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From |
Nick Cox <njcoxstata@gmail.com> |

To |
Francesco <k7br@gmx.fr> |

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 <k7br@gmx.fr> 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 <k7br@gmx.fr> 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 <njcoxstata@gmail.com> 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 <cariboupad@gmx.fr> 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... :) * * For searches and help try: * http://www.stata.com/help.cgi?search * http://www.stata.com/support/statalist/faq * http://www.ats.ucla.edu/stat/stata/

**Follow-Ups**:**Re: st: algorithmic question : running sum and computations***From:*Francesco <k7br@gmx.fr>

**References**:**st: algorithmic question : running sum and computations***From:*Francesco <cariboupad@gmx.fr>

**Re: st: algorithmic question : running sum and computations***From:*Nick Cox <njcoxstata@gmail.com>

**Re: st: algorithmic question : running sum and computations***From:*Francesco <k7br@gmx.fr>

**Re: st: algorithmic question : running sum and computations***From:*Francesco <k7br@gmx.fr>

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