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RE: st: Creating choice indicator!
From
<[email protected]>
To
<[email protected]>
Subject
RE: st: Creating choice indicator!
Date
Tue, 20 Mar 2012 14:37:25 -0400
Hello,
If I understand correctly, this will help Peter testing his choice model.
Expanding the data provides an environment similar to panel type data.
When programming his function in -ml-, only one observation "per panel" will contribute to the maximum likelihood.
In my experience, the long data structure or shape or form (when possible)
is much faster when programming choice model in -ml-.
Why?
Cheers,
Jean-François Bertrand
-----Original Message-----
From: [email protected] [mailto:[email protected]] On Behalf Of Nick Cox
Sent: March 20, 2012 2:24 PM
To: '[email protected]'
Subject: RE: st: Creating choice indicator!
Alternatively, look upwards!
ceil(5 * runiform())
Nick
[email protected]
Scott Merryman
For a random integer between 1 and 5 the code should be :
1 + floor(5*runiform())
On Tue, Mar 20, 2012 at 12:32 PM, Peter Grand <[email protected]> wrote:
> @ Scott: Exactly, thank you very much!
>
> @ Joerg: Sorry, if I have not been able to correctly specify my request.
>
> Again, thank you for your efforts!
>
Scott Merryman [[email protected]]
> Something like this?
>
>
> clear
> set obs 5
> gen id = _n
> expand 5
> sort id
>
> bys id: gen order = _n
> gen chosen = 0
> levelsof id, local(levels)
> foreach l of local levels {
> if runiform() < .5 { //probability that id is chosen
> local a = round(1 +4*runiform(),1)
> qui replace chosen= 1 if order == `a' & id == `l' //1 of 5
> alternatives is chosen
>
> }
> }
> drop order
> l
On Tue, Mar 20, 2012 at 11:51 AM, Peter Grand <[email protected]> wrote:
>> How can I create a random binary variable which indicates either only one chosen alternative or none of them within an id variable, e.g. 5 choices for each individual id :
>>
>>
>> id chosen
>> 1 0
>> 1 0
>> 1 1
>> 1 0
>> 1 0
>> 2 0
>> 2 0
>> 2 0
>> 2 0
>> 2 0
>> 3 0
>> 3 1
>> 3 0
>> 3 0
>> 3 0
>>
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