Re: st: generate variable versus define scalar, with conditional statement

[annoporci <annoporci@gmail.com>]

> scalar newscalar = open[mdy(1,3,2012)]
>
> is
>
> scalar newscalar = open[18995]


yes, I see, absolutely, I did try this also:

scalar newscalar = open[3jan2012]

but it didn't work, even after a tsset date, and there must be a logical  
explanation
for that too, naturally, but I guess I ran out of ideas and stopped  
thinking.

Your suggestion to:

su `var' if date == mdy(1,3,2012), meanonly
     scalar `var'_ini = r(min)

works inside my loop over the list of variables `var', so that's what I'm
using now

That's better than generating mostly-missing variables, I'm sure.

thanks Nick!





On Tue, 15 Jan 2013 22:07:42 +0800, Nick Cox <njcoxstata@gmail.com> wrote:

> It does work; it's just not what you want.
>
> Stata is perfectly logical here.
>
> scalar newscalar = open[mdy(1,3,2012)]
>
> is
>
> scalar newscalar = open[18995]
>
> which will be missing or some value, depending on the size and details
> of your dataset.
>
> Stata is very good at syntax, but it's lousy at semantics. You want
> Stata to be like a smart research assistant  and to look at
> mdy(1,3,2012) and to realise what you _mean_, which is to go and find
> the observation for that date, etc. Stata only notices what you _say_.
>
> But no; Stata is a robot and just evaluates -mdy(1,3,2012)- as a pure
> number, which is then treated as a subscript or observation number.
> Again, that is what you said as far as it is concerned.
>
> On Tue, Jan 15, 2013 at 1:47 PM, annoporci <annoporci@gmail.com> wrote:
> > you write faster than I think Nick!
>
> [...] I need to be
> > able to specify the date in a "human readable" way, so I tried:
> >
> >
> > scalar newscalar = open[mdy(1,3,2012)]
> >
> >
> > It doesn't work.
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--
Patrick Toche.
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