Efficiently Amend/Edit Character String Values

[NKormanik]

Please see following dataset snippet:

i_X
_20205_2
_21102a_5
_20103_5
_20902_2
_21405_3
_22805_3
_21506ab_3
_22304_5
_23003_4
_20903_4

The following is what I want:

i_X             iX
_20205_2        _20205
_21102a_5       _21102a
_20103_5        _20103
_20902_2        _20902
_21405_3        _21405
_22805_3        _22805
_21506ab_3      _21506ab
_22304_5        _22304
_23003_4        _23003
_20903_4        _20903

Here is partial code for bringing this about:

data want;
set have;
iX = i_X;
run;

Next iX values have to be edited, or amended.  The final two characters of each value must be deleted.  Example:  _2 in the first case.

Help with this greatly appreciated.

Nicholas Kormanik

[NKormanik]

Possible answer:

catx("_%", iX;

Maybe something like this?

[NKormanik]

Another possibility:

iX = scan(iX,-2);

I haven't tried the above two approaches.

Thinking I'd rather wait for the experts' advice.

[NKormanik]

Whoa!  And finally there's Ksharp's (Xia Keshan) answer to a similar question:

Perl Regular Expression is the most simple powerful flexible method for such question:

string=prxchange('s/\++$//',-1,strip(ur_string));

Good grief.  Help Mr. Wizzard.....

[Reeza]

Your strings are fixed length. Use substr. 

[NKormanik]

Reeza, unfortunately some of the strings are longer than others.  I'll change my example to show that.

If not fixed length, what's the best answer?

[Shmuel]

1) assuming you need truncate exactly last two caharcters:

       ix = substr(i_x,1,length(ix)-2);

2) assuming ix is made of 2 substrings separated by underscores and

     the 1st character should be an underscore:

      ix = '_' || scan(i_x,1,'_');

[NKormanik]

Wow.  So five or six ways of doing it, so far.

And others are likely to follow?

[Ksharp]

data have;
input i_X $20.;
string=prxchange('s/_\d+$//',-1,strip(i_x));
cards;
_20205_2
_21102a_5
_20103_5
_20902_2
_21405_3
_22805_3
_21506ab_3
_22304_5
_23003_4
_20903_4
;
run;

[NKormanik · Accepted solution]

Excellent contributions.  Surely useful for future reference, and searches for answers by others.

HOWEVER, in this present case, it appears the winner is as follows:

data want;
set have;
iX = scan(i_X,-2,"_");
run;

Easier to understand.