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Re: st: Looping across observations (forwards and backwards)


From   Nick Cox <njcoxstata@gmail.com>
To   statalist@hsphsun2.harvard.edu
Subject   Re: st: Looping across observations (forwards and backwards)
Date   Tue, 8 Nov 2011 22:01:32 +0000

Sorry, but I am going to back off from this. I've tried and failed to
understand this twice, and I don't have the inclination to try again.
Also, it does not seem that you have tried all my suggestions,
although they were only guesses, so I don't feel obliged to try again.

The real question for you is whether there is a completely different
way for you to explain all this. These rules come from somewhere and
it's possibly a context someone will recognise if you explain it
afresh. But pushing harder at the same shut door is unlikely to get
results. It would be nice if I were wrong about that.

Nick

On Tue, Nov 8, 2011 at 9:47 PM, Pedro Nakashima
<nakashimapedro@gmail.com> wrote:
> Thanks Nick, but it didn't work.
>
> Below I put a larger sample , a code that worked (for this small small
> sample) and, at the end, a description of what I want to do.
>
> clear
> input v269 v270 v271 ordem novaordem sinal
>         1        1986          10          96         -96           .
>         1        1988          50         148        -148           .
>         1        1986         100         187        -187           .
>         1        1986         100         513        -513           .
>         1        1985          20         743        -743           .
>         1        1985          40         944        -944           .
>         1        1985          40         945        -945           .
>         1        1988         100         954        -954           .
>         2        1985          40         966        -966           1
>         1        1986          40         971        -971           .
>         1        1986          40         992        -992           .
>         2        1985          20        1001       -1001           1
>         0        1985          20        1019       -1019           .
>         2        1985          20        1026       -1026          -1
>         0        1985          40        1032       -1032           .
>         1        1986         100        1034       -1034           .
>         0        1985          40        1035       -1035           .
>         0        1985          40        1045       -1045           .
>         2        1986          10        1053       -1053           1
>         0        1986          40        1054       -1054           .
>         2        1986         100        1056       -1056           1
>         2        1986          40        1062       -1062          -1
>         2        1985          20        1064       -1064          -1
>         2        1985          40        1065       -1065          -1
>         1        1986          45        1068       -1068           .
>         2        1986          45        1070       -1070           1
>         2        1986         100        1074       -1074           1
>         2        1988          10        1079       -1079           0
>         2        1988         100        1081       -1081           1
>         2        1988          50        1088       -1088           1
>         0        1988          50        1091       -1091           .
>         0        1988          50        1093       -1093           .
>         2        1988          70        1094       -1094           0
>         0        1988          50        1098       -1098           .
>         2        1988          50        1099       -1099          -1
>         0        1988          10        1102       -1102           .
>         2        1988          10        1103       -1103          -1
>         0        1988          50        1104       -1104           .
>         2        1988          10        1105       -1105          -1
>         2        1988          10        1107       -1107          -1
>         2        1988          10        1110       -1110          -1
>         0        1988          50        1113       -1113           .
>         2        1988          50        1115       -1115          -1
>         2        1988          10        1116       -1116          -1
>         2        1988          10        1118       -1118          -1
>         0        1988          10        1119       -1119           .
>         2        1988          10        1120       -1120          -1
>         0        1986          40        1124       -1124           .
>         2        1986          10        1127       -1127           1
>         2        1986          10        1131       -1131           1
>         2        1986          10        1135       -1135           1
> end
> sort time
> capture drop orde* sina*
> gen ordem = _n
> gen ordemnova = -_n
> sort ordemnova
> gen sinal2=.
>
> forvalues i=1/`=_N' {
>        if v269[`i']==2 {
>                local pr = v270[`i']
>                local qt = v271[`i']
>                local j=`i'+1
>                while ((v269[`j']==2) | (v270[`j']!=`pr' | v271[`j']!=`qt')) & (`j'<=`=_N') {
>                        local ++j
>                }
>                if v269[`j']==0 {
>                        local ordem = -1
>                }
>                else if v269[`j']==1 {
>                        local ordem = 1
>                }
>                else {
>                         local ordem = 0
>                }
>                quietly replace sinal2 = `ordem' in `i'
>        }
> }
> sort ordem
>
> Description:
> 1) The variable "sinal2" replicates de desired "sinal"
> 2) The first entry of v269 in which v269==2 has the pair v270=185 e v271=40.
> I want to put one of the 3 numbers (-1, 1 or 1) in the variable "sinal".
> What decides which one is the entry in v269 in other observation: the
> one that has the same values (v270==185 and v271==40).
> 3) To do that, I search backwards(in observations) for the pair
> v270==185 and v271==40, skiping observations that, even though they
> have the same pair v270, v271, have also v269==2. To conclude, I want
> to see the first observation that I find when looking backwards,
> starting from a observation in which v269==2, that have either v269==0
> or v269==1
> 4) For the first case in which v269==2 occurs, the looping go
> backwards 2 observations (2 observations before we have v269==1,
> v270==185 and v271==40). Seeing this v269==1, I store the value +1 in
> the local macro "ordem" and then put it in variable sinal.
>     For the second case in which v269==2 occurs, the looping go
> backwards 7 observations .
>     For the third case, the looping go backwards 2 observations.
> And so on..
>
> The problem is that when running this code in a dta-file that has
> 920,000 lines, time goes by and it seems the task will never end. And
> I think it's not normal.
>
> I wonder if a code without loopings, as you did first, would be able
> to do what I described, given that It's  perfect possible 1) that we
> can have consecutive observations v269==2 and, 2) the number of times
> the macro j is increased can overlap among v269==2 observations.
>
> I would thank if one could think with me of this problem. Also it
> might be usefull for other people..
>
> Best,
> Pedro.
>
> 2011/10/4 Nick Cox <njcoxstata@gmail.com>:
>> I have looked at this again. I am still not sure what you are trying
>> to do here, but this reproduces your first example:
>>
>> clear all
>> input v_269 v_270 v_271 desired_sinalt
>> 0 1.4 100 .
>> 1 1.5 100 .
>> 0 1.5 95 .
>> 0 1.4 100 .
>> 2 1.5 100 1
>> 1 1.7 98 .
>> 0 1.2 99 .
>> 2 1.5 95 -1
>> 0 1.8 101 .
>> end
>> gen long order = _n
>> gen start = v_269 == 2
>> gen block = sum(start)
>> bysort block (order) : ///
>>        gen match = sum(v_270 == v_270[1] | v_271 == v_271[1])
>> by block : ///
>>        replace match = sum(cond(inlist(v_269, 1, 0), v_269  * (match == 1),.))
>> by block : replace match = match[_N]
>> by block : gen sinalt = cond(match == 1, 1, cond(match == 0, -1, .)) if block
>>
>>
>>
>>
>> On Tue, Oct 4, 2011 at 3:32 PM, Nick Cox <n.j.cox@durham.ac.uk> wrote:
>>> I don't fully understand what you are trying to do here, but
>>>
>>> local ++j
>>>
>>> need not stop before
>>>
>>> v_270[`j']==v_270[`i'] | v_271[`j']==v_271[`i']
>>>
>>> and perhaps that is not guaranteed for all values of 2.
>>>
>>> so perhaps you need another condition to stop it, say that the next value of v_269 is 2.
>>>
>>> I think you need another approach. Evidently blocks start with some key values and then you count something within blocks. A few fragmentary suggestions
>>>
>>> gen start = v269 == 2
>>> gen block = sum(start)
>>> egen start_v269 = total(start * v269), by(block)
>>> egen start_v270 = total(start * v270), by(block)
>>> egen start_v271 = total(start * v271), by(block)
>>>
>>>
>>>
>>> Nick
>>> n.j.cox@durham.ac.uk
>>>
>>> -----Original Message-----
>>> From: owner-statalist@hsphsun2.harvard.edu [mailto:owner-statalist@hsphsun2.harvard.edu] On Behalf Of Pedro Nakashima
>>> Sent: 03 October 2011 20:39
>>> To: statalist@hsphsun2.harvard.edu
>>> Subject: Re: st: Looping across observations (forwards and backwards)
>>>
>>> Thanks, Nick
>>>
>>> When I applied you tip to the code:
>>>
>>> clear all
>>> input v_269 v_270 v_271 desired_sinalt
>>> 0 1.4 100 .
>>> 1 1.5 100 .
>>> 0 1.5 95 .
>>> 0 1.4 100 .
>>> 2 1.5 100 1
>>> 1 1.7 98 .
>>> 0 1.2 99 .
>>> 2 1.5 95 -1
>>> 0 1.8 101 .
>>> end
>>> gen order = _n
>>> gen neworder=-_n
>>> sort neworder
>>> gen sinalt=.
>>> set trace on
>>> forvalues i=1/`=_N' {
>>>        if v_269[`i']==2{
>>>                local j=`i'+1
>>>                while (v_270[`j']!=v_270[`i'] | v_271[`j']!=v_271[`i']) {
>>>                        local ++j
>>>                        }
>>>                if v_270[`j']==v_270[`i'] | v_271[`j']==v_271[`i'] {
>>>                        if v_269[`j']==1{
>>>                                local sinal=1
>>>                                }
>>>                        else if  v_269[`j']==0 {
>>>                                local sinal=-1
>>>                                }
>>>                        else {
>>>                                local sinal=.
>>>                                }
>>>                }
>>>                replace sinalt=`sinal' in `i'
>>>        }
>>> }
>>> set trace off
>>> sort order
>>>
>>> ,, it worked,
>>>
>>> But if I replace the third observation as follows:
>>> replace v_269 = 2 in 3
>>> replace v_271 = 100 in 3
>>>
>>> The looping never ends..
>>>
>>> Also, It's important to say that if the criterion matches v_269 and
>>> v_271 in observation number 3 (where v_269==2), as in the above
>>> example, I want to ignore it.
>>>
>>> Thanks in advance for the help.
>>>
>>> Best regards
>>> Pedro Nakashima.
>>>
>>> 2011/9/24 Nick Cox <njcoxstata@gmail.com>:
>>>> A different comment is that it is much easier to go forwards in Stata
>>>> than backwards. So, reversing the whole dataset, and defining spells
>>>> "started" in a certain way might be easier. When all is done you
>>>> reverse it again.
>>>>
>>>> Reversing is easy
>>>>
>>>> gen neworder = -_n
>>>> sort neworder
>>>>
>>>> On Sat, Sep 24, 2011 at 4:07 PM, Nick Cox <njcoxstata@gmail.com> wrote:
>>>>> When your program gets to
>>>>>
>>>>>      replace sinalt=`sinal' in `i'
>>>>>
>>>>> evidently `sinal' is undefined so Stata sees
>>>>>
>>>>>      replace sinalt= in `i'
>>>>>
>>>>> It tries first to interpret -in- as the name of a variable or scalar,
>>>>> fails, and aborts with error.
>>>>>
>>>>> Perhaps when you coded
>>>>>
>>>>>  if cod[j]==1 {
>>>>>
>>>>> you meant
>>>>>
>>>>>  if cod[`j']==1 {
>>>>>
>>>>> On Sat, Sep 24, 2011 at 3:28 PM, pedromfn <nakashimapedro@gmail.com> wrote:
>>>>>
>>>>>> My database looks like:
>>>>>>
>>>>>> obs cod pr qt sinalt
>>>>>> 1 1 1.4 100 .
>>>>>> 2 2 1.5 100 .
>>>>>> 3 1 1.5 95 .
>>>>>> 4 1 1.4 100 .
>>>>>> 5 3 1.5 100 .
>>>>>>
>>>>>> and I want to replace observations of sinalt in which cod==3, according to
>>>>>> the following rule:
>>>>>> 1) Go across observations looking for observations in which cod=3
>>>>>> 2) In the above example, the first observation is observation 5, in which
>>>>>> pr[5]=1.5 and qt[5]=100. Once that observation was found, go backwards
>>>>>> through observations looking for the first observation j in which
>>>>>> pr[j]==pr[5] & qt[j]==qt[5]. In the example, j=2.
>>>>>> 3) Replace sinalt[5]=`sinal' , where the macro sinal is defined as:
>>>>>>     if cod[j]==1, store in the local sinal the value 1
>>>>>>     if cod[j]==2, store in the local sinal the value -1
>>>>>> 4) Once last replace was done, look for the next observation in which cod==3
>>>>>> and do the same thing.
>>>>>>
>>>>>> I wrote the following do-file, but it didn't work:
>>>>>>
>>>>>> forvalues i=1/`=_N' {
>>>>>>        if cod[`i']==3{
>>>>>>                local j=`i'-1
>>>>>>                if pr[`j']==pr[`i'] & qt[`j']==qt[`i'] {
>>>>>>                        if cod[j]==1 {
>>>>>>                                local sinal 1
>>>>>>                        }
>>>>>>                        else if cod[`j']==2 {
>>>>>>                                local sinal -1
>>>>>>                        }
>>>>>>                        else {
>>>>>>                                local sinal
>>>>>>                        }
>>>>>>                }
>>>>>>                else {
>>>>>>                        while pr[`j']!=pr[`i'] | qt[`j']!=qt[`i'] {
>>>>>>                                local --j
>>>>>>                        }
>>>>>>                }
>>>>>>        replace sinalt=`sinal' in `i'
>>>>>>        }
>>>>>> }
>>>>>>
>>>>>> ERROR:
>>>>>> in not found
>>>>>> r(111);
>>>>>
>>>>

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