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# Open and closed intervals in the if statement

Is there a way to use open and/or closed numeric non-integer intervals with the IF statement?

e.g. X=1.26;
if X in (1:5) then Y=1; else Y=0; results in Y=0.
if X in (1.1:5.3) then Y=1; else Y=0; returns an error.

For programming reasons I can't use the syntax
if X >= 1 and X<=5 then Y=1; else Y=0;
as I need a single condition to follow the variable X without repeating the variable name.

I note that one can use the BETWEEN operator with a WHERE statement but not with an IF statement - does anyone know the rationale behind this?
Posts: 3,852

## Re: Open and closed intervals in the if statement

Posted in reply to deleted_user
It is not clear to me what values you expect from the range 1.1:5.3
What is the increment?

You can create an array of target values and use the IN operator on the array.

[pre]
1259 data _null_;
1260 array x[5];
1261 j = 1;
1262 do a=1.1 by 1 while(j le dim(x));
1263 x = a;
1264 j + 1;
1265 end;
1266 put x
• ;
1267 z = 2.1;
1268 z_in_x = z in x;
1269 put z_in_x=;
1270 run;

1.1 2.1 3.1 4.1 5.1
z_in_x=1
[/pre]
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Posts: 0

## Re: Open and closed intervals in the if statement

Posted in reply to data_null__
Hi, actually what I am looking for is the equivalent of the BETWEEN operator for the IF statement (for some reason BETWEEN can only be used with the WHERE statement). i.e. "IF X in range then do Y", where X is a non-integer numeric variable. My code can't use the syntax "IF X ge z1 and x le z2 then do Y", as there must be only a single condition following the variable X.
Super Contributor
Posts: 3,176

## Re: Open and closed intervals in the if statement

Posted in reply to deleted_user
Consider this "expression" structure:

IF ( 0.5 LE LE 1.5 ) then....;

Scott Barry
SBBWorks, Inc.

Suggested Google advanced search argument, this topic / post:

sas language data step if statement expression site:sas.com
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## Re: Open and closed intervals in the if statement

Thanks, I am looking for a condition that FOLLOWS the variable only as I am doing a lookup of a table that contains only the part after the variable. i.e. I am looking for "IF X then ...", where is kept in the lookup table. E.g. is , except you can't use BETWEEN with the IF statement.
Posts: 3,852

## Re: Open and closed intervals in the if statement

Posted in reply to deleted_user
Why do have such silly rules.

Perhaps this will meet the requirement. Check the bounds in different if statements.

[pre]
1487 data _null_;
1488 do x = 0.26 to 7.26 by 1;
1489 put 'NOTE: ' x=;
1490 if x ge 1.1 then if x le 5.3
1491 then put 'NOTE- ' x= 'is in the range 1.1 to 5.3';
1492 end;
1493 run;

NOTE: x=0.26
NOTE: x=1.26
x=1.26 is in the range 1.1 to 5.3
NOTE: x=2.26
x=2.26 is in the range 1.1 to 5.3
NOTE: x=3.26
x=3.26 is in the range 1.1 to 5.3
NOTE: x=4.26
x=4.26 is in the range 1.1 to 5.3
NOTE: x=5.26
x=5.26 is in the range 1.1 to 5.3
NOTE: x=6.26
NOTE: x=7.26
[/pre]
Super Contributor
Posts: 3,176

## Re: Open and closed intervals in the if statement

Posted in reply to data_null__
To the OP: did the IF example not make sense? Suggest you try it to test a value range. Otherwise explain why you don't think it will work (but without just repeating the question/post again).

Scott Barry
SBBWorks, Inc.
Regular Contributor
Posts: 241

## Re: Open and closed intervals in the if statement

Posted in reply to deleted_user
> Thanks, I am looking for a condition that FOLLOWS the
> variable only as I am doing a lookup of a table that
> contains only the part after the variable. i.e. I am
> looking for "IF X then ...", where
> is kept in the lookup table. E.g.
> is , except you
> can't use BETWEEN with the IF statement.
...
If the can contain the name of the variable (which will likely be the case when the lookup key is indeed the variable name), then you can do something like below. That is the part being:
[pre]
= ifn(x=0|missing(x), 999, ifn(1.3<=x<=10.7,x,-x))
[/pre]
hope this helps a bit.
[pre]
data _null_;
do x = ., 0, 1.26, 6.26, 10.71;
if x = ifn(x=0|missing(x), 999, ifn(1.3<=x<=10.7,x,-x)) then y = 1;
else y = 0;
put x= y=;
end;
run;
/* on log
x=. y=0
x=0 y=0
x=1.26 y=0
x=6.26 y=1
x=10.71 y=0
*/
[/pre]
Wait a minute. If you are *not* allowed to do a like:
[pre]
>= 1.3 and x <= 10.7
[/pre]
then, there is no hope. This cannot be done, as far as I know, that is.
Super Contributor
Posts: 3,176

## Re: Open and closed intervals in the if statement

Posted in reply to chang_y_chung_hotmail_com
Exactly how are you doing the "lookup table" invocation at SAS compilation time - that would be an interesting point to clarify? Also, please share the exact SAS code you are executing, preferably in a SAS -generated log, not just your pasted code-piece. Clearly your original post isn't quite consistent with whatever is being attempted to resolve "on the fly" a piece of your IF statement.

Scott Barry
SBBWorks, Inc.
Regular Contributor
Posts: 241

## Re: Open and closed intervals in the if statement

Scott wrote:
> Exactly how are you doing the "lookup table"
> invocation at SAS compilation time - that would be an
> interesting point to clarify?

I am not OP, so I can only guess. But it is quite easy to imagine a data step validating a lot of variables based on a lookup table with the variable name as the key that stores validating expression(s) for each variable. Below is an over-simplification (you should *not* rely on the global macro variables as the lookup table.), but illustrates the concept.
[pre]
/* validation macro looking up based on var name */
%macro inRange(var);
&&&var.cond
%mend inRange;

/* prepare lookup table */
data valid;
input var \$ cond & \$30.;
call symputx(catt(var,"cond"), cond, "g");
cards;
age 13 <= age <= 14
height 63 <= height <=65
;
run;

/* validation */
data _null_;
set sashelp.class;
ageOK = %inRange(age);
heightOK = %inRange(height);
if ageOK and heightOK then put name= age= height=;
run;
/* on log
Name=Henry Age=14 Height=63.5
Name=Judy Age=14 Height=64.3
*/
[/pre]
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Posts: 0

## Re: Open and closed intervals in the if statement

Posted in reply to chang_y_chung_hotmail_com
Thanks guys, the purpose was to see if there was an equivalent of the 'between' operator for the if statement (possibly even enhanced to allow for open, left-open and right-open as well as closed intervals) and this is clearly *not* the case. But I found the catx function and the symputx (as opposed to symput) call routine useful things I didn't know of.
Posts: 3,852

## Re: Open and closed intervals in the if statement

Posted in reply to deleted_user
Maybe I still don't understand what you're talking about, but then I don't think you do either.

To me the statements below confirm that BETWEEN is just a fancy way to write
(condition1 and condition2)

[pre]
2200 data class;
2201 set sashelp.class;
2202 where height between 59 and 65;
2203 run;

NOTE: There were 8 observations read from the data set SASHELP.CLASS.
WHERE (height>=59 and height<=65);
NOTE: The data set WORK.CLASS has 8 observations and 5 variables.

2204 data class;
2205 set sashelp.class;
2206 where 59 le height le 65;
2207 run;

NOTE: There were 8 observations read from the data set SASHELP.CLASS.
WHERE (height>=59 and height<=65);
NOTE: The data set WORK.CLASS has 8 observations and 5 variables.

2208
2209 data class;
2210 set sashelp.class;
2211 if 59 le height le 65;
2212 run;

NOTE: There were 19 observations read from the data set SASHELP.CLASS.
NOTE: The data set WORK.CLASS has 8 observations and 5 variables.
[/pre]
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## Re: Open and closed intervals in the if statement

Posted in reply to data_null__
I am physically writing out the code:

"IF "!! !! !! " then " !! !! " =1; else " !! !! " = 0;"

For this I need to have a valid value of that appears after only and does not need a repetition of to form a valid IF statement.

That's the clearest way I can put it.
Super Contributor
Posts: 3,176

## Re: Open and closed intervals in the if statement

Posted in reply to deleted_user
SAS does not support the syntax you are demonstrating -- of course you already know that which is why you are using a DATA step to generate your SAS code.

Once again, you are pounding the sand by suggesting that, without BETWEEN support, maybe there might be a suitable resolution for you --- and I provide you an example. Now you need to revise your DATA step that generates the code, the one with your condition values, and when there is a range-value consideration, you need to have a "low" and a "high" component code piece generated.

From your nomenclature, here's what you now need to generate for a range-value:

1) identify and determine type of comparison that must be used (LOW, HIGH, EQUAL, RANGE-VALUE, etc.).
2) based on #1, construct IF / THEN logic according to the type of comparison.
3) construct/append-to SAS variable containing code.
4) *repeat above as needed*
5) complete DATA step, generate macro variable with SYMPUT - end of DATA step.
6) execute macro variable containing generated SAS code.

Other techniques to achieve the same result would be to write (using PUT statement) to a "temporary" external sequential file in your main DATA step (no macro variable involved, so there is no implicit code-piece length limit, which no one has discussed here!), and then use %INCLUDE ; to cause your generated SAS code to get compiled and executed.

Scott Barry
SBBWorks, Inc.

Scott Barry
SBBWorks, Inc.
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Posts: 0

## Re: Open and closed intervals in the if statement

Thanks Scott, excellent thinking there. Actually I didn't know whether or not SAS supported the type of syntax I wanted, which is why I put the post up.

Because for any particular , can contain (on different rows) both conditions that don't need code modification and conditions that do need code modification, what I actually did since yesterday is:

1) text-search for " and ", " & ", " or ", " | " (with an implicit rule that there must be a space inserted in the condition cell as shown if any of these strings appears).
2) if 1) is false, proceed as given in previous post.
3) if 1) is true, break down the into three sub-strings: a) part before " and " or " or ", b) " and " or " or " c) part after " and " or "or ". Then modify the code to: "IF " " then " " =1; else " " =0;"

An example of a that doesn't need modification is "ge 5.2". An example of one that does need above modification is "ge 5.2 and le 9.1".

Cheers,
Steve
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