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
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;
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
;
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.
Thanks for both the answers.
Tom how can I put the answer in 3 seperate columns for each row?
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;
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.
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;
Thank you to everyone for your suggestions.
I used Tom and Art's suggestions.
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