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jerry898969
Pyrite | Level 9

I'm trying to get the individual numbers after the decimal place and i'm not sure how to do it.

Say I have .003  I want to pull out the first 0 in d1 and the second to d2 and 3 into d3.

Not sure if the mod function would be the way to go or not.

Thank you

1 ACCEPTED SOLUTION

Accepted Solutions
art297
Opal | Level 21

FriedEgg,  Tom's method correctly deals with negative numbers .. it simply keeps the minus sign.  If the minus sign isn't desired by the OP, a simple solution would be to add an abs function.  e.g.:

data want;

  input x;

  array digit(3);

  do place=1 to 3 ;

    digit(place)=abs(mod(int(x*10**place),10));

  end;

  put;

cards;

.003

.123

12.345

-12.345

;

proc print;

run;

View solution in original post

7 REPLIES 7
art297
Opal | Level 21

There may be an easier way, but you can always use brute force. e.g.,

data want;

  input x;

  d1=int(x*10);

  d2=int(x*100-(int(x*10)*10));

  d3=int(x*1000-(int(x*100)*10));

  cards;

.003

.123

;

Tom
Super User Tom
Super User

This is a math problem. Here is one way using powers of 10, INT and MOD functions.

data _null_;

  input x;

  put x= @;

  do place=1 to 3 ;

    digit=mod(int(x*10**place),10);

    put digit= @;

  end;

  put;

cards;

.003

.123

12.345

;

Try it starting from 0 or minus one.

jerry898969
Pyrite | Level 9

Thanks for both the answers. 

Tom how can I put the answer in 3 seperate columns for each row?

art297
Opal | Level 21

Using Tom's method is the following what you are looking for?

data want;

  input x;

  array digit(3);

  do place=1 to 3 ;

    digit(place)=mod(int(x*10**place),10);

  end;

  put;

cards;

.003

.123

12.345

;

proc print;

run;

FriedEgg
SAS Employee

If you have a negative number this should have one minor flaw.  I cannot test but I believe that if you have say -1.003 then using this method digit3=-3 which you probably do not want. You will probably want to account for this.

art297
Opal | Level 21

FriedEgg,  Tom's method correctly deals with negative numbers .. it simply keeps the minus sign.  If the minus sign isn't desired by the OP, a simple solution would be to add an abs function.  e.g.:

data want;

  input x;

  array digit(3);

  do place=1 to 3 ;

    digit(place)=abs(mod(int(x*10**place),10));

  end;

  put;

cards;

.003

.123

12.345

-12.345

;

proc print;

run;

jerry898969
Pyrite | Level 9

Thank you to everyone for your suggestions. 

I used Tom and Art's suggestions.

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