>
>>
>> Regards,
>>
>> Krishnaprasad
>>
>>
>> _______________________________________________
>> Help-octave mailing list
>> address@hidden
>> https://lists.gnu.org/mailman/listinfo/help-octave
>>
>
> Krishnaprasad
>
> If you want to use linear regression you do not need polyfit. The
> function regress indeed returns a model you can use, here a minimalist
> example
>
> X = randn(10,5);
> y = X*linspace(-1,1,5).' + 0.01*randn(10,1);
> B = regress (y, X);
> plot(y,'.;data;',X*B,'o;train;')
> X_new = randn(100,5);
> y_predict = X_new*B
>
> Just make sure you read the help of regress to understand its outputs
> and also check your theory on linear regression for more robust
> results.
>
> Hope this helps.
oh, check the model in B
B =
-0.9994197
-0.4920026
-0.0020738
0.5000850
1.0058200
and compare with linspace(-1,1,5).'