Re: st: FORVAL loop incomplete

[Sergiy Radyakin <serjradyakin@gmail.com>]

Rick

1) your original email mentions that the loop only runs once. Now we
know that the r(max) is
equal to 3 so we expect the loop to have three iterations. The only
way it can do 1 iteration
is if it breaks with an error. To avoid any confusion do the following
adjustment:

...
local limit=r(max)
forval i=1/`limit' {
   display "This is iteration `i' of `limit'"
...

2) Do you use fcf from here?
https://www.uni-goettingen.de/en/199539.html
If yes, note that it supports if condition. Instead of your "keep if
id==.." you can write "fcf if id==.."

3) Your i need to have a meaning. It is either ID or observation
number. This is not necessarily the same. It is definitely not the
same if you have multiple observations of the same subjects, which I
assume you do, since you have time series. This is hugely incompatible
with the -keep- statement, since it also affects the observation
number and hence the working of -in- modifier.

4) Whatever results you save with -replace- will be lost after you -restore-.

5) Make sure you understand how -append- works. It appends data to
memory from file, not the other way around. If you intend to save
results to file at each iteration look at postfile.

Sergiy


On Sat, Nov 9, 2013 at 8:41 AM, Francis, Richard N <rnfrancis@utep.edu> wrote:
> Hi Sergiy,
>
> Thank you for taking time out of your day for such a novice issue.
>
> The entire pgm looks like this:
>
>
> destring gvkey, replace
>
> egen id = group(gvkey), label
>
> gen date = yq(year, fqtr)
>
> tsset id date, quarterly
>
> su id, meanonly
>
> g Z14 = .
> g Z24 = .
> g Z34 = .
> g Z4t = .
> g Z3t = .
> g Z2t = .
> g Z1t = .
>
> forval i = 1/`r(max)' {
>     preserve
>     keep if id == `i'
>     sroot fcf
>         return list
>         ereturn list
>         replace Z14 = r(Z14) in `i'
>         replace Z24 = r(Z24) in `i'
>         replace Z34 = r(Z34) in `i'
>         replace Z4t = r(Z4t) in `i'
>         replace Z3t = r(Z3t) in `i'
>         replace Z2t = r(Z2t) in `i'
>         replace Z1t = r(Z1t) in `i'
>         append using new
>     restore
> }
>
> R(max) immediately prior to the FORVAL loop is three (3), but I'm not sure this is what you are looking for in your first comment.
>
> Does this help at all?
>
> Again, thank you!
>
> Rick Francis
>
> -----Original Message-----
> From: owner-statalist@hsphsun2.harvard.edu [mailto:owner-statalist@hsphsun2.harvard.edu] On Behalf Of Sergiy Radyakin
> Sent: Saturday, November 09, 2013 9:28 AM
> To: statalist@hsphsun2.harvard.edu
> Subject: Re: st: FORVAL loop incomplete
>
> Rick,
> 1) N may be 180, but what is r(max)? In Stata r(max) is the result of the previous command [of r-class]. You don't show us this command.
> Please do.
> 2) The combination of "append using new" and "restore" does not make sense, Whatever you append you immediately lose.
> 3) If append is meant to add one observation where you are saving the results (dataset new is a blank one observation of the same layout as the current dataset), consider using: expand in _N
> 4) If you want to accumulate the results of your tests, consider using a matrix to hold the results, and avoid modifying the data.
> Best, Sergiy
>
> On Sat, Nov 9, 2013 at 8:14 AM, Francis, Richard N <rnfrancis@utep.edu> wrote:
>> I'm sorry, N = 180 obs, currently cycles through the first 60 (3 groups of 60 =180 total obs).
>>
>> Thank you!!
>>
>> Rick Francis
>>
>> -----Original Message-----
>> From: owner-statalist@hsphsun2.harvard.edu
>> [mailto:owner-statalist@hsphsun2.harvard.edu] On Behalf Of Nick Cox
>> Sent: Saturday, November 09, 2013 9:08 AM
>> To: statalist@hsphsun2.harvard.edu
>> Subject: Re: st: FORVAL loop incomplete
>>
>> Tell us what r(max) is.
>> Nick
>> njcoxstata@gmail.com
>>
>>
>> On 9 November 2013 15:53, Francis, Richard N <rnfrancis@utep.edu> wrote:
>>> Hello Statalist,
>>>
>>> Have a simple FORVAL loop which should result in multiple cycles.
>>>
>>> However, she only cycles once. I'm sure the answer is obvious to experienced Stata users (which I am not).
>>>
>>> The loop is as follows:
>>>
>>>
>>> forval i = 1/`r(max)' {
>>>     preserve
>>>     keep if id == `i'
>>>     sroot fcf
>>>                return list
>>>                ereturn list
>>>                replace Z14 = r(Z14) in `i'
>>>                replace Z24 = r(Z24) in `i'
>>>                replace Z34 = r(Z34) in `i'
>>>                replace Z4t = r(Z4t) in `i'
>>>                replace Z3t = r(Z3t) in `i'
>>>                replace Z2t = r(Z2t) in `i'
>>>                replace Z1t = r(Z1t) in `i'
>>>                append using new
>>>     restore
>>> }
>>>
>>> I am grateful for any thoughts anyone may have.
>>>
>>> Thank you!!
>>>
>>> Rick Francis
>>>
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