How is the IF-THEN/ELSE condition working on this problem? - sas

I am a bit puzzled why the output is a certain way on the following program.
data a;
input name$ lv ##;
cards;
Frank 1 Joan 2 Sui 3 Burt 4 Kelly . Juan 1
;
data b;
set a;
if lv=.
then expertise='?';
else if lv=1
then expertise='L';
else if lv=2 or 3
then expertise ='M';
else expertise ='H';
run;
proc print data=b;
run;
In the program above, I am expecting that the output for the observation containing Burt has expertise of value H, but for some reason it is M.
I was thinking that the if statement should be lv=(2 or 3) but when I do that now those that I am thinking that the lv= 2 and lv=3 whose expertise should equal M now becomes H as well.
*This is probably explained by the fact that the syntax is inappropriate, and lv is never (2 or 3) /I am not sure why this does not cause an error/ thus the ELSE statement is executed.
I have a feeling that I am not understanding how the ELSE is really working.
However, according to that logic then if the lv=2 part is recognized but the others are not, I am not sure why the lv= 3 and lv=4 does not produce an H.
My goal is to understand why the program runs as if there is no syntax error and why the output is what it is.

Your mistake is here
If lv = 4 or 0
You think this means
If lv = 4 or lv = 0
But it doesn’t. What actually happens is
If (lv = 4) or 0
In SAS 0 is false and any number greater than 1 is true. Not sure about decimals. Anyways. The or condition makes this one false. But in your first example, it makes it always true.

Your syntax is incorrect, but not illegal, so SAS did what you asked for instead of what you wanted. The condition lv=2 or 3 is evaluated from left to right. So lv=2 will result in either 1 or 0 depending whether the value of lv is 2 or not. But both 1 or 3 and 0 or 3 will be true since 3 is always considered true.
What you really want is the in operator.
if missing(lv) then expertise='?';
else if lv=1 then expertise='L';
else if lv in (2 3) then expertise ='M';
else expertise ='H';

Related

Stata - Create within group ids

I have a data set of flows between locations, say they are 50 locations, but the number of pairs is not even because some locations do not have flows. I would like to create ids for each pair of observation (w_id and h_id)
Thank you.
Desired output
w_code h_code w_id h_id
295101011001003 291892204451015 1 1
295101011001003 295101011001003 1 2
295101011001003 291892202003011 1 3
295101011001025 295101021003001 2 1
295101011001025 295101011001025 2 2
295101011001026 291879507003038 3 1
295101011001026 190130007001013 3 2
295101011001026 295101105001027 3 3
295101011001026 291892126002008 3 4
295101011001026 291892126001005 3 5
295101011001029 291892199006006 4 1
295101011002007 295101011002015 5 1
295101011002014 295101011002016 6 1
295101011002014 295101011001003 6 2
295101011002016 295101011001007 7 1
295101011002030 295101255001008 8 1
Documentation accessible through Stata includes this paper on composite categorical variables and this paper on handling dyadic data. The Stata command search would have led to these papers, except that the art to finding as well as searching is thinking of the right keywords.
In your case the natural question arises whether for example the pair (1, 2) is really the same as (2, 1) and for flows my guess is No. In mathematics, abstraction is often the key to solving a problem; in statistical computing some concreteness may make a problem clearer. Perhaps h means husband and w means wife, and perhaps not. Assuming that (1, 2) and (2, 1) are quite different, a joint identifier is immediately obtained by
egen newid = group(w_id h_id)
and for a small number of identifiers -- you mention 50 -- there is no pain in asking for values to be labelled, so that with
egen newid = group(w_id h_id), label
the pair (1, 1) would be mapped to the value 1 and the value label 1 1.
As this solution was not immediately obvious, it is likely that a study of help egen will reveal a bunch of tools likely to be useful in data management; some are directly statistical.
For pairs of identifiers where Billy, Bob is to be treated like Bob, Billy see the second paper linked above. Whether this is true for the OP is a little unclear, but it is likely to be true for some others reading this in the future.

Impute missing covariates at random in Stata

I am trying to randomly impute missing data for several covariates using Stata. I have never done this, and I am trying to use this code from a former employee:
local covarall calc_age educcat ipovcat_bl US_born alc_yn2 drug_yn lnlpcbsum tot_iod
local num = 0
foreach j of local covarall {
gen iflag_`j'=0
replace iflag_`j'=1 if `j'==.
local num = `num'+1000
forvalues i = 1/476 {
sort `j'
count if `j'==.
di r(N)
local num2 = `num'+`i'
set seed `num2'
replace `j' in `i'=`j'[1+int((400-r(N))*runiform())] if iflag_`j'[`i']==1
}
}
When I run this, Stata just gives me this over and over forever:
(0 real changes made)
0
0
What am I doing wrong?
The three messages seem interpretable as follows:
replace iflag_`j' = 1 if `j' == .
will lead to a message (0 real changes made) whenever that is so, meaning that the variable in question is never equal to system missing, the requirement for replacement.
count if `j' == .
will lead to the display of 0 in the same circumstance.
di r(N)
ditto. count shows a result by default and then the code insists that it be shown again. Strange style, but not a bug.
All that said the line
replace `j' in `i'=`j'[1+int((400-r(N))*runiform())] if iflag_`j'[`i'] == 1
is quite illegal. My best guess is that you have copied it incorrectly somehow and that it should have been
replace `j' =`j'[1+int((400-r(N))*runiform())] in `i' if iflag_`j'[`i'] == 1
but this too should produce the same message as the first if a value is not missing.
I add that it is utterly pointless to enter the innermost loop if there are no missing values in a variable: there is then nothing to impute.
Changing the seed every time a change is made is strange, but that is partly a matter of taste.

How to write loop across Hierarchical Data (household-individual) in stata?

I'm now working on a household survey data set and I'd like to give certain members extra IDs according to their relationship to the household head. More specifically, I need to identify the adult children of household head and his/her spouse, if married, and assign them "sub-household IDs".
The variables are: hhid - household ID; pid -individual ID; relhead - relationship with head.
Regarding relhead, a 1 represents the head, a 6 represents a child, and a 7 represents a child-in-law. Below some example data, including in the last column the desired outcome. I assume that whenever a 6 is followed by a 7, they constitute a couple and belong to the same sub-household.
hhid pid relhead sub_hhid(desired)
50 1 1 1
50 2 3 1
50 3 6 2
50 4 6 3
50 5 7 3
-----------------------------------------------
67 1 1 1
67 3 6 2
67 4 7 2
Here are some thoughts:
There may be married and unmarried adult children within one household, the family structure is a little bit complicated, so I want to write some loop across the members in a household.
The basic idea is in the outer loop we identify the children staying-at-home and then check if there's a spouse presented, if there is, then we give the couple an indicator, if not, we continue and give the single stay_chil other indicator. After walking through all the possible members within a household, we get a series of within-household IDs. To facilitate further analysis , I need some kind of external ID variable to separate the sub-families.
* Define N as the total number of household, n as number of individual household size
* sty_chil is indicator for adult child who living with parents(head)
* sty_chil_sp is adult child's spouse
* "hid" and "ind_id" are local macros
forvalue hid=1/N {
forvalue ind_id= 1/n {
if sty_chil[`ind_id']==1 {
check if sty_chil_sp[`ind_id+1']==1 {
if yes then assign sub_hhid to this couples *a 6-7 pairs,identifid as couple
}
else { * single 6 identifid as single child
assign sub_hhid to this child
}
else { *Other relationships rather than 6, move forward
++ind_id the members within a household
}
++hid *move forward across households
}
The built-in stata by,sort: is pretty powerful but here I want to treat part of family members who fall into certain criterion and leave other untouched, so a if-else type loop is more natural for me (even by: may achieve my goal,it's always too tactful when situation become not so simple,and we cannot exhaust all the possible pattern of household pattern).
An immediate problem is that I don't know how to write loop across house IDs and individual IDs, because I used to acquire the household size (increment of outer loop) using by command (I'm not sure in this case it's 1 or the numerber of family members), and I'm not sure if mix up the by and if loops is a good programming practice, I favor write a "full loop" in this case. Please give me some clues how to achieve my goal and provide (illustrate)pseudo code for me.
An extra question is I cannot find the ado file which contains the content of by command, does it exist?
I will abstract from the issue of whether the assumption used to create matches is a sensible one or not. Rather, let this be an example of reaching the desired results without using explicit loops. Some logic and the use of subscripting (see help subscripting) can get you far.
clear
set more off
*----- example data -----
input ///
hhid pid relhead sub_hhid
50 1 1 1
50 3 6 2
50 4 6 3
50 5 7 3
67 1 1 1
67 3 6 2
67 4 7 2
67 5 6 3
end
list, sepby(hhid)
*----- what you want -----
bysort hhid (pid): gen hhid2 = sum( !(relhead == 7 & relhead[_n-1] == 6) )
list, sepby(hhid)
As you can see, one line of code gets you there. The reasoning is the following:
sum() gives the running sum. The arguments to sum(), being conditions, can either be True or False. The ! denotes the logical not (see help operators).
If it is not the case that the relationship is daughter/son-in-law AND the previous relationship is daughter/son, the condition evaluates to True and takes on the value of 1, increasing the running sum by 1. If it evaluates to False, meaning that the relationship is daughter/son-in-law AND the previous relationship is daughter/son, then it takes on the value of 0 and the running sum will not increase. This gives the result you seek.
You do this using the by: prefix, since you want to check each original household independently, so to speak.
For the the first observation of each original household, the condition always evaluates to True. This is because there exist no "previous" observation (relationship), and Stata considers relhead to be missing (., a very large number) and therefore, not equal to 6. This takes the running sum from 0 to 1 for the first observation of each sub-group, and so on.
Bottom line: learn how to use by: and take advantage of the features offered by Stata. Do not swim against the current; not here.
Edit
Please note that instead of progressively changing your example data set, you should provide a representative example from the beginning. Not doing so can render answers that are initially OK, completely inadequate.
For your modified example, add:
replace hhid2 = 1 if !inlist(relhead,6,7)
That will simply assign anyone not 6 or 7 to the same household as the head. The head is assumed to always have hhid2 == 1. If the head can have hhid2 != 1, then
bysort hhid (relhead): replace hhid2 = hhid2[1] if !inlist(relhead,6,7)
should work.
You can follow with:
bysort hhid (pid): replace hhid2 = hhid2[_n-1] + 1 if hhid2 != hhid2[_n-1] & _n > 1
but because they are IDs, it's not really necessary.
Finally, use:
gen hhid3 = string(hhid) + "_" + string(hhid2)
to create IDs with the form 50_1, 50_2, 50_3, etc.
Like I said before, if your data presents more complications, you should present a relevant example.

How do I calculate the maximum or minimum seen so far in a sequence, and its associated id?

From this Stata FAQ, I know the answer to the first part of my question. But here I'd like to go a step further. Suppose I have the following data (already sorted by a variable not shown):
id v1
A 9
B 8
C 7
B 7
A 5
C 4
A 3
A 2
To calculate the minimum in this sequence, I do
generate minsofar = v1 if _n==1
replace minsofar = min(v1[_n-1], minsofar[_n-1]) if missing(minsofar)
To get
id v1 minsofar
A 9 9
B 8 9
C 7 8
B 7 7
A 5 7
C 4 5
A 3 4
A 2 3
Now I'd like to generate a variable, call it id_min that gives me the ID associated with minsofar, so something like
id v1 minsofar id_min
A 9 9 A
B 8 9 A
C 7 8 B
B 7 7 C
A 5 7 C
C 4 5 A
A 3 4 C
A 2 3 A
Note that C is associated with 7, because 7 is first associated with C in the current sorting. And just to be clear, my ID variable here shows as a string variable just for the sake of readability -- it's actually numeric.
Ideas?
EDIT:
I suppose
gen id_min = id if _n<=2
replace id_min = id[_n-1] if v1[_n-1]<minsofar[_n-1] & missing(id_min)
replace id_min = id_min[_n-1] if missing(id_min)
does the job at least for the data in this example. Don't know if it would work for more complex cases.
This works for your example. It uses the user-written command vlookup, which you can install running findit vlookup and following through the link that appears.
clear
set more off
input ///
str1 id v1
A 9
B 8
C 7
B 7
A 5
C 4
A 3
A 2
end
encode id, gen(id2)
order id2
drop id
list
*----- what you want -----
// your code
generate minsofar = v1 if _n==1
replace minsofar = min(v1[_n-1], minsofar[_n-1]) if missing(minsofar)
// save original sort
gen osort = _n
// group values of v1 but respecting original sort so values of
// id2 don't jump around
sort v1 osort
// set obs after first as missing so id2 is unique within v1
gen v2 = v1
by v1: replace v2 = . if _n > 1
// lookup
vlookup minsofar, gen(idmin) key(v2) value(id2)
// list
sort osort
drop osort v2
list, sep(0)
Your code has generate minsofar = v1 if _n==1 which is better coded as generate minsofar = v1 in 1, because it is more efficient.
Your minsofar variable is just a displaced copy of v1, so if this is always the case, there should be simpler ways of handling your problem. I suspect your problem is easier than you have acknowledged until now, and that has come through your post. Perhaps giving more context, expanded example data, etc. could get you better advice.
This is both easier and a little more challenging than implied so far. Given value (a little more evocative than the OP's v1) and a desire to keep track of minimum so far, that's for example
generate min_so_far = value[1]
replace min_so_far = value if value < min_so_far[_n-1] in 2/L
where the second statement exploits the unsurprising fact that Stata replaces in the current order of observations. [_n-1] is the index of the previous observation and in 2/L implies a loop over all observations from the second to the last.
Note that the OP's version is buggy: by always looking at the previous observation, the code never looks at the very last value and will overlook that if it is a new minimum. It may be that the OP really wants "minimum before now" but that is not what I understand by "minimum so far".
If we have missing values in value they will not enter the comparison in any malign way: missing is always regarded as arbitrarily large by Stata, so missings will be recorded if and only if no non-missings are present so far, which is as it should be.
The identifier of that minimum at first sight yields to the same logic
generate min_so_far = value[1]
gen id_min = id[1]
replace min_so_far = value if value < min_so_far[_n-1] in 2/L
replace id_min = id if value < min_so_far[_n-1] in 2/L
There are at least two twists that might bite. The OP mentions a possibility that the identifier might be missing so that we might have a new minimum but not know its identifier. The code just given will use a missing identifier, but if the desire is to keep separate track of the identifier of the minimum value with known identifiers, different code is needed.
A twist not mentioned to date is that observations with different identifier might all have the same minimum so far. The code above replaces the identifier only the first time a particular minimum is seen; if the desire is to record the identifier of the last occurrence the < in the last code line above should be replaced with <=. If the desire is to keep track of the all the identifiers of the minimum so far, then a string variable is needed to concatenate all the identifiers.
With a structure of panel or longitudinal data the whole thing is done under the aegis of by:.
I can't see a need to resort to user-written extensions here.

Perform Fisher Exact Test from aggregated using Stata

I have a set of data like below:
A B C D
1 2 3 4
2 3 4 5
They are aggregated data which ABCD constitutes a 2x2 table, and I need to do Fisher exact test on each row, and add a new column for the p-value of the Fisher exact test for that row.
I can use fisher.exact and loop to do it in R, but I can't find a command in Stata for Fisher exact test.
You are thinking in R terms, and that is often fruitless in Stata (just as it is impossible for a Stata guy to figure out how to do by ... : regress in R; every package has its own paradigm and its own strengths).
There are no objects to add columns to. May be you could say a little bit more as to what you need to do, eventually, with your p-values, so as to find an appropriate solution that your Stata collaborators would sympathize with.
If you really want to add a new column (generate a new variable, speaking Stata), then you might want to look at tabulate and its returned values:
clear
input x y f1 f2
0 0 5 10
0 1 7 12
1 0 3 8
1 1 9 5
end
I assume that your A B C D stand for two binary variables, and the numbers are frequencies in the data. You have to clear the memory, as Stata thinks about one data set at a time.
Then you could tabulate the results and generate new variables containing p-values, although that would be a major waste of memory to create variables that contain a constant value:
tabulate x y [fw=f1], exact
return list
generate p1 = r(p_exact)
tabulate x y [fw=f2], exact
generate p2 = r(p_exact)
Here, [fw=variable] is a way to specify frequency weights; I typed return list to find out what kind of information Stata stores as the result of the procedure. THAT'S the object-like thing Stata works with. R would return the test results in the fisher.test()$p.value component, and Stata creates returned values, r(component) for simple commands and e(component) for estimation commands.
If you want a loop solution (if you have many sets), you can do this:
forvalues k=1/2 {
tabulate x y [fw=f`k'], exact
generate p`k' = r(p_exact)
}
That's the scripting capacity in which Stata, IMHO, is way stronger than R (although it can be argued that this is an extremely dirty programming trick). The local macro k takes values from 1 to 2, and this macro is substituted as ``k'` everywhere in the curly bracketed piece of code.
Alternatively, you can keep the results in Stata short term memory as scalars:
tabulate x y [fw=f1], exact
scalar p1 = r(p_exact)
tabulate x y [fw=f2], exact
scalar p2 = r(p_exact)
However, the scalars are not associated with the data set, so you cannot save them with the
data.
The immediate commands like cci suggested here would also have returned values that you can similarly retrieve.
HTH, Stas
Have a look the cci command with the exact option:
cci 10 15 30 10, exact
It is part of the so-called "immediate" commands. They allow you to do computations directly from the arguments rather than from data stored in memory. Have a look at help immediate
Each observation in the poster's original question apparently consisted of the four counts in one traditional 2 x 2 table. Stas's code applied to data of individual observations. Nick pointed out that -cci- can analyze a b c d data. Here's code that applies -cci to each table and, like Stas's code, adds the p-values to the data set. The forvalues i = 1/`=_N' statement tells Stata to run the loop from the first to the last observation. a[`i'] refers to the the value of the variable `a' in the i-th observation.
clear
input a b c d
10 2 8 4
5 8 2 1
end
gen exactp1 = .
gen exactp2 =.
label var exactp1 "1-sided exact p"
label var exactp2 "2-sided exact p"
forvalues i = 1/`=_N'{
local a = a[`i']
local b = b[`i']
local c = c[`i']
local d = d[`i']
qui cci `a' `b' `c' `d', exact
replace exactp1 = r(p1_exact) in `i'
replace exactp2 = r(p_exact) in `i'
}
list
Note that there is no problem in giving a local macro the same name as a variable.