## arithmetic operations within a variable

Solved
Occasional Contributor
Posts: 7

# arithmetic operations within a variable

I have multiple observations for each ID. I also have a variable called "score". I want to make a new variable "range" based on the values in "score" variable. Range is the difference between maximum and minimum values for score for each id. Below is what I have and what I want.

What I have:

ID      score

1         10

1         20

1         30

2         20

2         30

3         50

3         60

What I want;

ID      Score     average (this variable is the difference between the last score for an ID and the first score for the same ID)

1         10              20

1         20              .

1         30              .

2         20              10

2         30              .

3         50              10

3         60              .

Accepted Solutions
Solution
‎02-11-2018 03:33 PM
Super User
Posts: 6,624

## Re: arithmetic operations within a variable

[ Edited ]

Many things don't make sense.  Why should the difference between the first and last be the same as the difference between the min and the max?  Why should the result appear on just the first observation for each ID?  Why should the new field be named AVERAGE?

At any rate, here's a program that takes care of one interpretation of the question:

data want;

do until (last.id);

set have;

by id;

min_val = min(min_val, score);

max_val = max(max_val, score);

end;

average = max_val - min_val;

do until (last.id);

set have;

by id;

output;

average= .;

end;

drop min_val max_val;

run;

All Replies
Posts: 2,802

## Re: arithmetic operations within a variable

Use PROC SUMMARY, which computes the RANGE for you.

``````proc summary nway data=have;
class id;
var score;
output out=_ranges_ range=range;
run;``````

Then you can merge this back to the original data set using BY ID;

--
Paige Miller
Solution
‎02-11-2018 03:33 PM
Super User
Posts: 6,624

## Re: arithmetic operations within a variable

[ Edited ]

Many things don't make sense.  Why should the difference between the first and last be the same as the difference between the min and the max?  Why should the result appear on just the first observation for each ID?  Why should the new field be named AVERAGE?

At any rate, here's a program that takes care of one interpretation of the question:

data want;

do until (last.id);

set have;

by id;

min_val = min(min_val, score);

max_val = max(max_val, score);

end;

average = max_val - min_val;

do until (last.id);

set have;

by id;

output;

average= .;

end;

drop min_val max_val;

run;

PROC Star
Posts: 1,555

## Re: arithmetic operations within a variable

``````data have;
input ID      score;
datalines;
1         10
1         20
1         30
2         20
2         30
3         50
3         60
;

proc sql;
create table want as
select *,case when score = min(score) then range(score) else .  end as range
from have
group by id
order by 1,2;
quit;``````
Posts: 5,474

## Re: arithmetic operations within a variable

Nice code @novinosrin. But, for the sake of clarity, I would prefer if you didn't use numbers in the order by clause, especially when referring to * as a column list. It forces the reader to go back to the definition of the dataset to identify the variables.

PG
PROC Star
Posts: 1,555

## Re: arithmetic operations within a variable

Thank you Sir @PGStats and my apologies to OP @Sinakian1  please see the revised:

``````data have;
input ID      score;
datalines;
1         10
1         20
1         30
2         20
2         30
3         50
3         60
;

proc sql;
create table want as
select *,case when score = min(score) then range(score) else .  end as range
from have
group by id
order by id,score;
quit;``````
Super User
Posts: 23,228

## Re: arithmetic operations within a variable

It doesn't seem useful to have the RANGE in a single value. That's usually done when you want to use it later on in a calculation with most values.

Here's an example of how you do it with average, change it to RANGE to have it work.

☑ This topic is solved.

Discussion stats
• 6 replies
• 172 views
• 8 likes
• 6 in conversation