Lags and Lead

[yp2609]

I want to divide 2nd row of j_top by the 1st row of j_denom; divide 3rd row of j_top by 2nd rows of j_denom; so on and so forth. 

lag function is not working as I want it to be. 

Can anyone provide code for this? Thank you so much!

[PaigeMiller]

Show us your code. Show us the log. Explain what is wrong.

[yp2609]

Nothing is wrong. I just have no clue how to perform this code. 

[ballardw]

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@yp2609 wrote:

I want to divide 2nd row of j_top by the 1st row of j_denom; divide 3rd row of j_top by 2nd rows of j_denom; so on and so forth. 

lag function is not working as I want it to be. 

Can anyone provide code for this? Thank you so much!

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When you say code isn't working you should show the code and possibly the LOG if you are getting unexpected notes in the log.

Because Lag, and its partner Dif, function is a queue function then using it conditionally is the main cause of unexpected results as the "lag" then becomes the last time the condition was true.

[Tom · Accepted solution]

divide 2nd row of j_top by the 1st row of j_denom;

data want;
  set have;
  lag_j_denom=lag(j_denom);
  if _n_ =1 then want = j_top;
 else want=j_top/lag_j_denom;
run;

[yp2609]

Thank you so much!

[sbxkoenk]

Hello @yp2609 ,

Your topic title (title of your post) is misleading.

But if you don't know how a data step works, I agree you have difficulties in knowing whether you need lag or lead.

I hope it's clear by now you only need lag.

When processing observation k you need info that is from observation (k-1).

That's looking back.

Looking ahead is not needed here.

BR,

Koen