data step execution

[Tom]

Since it is the first time that it executed the result of any LAG() function calls in the ELSE part on the third iteration will be missing values.

[Tom]

And each place in the code where you have a LAG function will keep its own queue.

Look at this example.

data x;
  do i=1 to 10 ;
    lag1=lag(i) ;
    if i ne 4 then lag2=lag(i);
    put (i lag1 lag2) (=);
  end;
run;
i=1 lag1=. lag2=.
i=2 lag1=1 lag2=1
i=3 lag1=2 lag2=2
i=4 lag1=3 lag2=2
i=5 lag1=4 lag2=3
i=6 lag1=5 lag2=5
i=7 lag1=6 lag2=6
i=8 lag1=7 lag2=7
i=9 lag1=8 lag2=8
i=10 lag1=9 lag2=9

[ender_akpil]

Thanks Tom, this is very nice code..

I have a question about lag2 variable, if lag's in the data step follw their own que then lag2 values should be:

lag2=.

lag2=1

lag2=2

lag2=2

lag2=3

lag2=4

lag2=5

lag2=6

lag2=7

lag2=8

lag2=9

but according to ypur result in line 6 in turns to normal flow?

[Tom]

When LAG() is called the top of the stack is returned and the current value is pushed onto the stack.

When I=1 then missing is returned and 1 is pushed.

When I=2 then 1 is returned and 2 is pushed.

When I=3 then 2 is returned and 3 is pushed.

When I=4 it is skipped so LAG2 stays as 2.

When I=5 then 3 is returned and 5 is pushed.

When I=6 then 5 is returned and 6 is pushed.

....

[ender_akpil]

Thanks Tom..

Still i want to ask, should lag function should use same que? i mean this is more meaningfull isnt it?

I come across with this problem while i was searching equivalent of oracle partitioned analytic query ( row_number over (partition by a order by b asc) )

Is there a anoher way to do this?