eagles_dare13 Tracker

What can a stored process do that a normal SAS program cannot

eagles_dare13 — Fri, 28 Aug 2015 20:58:14 GMT

What is the use of a SAS stored process? What can a stored process do that a normal sas program / macro cannot? It seems like a stored process is an unnecessary wrapper around a simple SAS program.

In my company, the stored process is the first layer on the server side that intercepts web requests and sends back response in the form of JSON.

Can a web request not directly interact with a SAS server side macro? IS that what the stored process is for? Some kind of a Mid Tier controller that acts as the gateway between web client and the SAS server side code?

What ML algorithms are part of base SAS

eagles_dare13 — Fri, 28 Aug 2015 19:07:18 GMT

Am I correct to say that simple learning algorithms like linear and logistic regression can be done in sas enterprise guide using proc reg, but advanced algorithms like SVM and neural networks require enterprise miner?

SAS global macro variable not being resolved

eagles_dare13 — Wed, 05 Aug 2015 19:46:54 GMT

I have 2 jobs in DI studio. In job 1, I need to define some variables that I need to use in job2.

So in job 1, I do:

proc sql ;

select col1 into :new_col from table1 where 1 = 1;

quit;

%global new_col;

In job 2:

I try to use the macro variable new_col, but the variable name cannot be resolved.

How to solve this?

How to use a sql statement ina DATA step

eagles_dare13 — Wed, 24 Jun 2015 18:32:43 GMT

How to achieve this in SAS?

DATA TEST;

A=1;

B=2;

IF A IN (LIST_OF_VALUES) THEN B = 5;

OUTPUT;

RUN;

LIST_OF_VALUES is a column in a lookup table.

Ideally, Iwould like something like:

DATA TEST;

A=1;

B=2;

IF A IN (select LIST_OF_VALUES from TABLE_TEST) THEN B = 5;

OUTPUT;

RUN;

But this syntax is not allowed.

How to achieve this?

Re: How to simulate a Weiner process in SAS

eagles_dare13 — Thu, 11 Jun 2015 18:54:12 GMT

Thank you. I copied your code but am getting:

ERROR: Invocation of unresolved module RANDFUN

Re: How to generate a list of random numbers in SAS

eagles_dare13 — Thu, 11 Jun 2015 16:46:50 GMT

Thanks. How to get the plot of the distribution? I am looking for something like xyplot(x, p);

How to simulate a Weiner process in SAS

eagles_dare13 — Thu, 11 Jun 2015 13:36:30 GMT

Is there a way to simulate data points that have been drawn from a Geometric Brownian Motion (Weiner Process)?

I am looking at this function:SAS(R) 9.2 Language Reference: Dictionary, Fourth Edition

But I don't see an option for a Weiner process.

I know the Black Scholes internally simulates a Weiner process but is it possible to see what a set of 10,000 numbers drawn from a Weiner process looks like?

Re: How to generate a list of random numbers in SAS

eagles_dare13 — Thu, 11 Jun 2015 13:27:25 GMT

Thanks. How come the probplot does not represent a bell curve but more like a 45 degree straight line?

How to generate a list of random numbers in SAS

eagles_dare13 — Wed, 10 Jun 2015 18:53:13 GMT

I wish to simulate(generate) a list of random numbers, where I specify the number of observations that I wish, the underlying distribution(normal,lognormal), the mean, the std. deviation and I want as output a list of random numbers from this distribution.

Further I wish to draw a probability plot of these numbers, conforming to stathe distribution that they were drawn from.

How to do it in SAS?

Re: How to convert to char from numeric without losing decimals

eagles_dare13 — Fri, 06 Mar 2015 18:48:49 GMT

Sorry, but 3.10 is coming as 3.1, the last 0 is getting dropped.

data have;

num=3; output;

num=3.01; output;

num=3.10; output;

run;

data want;

set have;

length text $10.;

text=strip(put(num,best7.2));

run;

How to convert to char from numeric without losing decimals

eagles_dare13 — Fri, 06 Mar 2015 17:01:48 GMT

Hello,

I have a numeric variable x and i tried to convert to character using: put (x, best32.).

Now, x can have values like 3, 3.0, 3.02 etc...upto 2 decimal places.

How to keep the format as is..i.e 3 -->3, 3.0 --> 3.0 and so on?

Currently, with my code, it is outputting everything as 1 decimal places...and dropping the second decimal.

Thanks very much

Re: drop variables from a data set conditionally

eagles_dare13 — Fri, 30 Jan 2015 20:25:53 GMT

Thanks.

This seems to work, but is it correct?

In SAS, how can I do the equivalent of:

data a;

set b;

if (somecondition = true) then do;

drop col2;

end;/*col2 is a column included in b*/

run;

drop variables from a data set conditionally

eagles_dare13 — Fri, 30 Jan 2015 20:20:39 GMT

In SAS, how can I do the equivalent of:

data a;

set b;

if (somecondition = true)

drop col2; /*col2 is a column included in b*/

run;

Re: sas if else not working when using parentheses

eagles_dare13 — Thu, 29 Jan 2015 20:38:52 GMT

Thanks. Can you show with example what you mean by recoding with SELECT statement.

sas if else not working when using parentheses

eagles_dare13 — Thu, 29 Jan 2015 20:29:03 GMT

Why is this code giving error:

data test;

length a: $5;
a='2';
if a = '1' then a ='a' ;
else (
a= 'b');
run;

ERROR: Undeclared array referenced: else.

I realise removing the parentheses will solve the issue, but i want to keep the parentheses for ease of understanding. Ultimately the else box will have many if else if conditions and the parentheses helps greatly.

SAS HLO= O not working as expected

eagles_dare13 — Mon, 26 Jan 2015 18:58:16 GMT

I am creating a data step as follows:

data TEST;

INFILE DATALINES DSD;

INPUT CODE: $4. FLAG_IND: $1.;

DATALINES;

0111,Y

0115,N

;

if last then do;

hlo ='O';

FLAG_IND ='U';

output;

end;

run;

My desired output is:

CODE FLAG_IND HLO

0111 Y

0115 N

U O

But I get:

Statement is not valid or it is used out of proper order.

Is my syntax not right? I was following this:

http://www.pharmasug.org/proceedings/2011/CC/PharmaSUG-2011-CC19.pdf

Page 2

Find variable names that are in one dataset but not in another

eagles_dare13 — Mon, 05 Jan 2015 17:09:56 GMT

If I have 2 data sets:

data investment;

begin='01JAN1990'd;

end='31DEC2009'd;

run;

data investment_2;

begin='01JAN1990'd;

run;

I wish to write a program that will give me list of all columns that are in investment but not in investment_2. So in this case, the variable 'end'.

How to do this? I dont want to list out the column names using proc contents on eaqch dataset and then manually eyeball it and check for differences.

Re: A character operand was found in the %EVAL function or %IF condition where a numeric operand is required

eagles_dare13 — Wed, 31 Dec 2014 21:14:28 GMT

Thank you. You can replicate the error like this:

%macro test();

%if &macro_var =%str() %then

%do;

%put ERROR: The macro_var is not set.;

%return;

%end;

%else %do;

%put macro_var was set.;

%return;

%end;

%mend test;

%test();

A character operand was found in the %EVAL function or %IF condition where a numeric operand is required

eagles_dare13 — Wed, 31 Dec 2014 20:24:28 GMT

I have this code inside a macro:

%if &macro_var =%str() %then

%do;

%put ERROR: The macro_var is not set.;

%return;

%end;

I am getting error:

WARNING: Apparent symbolic reference macro_var not resolved.

ERROR: A character operand was found in the %EVAL function or %IF condition where a numeric operand is required. The condition was: &macro_var=

The WARNING is ok..because the macro_var has not been set. But Why is it throwing an error? Why is it looking for a numeric operand? How to resolve this? Basically I am trying to check if a macro variable is null or has been set with a proper value.

Macro variable getting set to the name of the macro

eagles_dare13 — Wed, 31 Dec 2014 19:09:48 GMT

When I run this progrma:

data _null_;
   set info1;
   if compress(upcase(a),'','cops') eq 'abc' then call symputx('macro_var1',b,'g'); /*if condition evaluates to false*/

run;

    %put Macro var1 is: &macro_var1;

I get Output:
Macro var1 is: &macro_var1

I was expecting output:
Macro var1 is:

i.e. the macro variable should be set to a null (blank) value. Why did the macro variable value get set to the name of the macro?
The dataset info1 looks like (2 by 2 matrix):

Name user12
Salary 100