topic Re: A division with a denominator which is the sum of values of a variable for two days before and a in SAS Programming

A division with a denominator which is the sum of values of a variable for two days before and after

aminkarimid — Sun, 10 Dec 2017 20:05:36 GMT

Hello everybody;

I want to calculate a division which the nominator is intraday variable and the denominator is the aggregate of daily values of two days before and after that specific date and name. This new variable is called adjusted as shown in table below;

table1:

name	date	time	Intraday	daily	adjusted
A	12/1/2013	9:30:00	1
A	12/1/2013	10:00:00	2	3
A	12/2/2013	10:30:00	3
A	12/2/2013	11:00:00	2	5
A	12/3/2013	10:00:00	2		0.0952381
A	12/3/2013	10:30:00	1		0.04761905
A	12/3/2013	11:00:00	1	4	0.04761905
A	12/4/2013	11:30:00	1		0.04347826
A	12/4/2013	12:00:00	3		0.13043478
A	12/4/2013	12:30:00	3	7	0.13043478
A	12/7/2013	9:30:00	1		0.05
A	12/7/2013	10:00:00	1	2	0.05
A	12/8/2013	10:30:00	2		0.11111111
A	12/8/2013	11:00:00	3	5	0.16666667
A	12/17/2013	11:30:00	1		0.07692308
A	12/17/2013	12:00:00	1	2	0.07692308
A	12/18/2013	12:30:00	2	2
A	12/28/2013	9:30:00	1
A	12/28/2013	10:00:00	1	2
B	12/8/2013	10:30:00	2
B	12/8/2013	11:30:00	2
B	12/8/2013	12:30:00	1	5
B	12/14/2013	9:30:00	1
B	12/14/2013	10:30:00	1	2
B	12/15/2013	12:00:00	2		0.13333333
B	12/15/2013	12:30:00	1	3	0.06666667
B	12/24/2013	9:30:00	1	1	0.08333333
B	12/25/2013	10:00:00	2		0.125
B	12/25/2013	10:30:00	2	4	0.125
B	12/26/2013	12:00:00	2	2	0.125
B	12/30/2013	9:30:00	3
B	12/30/2013	10:00:00	2
B	12/30/2013	10:30:00	1	6
B	1/3/2014	11:00:00	1
B	1/3/2014	11:30:00	1
B	1/13/2014	12:00:00	1	3

Now, let me give you an example. the value '0.0952381' at date '12/3/2013' in adjusted column is (2/(3+5+4+7+2)). the value '2' in nominator is intraday value at '12/3/2013' and the denominator is the sum of daily values for two days before and after plus daily volume of that specific day. For another instance, the adjusted value '0.04347826' at date '12/4/2013' is (1/(3+5+4+7+2)).

How can I create the variable adjusted in table 1?

Thanks in advance.

Re: A division with a denominator which is the sum of values of a variable for two days before and a

novinosrin — Sun, 10 Dec 2017 20:32:50 GMT

The question is indeed interesting however the dates in your convert logic explanation seem to fall outside of range or perhaps I am the one who isn't understanding well. Please clarify:

Denominator: aggregate of daily values of two days before and after that specific date and name

Eg:

A	12/3/2013 --> 3,5, 4 ,7makes sense to me as those are within two days before and after, but what about 2 that falls on 12/7/2013

I am sure most others may gauge your requirement thinking out of the box, but I need more comprehensive explanation. Thank you!

Re: A division with a denominator which is the sum of values of a variable for two days before and a

aminkarimid — Sun, 10 Dec 2017 21:28:09 GMT

Thanks @novinosrin;
values of two days before are '3' at '12/1/2013' and '5' at '12/2/2013';
The value of current day is '4' at '12/3/2013';
Values of two days after are '7' at '12/4/2013' and 2 at '12/7/2013'. So, the denominator at date '12/3/2013' is (3+5+4+7+2).

Re: A division with a denominator which is the sum of values of a variable for two days before and a

art297 — Sun, 10 Dec 2017 23:46:29 GMT

It would help if you added a column for each row in your example explain either (1) which rows were selected or (2) why a value wasn't computed.

Art, CEO, AnalystFinder.com

Re: A division with a denominator which is the sum of values of a variable for two days before and a

novinosrin — Mon, 11 Dec 2017 03:01:20 GMT

For an ordinary bloke like me, that explanation doesn't quite suffice. But here you, my code works for your sample:

data have;

infile datalines expandtabs truncover ;

input name $ date :mmddyy10. time : time10. Intraday daily ;

format date mmddyy10. time time10.;

datalines;

A 12/1/2013 9:30:00 1

A 12/1/2013 10:00:00 2 3

A 12/2/2013 10:30:00 3

A 12/2/2013 11:00:00 2 5

A 12/3/2013 10:00:00 2

A 12/3/2013 10:30:00 1

A 12/3/2013 11:00:00 1 4

A 12/4/2013 11:30:00 1

A 12/4/2013 12:00:00 3

A 12/4/2013 12:30:00 3 7

A 12/7/2013 9:30:00 1

A 12/7/2013 10:00:00 1 2

A 12/8/2013 10:30:00 2

A 12/8/2013 11:00:00 3 5

A 12/17/2013 11:30:00 1

A 12/17/2013 12:00:00 1 2

A 12/18/2013 12:30:00 2 2

A 12/28/2013 9:30:00 1

A 12/28/2013 10:00:00 1 2

B 12/8/2013 10:30:00 2

B 12/8/2013 11:30:00 2

B 12/8/2013 12:30:00 1 5

B 12/14/2013 9:30:00 1

B 12/14/2013 10:30:00 1 2

B 12/15/2013 12:00:00 2

B 12/15/2013 12:30:00 1 3

B 12/24/2013 9:30:00 1 1

B 12/25/2013 10:00:00 2

B 12/25/2013 10:30:00 2 4

B 12/26/2013 12:00:00 2 2

B 12/30/2013 9:30:00 3

B 12/30/2013 10:00:00 2

B 12/30/2013 10:30:00 1 6

B 1/3/2014 11:00:00 1

B 1/3/2014 11:30:00 1

B 1/13/2014 12:00:00 1 3

;

proc sort data= have(where=(daily ne .)) out=_have nodupkey;

by name date;

run;

data want;

if (_n_ = 1) then do;

declare hash h(dataset: "_have", multidata: "yes");

h.definekey('name','date');

h.definedone();

end;

call missing(max_date);

do n=1 by 1 until(last.name);

set _have;

by name date;

max_date=max(date,max_date);

end;

array dt(*) prev_date1 prev_date2 after_date1 after_date2;

do n1=1 by 1 until(last.name);

set _have;

by name date;

if first.name then call missing(of dt(*));

curr_date=date;

curr_date_daily=daily;

prev_date1=lag1(date);

prev_date2=lag2(date);

if n1>2 and n1<=n-2 then

do;

if not missing(prev_date1) and not missing(prev_date2) then

do;

c=0;

do t=date+1 to max_date;

rc=h.check(key:name, key:t);

if rc=0 then

do;

c+1;

if c=1 then after_date1=t;

else if c=2 then

do;

after_date2=t;

leave;

end;

if cmiss(of dt(*))=0 then output;

end;

format t max_date prev_date1 prev_date2 after_date1 after_date2 curr_date mmddyy10.;

keep name date prev_date1 prev_date2 after_date1 after_date2 curr_date_daily;

run;

data final_WANT;

keep name _date time intraday _daily adjusted;

retain name _date time intraday _daily adjusted;

if (_n_ = 1) then do;

if 0 then set have;

if 0 then set want;

declare hash h(dataset: "have(where=(daily ne .))", multidata: "yes",ordered:'y');

h.definekey('name','date');

h.definedata('name','date','daily');

h.definedone();

declare hash h1(dataset: "want", multidata: "yes",ordered:'y');

h1.definekey('name','date');

h1.definedata(ALL: 'YES' );

h1.definedone();

end;

array t(*) prev_date1 prev_date2 after_date1 after_date2;

array t1(*) prev_date1_daily prev_date2_daily after_date1_daily after_date2_daily;

do until(last.name);

set have;

by name date;

_daily=daily;

_date=date;

rc=h1.find();

if rc=0 then

do;

do i= 1 to dim(t);

temp= t(i);

rc2=h.find(key:name, key:temp);

if rc2=0 then t1(i)=daily;

end;

adjusted=divide(intraday,sum(of t1(*),curr_date_daily));

end;

else do;

call missing(of t(*));

call missing(of t1(*),adjusted);

end;

output;

end;

format _date mmddyy10.;

run;

Re: A division with a denominator which is the sum of values of a variable for two days before and a

Reeza — Mon, 11 Dec 2017 04:35:05 GMT

Is it +/- 2 BUSINESS days, calendar days, event days, ....random days?

values of two days before are '3' at '12/1/2013' and '5' at '12/2/2013';
The value of current day is '4' at '12/3/2013';
Values of two days after are '7' at '12/4/2013' and 2 at '12/7/2013'. So, the denominator at date '12/3/2013' is (3+5+4+7+2).

December 7 is 4 days past December 3rd NOT 2 days so why/how is it included?

Business days and/or calendar days don't make sense, so I'm guessing event days of some kind? But then those are days and that's the confusion with how you've phrased your question. Please take more time in detailing your questions.

If it is event days, create an event day variable and then use a SQL join with itself and a margin of +/- 2 days. You've asked previously solved questions that demonstrate this type of logic.

Re: A division with a denominator which is the sum of values of a variable for two days before and a

aminkarimid — Mon, 11 Dec 2017 09:26:08 GMT

As you said, there is no same length between days. So, the better explanation is two rows after and before of current row.