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# Re: st: RE: How to define shortest possible period with 95% of observations

 From Daniel Mueller To statalist@hsphsun2.harvard.edu Subject Re: st: RE: How to define shortest possible period with 95% of observations Date Thu, 13 May 2010 02:51:35 +0700

Well, we have thought about that. It is surely not a problem as the fire season in our target area is relatively short (~3 month) and the 95% are in a much shorter interval. Therefore, it does not matter where the max is - we will capture the relevant period. We have other good reasons to use the max for this definition.
```
```
But I bow in and use -egen- with the function -dayofyear(daily_date_variable)- and map the day to the fire season accordingly. I guess that was Nick's suggestion. It saves two lines of code and maybe half a second of computing time. Are there other advantages?
```
Thank you.
Daniel

Steve Samuels wrote on 5/13/2010 2:33 AM:
```
```What you are asking is not only contrary to Nick's recommendation
(and, now, mine), it is a mistake. You are assuming that the peak day
for fires occurs exactly in the middle of a "fire year".  Of course
the  peak days will differ from year to year, and you will wind up
with overlapping periods.

Steve

On Wed, May 12, 2010 at 3:13 PM, Nick Cox<n.j.cox@durham.ac.uk>  wrote:
```
```What you outline is not what I was recommending. It's an awkward
half-way house.

But in terms of your new question, see

SJ-6-4  dm0025  . . . . . . . . . .  Stata tip 36: Which observations?
Erratum
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  N.
J. Cox
Q4/06   SJ 6(4):596                              (no commands)
correction of example code for Stata tip 36

SJ-6-3  dm0025  . . . . . . . . . . . . . .  Stata tip 36: Which
observations?
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  N.
J. Cox
Q3/06   SJ 6(3):430--432                                 (no
commands)
tip for identifying which observations satisfy some
specified condition

Nick
n.j.cox@durham.ac.uk

Daniel Mueller

Thanks.

In the code below I did my best to think in fire years by defining the
peak fire day (these are always unique) as the middle of the year. Then
I'd like to place the maximum into a local macro (where I fail miserably

in line 2 as Steve rightly pointed out):

qui su no_fire_day
loc yearstart = Day[r(max)] - 183
loc yearend = `yearstart' + 365

My simple question is how I can place the Day where no_fires_day =
r(max) into a local macro?

(I ignore leap years for the sake of simplicity..)

Nick Cox wrote on 5/12/2010 10:44 PM:
```
```Without looking at this in detail, it seems to me that you might
```
```benefit
```
```from thinking in terms of fire years, rather than calendar years,
starting on some day other than January 1.  After all, all sorts of
different sciences, not to mention religions, have years that don't
coincide with conventional Western calendar years: fiscal years, water

Several pertinent -egen- functions are included in -egenmore- on SSC.

In other words, define the time scale in terms of those fire years;
```
```then
```
```Robert's code will probably not need any complicated adjustments.

Nick
n.j.cox@durham.ac.uk

Daniel Mueller

Robert, this works like charm!!! Thanks a bunch for this neat code.
```
```Also
```
```
thanks to Nick for pointing me to -shorth- which I will certainly
explore in more detail after having sipped through the extensive
reference list.

Using Roberts code I can seamlessly loop over the nine years of data
```
```and
```
```
generate the shortest fire season per year with 95% of obs. The
```
```results
```
```suggested an additional complication.. For some subsets the shortest
possible period likely starts a couple of days before Jan 1st, at the
end of the preceding year.

I tweaked Roberts code a little to loop over years and defined the
middle of a year as the peak fire day. The code runs through, yet sets
the start of the fire season for some subsets to Jan 1st, while my
educated guess is that it should be somewhere around mid to end of
December. Something went wrong, but I can't spot the glitch in the
```
```code
```
```below. Can someone please help?

Thanks a lot in advance and best regards,
Daniel

*** start
forv y = `yearfirst'/`yearlast' {

* keep previous year
if `y' != `yearfirst' {
keep if Year == `y' | Year == (`y'-1)
}
bys Day: g no_fire_day = _N
qui su no_fire_day

* define year to start 183 days before peak fire day
loc yearstart = Day[r(max)] - 183
loc yearend = `yearstart' + 365
keep if Day>    `yearstart'&    Day<    `yearend' // or with
```
```egen->rotate?
```
```    bys Day: keep if _n == _N
g nobs = _n

* the target is a continuous run that includes 95% of all fires
sum no_fire_day, meanonly
scalar target = .95 * r(sum)

scalar shortlen = .
gen arun = .
gen bestrun = .

* at each pass, create a run that starts at nobs == `i'
* and identify the nobs where the number of fires>= 95%
local more 1
local i = 0
while `more' {
local i = `i' + 1
qui replace arun = sum(no_fire_day * (nobs>= `i'))
sum nobs if arun>= target, meanonly
if r(N) == 0 local more 0
else if (Day[r(min)] - Day[`i'])<    shortlen {
scalar shortlen = Day[r(min)] - Day[`i']
qui replace bestrun = arun
qui replace bestrun = . if nobs>    r(min) | nobs<    `i'
}
}
qui drop if bestrun == .
drop bestrun arun
save fires_`y', replace
}
*** end

Robert Picard wrote on 5/11/2010 3:28 AM:
```
```Here is how I would approach this problem. I would do each year
separately; it could be done all at once but it would complicate the
code unnecessarily. If the fire data is one observation per fire, I
would -collapse- it to one observation per day. Each observation
```
```would
```
```contain the number of fires that day. The following code will
```
```identify
```
```the first instance of the shortest run of days that includes 95% of
fires for the year.

Note that the following code will work, even if there are days
```
```without
```
```fires (and thus no observation for that day).

*--------------------------- begin example -----------------------
version 11

* daily fire counts; with some days without fires
clear all
set seed 123
set obs 365
gen day = _n
drop if uniform()<     .1
gen nobs = _n
gen nfires = round(uniform() * 10)

* the target is a continuous run that includes 95% of all fires
sum nfires, meanonly
scalar target = .95 * r(sum)
dis target

scalar shortlen = .
gen arun = .
gen bestrun = .

* at each pass, create a run that starts at nobs == `i'
* and identify the nobs where the number of fires>= 95%
local more 1
local i 0
while `more' {
local i = `i' + 1
qui replace arun = sum(nfires * (nobs>=`i'))
sum nobs if arun>= target, meanonly
if r(N) == 0 local more 0
else if (day[r(min)] - day[`i'])<     shortlen {
scalar shortlen = day[r(min)] - day[`i']
qui replace bestrun = arun
qui replace bestrun = . if nobs>     r(min) | nobs<     `i'
}
}

*--------------------- end example --------------------------

Hope this help,

Robert

On Mon, May 10, 2010 at 6:19 AM, Nick Cox<n.j.cox@durham.ac.uk>
```
```wrote:
```
```I don't think any trick is possible unless you know in advance the
precise distribution, e.g. that it is Gaussian, or exponential, or
whatever, which here is not the case.

So, you need to look at all the possibilities from the interval
```
```starting
```
```at the minimum to the interval starting at the 5% point of the fire
number distribution in each year.

However, this may all be achievable using -shorth- (SSC). Look at
```
```the
```
```-proportion()- option, but you would need to -expand- first to get a
separate observation for each fire. If that's not practicable, look
inside the code of -shorth- to get ideas on how to proceed. Note
```
```that
```
```no
```
```looping is necessary: the whole problem will reduce to use of -by:-
```
```and
```
```subscripts.

Nick
n.j.cox@durham.ac.uk

Daniel Mueller

I have a strongly unbalanced panel with 100,000 observations (=fire
occurrences per day) that contain between none (no fire) and 3,000
```
```fires
```
```
per day for 8 years. The fire events peak in March and April with
```
```about
```
```85-90% of the yearly total.

My question is how I can define the shortest possible continuous
```
```period
```
```of days for each year that contains 95% of all yearly fires. The
```
```length
```
```and width of the periods may slightly differ across the years due to
climate and other parameters.

I am sure there is a neat trick in Stata for this, yet I have not
spotted it. Any suggestions would be appreciated.
```
```
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```

```
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