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st: RE: Determining the width of a formatted value


From   "Nick Cox" <n.j.cox@durham.ac.uk>
To   <statalist@hsphsun2.harvard.edu>
Subject   st: RE: Determining the width of a formatted value
Date   Fri, 4 Apr 2003 18:52:05 +0100

Sayer, Bryan
>
> I am a bit confused over how Stata determines what can fit
> into the width of
> a formatted field (the w in w.d)  All references are Stata
> 8 on Windows 95.
>
> I've generally used a comma 12.0 type format to display
> U.S. population
> counts.  But I was using %12.0gc and I got
>
> . display %12.0gc 123456789
>    123456789
>
> I've always assumed that 12 was enough spaces for hundreds
> of millions
> including commas.  But apparently not with the %gc format.
>
> In comparison %fc is fine,
>
> . display %12.0fc 123456789
>  123,456,789
>
> The reference manual (A-F) page 394 states that %11.0gc is
> too small for
> 23,667,902 to display with commas, which does appear to be the case:
> . display %11.0gc 23667902
>    23667902
> But %11.0fc is fine:
> . display %11.0fc 23667902
>  23,667,902
>
> The User's manual on page 121 states that 10,000,000 with a
> %9.2fc format
> will display
> 10,000,000.00
> which is clearly incorrect
> . display %9.2fc 10000000
>  1.00e+07
>
> It also states (page 121) that the f format will always
> display the same
> number of decimal places , even if there is a loss in
> displayed precision.
>
> So does the gc format keep a space for the decimal, while
> fc does not or
> what?  %11.0gc should be fine for 23667902 in my mind, as
> that is only TEN
> required spaces as far as I can tell.  How do I calculate
> the number of
> digits that will display with the gc format?

I can't add much here, except one tip:

. di length("`: di %11.0gc 123456789'")

shows how much space Stata thinks is consumed
by the innermost expression, and the answer is

11

-- which presumably includes two leading spaces.

This one-liner is a contraction of

local foo : di %11.0gc 123456789
di length("`foo'")

Nick
n.j.cox@durham.ac.uk

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