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Re: st: Looping across observations (forwards and backwards)
From
Pedro Nakashima <[email protected]>
To
[email protected]
Subject
Re: st: Looping across observations (forwards and backwards)
Date
Wed, 9 Nov 2011 09:53:06 -0200
Actually, I did try.
In this case, bysort operations within groups seems too much unflexible.
I created a counter (in percentage) and it's showing me that the code
is actually doing it's job (consequently not doing an infinite loop),
although not so fast.
Later I will post comments about the results.
Best regards,
Pedro Nakashima.
2011/11/8 Nick Cox <[email protected]>:
> 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
> <[email protected]> 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 <[email protected]>:
>>> 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 <[email protected]> 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
>>>> [email protected]
>>>>
>>>> -----Original Message-----
>>>> From: [email protected] [mailto:[email protected]] On Behalf Of Pedro Nakashima
>>>> Sent: 03 October 2011 20:39
>>>> To: [email protected]
>>>> 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 <[email protected]>:
>>>>> 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 <[email protected]> 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 <[email protected]> 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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