data l;
input id $ 1-20;
cards;
aaaa1234a
bbbb1234b
ccccx1234c
asdf1234b
run;
I dont want to do by substr(1,1)=substr(2,1)
i want the output if the first character and last are same
output :
ccccx1234c
asdf1234b
Your output example seems inconsistent with your description, I will just go with the latter:
data l;
input id $ 1-20;
cards;
aaaa1234a
bbbb1234b
ccccx1234c
asdf1234b
run;
data want;
set l;
if first(id)=:left(reverse(id));
run;
proc print;run;
Haikuo
hi ... another idea ...
data want;
set l;
where char(id,1) eq char(right(id),20);
run;
OR You like Perl Regular Expression :
data l; input id $ 1-20; cards; aaaa1234a bbbb1234b ccccx1234c asdf1234b ; run; data want; set l; where id=prxchange('s/(\S)(.*)(\S)/\3\2\1/o',-1,id); run;
Ksharp
Sory i have not framed my question properly
this is the req below
data l;
input id $ 1-20;
cards;
aaaa1234a
bbbb1234b
ccccx1234c
asdf1234b
run;
I dont want to do by substr(1,1)=substr(2,1)
i want the output if the first,second,third,foruth,fifth and last character and are not same.
output :
ccccx1234c
asdf1234b
My_SAS wrote:
i want the output if the first,second,third,foruth,fifth and last character and are same.output :
ccccx1234c
asdf1234b
But in the case of 'ccccx1234c"
the first, -> c
second,-> c
third,-> c
foruth-> c
,fifth-> x
last character-> c
They are different . Any other logic are you using ?
Ksharp
i want the output if the first,second,third,foruth,fifth and last character and are NOT same.
output :
ccccx1234c
asdf1234b
Sorry . overlook "NOT"
Oh, I undertand what you mean now.
Ksharp
should check only the characters are same not the numerics
bbbbb1234b
bbbbb b (all the b characters are same in 1,2,3,4,5 and last obs are same)
How about:
data l; input id $ 1-20; cards; aaaa1234a bbbb1234b ccccx1234c asdf1234b ; run; data want(drop=x); set l; x=substrn(left(id),1,1); if not missing (compress(id,x,'d')) ; run;
Ksharp
Message was edited by: xia keshan
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