About gamotte

gamotte · ‎06-29-2016

As you can see, the array is declared as array a_mngrs {*} a--c; That is all columns from a to c. If you have more managers, just indicate the first and the last instead of a and c.

gamotte · ‎06-29-2016

I don't understand exactly what You are trying to do. But when the interpreter meets the end of the "do i" loop, i will be reset to the next value of the list V7200,V7210,V7211,V7212,V7213,V7240,V7250,V7260,V7280. So the only use of the new i value You computed is to determine the label. In that case, You can use another variable than i or directly origtillg : LABEL origtillg=CATS('orig',origtillg); Why do You have to modify the looping variable i and what effect do You expect ?

gamotte · ‎06-29-2016

Hello, You are modifying the value of i inside the do loop on the different values of i.

gamotte · ‎06-29-2016

data test; input accounts; datalines; 12 2 32 4 5 6 7 8 9 50 11 46 13 14 15 ; run; data test2; input managers$ count; datalines; a 48 b 45 c 49 ; run; proc transpose data=test2 out=tr_test2; id managers; idlabel managers; run; proc sql; SELECT count(*) INTO :n_accounts FROM test; quit; data test3; set tr_test2; keep managers; array a_mngrs {*} a--c; if _N_=1 then index=1; do i=1 to &n_accounts.; if a_mngrs{index} lt 50 then do; a_mngrs{index}=a_mngrs{index}+1; managers=vname(a_mngrs{index}); output; end; /* Probably a better way to do this */ index=mod(index+1,dim(a_mngrs)); if index=0 then index=dim(a_mngrs); end; run; data want; merge test test3 (in=a); if a; run;

gamotte · ‎06-29-2016

There was a proble with the accounts column. Here is a corrected version : data want; set tr_test2 (drop=_NAME_); array a_mngrs {*}_NUMERIC_; keep accounts managers; if _N_=1 then index=1; accounts=0; do i=1 to 15; if a_mngrs{index} lt 50 then do; a_mngrs{index}=a_mngrs{index}+1; managers=vname(a_mngrs{index}); accounts+1; output; end; /* Probably a better way to do this */ index=mod(index+1,dim(a_mngrs)); if index=0 then index=dim(a_mngrs); end; run;

gamotte · ‎06-29-2016

It doesn't, sorry 😕

gamotte · ‎06-29-2016

Hello, I think this should work : proc transpose data=test2 out=tr_test2; id managers; idlabel managers; run; data want; set tr_test2 (drop=_NAME_); array a_mngrs {*}_NUMERIC_; if _N_=1 then index=1; do accounts=1 to 15; if a_mngrs{index} lt 50 then do; put index=; a_mngrs{index}=a_mngrs{index}+1; managers=vname(a_mngrs{index}); output; end; /* Probably a better way to do this */ index=mod(index+1,3); if index=0 then index=3; end; run;

gamotte · ‎06-28-2016

Hello, Data step processes the input dataset row by row. If there is no explicit output instruction in the code, the currently processed row (*) is output when the interpreter hits the run instruction. If there remains rows to process, there is an implicit loop and the instructions in the data step are executed again for the next row of the input dataset. (*) More precisely the program data vector (PDV) wich is SAS terminology to indicate the data being processed (row extracted from the input dataset + newly created variables). On the other hand, when the code in the data step contains some output instructions, those determine when to output new lines in the created dataset and the run instruction does not implicitely output data anymore. So in your code, for each input row where Var1='A', the instruction output WORK.ONE creates a new row in table ONE. The following output instruction creates a new row in both output datasets. With 3 rows having Var1='A' and two rows where Var1 <> 'A', the ONE dataset will have : 3x2+2=8 rows.

gamotte · ‎06-28-2016

Hello, Does the following program answer your problem ? data new; set sashelp.class; length Total $1000.; retain total; if _N_=1 then Total=Name; else Total=catx(',',Name,Total); run;

gamotte · ‎06-24-2016

%macro handimp; /* Declares a macro function named handimp */ %let var1 = Intake; /* declares a macrovariable var1 that takes the value "Intake" (wo quotes) */ %let var2 = Discharge; /* declares a macrovariable var2 that takes the value "Discharge" (wo quotes) */ %do k=1 %to 2; /* Loops on k from 1 to 2 */ /* Imports data from an excel file, the name of which depends on k */ proc import out= hand_&&var&k.._paper /* &&var&k. first, Intake then Discharge */ /* var&k => var1 or var2 | &&var&k => &var1 or &var2 */ datafile= "&fpath\Hand_Wrist.xlsx" /* macrovariable fpath should be initialized somewhere else */ dbms=xlsx replace; getnames=yes; sheet = "&&var&k"; run; /* Sorts the newly created SAS dataset by client_id */ proc sort data = hand_&&var&k.._paper; by client_id; run; %end; /* End of loop */ %mend; /* End of macro definition */ %handimp; /* Calls the macro handimp */

gamotte · ‎06-10-2016

Thanks. Using findw is the goal here, the string always being "U Z E" in the execution context where the problem occur (the U is always isolated).

gamotte · ‎06-10-2016

RW9, Thanks for your answer. The program in question is a stored process from a web application. The "U" is in fact the value of a macrovariable that results from the user's choices in a form. We use macro code to take users inputs into account and determine the actions to take. I was unaware of the index function so we could replace the test by : %if %sysfunc(index(&string.,&element))>0 %then %do; .... I will try this. Yet, I am still wondering how this macro can give random results.

gamotte · ‎06-10-2016

I have translated the given code from french to english and missed some elements. The first arguments to findw adn find should be &string. not &chaine. There is no such problem is the original source code.

gamotte · ‎06-10-2016

Hi, I have a macro "isIn" which is supposed to test if an element is in a given (macro) string. The code is as follows : %macro rank(string,element,sep=); /* Returns element's index in a list of elements */ %if &sep ne %then %do; %sysfunc(findw(&chaine.,&element.,&sep,E)) %end; %else %do; %sysfunc(find(&chaine.,&element.)) %end; %mend; %macro isIn(string,element); /* Tests whether an element is in a given string */ %eval(%rank(&string.,&element.,sep=' ')>0) %mend; Somewhere in the program, &mystring. is initialized to "U Z E" (without the quotes) and elsewhere the following test occur : %if %isIn(&mystring,U) %then ... The test is expected to return 1 since "U" is in "U Z E". Several execution of the program give different results for the above test, yet the input is always the same. It shoudn't be too complicated to modify the code and avoid using the findw function but I am just wondering what is wrong exactly and why the macro behavior seems random. Do you see anything wrong with the above code ? We use SAS 9.4. Thanks

gamotte · ‎06-07-2016

Hello, Are you asking how to import some available data in a sas dataset for further processing ? If so, you will probably get more feedback if you post in a more general forum (Base SAS programming for instance) and avoid technical terms (behavioral model, logistic regression, ...) that are not strictly relevant to your problem. If your problem really concerns importing data in SAS, you should also give some precisions on the data at your disposal (file type, structure,...) If I completely miss your point, sorry for the irrelevant advice.

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