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08-13-2016 06:33 AM

Dear community,

I would be glad, if you could help me with the following problem.

What I am aiming at implementing is the following:

In the __first row__, I calculate a specific variable, e.g., y = x*a;

Said y shall be carried forward to the __second row__ a play the role as the new x. The same calculation is performed, i.e., y=x*a, whereby the x for the second row is the y from the first row;

That is the input

and output

I am considering.

It would be kind, if you could give me a hint, on how to I tackle this kind of problem?

As SAS "work" row to row (as I have understood), it should give a "trick"?

Otherwise, for each step, I would have to create an index, separate the y, merge the y to the next observation, replace x with the carried-forward-y and calculate once again.

This would result in a huge amount of data steps.

```
data have;
input x a;
datalines;
1 2
. 2
. 2
. 2
. 2
. 2
. 2
. 2
. 2
. 2
;
run;
```

Accepted Solutions

Solution

08-14-2016
08:57 AM

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08-13-2016 09:39 PM

A good example of 'RETAIN';

```
data want;
set have;
retain y;
if _n_ ne 1 then x=y;
y=x*a;
run;
```

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08-13-2016 07:56 AM

Here is a way by data step. Replace x by everytime before the next iteration like:

```
data have;
a = 2;
do i = 1 to 10;
if i = 1 then x = 1;
y = x * a;
output;
x = y;
end;
drop i;
run;
proc print; run;
```

Hope this is an acceptable solution to you.

Solution

08-14-2016
08:57 AM

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08-13-2016 09:39 PM

A good example of 'RETAIN';

```
data want;
set have;
retain y;
if _n_ ne 1 then x=y;
y=x*a;
run;
```

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08-14-2016 09:04 AM

Hello,

this is enormous.

I really had allready implemented it via a %macro %Do Loop.

The amount of time if do save is astonishing.

Thank you so much for this - I need this very procedure "over an over" so this really changes the game.

As I do have to implement it for more than one variable, would you mind considering if I have transferred it correctly to a setting more closely to the actual one?

```
data want;
set have;
retain x_end y_end z_end;
if _n_ ne 1 then x=x_end;
if _n_ ne 1 then y=y_end;
if _n_ ne 1 then z=z_end;
x_end = intercept1
+ a*x
+ b*y
+ c*z
;
y_end = intercept2
+ d*x
+ e*y
+ f*z
;
z_end = intercept3
+ g*x
+ h*y
+ i*z
;
run;
```

So, former y is now "x_end";

Especially, the three ifs are a concern of mine - could I achieve a better solution?

But, nonetheless, your very answer means a huge leap indeed for me.

Once again, thank you so much.