topic Re: Ascending day counter in SAS Programming

Ascending day counter

jorentergesen — Sun, 07 May 2017 23:52:01 GMT

Hi SAS Community,

I am trying to write what should be a simple program to capture data with ascending days, eg 01/01/2012 to 01/02/2012.

I have done it somewhat, I have created a variable COUNT, which shows me when I have ascending days, however it doesn't give me the origin day eg 12/12/2011 count=0 01/01/2012 count=0 01/02/2012 count=1. As well, I would like to come up with how many ascending days I have consecutively.

I know there must be a loop function to create this but I am not very strong at sas.

Any advice would be greatly appreciated.

data wcourt.event;

set wcourt.PM25_high;

/*count2=(consec+1);*/
 
tempdate=lag(&start);

leaddate=sum(tempdate,1);

if date=tempdate+1 then count=1;
else count=0;

if leaddate=Date then count2=1;
else count2=0;


/*if count2=1 and count3=*/

format tempdate date9. leaddate date9.;

run;quit;

Re: Ascending day counter

Reeza — Mon, 08 May 2017 00:11:50 GMT

Look at DIF function.

And the RETAIN statement to increment counter. It's unclear what you want as output.

You'll get better answers if you post your data as a data step and show what you want as output. Include it directly in the forum - not as an attachment.

Re: Ascending day counter

art297 — Mon, 08 May 2017 00:24:27 GMT

Early in your code you use:

tempdate=lag(&start);

Where do you create the macro variable start and what is its value?

Art, CEO, AnalystFinder.com

Re: Ascending day counter

jorentergesen — Mon, 08 May 2017 00:27:34 GMT

Oh sorry I thought I took that out. &start=Date




data wcourt.event;

set wcourt.PM25_high;

/*count2=(consec+1);*/
 
tempdate=lag(Date);

leaddate=sum(tempdate,1);

if date=tempdate+1 then count=1;
else count=0;
retain count;

if leaddate=Date then count2=1;
else count2=0;


/*if count2=1 and count3=*/

format tempdate date9. leaddate date9.;

run;quit;

Re: Ascending day counter

art297 — Mon, 08 May 2017 00:33:22 GMT

Post enough records of what you want as a result .. enough defined as a sufficient amount that we can understand what you are trying to do.

Art, CEO, AnalystFinder.com

Re: Ascending day counter

jorentergesen — Mon, 08 May 2017 00:45:44 GMT

The data looks like this

Date,Mean

01/01/2012,xx

01/05/2012,xx

01/08/2012,xx

01/09/2012,xx

01/11/2012,xx

01/23/2012,xx

01/24/2012,xx

01/25/2012,xx

02/01/2013,xx

02/09/2013,xx

01/01/2014,xx

I want my data output to have this: The date, the mean, if the day is apart of ascending days in order, the numbers of days ascending

Date,Mean,Ascending_days_occurence,#_of_ascending_days

01/01/2012,xx,0,0

01/05/2012,xx,0,0

01/08/2012,xx,1,1

01/09/2012,xx,1,2

01/11/2012,xx0,0

01/23/2012,xx1,1

01/24/2012,xx1,2

01/25/2012,xx1,3

02/01/2013,xx,0,0

02/09/2013,xx,0,0

01/01/2014,xx,0,0

Re: Ascending day counter

art297 — Mon, 08 May 2017 01:14:47 GMT

If all you are looking for is a consecutive days counter then the following would work:

data wcourt.event (drop=date2);
  merge wcourt.PM25_high wcourt.PM25_high (firstobs=2 keep=date rename=(date=date2));
  if _n_ eq 1 then count=0;
  if dif(date) eq 1 or date+1 eq date2 then count+1;
  else count=0;
  output;
  if date+1 ne date2 then count=0;
run;

Art, CEO, AnalystFinder.com

Re: Ascending day counter

jorentergesen — Mon, 08 May 2017 01:46:40 GMT

I know this may be annoying but do you mind explaining what "_n_" does?

Re: Ascending day counter

art297 — Mon, 08 May 2017 04:53:57 GMT

_n_, in a datastep, is the automatic record counter. Thus, for record 1, _n_ equals 1, for record 2 it equals 2, etc.

Art, CEO, AnalystFinder.com