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Re: st: algorithmic question : running sum and computations
From
Nick Cox <[email protected]>
To
Francesco <[email protected]>
Subject
Re: st: algorithmic question : running sum and computations
Date
Fri, 17 Aug 2012 13:55:23 +0100
I don't have easy advice on this. As I understand it sorting on
id product (date)
can't distinguish between
id 1 product A date 42 quantity 12
id 1 product A date 42 quantity -12
id 1 product A date 42 quantity 21
id 1 product A date 42 quantity -21
and
id 1 product A date 42 quantity 12
id 1 product A date 42 quantity -21
id 1 product A date 42 quantity 21
id 1 product A date 42 quantity -12
In the first case you have two spells to 0, and in the second one
spell to 0. Your example shows that spells need not be two
observations long, so I don't know what to suggest.
Nick
On Fri, Aug 17, 2012 at 1:45 PM, Francesco <[email protected]> wrote:
> Actually Nick there is only a slight problem : dates could be repeated
> for the same individual AND the same product : for example there
> could be several round trips during the same day for the same
> product... In that case I would consider that there are as many
> delta_Date equal to zero as different round trips during the day for a
> particular product... My apologies I did not think of this particular
> and important case...
>
> Could the trick egen panelid = group(id product) be adapted in that case ?
>
> Many thanks
> Best Regards
>
> On 17 August 2012 13:58, Francesco <[email protected]> wrote:
>> Many, Many thanks Nick and Scott for your kind and very precise
>> answers! Spells is indeed what I needed ;-)
>>
>>
>> On 17 August 2012 13:43, Nick Cox <[email protected]> wrote:
>>> Using your data as a sandpit
>>>
>>> . clear
>>>
>>> . input id date str1 product quantity
>>>
>>> id date product quantity
>>> 1. 1 1 A 10
>>> 2. 1 2 A -10
>>> 3. 1 1 B 100
>>> 4. 1 2 B -50
>>> 5. 1 4 C 15
>>> 6. 1 8 C 100
>>> 7. 1 9 C -115
>>> 8. 1 10 C 10
>>> 9. 1 11 C -10
>>> 10. end
>>>
>>> it seems that we are interested in the length of time it takes for
>>> cumulative quantity to return to 0. -sum()- is there for cumulative
>>> sums:
>>>
>>> . bysort id product (date) : gen cumq = sum(q)
>>>
>>> In one jargon, we are interested in "spells" defined by the fact that
>>> they end in 0s for cumulative quantity. In Stata it is easiest to work
>>> with initial conditions defining spells, so we negate the date
>>> variable to reverse time:
>>>
>>> . gen negdate = -date
>>>
>>> As dates can be repeated for the same individual, treating data as
>>> panel data requires another fiction, that panels are defined by
>>> individuals and products:
>>>
>>> . egen panelid = group(id product)
>>>
>>> Now we can -tsset- the data:
>>>
>>> . tsset panelid negdate
>>> panel variable: panelid (unbalanced)
>>> time variable: negdate, -11 to -1, but with a gap
>>> delta: 1 unit
>>>
>>> -tsspell- from SSC, which you must install, is a tool for handling
>>> spells. It requires -tsset- data; the great benefit of that is that it
>>> handles panels automatically. (In fact almost all the credit belongs
>>> to StataCorp.) Here the criterion is that a spell is defined by
>>> starting with -cumq == 0-
>>>
>>> . tsspell, fcond(cumq == 0)
>>>
>>> -tsspell- creates three variables with names by default _spell _seq
>>> _end. _end is especially useful: it is an indicator variable for end
>>> of spells (beginning of spells when time is reversed). You can read
>>> more in the help for -tsspell-.
>>>
>>> . sort id product date
>>>
>>> . l id product date cumq _*
>>>
>>> +---------------------------------------------------+
>>> | id product date cumq _spell _seq _end |
>>> |---------------------------------------------------|
>>> 1. | 1 A 1 10 1 2 1 |
>>> 2. | 1 A 2 0 1 1 0 |
>>> 3. | 1 B 1 100 0 0 0 |
>>> 4. | 1 B 2 50 0 0 0 |
>>> 5. | 1 C 4 15 2 3 1 |
>>> |---------------------------------------------------|
>>> 6. | 1 C 8 115 2 2 0 |
>>> 7. | 1 C 9 0 2 1 0 |
>>> 8. | 1 C 10 10 1 2 1 |
>>> 9. | 1 C 11 0 1 1 0 |
>>> +---------------------------------------------------+
>>>
>>> You want the mean length of completed spells. Completed spells are
>>> tagged by _end == 1 or cumq == 0
>>>
>>> . egen meanlength = mean(_seq/ _end), by(id)
>>>
>>> This is my favourite division trick: _seq / _end is _seq if _end is 1
>>> and missing if _end is 0; missings are ignored by -egen-'s -mean()-
>>> function, so you get the mean length for each individual. It is
>>> repeated for each observation for each individual so you could go
>>>
>>> . egen tag = tag(id)
>>> . l id meanlength if tag
>>>
>>> I wrote a tutorial on spells.
>>>
>>> SJ-7-2 dm0029 . . . . . . . . . . . . . . Speaking Stata: Identifying spells
>>> . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . N. J. Cox
>>> Q2/07 SJ 7(2):249--265 (no commands)
>>> shows how to handle spells with complete control over
>>> spell specification
>>>
>>> which is accessible at
>>> http://www.stata-journal.com/sjpdf.html?articlenum=dm0029
>>>
>>> Its principles underlie -tsspell-, but -tsspell- is not even
>>> mentioned, for which there is a mundane explanation. Explaining some
>>> basics as clearly and carefully as I could produced a paper that was
>>> already long and detailed, and adding detail on -tsspell- would just
>>> have made that worse.
>>>
>>> For more on spells, see Rowling (1997, 1998, 1999, etc.).
>>>
>>> Nick
>>>
>>> On Fri, Aug 17, 2012 at 11:30 AM, Francesco <[email protected]> wrote:
>>>> Dear Statalist,
>>>>
>>>> I am stuck with a little algorithmic problem and I cannot find an
>>>> simple (or elegant) solution...
>>>>
>>>> I have a panel dataset as (date in days) :
>>>>
>>>> ID DATE PRODUCT QUANTITY
>>>> 1 1 A 10
>>>> 1 2 A -10
>>>>
>>>> 1 1 B 100
>>>> 1 2 B -50
>>>>
>>>> 1 4 C 15
>>>> 1 8 C 100
>>>> 1 9 C -115
>>>>
>>>> 1 10 C 10
>>>> 1 11 C -10
>>>>
>>>>
>>>>
>>>> and I would like to know the average time (in days) it takes for an
>>>> individual in order to complete a full round trip (the variation in
>>>> quantity is zero)
>>>> For example, for the first id we can see that there we have
>>>>
>>>> ID PRODUCT delta_DATE delta_QUANTITY
>>>> 1 A 1=2-1 0=10-10
>>>> 1 C 5=4-9 0=15+100-115
>>>> 1 C 1=11-10 0=10-10
>>>>
>>>> so on average individual 1 takes (1+5+1)/3=2.3 days to complete a full
>>>> round trip. Indeed I can discard product B because there is no round
>>>> trip, that is 100-50 is not equal to zero.
>>>>
>>>> My question is therefore ... do you have an idea obtain this simply in
>>>> Stata ? I have to average across thousands of individuals... :)
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