RE: st: passing indefinite no of arguments

[tashi lama <ltashi32@hotmail.com>]

Hello Nick,

   Sorry for not being clear enough. The problem with this code is you have to edit the code everytime you want to change the variables to be ignored. That is exactly why I wanted to pass those variables as arguments. So, again I wanted to be able to say 

 

do filename var2 var3    and expect to get the sum of the rest. 

 

What I am really trying to do is understand how to do this and use it for the actual program that I am working. I didn't write the actual program because it is too long and I am afraid it might bore you guys. I would be more than happy to put the actual program and actual problem I am facing if you are interested. But, I am going to be using similar concept somewhere in my actual program. 

 

Thank you very much for your time, 

Tashi 
 


> Date: Sat, 31 Mar 2012 17:47:55 +0100
> Subject: Re: st: passing indefinite no of arguments
> From: njcoxstata@gmail.com
> To: statalist@hsphsun2.harvard.edu
> 
> Frankly, this is very confused. You need to keep separate macro
> manipulations and variable calculations.
> 
> Also, thisn't really a problem that needs a program or a do-file at all.
> 
> local myvars : var2 var3
> unab all : var*
> local rest : list all - myvars
> egen rest_total = rowtotal(`rest')
> 
> I advise trying simpler problems and studying the User's manual much
> more slowly and carefully.
> 
> Nick
> 
> On Sat, Mar 31, 2012 at 2:10 PM, tashi lama <ltashi32@hotmail.com> wrote:
> 
> > That sounds great. Here is a slightly diff problem but with a similar flavor..Any idea?
> >
> > I have a dataset as follows
> >
> > var1 var2 var3 va4 var5
> >
> > 4 5 6 3 3
> >
> > I would like to enter variables as arugments and find the sum of the rest. Here is my attempt
> >
> > *sum of the selected vars
> >
> > tokenize `0'
> >
> > local length: word count `0'
> >
> > local varlist var*
> >
> > forvalues i=1/length {
> >
> > local var_new=varlist-`i' /* i am not so sure I can do this
> > replace varlist=`var_new'
> > drop var_new
> > }
> >
> > egen total=rowtotal(`varlist')
> >
> 
> > so if I run,
> >
> > do filename var2 var3 , I expect to get the sum of var1 var4 and var5.
> >
> > I am thinkin I could also do sth like this,
> >
> > tokenize `0'
> > local length:word count `0'
> >
> > unab varlist:var*
> >
> > forvalues i=1/length {
> >
> > local var_new:list varlist- `i'
> >
> > }
> >
> > egen total=rowtotal(`var_new')
> >
> > Essentially I am trying to learn to subtract certain variables from the list of variables or a string. Any help will be highly appreciated...
> >
> > ----------------------------------------
> >> Date: Sat, 31 Mar 2012 00:48:50 +0100
> >> Subject: Re: st: passing indefinite no of arguments
> >> From: njcoxstata@gmail.com
> >> To: statalist@hsphsun2.harvard.edu
> >>
> >> di `: list sizeof 0 '
> >>
> >> Nick
> >>
> >> On Fri, Mar 30, 2012 at 8:52 PM, Maarten Buis <maartenlbuis@gmail.com> wrote:
> >> > On Fri, Mar 30, 2012 at 8:43 PM, tashi lama wrote:
> >> >> So, say I have a do file which will add the arguments and display
> >> >>
> >> >> *sum
> >> >>
> >> >> args a b c d
> >> >>
> >> >> di `a' +`b'+`c'+`d'
> >> >>
> >> >> This one is easy since I could pass 4 arguments a!
> > , b, c and d and my do file will sum them. What if I don't know the no of arguments I am passing? I could have passed 2, 3, 5 or even 1 and find their sum.
> >> >
> >> > Here is a quick solution. The logic is that within a program a number
> >> > of locals are created. The local `0' contains all arguments, and the
> >> > local `1' contains the first argument, `2' the second, etc. This is
> >> > not very stable, e.g. when you call -tokenize- the locals `1', `2'
> >> > etc. will be replaced, but in a small program like the one below it is
> >> > an easy enough solution.
> >> >
> >> > *---------- begin example ------------
> >> > program drop _all
> >> > program define sumargs
> >> > // count the number of arguments
> >> > local k : word count!
> > `0'
> >> >
> >> > // add them&#13
> >
> > ;
> >> > tempname sum
> >> > scalar `sum' = 0
> >> > forvalues i = 1/`k' {
> >> > scalar `sum' = `sum' + ``i''
> >> > }
> >> >
> >> > // display the result
> >> > di `sum'
> >> > end
> >> >
> >> > sumargs 2 3 7
> >> > sumargs 4
> >> > sumargs 2 3 4 5 6 7 8
> >> > *---------- end example -------------
> >> >
> 
> *
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