Hello
I have a data set with char field that contain 6 figures 1/0.
I want to identify the following pattern (If this pattern then Indicator=1 else 0).
Block A- block of '1's in the right side
In the right side there will be continuous of one or several '1's and last number must be 1
Block B- block of '0's in the left side of block A
There can be one zero or multiple zeros in consecutive
Block C- block of '1's in the left side of block A
There can be one '1' or multiple '1's in consecutive
examples:
010101 -This pattern meet the criteria because Block A-1 Block B-0 Block C-1
000001--This pattern doesn't meet the criteria
000011--This pattern doesn't meet the criteria
101101-This pattern meet the criteria because Block A-1 Block B-0 Block C-11
101001-This pattern meet the criteria because Block A-1 Block B-00 Block C-1
010011-This pattern meet the criteria because Block A-11 Block B-00 Block C-1
What is the way to recognize this pattern?
Data have;
input X $;
cards;
010101
000001
000011
000111
001111
011111
111111
110011
110111
111001
100001
001001
001011
010011
101001
101101
;
Run;
Data have;
input X $;
cards;
010101
000001
000011
000111
001111
011111
111111
110011
110111
111001
100001
001001
001011
010011
101001
101101
;
Run;
data want;
set have;
y=prxchange('s/0+/0/',-1,x);
z=prxchange('s/1+/1/',-1,y);
Indicator=ifn(find(z,'101'),1,0);
drop y z;
run;
Data have;
input X $;
cards;
010101
000001
000011
000111
001111
011111
111111
110011
110111
111001
100001
001001
001011
010011
101001
101101
;
Run;
data want;
set have;
y=prxchange('s/0+/0/',-1,x);
z=prxchange('s/1+/1/',-1,y);
Indicator=ifn(find(z,'101'),1,0);
drop y z;
run;
Thanks
May you please explain
y=prxchange('s/0+/0/',-1,x);
Hi,
Thanks for supplying the input data in a data step. Please supply the expected output data in a data step too.
Please also clarify the following if the examples "000001" & "000011" don't meet the criteria because they do not satisfy Block C or is it something else?
Thanks & kind regards,
Amir.
/*
Sorry.
I found a problem in my code.
if you want 010100 be NOT matched, try this one:
*/
Data have;
input X $;
cards;
010101
010100
;
Run;
data want;
set have;
y=prxchange('s/0+/0/',-1,x);
z=prxchange('s/1+/1/',-1,y);
Indicator=ifn(prxmatch('/101$/',strip(z)),1,0);
drop y z;
run;
You already have an answer but, just for fun, one more:
Data have;
input X $;
cards;
010101
000001
000011
000111
001111
011111
111111
110011
110111
111001
100001
001001
001011
010011
101001
101101
101100
;
Run;
data want;
set have;
marker =
mod(X,2) /* not ends with 0 */
*
(countw(X,"0","B") > 1) /* has at least 2 sequences of 1s (counting from right) */
;
run;
proc print;
run;
{EDIT:}
Of course if zeros on right aren't a problem (e.g., 010110 should be marked) then it reduces just to:
marker = (countw(X,"0","B") > 1);
{EDIT 2:}
To avoid ugly "types conversion" note in the log, the MOD() can be replaced with CHAR() function:
marker =
char(X,6) = "1" /* not ends with 0 */
*
(countw(X,"0","B") > 1) /* has at least 2 sequences of 1s (counting from right) */
;
Bart
ok, so it looks like my code does what you need.
Both codes provide same result which is perfect:)
Data have;
input X $ 20.;;
cards;
010101
000001
000011
000111
001111
011111
111111
110011
110111
111001
100001
001001
001011
010011
101001
101101
000000000001
000000000101
000000000011
000001011111
100000000001
101010101011
;
Run;
data want1;
set have;
y=prxchange('s/0+/0/',-1,x);
z=prxchange('s/1+/1/',-1,y);
Indicator1=ifn(find(z,'101'),1,0);
drop y z;
run;
data want2;
set have;
Indicator2 = mod(X,2) /* not ends with 0 */ * (countw(X,"0","B") > 1) /* has at least 2 sequences of 1s (counting from right) */;
run;
proc sql;
create table compare_results(Where=(Indicator1 ne Indicator2)) as
select a.*,b.Indicator2
from want1 as a
left join want2 as b
on a.X=b.X
;
quit;
You didn't read what I wrote in my answer about "if zeros at the end are not important"...
Actually cannot have zero at end because the check is for customers that exist today (value 1)
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