Sorry that title is confusing. Hopefully it's clear below.
I'm using Stata and I'd like to assign the value 1 to a variable that depends on the value within a different variable. I have 20 order variables and also 20 corresponding variables. For example if order1 = 3, I'd like to assign variable3 = 1. Below is a snippet of what the final dataset would look like if I had only 3 of each variable.
Right now I'm doing this with two loops but I have to another loop around this that goes through this 9 more times plus I'd doing this for a couple hundred data files. I'd like to make it more efficient.
forvalues i = 1/20 {
forvalues j = 1/20 {
replace variable`j' = 1 if order`i'==`j'
}
}
Is it possible to use the value of order'i' to assign the variable[order`i'VALUE] directly? Then I can get rid of the j loop above. Something like this.
forvalues i = 1/20 {
replace variable[`order`i'value] = 1
}
Thanks for your help!
***** CLARIFICATION ADDED Feb 2nd.**
I simplified my problem and the dataset too much bc the solutions suggested work for what I presented but, are not getting at what I'm really attempting to do. Thank you three for your solutions though. I was not clear enough in my post.
To clarify, my data doesn't have a one to one correspondence of each order# assigning variable# a 1 if it's not missing. For example, the first observation for order1=3, variable1 isn't supposed to get a 1, variable3 should get a 1. What I didn't include in my original post is that I'm actually checking for other conditions to set it equal to 1.
For more background, I'm counting up births of women by birth order(1st child, 2nd child, etc) that occurred at different ages of mothers. So in the data, each row is a woman, each order# is the number birth (order1=3, it's her third child). The corresponding variable#s are the counts (variable# means the woman has a child of birth order #). I mentioned in the post, that I do this 9 times bc I'm doing it for 5 year age groups (15-19; 20-24; etc). So the first set of variable# would be counts of birth by order when women were ages 15-19; the second set of variable# would be counts of births by order when women were 20-24. etc etc. After this, I sum up the counts in different ways (by woman's education, geography, etc).
So with the additional loop what I do is something more like this
forvalues k = 1/9{
forvalues i = 1/20 {
forvalues j = 1/20 {
replace variable`k'_`j' = 1 if order`i'==`j' & age`i'==`k' & birth_age`i'<36
}
}
}
Not sure if it's possible, but I wanted to simplify so I only need to cycle through each child once, without cycling through the birth orders and directly use the value in order# to assign a 1 to the correct variable. So if order1=3 and the woman had the child at the specific age group, assign variable[agegroup][3]=1; if order1=2, then variable[agegroup][2] should get a 1.
forvalues k=1/9{
forvalues i = 1/20 {
replace variable`k'_[`order`i'value] = 1 if age`i'==`k' & birth_age`i'<36
}
}
I would reshape twice. First reshape to long, then condition variable on !missing(order), then reshape back to wide.
* generate your data
clear
set obs 3
forvalues i = 1/3 {
generate order`i' = .
local k = (3 - `i' + 1)
forvalues j = 1/`k' {
replace order`i' = (`k' - `j' + 1) if (_n == `j')
}
}
list
*. list
*
* +--------------------------+
* | order1 order2 order3 |
* |--------------------------|
* 1. | 3 2 1 |
* 2. | 2 1 . |
* 3. | 1 . . |
* +--------------------------+
* I would rehsape to long, then back to wide
generate id = _n
reshape long order, i(id)
generate variable = !missing(order)
reshape wide order variable, i(id) j(_j)
order order* variable*
drop id
list
*. list
*
* +-----------------------------------------------------------+
* | order1 order2 order3 variab~1 variab~2 variab~3 |
* |-----------------------------------------------------------|
* 1. | 3 2 1 1 1 1 |
* 2. | 2 1 . 1 1 0 |
* 3. | 1 . . 1 0 0 |
* +-----------------------------------------------------------+
Using a simple forvalues loop with generate and missing() is orders of magnitude faster than other proposed solutions (until now). For this problem you need only one loop to traverse the complete list of variables, not two, as in the original post. Below some code that shows both points.
*----------------- generate some data ----------------------
clear all
set more off
local numobs 60
set obs `numobs'
quietly {
forvalues i = 1/`numobs' {
generate order`i' = .
local k = (`numobs' - `i' + 1)
forvalues j = 1/`k' {
replace order`i' = (`k' - `j' + 1) if (_n == `j')
}
}
}
timer clear
*------------- method 1 (gen + missing()) ------------------
timer on 1
quietly {
forvalues i = 1/`numobs' {
generate variable`i' = !missing(order`i')
}
}
timer off 1
* ----------- method 2 (reshape + missing()) ---------------
drop variable*
timer on 2
quietly {
generate id = _n
reshape long order, i(id)
generate variable = !missing(order)
reshape wide order variable, i(id) j(_j)
}
timer off 2
*--------------- method 3 (egen, rowmax()) -----------------
drop variable*
timer on 3
quietly {
// loop over the order variables creating dummies
forvalues v=1/`numobs' {
tab order`v', gen(var`v'_)
}
// loop over the domain of the order variables
// (may need to change)
forvalues l=1/`numobs' {
egen variable`l' = rmax(var*_`l')
drop var*_`l'
}
}
timer off 3
*----------------- method 4 (original post) ----------------
drop variable*
timer on 4
quietly {
forvalues i = 1/`numobs' {
gen variable`i' = 0
forvalues j = 1/`numobs' {
replace variable`i' = 1 if order`i'==`j'
}
}
}
timer off 4
*-----------------------------------------------------------
timer list
The timed procedures give
. timer list
1: 0.00 / 1 = 0.0010
2: 0.30 / 1 = 0.3000
3: 0.34 / 1 = 0.3390
4: 0.07 / 1 = 0.0700
where timer 1 is the simple gen, timer 2 the reshape, timer 3 the egen, rowmax(), and timer 4 the original post.
The reason you need only one loop is that Stata's approach is to execute the command for all observations in the database, from top (first observation) to bottom (last observation). For example, variable1 is generated but according to whether order1 is missing or not; this is done for all observations of both variables, without an explicit loop.
I wonder if you actually need to do this. For future questions, if you have a further goal in mind, I think a good strategy is to mention it in your post.
Note: I've reused code from other posters' answers.
Here's a simpler way to do it (that still requires 2 loops):
// loop over the order variables creating dummies
forvalues v=1/20 {
tab order`v', gen(var`v'_)
}
// loop over the domain of the order variables (may need to change)
forvalues l=1/3 {
egen variable`l' = rmax(var*_`l')
drop var*_`l'
}
Related
I am attempting to reproduce the following in stata. This is a scatter plot of average portfolio returns (y axis) and predicted retruns (x axis).
To do so, I need your help on how I can extract the intercepts from 25 regressions into one variable? I am currently running the 25 portfolio regressions as follows. I have seen that parmest can potentially do this but can't get it to work with the forval. Many thanks
forval s = 1 / 5 {
forval h = 1 / 5 {
reg S`s'H`h' Mkt_Rf SMB HML
}
}
I don't know what your data look like, but maybe something like this will work:
gen intercepts = .
local i = 1
forval s = 1 / 5 {
forval h = 1 / 5 {
reg S`s'H`h' Mkt_Rf SMB HML
// assign the ith observation of intercepts
// equal to the regression constant
replace intercepts = _b[_cons] if _n == `i'
// increment i
local ++i
}
}
The postfile series of commands can be very helpful in a situation like this. The commands allows you to store results in a separate data set without losing the data in memory.
You can start with this as a simple example. This code will produce a Stata data set called "results.dta" with the variables s h and constant with a record of each regression.
cap postclose results
postfile results s h constant using results.dta, replace
forval s = 1 / 5 {
forval h = 1 / 5 {
reg S`s'H`h' Mkt_Rf SMB HML
loc c = _b[_cons]
post results (`s') (`h') (`c')
}
}
postclose results
use results, clear
I have some panel data of the form...
id | amount
-----------
1 | 10
1 | 10
1 | 100
2 | 10
2 | 15
2 | 10
3 | 100
What I'm looking to do seems like it should be fairly simple, but my experience with Stata is limited and I'm used to programming in languages similar to C/Java. Essentially, I want to drop an entire person (id) if any of their individual observations ever exceed a certain amount. So let's say I set this amount to 50, I want to drop all the observations from id 1 and id 3 such that the data will then only contain observations from id 2.
The pseudo-code in Java would be fairly straightforward...
for(int i = 0; i < dataset_length; i++) {
if dataset[i].amount > 50 {
int drop_id = dataset[i].id;
for(int j = 0; j < dataset_length; j++) {
if dataset[j].id == drop_id {
delete observation
}
}
}
}
What would the Stata equivalent of something akin to this be? I'm surely missing something and making it more complicated than it ought to be, but I cannot figure it out.
If there are no missings on amount this is just
bysort id (amount) : drop if amount[_N] > 50
If there are missings, then
gen ismissing = -missing(amount)
bysort id (ismissing amount): drop if amount[_N] > 50 & amount[_N] < .
would be one kind of check, although it's hard to see how the missings could be interesting or useful.
The machinery here (for one introduction see here) in effect builds in a loop over identifiers, and over the observations for each identifier. Literal translation using models from mainstream programming languages could only result in lengthier and less efficient code.
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 use Stata for estimating rolling standard deviation of ROA (using 4 window in previous year). Now, I would like to keep only those rolling standard deviation that has at least 3 observation (out of 4) in the ROA. How can I do this using Stata?
ROA roa_sd
. .
. .
. .
.0108869 .
.0033411 .
.0032814 .0053356 (this value should be missing as it was calculated using only 2 valid value)
.0030827 .0043739
.0029793 .0038275
Your question is answered on the blog post I link to above in the comments. You can use rolling and then add an additional screen to discard sigma when the number of observations doesn't meet your threshold.
But for simple calculations like sigma and beta (i.e., standard deviation and univariate regression coefficient) you can do much better with a more manual approach. Compare the rolling solution with my manual solution.
/* generate panel by adpating the linked code */
clear
set obs 20000
gen date = _n
gen id = floor((_n - 1) / 20) + 1
gen roa = int((100) * runiform())
replace roa = . in 1/4
replace roa = . in 10/12
replace roa = . in 18/20
/* solution with rolling */
/* http://statadaily.wordpress.com/2014/03/31/rolling-standard-deviations-and-missing-observations/ */
timer on 1
xtset id date
rolling sd2 = r(sd), window(4) keep(date) saving(f2, replace): sum roa
merge 1:1 date using f2, nogenerate keepusing(sd2)
xtset id date
gen tag = missing(l3.roa) + missing(l2.roa) + missing(l1.roa) + missing(roa) > 1
gen sd = sd2 if (tag == 0)
timer off 1
/* my solution */
timer on 2
rolling_sd roa, window(4) minimum(3)
timer off 2
/* compare */
timer list
list in 1/50
I show the manual solution is much faster.
. /* compare */
. timer list
1: 132.38 / 1 = 132.3830
2: 0.10 / 1 = 0.0990
Save the following as rolling_sd.ado in your personal ado file directory (or in your current working directory). I'm sure that someone could further streamline this code. Note that this code has the additional advantage of meeting the minimum data requirements at the front edge of the window (i.e., calculates sigma with first three observations, rather than waiting for all four).
*! 0.2 Richard Herron 3/30/14
* added minimum data requirement
*! 0.1 Richard Herron 1/12/12
program rolling_sd
version 11.2
syntax varlist(numeric), window(int) minimum(int)
* get dependent and indpendent vars from varlist
tempvar n miss xs x2s nonmiss1 nonmiss2 sigma1 sigma2
local w = `window'
local m = `minimum'
* generate cumulative sums and missing values
xtset
bysort `r(panelvar)' (`timevar'): generate `n' = _n
by `r(panelvar)': generate `miss' = sum(missing(`varlist'))
by `r(panelvar)': generate `xs' = sum(`varlist')
by `r(panelvar)': generate `x2s' = sum(`varlist' * `varlist')
* generate variance 1 (front of window)
generate `nonmiss1' = `n' - `miss'
generate `sigma1' = sqrt((`x2s' - `xs'*`xs'/`nonmiss1')/(`nonmiss1' - 1)) if inrange(`nonmiss1', `m', `w') & !missing(`nonmiss1')
* generate variance 2 (back of window, main part)
generate `nonmiss2' = `w' - s`w'.`miss'
generate `sigma2' = sqrt((s`w'.`x2s' - s`w'.`xs'*s`w'.`xs'/`nonmiss2')/(`nonmiss2' - 1)) if inrange(`nonmiss2', `m', `w') & !missing(`nonmiss2')
* return standard deviation
egen sigma = rowfirst(`sigma2' `sigma1')
end
Is the modified version of kappa proposed by Conger (1980) available in Stata? Tried to google it to no avail.
This is an old question, but in case anyone is still looking--the SSC package kappaetc now calculates that, along with every other inter-rater statistic you could ever want.
Since no one has responded with a Stata solution, I developed some code to calculate Conger's kappa using the formulas provided in Gwet, K. L. (2012). Handbook of Inter-Rater Reliability (3rd ed.), Gaithersburg, MD: Advanced Analytics, LLC. See especially pp. 34-35.
My code is undoubtedly not as efficient as others could write, and I would welcome any improvements to the code or to the program format that others wish to make.
cap prog drop congerkappa
prog def congerkappa
* This program has only been tested with Stata 11.2, 12.1, and 13.0.
preserve
* Number of judges
scalar judgesnum = _N
* Subject IDs
quietly ds
local vlist `r(varlist)'
local removeit = word("`vlist'",1)
local targets: list vlist - removeit
* Sums of ratings by each judge
egen judgesum = rowtotal(`targets')
* Sum of each target's ratings
foreach i in `targets' {
quietly summarize `i', meanonly
scalar mean`i' = r(mean)
}
* % each target rating of all target ratings
foreach i in `targets' {
gen `i'2 = `i'/judgesum
}
* Variance of each target's % ratings
foreach i in `targets' {
quietly summarize `i'2
scalar s2`i'2 = r(Var)
}
* Mean variance of each target's % ratings
foreach i in `targets' {
quietly summarize `i'2, meanonly
scalar mean`i'2 = r(mean)
}
* Square of mean of each target's % ratings
foreach i in `targets' {
scalar mean`i'2sq = mean`i'2^2
}
* Sum of variances of each target's % ratings
scalar sumvar = 0
foreach i in `targets' {
scalar sumvar = sumvar + s2`i'2
}
* Sum of means of each target's % ratings
scalar summeans = 0
foreach i in `targets' {
scalar summeans = summeans + mean`i'2
}
* Sum of meansquares of each target's % ratings
scalar summeansqs = 0
foreach i in `targets' {
scalar summeansqs = summeansqs + mean`i'2sq
}
* Conger's kappa
scalar conkappa = summeansqs -(sumvar/judgesnum)
di _n "Conger's kappa = " conkappa
restore
end
The data structure required by the program is shown below. The variable names are not fixed, but the judge/rater variable must be in the first position in the data set. The data set should not include any variables other than the judge/rater and targets/ratings.
Judge S1 S2 S3 S4 S5 S6
Rater1 2 4 2 1 1 4
Rater2 2 3 2 2 2 3
Rater3 2 5 3 3 3 5
Rater4 3 3 2 3 2 3
If you would like to run this against a test data set, you can use the judges data set from StataCorp and reshape it as shown.
use http://www.stata-press.com/data/r12/judges.dta, clear
sort judge
list, sepby(judge)
reshape wide rating, i(judge) j(target)
rename rating* S*
list, noobs
* Run congerkappa program on demo data set in memory
congerkappa
I have run only a single validation test of this code against the data in Table 2.16 in Gwet (p. 35) and have replicated the Conger's kappa = .23343 as calculated by Gwet on p. 34. Please test this code on other data with known Conger's kappas before relying on it.
I don't know if Conger's kappa for multiple raters is available in Stata, but it is available in R via the irr package, using the kappam.fleiss function and specifying the exact option. For information on the irr package in R, see http://cran.r-project.org/web/packages/irr/irr.pdf#page.12 .
After installing and loading the irr package in R, you can view a demo data set and Conger's kappa calculation using the following code.
data(diagnoses)
print(diagnoses)
kappam.fleiss(diagnoses, exact=TRUE)
I hope someone else here can help with a Stata solution, as you requested, but this may at least provide a solution if you can't find it in Stata.
In response to Dimitriy's comment below, I believe Stata's native kappa command applies either to two unique raters or to more than two non-unique raters.
The original poster may also want to consider the icc command in Stata, which allows for multiple unique raters.