topic Re: return over a 5 minute/ second period in SAS Procedures

return over a 5 minute/ second period

lakshmanmv — Sat, 19 Jul 2014 03:22:51 GMT

I am trying to find the return over a 5 minute period by using proc expand in SAS. I am working on financial time series data. The time data is in hr:min:sec. Sample data is given below:

date Co time price vol

01JAN2012 ABC 9:55:59 166.00 100

01JAN2012 ABC 9:56:02 165.40 50

01JAN2012 ABC 9:56:02 165.40 25

01JAN2012 ABC 9:56:03 166.45 700

01JAN2012 ABC 9:56:03 166.80 20

01JAN2012 ABC 9:56:03 167.00 280

01JAN2012 ABC 9:56:06 166.45 1

01JAN2012 ABC 9:56:06 166.05 24

01JAN2012 ABC 9:56:15 166.50 500

01JAN2012 ABC 9:56:16 166.50 25

01JAN2012 ABC 9:56:17 166.50 475

01JAN2012 ABC 9:56:22 166.25 10

01JAN2012 ABC 9:56:26 166.05 50

01JAN2012 ABC 9:56:30 166.25 20

01JAN2012 ABC 9:56:38 166.05 126

01JAN2012 ABC 9:56:40 166.00 100

01JAN2012 ABC 9:56:40 166.00 1

01JAN2012 ABC 9:56:43 166.00 1

01JAN2012 ABC 9:56:46 166.00 2

But the output throws up 'time' as 0,1,2,3.... when I run the following code:

data temp;set alldata_sas7bdat; proc expand data=temp out=lagtst method=none; convert time=time_lag1 / transform=(lag 1); convert price=price_lag1 / transform=(lag 1); by date; run;

proc print data=lagtst (obs=5000); run;

proc delete data=temp; data lagtst_2;set lagtst; timegap = time-time_lag1; if timegap > 0 and timegap < 5 then return=log(price/price_lag1); else return=.; run;

While I have tried to calculate return over a '5 minute' period, the data I have posted here is not enough. I can provide more data but I wonder if it would be too cumbersome here. So please suggest a method for finding out return of a '5 second' period. I will then make changes to find return over a 5-minute period. Thanks.

Re: return over a 5 minute/ second period

Ksharp — Sat, 19 Jul 2014 12:17:35 GMT

You can also achieve it by using sql. Here is for 5-seconds period .

data have;
input date :date9. Co $ time : time9. price vol     ;
format date date9. time time.;
cards;
01JAN2012 ABC 9:55:59 166.00 100
01JAN2012 ABC 9:56:02 165.40 50
01JAN2012 ABC 9:56:02 165.40 25
01JAN2012 ABC 9:56:03 166.45 700
01JAN2012 ABC 9:56:03 166.80 20
01JAN2012 ABC 9:56:03 167.00 280
01JAN2012 ABC 9:56:06 166.45 1
01JAN2012 ABC 9:56:06 166.05 24
01JAN2012 ABC 9:56:15 166.50 500
01JAN2012 ABC 9:56:16 166.50 25
01JAN2012 ABC 9:56:17 166.50 475
01JAN2012 ABC 9:56:22 166.25 10
01JAN2012 ABC 9:56:26 166.05 50
01JAN2012 ABC 9:56:30 166.25 20
01JAN2012 ABC 9:56:38 166.05 126
01JAN2012 ABC 9:56:40 166.00 100
01JAN2012 ABC 9:56:40 166.00 1
01JAN2012 ABC 9:56:40 166.00 1
01JAN2012 ABC 9:56:43 166.00 1
01JAN2012 ABC 9:56:46 166.00 2
;
run;
proc sql;
create table want as
 select *,log(price/(select price from have where date=a.date and time+5=a.time)) as return
  from have as a;
quit;

Xia Keshan

Re: return over a 5 minute/ second period

lakshmanmv — Sun, 20 Jul 2014 15:15:29 GMT

Thank you very much..

However, the program gives a blank 'return'.

date stk time price vol return

01JAN2003 ABC 9:55:59 166.00 100 .

I have used the program giving the path to the datafiles.

libname in 'c:\Users\VLM\Stk data';

data temp;set in.ABC_sas7bdat;

run;

proc sql;

create table want as

select *,log(price/(select price from temp where date=temp.date and time+5=temp.time)) as return

from temp;

quit;

Re: return over a 5 minute/ second period

Haikuo — Sun, 20 Jul 2014 17:27:55 GMT

's code assumes that you will have one unique data entry per second, obviously it is not the case, but the logic of his code still holds:

data have;

input date :date9. Co $ time : time9. price vol;

format date date9. time time.;

cards;

01JAN2012 ABC 9:55:59 166.00 100

01JAN2012 ABC 9:56:02 165.40 50

01JAN2012 ABC 9:56:02 165.40 25

01JAN2012 ABC 9:56:03 166.45 700

01JAN2012 ABC 9:56:03 166.80 20

01JAN2012 ABC 9:56:03 167.00 280

01JAN2012 ABC 9:56:06 166.45 1

01JAN2012 ABC 9:56:06 166.05 24

01JAN2012 ABC 9:56:15 166.50 500

01JAN2012 ABC 9:56:16 166.50 25

01JAN2012 ABC 9:56:17 166.50 475

01JAN2012 ABC 9:56:22 166.25 10

01JAN2012 ABC 9:56:26 166.05 50

01JAN2012 ABC 9:56:30 166.25 20

01JAN2012 ABC 9:56:38 166.05 126

01JAN2012 ABC 9:56:40 166.00 100

01JAN2012 ABC 9:56:40 166.00 1

01JAN2012 ABC 9:56:43 166.00 1

01JAN2012 ABC 9:56:46 166.00 2

;

run;

proc sql;

create table want as

select *,log(price/(select avg(price) from (select price from have where date=a.date and co=a.co and time between a.time and a.time-5 group by co, date having time=min(time) ))) as return

from have as a;

quit;

Above code takes mean value if you have multiple "price" for the same "second", alternatively, you can also choose to use Max or Min price.

Haikuo

Re: return over a 5 minute/ second period

Ksharp — Mon, 21 Jul 2014 12:51:49 GMT

That is because there is no price at that time . haikuo bian has already pointed out it .

I make a TEMP dataset to hold every second price value.

data have;
input date :date9. Co $ time : time9. price vol     ;
format date date9. time time.;
cards;
01JAN2012 ABC 9:55:59 166.00 100
01JAN2012 ABC 9:56:02 165.40 50
01JAN2012 ABC 9:56:02 165.40 25
01JAN2012 ABC 9:56:03 166.45 700
01JAN2012 ABC 9:56:03 166.80 20
01JAN2012 ABC 9:56:03 167.00 280
01JAN2012 ABC 9:56:06 166.45 1
01JAN2012 ABC 9:56:06 166.05 24
01JAN2012 ABC 9:56:15 166.50 500
01JAN2012 ABC 9:56:16 166.50 25
01JAN2012 ABC 9:56:17 166.50 475
01JAN2012 ABC 9:56:22 166.25 10
01JAN2012 ABC 9:56:26 166.05 50
01JAN2012 ABC 9:56:30 166.25 20
01JAN2012 ABC 9:56:38 166.05 126
01JAN2012 ABC 9:56:40 166.00 100
01JAN2012 ABC 9:56:40 166.00 1
01JAN2012 ABC 9:56:40 166.00 1
01JAN2012 ABC 9:56:43 166.00 1
01JAN2012 ABC 9:56:46 166.00 2
;
run;
data temp;
 merge have have(firstobs=2 keep=date time rename=(date=_date time=_time));
 output;
 if date eq _date then do;
  do i=time+1 to _time-1;
   time=i;output;
  end;
end;
run;


proc sql;
create table want as
 select *,log(price/(select price from temp where date=a.date and time+5=a.time)) as return
  from have as a;
quit;

Xia Keshan