As you are pushing me, my guess is yes,
it can be done, but I recommend collaboration
with someone with more Stata experience.
I'm surprised at the implication that -predict-
is irrelevant here.
Nick
[email protected]
Patrick McElduff
> Dear Nick
>
> Unless I am mistaken 'predict' doesn't give the information I
> need and therefore your reply isn't particulary helpful. The
> code by Joe Hibe takes 3 arguements and does work. His code
> can easily be modified to take two arguements as one of the
> first lines of his code combines two of the arguements into
> one variable, but as you say it isn't suitable for conversion
> into a function because it drops all the variables and then
> creates a different structure for the dataset. My code does
> the calculation correctly and will provide the answer if the
> dataset is structured properly. But I know that it is simple
> and incorrect.
>
> My question is, can the program be modified into a function
> that takes two arguements from each observation and returns a
> value to a new variable? If so then I can spend time working
> on correcting the program.
> >>> [email protected] 21/12/2005 10:23:11 am >>>
> Joe Hilbe has used the name -cpoisson- for
> two quite different programs, for cumulative
> and for censored Poisson.
>
> The first, published in STB-1, sounds closer to what
> you want, but it long predates
> the -poisson- command, and it is not compatible
> with that. Also, it -drop-s all data in memory
> without warning!!!
>
> My guess is that your own program needs a lot of
> changes before it will work. You are putting constants in variables
> (not illegal, but inefficient) and variables in constants
> (usually fatal). The line
>
> local cum = `cum' = ((exp(-poi_pred)*(poi_pred^`i'))/`fact')
>
> even after correcting the second = to + is not
> going to work the way you want it to.
>
> I recommend keeping much closer to what -predict-
> can supply for you.
>
> Nick
> [email protected]
>
> Patrick McElduff
>
> > Hi Nick
> >
> > What I am attempting to do (and there may be a much more
> > simple solution) is to calculate the exact probability of
> > observing a value of x or greater given that I expect to
> > observe mu. The value of mu comes from a poisson regression
> > and is potentially different for each observed value.
> >
> > The code that I believe works is:
> >
> > generate temp = 0
> >
> > local i = 0
> > local cum = 0
> > generate prob = 0
> >
> > while `i' <= cases {
> > local j = 1
> > local fact = 1
> >
> > while `j' <= `i' {
> > local fact = `fact' * `j'
> > local j = `j' + 1
> > }
> >
> > local cum = `cum' = ((exp(-poi_pred)*(poi_pred^`i'))/`fact')
> > local i = `i' + 1
> > replace prob = 1 - `cum' if `i' == cases & cases ~= 0
> > replace prob = 1 if cases == 0
> > }
> >
> > There is also a program called cpoisson by Joseph Hilbe that
> > can by easily modified to work for my scenario but it doesn't
> > seem to return the answer the way I need it to.
>
>
> > >>> [email protected] 21/12/2005 8:46:33 am >>>
> > You can't write a program with exactly
> > this syntax. You could write an -egen-
> > function.
> >
> > If you want a new variable as output,
> > why this question?
> >
> > I could start speculating on what you might mean,
> > but better that you should tell us much more.
> >
> > Indeed, please show us your program.
> >
> > Nick
> > [email protected]
> >
> > Patrick McElduff
> >
> > > I want to write a program that has two arguements and
> > > returns a value so that I can use it in the form:
> > >
> > > generate temp = program(var1 var2)
> > >
> > > I have written the guts of the program to do the work but I
> > > don't know how to get the input and output in the form I want.
> > >
> > > Could you please tell me where I should start looking.
> >
>
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>
>
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