Help in substring function

[Kalai2008]

data want;
set have (firstobs=1 obs=10);
id1=SUBSTRN(id,13,10);
RUN;

given data:

ABC00000012365948952321
ABC00000000000948952321
ABC00000000025369214558

Output want: 

12365948952321

 948952321

25369214558

I have set of ID's that consist of 23 digits.
First 3 letters starts with alphabets and the rest are numbers.
I want to pull only the last numbers that starts after zero.
Since it's not constant position not sure how to use the function?

[Reeza]

Use COMPRESS to remove any characters that are letters.

Use INPUT to read the remaining text as a number.

[novinosrin]

data have;
input var $25.;
datalines;
ABC00000012365948952321
ABC00000000000948952321
ABC00000000025369214558
;

data want;
set have;
new_var=compress(var,'abcdefghijklmnopqrstuvwxyz0');
run;

Regards,

Naveen Srinivasan

[Reeza]

@novinosrin It's not shown in the OP's example, but what if there was a 0 in the middle of the number? I think it's highly likely that's a possibility. 

ABC00000012365948052321
ABC00000000000948952321
ABC00000000025369214008

[Kalai2008]

Thank you, I tried the compress function, but as you said its eliminating the 0 .

example:

ABC000000000025225250

ABC000000000025205250

 its eliminating the last zero and middle zero  as well.

[Reeza]

---

@Kalai2008 wrote:

Thank you, I tried the compress function, but as you said its eliminating the 0 .

example:

ABC000000000025225250

ABC000000000025205250

 its eliminating the last zero and middle zero  as well.

---

POST YOUR CODE. You can specify digits or characters using the 3rd or 4th parameter of the COMPRESS function, you don't have to list them at all. 

[Kalai2008]

data chk;
set test1.CHK2(firstobs=1 obs=10);
acct=compress(acct_id,'ABC0');
RUN;

DATA:

ABC000000000014256320

ABC000000000140252301

Output Results :

1425632

1425231

[kiranv_ · Accepted solution]

Something like this. What we are doing is making three groups. Anything mentioned in parentheis is group.1 st group is alphabets 2nd group is anything with 0 i.e. in midddle third group is from 0-9. We are replacing everything with last group that is third group mentioned as $3

data abc;
input var $50.;
datalines;
ABC00000012365948952321
ABC00000000000948952321
ABC00000000025369214558
ABC000000000253692145580
;


data bde;
set abc;
var1 =prxchange('s/([A-Z]+)(0+)([0-9]+)/$3/i', -1, var);
run;

. 

[Kalai2008]

Wow..It worked..Thanks a lot! Saved my time! Good & clear  Explanation.

[Reeza]

PRX is a great solution, but which one of these can you understand and modify if necessary?

data want;
set abc;

*Keep only digits in the string;
x=compress(var, , 'kd');

*convert to numeric;
y=input(x, best32.);

*use substring to extract all text after 3/4th character and convert to number;
z=input(substr(var, 4), best32.);

*PRX formula;
w=prxchange('s/([A-Z]+)(0+)([0-9]+)/$3/i', -1, var);

format  y z  best32.;
format x w $32.;

run;

[Kalai2008]

Thank you Reeza. Let me try everything

[Kalai2008]

Somehow the code didn't picked this id,
It excluded the last zero.
ABCD0000000011423744160
ABCD0000000011129735160..

Anyways Thank you Reeza, will try to modify as per your solution.

[Reeza]

---

@Kalai2008 wrote:
Somehow the code didn't picked this id,
It excluded the last zero.
ABCD0000000011423744160
ABCD0000000011129735160..

Anyways Thank you Reeza, will try to modify as per your solution.

---

Note that I don't understand PRX, so the function used above is as the original poster posted.  It's a perfectly valid solution, just not easy to modify especially as a beginner. 

[Kalai2008]

I am so sorry, the code worked . I did a mistake in the input function ( char to num). Thank you!

[Ksharp]

data abc;
input var $50.;
datalines;
ABC00000012365948952321
ABC00000000000948952321
ABC00000000025369214558
ABC000000000253692145580
;


data bde;
set abc;
var1 =prxchange('s/^[A-Z0]+//i', -1, var);
run;