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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 Fri, 14 May 2010 02:21:13 +0700

Robert, your code works just perfect for me and I learned a lot from it. Thank you so much, and also thanks to other contributors!
```
Daniel

Robert Picard wrote on 5/13/2010 3:03 AM:
```
```Apologies for a poor choice of variable name. Instead of day, I should
have called it firedate. Since Stata dates are actually days since
01jan1960, you can do date arithmetics with them. My code should work
for any time period in memory, as long as the firedate variable
contains a Stata date and the records are sorted by date. You can
target 95% of a 6 month span or 3 years, it does not matter.

As to choosing a year of observation centered around the day with the
most fire, I think you were on the right tract Daniel,
and it gives an opportunity to show again how to identify an
observation that matches a certain condition. All you need to do is to
have a variable set to the number of observation (_n). In the updated
example below, you do a summarize to identify the maximum number of
fires and then you do another summarize on n if nfires == r(max). The
r(min) returned from the second sum indicates the number of the first
observation that matches the max number of fires.

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

* generate two years of data
clear all
set seed 12345
set obs 730
gen firedate = mdy(1,1,2001) + _n - 1
format %d firedate
sum, format

* generate more fires arount May; some days without fires
gen fireyear = year(firedate)
sort fireyear firedate
by fireyear: gen nfires = round(uniform() * 1000 / abs(_n-140.1)^.5)
drop if uniform()<  .1
line nfires firedate

* Identify peak fire date; choose first if more than one peak date
gen n = _n
sum nfires if fireyear == 2002
sum n if nfires == r(max)
local peakdate = firedate[r(min)]
list if firedate == `peakdate'

* select a year around peak fire date
drop if firedate<  `peakdate' - 183
drop if firedate>  firedate[1] + 364
sum, format

* the target is a continuous run that includes 95% of all fires
sort firedate
gen nobs = _n
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 (firedate[r(min)] - firedate[`i'])<  shortlen {
scalar shortlen = firedate[r(min)] - firedate[`i']
qui replace bestrun = arun
qui replace bestrun = . if nobs>  r(min) | nobs<  `i'
}
}

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

Hope this helps,

Robert

I

On Wed, May 12, 2010 at 11:44 AM, Nick Cox<n.j.cox@durham.ac.uk>  wrote:
```
```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

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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```
```
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```