On May 3, 11:17 am, art...@NETSCAPE.NET (Arthur Tabachneck) wrote: > Sorry, > > I never provided my original test code and, in the interim, discovered the > answer. > > First, the test code was: > > data test1; > input date permno x; > lx=lag(x); > put _all_; > lx=lag(x); > put _all_; > datalines; > . 4 2 > 1 4 . > 2 4 1 > ; > > data test2; > input date permno x; > do i=1 to 2; > lx=lag(x); > put _all_; > end; > datalines; > . 4 2 > 1 4 . > 2 4 1 > ; > > I found the reason for the observed difference in the documentation, > namely: "Each occurrence of a LAGn function in a program generates its own > queue of values." > > Art > ------- > On Sun, 3 May 2009 10:16:54 -0400, Arthur Tabachneck <art...@NETSCAPE.NET> > wrote: > > >List, > > >I was going to answer Barry's question by directing him toward Dan's nice > >response from a couple of years ago, namely: > > >http://www.listserv.uga.edu/cgi-bin/wa?A2=ind0511e&L=sas-l&D=0&P=19479 > > >Yes, of course the problem is that lag is que rather than observation > >specific. > > >However, before posting my response, I wrote a short program in order to > >show what was happening and realized that I don't understand all of the > >specifics. > > >Can anyone provide an explanation for the following differences: > > >145 data test1; > >146 input date permno x; > >147 lx=lag(x); > >148 put _all_; > >149 lx=lag(x); > >150 put _all_; > >151 datalines; > > >date=. permno=4 x=2 lx=. _ERROR_=0 _N_=1 > >date=. permno=4 x=2 lx=. _ERROR_=0 _N_=1 > >date=1 permno=4 x=. lx=2 _ERROR_=0 _N_=2 > >date=1 permno=4 x=. lx=2 _ERROR_=0 _N_=2 > >date=2 permno=4 x=1 lx=. _ERROR_=0 _N_=3 > >date=2 permno=4 x=1 lx=. _ERROR_=0 _N_=3 > >NOTE: The data set WORK.TEST has 3 observations and 4 variables. > >NOTE: DATA statement used (Total process time): > > real time 0.01 seconds > > cpu time 0.01 seconds > > >155 ; > > >156 data test2; > >157 input date permno x; > >158 do i=1 to 2; > >159 lx=lag(x); > >160 put _all_; > >161 end; > >162 datalines; > > >date=. permno=4 x=2 i=1 lx=. _ERROR_=0 _N_=1 > >date=. permno=4 x=2 i=2 lx=2 _ERROR_=0 _N_=1 > >date=1 permno=4 x=. i=1 lx=2 _ERROR_=0 _N_=2 > >date=1 permno=4 x=. i=2 lx=. _ERROR_=0 _N_=2 > >date=2 permno=4 x=1 i=1 lx=. _ERROR_=0 _N_=3 > >date=2 permno=4 x=1 i=2 lx=1 _ERROR_=0 _N_=3 > >NOTE: The data set WORK.TEST has 3 observations and 5 variables. > >NOTE: DATA statement used (Total process time): > > real time 0.01 seconds > > cpu time 0.01 seconds > > >166 ; > > >Art > >--------- > >On Sat, 2 May 2009 15:35:59 -0700, barry.brian.barr...@GMAIL.COM wrote: > > >>I was trying to understand the code by looking at one and then data > >>set two. > > >>The real code is to produce data set three. I am not really sure how > >>the code is working. > >>When I look at the output of lx, ly, and lz for data set three, it > >>doesn't make sense that the all the columns are filled up, since there > >>are missing values in the corresponding adjacent lx, ly, and lz row > >>variables. Let me know how this code works. It looks like magic to me, > >>LOL. I have programmed in SAS for a year. Let me know what's the best > >>way to learn SAS even more. Are there good reference books, a must > >>buy. I just used the online SAS reference guide and google. > >>I know that the SAS Institute publishes Tutorial books as well. > > >>Below is the code ( just asking what is the logic, trying to figure it > >>out for days. There are some things in SAS that just doesn't make > >>sense how it works LOL but it just does LOL.): > > >>/****CODE BEGINS******/ > > >>libname home '/home/mit/bcubeb3'; > >>data home.one; > >>input date permno x y z w; > >>datalines; > >>. 4 . . . . > >>1 4 . . . . > >>2 4 10 3 3990 . > >>3 4 . . . 3680 > >>4 4 . . . . > >>5 4 . . . . > >>6 4 . . . . > >>7 4 . . . . > >>8 4 . . . . > >>9 4 . . . . > >>10 4 . . . 3680 > >>11 4 . . . . > >>12 4 . . . . > >>13 4 . . . . > >>14 4 . . . . > >>15 4 . . . . > >>16 4 . . . 3793 > >>17 4 . . . . > >>18 4 . . . . > >>19 4 . . . . > >>20 5 . . . 3843 > >>21 5 . . . . > >>22 5 20 2 4000 . > >>23 5 . . . . > >>24 5 . . . . > >>; > >>run; > > >>data home.two; > >>set home.one; > >>by permno date; > >>i=0; > >>do while(i<1); > >>lx=lag(x); > >>ly=lag(y); > >>lz=lag(z); > >>lw=lag(w); > >>if permno=lag(permno) and x=. then x=lx; > >>if permno=lag(permno) and y=. then y=ly; > >>if permno=lag(permno) and z=. then z=lz; > >>if permno=lag(permno) and w=. then w=lw; > >>i+1; > >>end; > >>run; > > >>data home.three; > >>set home.one; > >>by permno date; > >>i=0; > >>do while(i<2); > >>lx=lag(x); > >>ly=lag(y); > >>lz=lag(z); > >>lw=lag(w); > >>if permno=lag(permno) and x=. then x=lx; > >>if permno=lag(permno) and y=. then y=ly; > >>if permno=lag(permno) and z=. then z=lz; > >>if permno=lag(permno) and w=. then w=lw; > >>i+1; > >>end; > >>run;

I was reading the link you sent me. From my understanding the lag function follows a queue specifically FIFO. Now initially the data as you have it is: data permno x . 4 2 1 4 . 2 4 1

then when you do the do i=1 to 2. It executes the lag function so for i=1 data permno x lx . 4 2 . 1 4 . 2 2 4 1 . 3 N/A N/A 1

The 3 is there for illustrations purposes

so when you apply the lag function it shifts up by one as below for i=2 data permno x lx . 4 2 2 1 4 . . 2 4 1 1

I am guessing if you do i=1 to 3. you will get the following: data permno x lx . 4 2 . 1 4 . 1 2 4 1 .

but instead you get i=3 data permno x lx . 4 2 2 1 4 . . 2 4 1 1

I thought it was following a queue which to mean all it means there is a shift.

How does that apply to my case, where I don't see how the last data set (THREE) have all observations filled when using the code. I was thinking about FIFO but it didn't make sense to me still.

For my case in my initial post:

Dataset ONE:

date permno x y z w; date permno x y z w . 4 . . . . 1 4 . . . . 2 4 10 3 3990 . 3 4 . . . 3680 4 4 . . . . 5 4 . . . . 6 4 . . . . 7 4 . . . . 8 4 . . . . 9 4 . . . . 10 4 . . . 3680 11 4 . . . . 12 4 . . . . 13 4 . . . . 14 4 . . . . 15 4 . . . . 16 4 . . . . 17 4 . . . . 18 4 . . . . 19 4 . . . . 20 4 . . . . 21 5 . . . . 22 5 20 2 4000 . 23 5 . . . . 24 5 . . . .

Dataset TWO: date permno x y z w i lx ly lz lw . 4 . . . . 1 . . . . 1 4 . . . . 1 . . . . 2 4 10 3 3990 . 1 . . . . 3 4 10 3 3990 3680 1 10 3 3990 . 4 4 . . . 3680 1 . . . 3680 5 4 . . . . 1 . . . . 6 4 . . . . 1 . . . . 7 4 . . . . 1 . . . . 8 4 . . . . 1 . . . . 9 4 . . . . 1 . . . . 10 4 . . . 3680 1 . . . . 11 4 . . . 3680 1 . . . 3680 12 4 . . . . 1 . . . . 13 4 . . . . 1 . . . . 14 4 . . . . 1 . . . . 15 4 . . . . 1 . . . . 16 4 . . . . 1 . . . . 17 4 . . . . 1 . . . . 18 4 . . . . 1 . . . . 19 4 . . . . 1 . . . . 20 5 . . . . 1 . . . . 21 5 . . . . 1 . . . . 22 5 20 2 4000 . 1 . . . . 23 5 20 2 4000 . 1 20 2 4000 . 24 5 . . . . 1 . . . .

data set THREE:

date permno x y z w i lx ly lz lw . 4 . . . . 2 . . . . 1 4 . . . . 2 . . . . 2 4 10 3 3990 . 2 10 3 3990 . 3 4 10 3 3990 3680 2 . . . 3680 4 4 10 3 3990 3680 2 . . . . 5 4 10 3 3990 3680 2 . . . . 6 4 10 3 3990 3680 2 . . . . 7 4 10 3 3990 3680 2 . . . . 8 4 10 3 3990 3680 2 . . . . 9 4 10 3 3990 3680 2 . . . . 10 4 10 3 3990 3680 2 . . . 3680 11 4 10 3 3990 3680 2 . . . . 12 4 10 3 3990 3680 2 . . . . 13 4 10 3 3990 3680 2 . . . . 14 4 10 3 3990 3680 2 . . . . 15 4 10 3 3990 3680 2 . . . . 16 4 10 3 3990 3680 2 . . . . 17 4 10 3 3990 3680 2 . . . . 18 4 10 3 3990 3680 2 . . . . 19 4 10 3 3990 3680 2 . . . . 20 5 . . . . 2 . . . . 21 5 . . . . 2 . . . . 22 5 20 2 4000 . 2 20 2 4000 . 23 5 20 2 4000 . 2 . . . . 24 5 20 2 4000 . 2 . . . .