I am trying to create a school-level learning inequality index based on the difference in learning outcomes between the top 10% and the bottom 50% of students in each school. The data are student-level (below is my attempt) but evidently, I am not creating the deciles at the school level. I imagine I need to use a foreach loop, but that is not yet something I know am fluent at.
My attempt:
bysort IDSCHOOL: egen pv__std_deciles = std(avg_pv)
xtile pv_deciles=avg_pv, nq(10)
bysort IDSCHOOL: egen school_pv_5 = mean(avg_pv) if pv_deciles<6
bysort IDSCHOOL: egen school_pv_10 = mean(avg_pv) if pv_deciles==10
egen max_sp_10=max(school_pv_10), by(IDSCHOOL)
egen max_sp_5=max(school_pv_5), by(IDSCHOOL)
gen school_pv_diff= max_sp_10 - max_sp_5
sort IDSCHOOL
quietly by IDSCHOOL: gen dup = cond(_N == 1, 0, _n)
drop if dup > 1
isid IDSCHOOL
Related
In a previous question, I got an efficient solution to generate a variable and at the same time order it:
sysuse auto, clear
generate random = runiform(), before(make)
This solution does not seem to work if the egen command is used:
egen avgprice = mean(price), before(make)
option before() not allowed
r(198);
Is it possible to generate a variable and at the same time order it when using egen?
The egen command does not have an option similar to the before() option of generate.
However, you can accomplish what you want by writing a small program:
program define egen2
unab allvars : *
gettoken firstvar : allvars
tempname var
gettoken firstarg 0 : 0, parse("=")
egen `var' `0'
generate `firstarg' = `var', before(`firstvar')
end
You could then do the following:
sysuse auto, clear
egen2 foo = mean(price)
EDIT:
The program can be reduced to the following if you do not want to completely avoid order:
program define egen2
gettoken firstarg 0 : 0, parse("=")
egen `firstarg' `0'
order `firstarg'
end
I have a dataset with multiple subgroups (variable economist) and dates (variable temps99).
I want to run a tabsplit command that does not accept bysort or by prefixes. So I created a macro to apply my tabsplit command to each of my subgroups within my data.
For example:
levelsof economist, local(liste)
foreach gars of local liste {
display "`gars'"
tabsplit SubjectCategory if economist=="`gars'", p(;) sort
return list
replace nbcateco = r(r) if economist == "`gars'"
}
For each subgroup, Stata runs the tabsplit command and I use the variable nbcateco to store count results.
I did the same for the date so I can have the evolution of r(r) over time:
levelsof temps99, local(liste23)
foreach time of local liste23 {
display "`time'"
tabsplit SubjectCategory if temps99 == "`time'", p(;) sort
return list
replace nbcattime = r(r) if temps99 == "`time'"
}
Now I want to do it on each subgroups economist by date temps99. I tried multiple combination but I am not very good with macros (yet?).
What I want is to be able to have my r(r) for each of my subgroups over time.
Here's a solution that shows how to calculate the number of distinct publication categories within each by-group. This uses runby (from SSC). runby loops over each by-group, each time replacing the data in memory with the data from the current by-group. For each by-group, the commands contained in the user's program are executed. Whatever is left in memory when the user's program terminates is considered results and accumulates. Once all the groups have been processed, these results replace the data in memory.
I used the verbose option because I wanted to present the results for each by-group using nice formatting. The derivation of the list of distinct categories is done by splitting each list, converting to a long layout, and reducing to one observation per distinct value. The distinct_categories program generates one variable that contains the final count of distinct categories for the by-group.
* create a demontration dataset
* ------------------------------------------------------------------------------
clear all
set seed 12345
* Example generated by -dataex-. To install: ssc install dataex
clear
input str19 economist
"Carmen M. Reinhart"
"Janet Currie"
"Asli Demirguc-Kunt"
"Esther Duflo"
"Marianne Bertrand"
"Claudia Goldin"
"Bronwyn Hughes Hall"
"Serena Ng"
"Anne Case"
"Valerie Ann Ramey"
end
expand 20
bysort economist: gen temps99 = 1998 + _n
gen pubs = runiformint(1,10)
expand pubs
sort economist temps99
gen pubid = _n
local nep NEP-AGR NEP-CBA NEP-COM NEP-DEV NEP-DGE NEP-ECM NEP-EEC NEP-ENE ///
NEP-ENV NEP-HIS NEP-INO NEP-INT NEP-LAB NEP-MAC NEP-MIC NEP-MON ///
NEP-PBE NEP-TRA NEP-URE
gen SubjectCategory = ""
forvalues i=1/19 {
replace SubjectCategory = SubjectCategory + " " + word("`nep'",`i') ///
if runiform() < .1
}
replace SubjectCategory = subinstr(trim(SubjectCategory)," ",";",.)
leftalign // from SSC
* ------------------------------------------------------------------------------
program distinct_categories
dis _n _n _dup(80) "-"
dis as txt "fille = " as res economist[1] as txt _col(68) " temps = " as res temps99[1]
// if there are no subjects for the group, exit now to avoid a no obs error
qui count if !mi(trim(SubjectCategory))
if r(N) == 0 exit
// split categories, reshape to a long layout, and reduce to unique values
preserve
keep pubid SubjectCategory
quietly {
split SubjectCategory, parse(;) gen(cat)
reshape long cat, i(pubid)
bysort cat: keep if _n == 1
drop if mi(cat)
}
// show results and generate the wanted variable
list cat
local distinct = _N
dis _n as txt "distinct = " as res `distinct'
restore
gen wanted = `distinct'
end
runby distinct_categories, by(economist temps99) verbose
This is an example of the XY problem, I think. See http://xyproblem.info/
tabsplit is a command in the package tab_chi from SSC. I have no negative feelings about it, as I wrote it, but it seems quite unnecessary here.
You want to count categories in a string variable: semi-colons are your separators. So count semi-colons and add 1.
local SC SubjectCategory
gen NCategory = 1 + length(`SC') - length(subinstr(`SC', ";", "", .))
Then (e.g.) table or tabstat will let you explore further by groups of interest.
To see the counting idea, consider 3 categories with 2 semi-colons.
. display length("frog;toad;newt")
14
. display length(subinstr("frog;toad;newt", ";", "", .))
12
If we replace each semi-colon with an empty string, the change in length is the number of semi-colons deleted. Note that we don't have to change the variable to do this. Then add 1. See also this paper.
That said, a way to extend your approach might be
egen class = group(economist temps99), label
su class, meanonly
local nclass = r(N)
gen result = .
forval i = 1/`nclass' {
di "`: label (class) `i''"
tabsplit SubjectCategory if class == `i', p(;) sort
return list
replace result = r(r) if class == `i'
}
Using statsby would be even better. See also this FAQ.
I need to generate a variable representing household income. I have each individual's income information and have identified the head (head == 1) and spouse (spouse == 1). Now I decide to define the household income as head's income plus their spouse's income if the head is married or the head's income if unmarried. In this case, I cannot simply use some command like
bys hhid : egen hhincome = total(income)
because there may be other members in the household who receiveincome, such as adult children living with their parents. So how can I achieve my goal, either using an egen function or other approach?
Thanks,Nick,it's really a tactful solution. Another a little bit lumpy solution come to my mind later on : Given I have used bysort prefix to generate the two dummy,head and spouse, the head and his/her spouse should appear as the first two members in each household group,then I can write something like bys hhid:gen hhincome=sum(income) in 1/2 if（pid==2&spouse==1）for married head and bys hhid:gen hhincome=income if(（pid==2&spouse==.）| hsize==1)for unmarried head or single-member household,where pidstands for individual's id and hsizeis a previously created dummy for household size
bysort hhid : egen hhincome = total(cond(head == 1, income, 0) + cond(spouse == 1, income, 0))
or
bysort hhid : egen hhincome = total((head == 1) * income + (spouse == 1) * income)
I have a large data set. I have to subset the data set (Big_data) by using values stored in other dta file (Criteria_data). I will show you the problem first:
**Big_data** **Criteria_data**
==================== ================================================
lon lat 4_digit_id minlon maxlon minlat maxlat
-76.22 44.27 0765 -78.44 -77.22 34.324 35.011
-67.55 33.19 6161 -66.11 -65.93 40.32 41.88
....... ........
(over 1 million obs) (271 observations)
==================== ================================================
I have to subset the bid data as follows:
use Big_data
preserve
keep if (-78.44<lon<-77.22) & (34.324<lat<35.011)
save data_0765, replace
restore
preserve
keep if (-66.11<lon<-65.93) & (40.32<lat<41.88)
save data_6161, replace
restore
....
(1) What should be the efficient programming for the subsetting in Stata? (2) Are the inequality expressions correctly written?
1) Subsetting data
With 400,000 observations in the main file and 300 in the reference file, it takes about 1.5 minutes. I can't test this with double the observations in the main file because the lack of RAM takes my computer to a crawl.
The strategy involves creating as many variables as needed to hold the reference latitudes and longitudes (271*4 = 1084 in the OP's case; Stata IC and up can handle this. See help limits). This requires some reshaping and appending. Then we check for those observations of the big data file that meet the conditions.
clear all
set more off
*----- create example databases -----
tempfile bigdata reference
input ///
lon lat
-76.22 44.27
-66.0 40.85 // meets conditions
-77.10 34.8 // meets conditions
-66.00 42.0
end
expand 100000
save "`bigdata'"
*list
clear all
input ///
str4 id minlon maxlon minlat maxlat
"0765" -78.44 -75.22 34.324 35.011
"6161" -66.11 -65.93 40.32 41.88
end
drop id
expand 150
gen id = _n
save "`reference'"
*list
*----- reshape original reference file -----
use "`reference'", clear
tempfile reference2
destring id, replace
levelsof id, local(lev)
gen i = 1
reshape wide minlon maxlon minlat maxlat, i(i) j(id)
gen lat = .
gen lon = .
save "`reference2'"
*----- create working database -----
use "`bigdata'"
timer on 1
quietly {
forvalues num = 1/300 {
gen minlon`num' = .
gen maxlon`num' = .
gen minlat`num' = .
gen maxlat`num' = .
}
}
timer off 1
timer on 2
append using "`reference2'"
drop i
timer off 2
*----- flag observations for which conditions are met -----
timer on 3
gen byte flag = 0
foreach le of local lev {
quietly replace flag = 1 if inrange(lon, minlon`le'[_N], maxlon`le'[_N]) & inrange(lat, minlat`le'[_N], maxlat`le'[_N])
}
timer off 3
*keep if flag
*keep lon lat
*list
timer list
The inrange() function implies that the minimums and maximums must be adjusted beforehand to satisfy the OP's strict inequalities (the function tests <=, >=).
Probably some expansion using expand, use of correlatives and by (so data is in long form) could speed things up. It's not totally clear for me right now. I'm sure there are better ways in plain Stata mode. Mata may be even better.
(joinby was also tested but again RAM was a problem.)
Edit
Doing computations in chunks rather than for the complete database, significantly improves the RAM issue. Using a main file with 1.2 million observations and a reference file with 300 observations, the following code does all the work in about 1.5 minutes:
set more off
*----- create example big data -----
clear all
set obs 1200000
set seed 13056
gen lat = runiform()*100
gen lon = runiform()*100
local sizebd `=_N' // to be used in computations
tempfile bigdata
save "`bigdata'"
*----- create example reference data -----
clear all
set obs 300
set seed 97532
gen minlat = runiform()*100
gen maxlat = minlat + runiform()*5
gen minlon = runiform()*100
gen maxlon = minlon + runiform()*5
gen id = _n
tempfile reference
save "`reference'"
*----- reshape original reference file -----
use "`reference'", clear
destring id, replace
levelsof id, local(lev)
gen i = 1
reshape wide minlon maxlon minlat maxlat, i(i) j(id)
drop i
tempfile reference2
save "`reference2'"
*----- create file to save results -----
tempfile results
clear all
set obs 0
gen lon = .
gen lat = .
save "`results'"
*----- start computations -----
clear all
* local that controls # of observations in intermediate files
local step = 5000 // can't be larger than sizedb
timer clear
timer on 99
forvalues en = `step'(`step')`sizebd' {
* load observations and join with references
timer on 1
local start = `en' - (`step' - 1)
use in `start'/`en' using "`bigdata'", clear
timer off 1
timer on 2
append using "`reference2'"
timer off 2
* flag observations that meet conditions
timer on 3
gen byte flag = 0
foreach le of local lev {
quietly replace flag = 1 if inrange(lon, minlon`le'[_N], maxlon`le'[_N]) & inrange(lat, minlat`le'[_N], maxlat`le'[_N])
}
timer off 3
* append to result database
timer on 4
quietly {
keep if flag
keep lon lat
append using "`results'"
save "`results'", replace
}
timer off 4
}
timer off 99
timer list
display "total time is " `r(t99)'/60 " minutes"
use "`results'"
browse
2) Inequalities
You ask if your inequalities are correct. They are in fact legal, meaning that Stata will not complain, but the result is probably unexpected.
The following result may seem surprising:
. display (66.11 < 100 < 67.93)
1
How is it the case that the expression evaluates to true (i.e. 1) ? Stata first evaluates 66.11 < 100 which is true, and then sees 1 < 67.93 which is also true, of course.
The intended expression was (and Stata will now do what you want):
. display (66.11 < 100) & (100 < 67.93)
0
You can also rely on the function inrange().
The following example is consistent with the previous explanation:
. display (66.11 < 100 < 0)
0
Stata sees 66.11 < 100 which is true (i.e. 1) and follows up with 1 < 0, which is false (i.e. 0).
This uses Roberto's data setup:
clear all
set obs 1200000
set seed 13056
gen lat = runiform()*100
gen lon = runiform()*100
local sizebd `=_N' // to be used in computations
tempfile bigdata
save "`bigdata'"
*----- create example reference data -----
clear all
set obs 300
set seed 97532
gen minlat = runiform()*100
gen maxlat = minlat + runiform()*5
gen minlon = runiform()*100
gen maxlon = minlon + runiform()*5
gen id = _n
tempfile reference
save "`reference'"
timer on 1
levelsof id, local(id_list)
foreach id of local id_list {
sum minlat if id==`id', meanonly
local minlat = r(min)
sum maxlat if id==`id', meanonly
local maxlat = r(max)
sum minlon if id==`id', meanonly
local minlon = r(min)
sum maxlon if id==`id', meanonly
local maxlon = r(max)
preserve
use if (inrange(lon,`minlon',`maxlon') & inrange(lat,`minlat',`maxlat')) using "`bigdata'", clear
qui save data_`id', replace
restore
}
timer off 1
I would try to avoid preserveing and restoreing the "big" file, and doing so is possible, but at the expense of losing Stata format.
Using the same set up as Roberto and Dimitriy did,
set more off
use `bigdata', clear
merge 1:1 _n using `reference'
* check for data consistency:
* minlat, maxlat, minlon, maxlon are either all defined or all missing
assert inlist( mi(minlat) + mi(maxlat) + mi(minlon) + mi(maxlon), 0, 4)
* this will come handy later
gen byte touse = 0
* set up and cycle over the reference data
count if !missing(minlat)
forvalues n=1/`=r(N)' {
replace touse = inrange(lat,minlat[`n'],maxlat[`n']) & inrange(lon,minlon[`n'],maxlon[`n'])
local thisid = id[`n']
outfile lat lon if touse using data_`thisid'.csv, replace comma
}
Time it on your machine. You could avoid touse and thisid and only have the single outfile within the cycle, but it would be less readable.
You can then infile lat lon using data_###.csv, clear later. If you really need the Stata files proper, you can convert that swarm of CSV files with
clear
local allcsv : dir . files "*.csv"
foreach f of local allcsv {
* change the filename
local dtaname = subinstr(`"`f'"',".csv",".dta",.)
infile lat lon using `"`f'"', clear
if _N>0 save `"`dtaname'"', replace
}
Time it, too. I protected the save as some of the simulated data sets were empty. I think this was faster than 1.5 min on my machine, including the conversion.
I am trying to create a variable for each year in my data based on mathematical expressions of other variables (I have annual data and used "..." to avoid writing each year). I am using the summarize command in Stata to extract the standard deviation but Stata does not recognize the frac variable. I have tried to use egen but that results in an unknown function error. Using gen results in an already defined variable. I would appreciate anyone helping with the following code or pointing me to a link where this issue has been discussed.
foreach yr of numlist 1995...2012 {
local row = `yr' - 1994
local numerator = 100*(income - L1.income)
local denominator = ((abs(income) + abs(L1.income)) / 2)
local frac = (`numerator' / `denominator')
summarize frac
local sdfrac = r(sd)
matrix C[`row', 1] = `numerator'
matrix C[`row', 2] = `denominator'
matrix C[`row', 3] = `sdfrac'
}
If I am understanding your question right, maybe you don't need to use a loop until the end and then you can post the results to a postfile:
This is just a thought:
tempname memhold
tempfile filename
postfile `memhold' year sdfrac using `filename'
gen row=year-1994
gen numerator=100*(income-L1.income)
gen denominator=((abs(income)+abs(L1.income))/2)
gen frac=numerator/denominator
foreach yr of numlist 1995...2012 {
summarize frac if year=`yr'
local sdfrac=r(sd)
post `memhold' (year) (`sdfrac')
}
postclose `memhold'
clear all
use `filename'
*View Results
list
This code should get you a data set with the name of the year and the standard deviation of the frac variable as variables.
In a comment, OP added a question about code similar to this (but ignored the request to post it in a more civilised form). Note that backticks or left quotation marks in Stata clash with SO mark-up codes in comments. Presumably some
tempname memhold
definition preceded this.
postfile `memhold' year sdfrac sex race using myresults
levels of sex, local (s)
levelsof race, local (r)
foreach a of local s {
foreach b of local r {
forval yr = 1995/2012 {
summarize frac if year == `yr' & sex == `a' & race == `b'
post `memhold' (`yr') (`r(sd)') (`sex') (`race')
}
}
}
Let's focus on what the problem is. You want the standard deviations of frac for all combinations of sex, race and year in a separate file. That's one line
collapse (sd) frac, by(year sex race)
If you want to see a table alongside the data, consider
egen group = group(sex race year), label
and then
tab group, su(frac)
or
tabstat frac, by(group) stat(sd)
This code modifies that by #Pcarlitz, mostly by simplifying it. I can't check with your data, which I don't have.
It's too long to fit into a comment.
I would not use a temporary file as you want to save these results, it seems.
tempname memhold
postfile `memhold' year sdfrac using myresults
gen frac = (100*(income - L1.income))/((abs(income) + abs(L1.income))/2)
forval yr = 1995/2012 {
summarize frac if year==`yr'
post `memhold' (`yr') (`r(sd)')
}
postclose `memhold'
use myresults
list
UPDATE As in a later answer, consider collapse as a much simpler direct alternative here.