Other way to solve this problem is by defining a loop which keep track of pattern counter. For example, If the loop operates twice then it will define b2=1 and if there will be another pattern then it will add counter = 1 in b2 value.

I am not sure if this works for all the possibilities of observations (2**5 =32) but this code should work.

Vishal.

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*List data. DATA LIST LIST/var1 TO var5 (5 F8.0). BEGIN DATA 1 1 1 1 1 1 0 1 1 0 1 0 1 1 1 1 1 0 1 1 1 1 1 1 0 1 0 0 0 1 0 0 0 1 0 0 1 0 1 1 END DATA. SET MPRINT = ON PRINTBACK ON.

*Define the vector to compare each variable. DEFINE !vars(nbvars=!TOKENS(1) /v1=!TOKENS(1) /v2=!TOKENS(1)) VECTOR v=!v1 TO !v2 /#val(!nbvars). NUMERIC b1 TO b5. VECTOR b=b1 TO b5. LOOP #cnt=1 TO !nbvars. - COMPUTE #val(#cnt) = v(#cnt). - COMPUTE b(#cnt) = 0. END LOOP.

*Try to look for the continuous patten and add the counter in loop statement to define pattern type (ie. b1, b2, b3...). LOOP #i = 1 TO !nbvars-1. DO IF (#i=1). - COMPUTE #k=1. - COMPUTE #l=1. - LOOP #j = #i +1 TO !nbvars IF (#val(#i) EQ v(#j)). - COMPUTE #k = #k +1. - COMPUTE #l = #j. - END LOOP. - COMPUTE b(#k) = b(#k) + 1. ELSE IF((#val(#i) NE v(#i+1)) OR (#val(#i) NE v(#i-1))). - COMPUTE #k=1. - COMPUTE #l=1. - LOOP #j = #i +1 TO !nbvars IF (#val(#i) EQ v(#j)). - COMPUTE #k = #k +1. - COMPUTE #l = #j. - END LOOP. - COMPUTE b(#k) = b(#k) + 1. END IF. END LOOP IF (#l EQ !nbvars).

*b1 won't be the correct value from the counter variable but rest of the patterns are correct one. *calculate b1 from rest of the values of b(i). COMPUTE #m=0. LOOP #i=2 TO !nbvars. - COMPUTE #m = #m + #i*b(#i). END LOOP. COMPUTE #cnt2=1. COMPUTE b(#cnt2) = !nbvars - #m. !ENDDEFINE.

*Call Macro !vars nbvars=5 v1=var1 v2=var5. EXECUTE. *******************************************************************

VD> -----Original Message----- VD> From: SPSSX(r) Discussion [mailto:SPSSX-L@LISTSERV.UGA.EDU] On Behalf Of VD> Marks, Jim VD> Sent: Friday, December 02, 2005 10:26 AM VD> To: SPSSX-L@LISTSERV.UGA.EDU VD> Subject: Re: COUNTING CONSECUTIVES NUMBERS IN A SET OF VARIABLES VD> VD> Here is a solution that can be extended to more variables (I am working VD> on a similar problem, but I need the length of all sequences). VD> VD> To extend the number of variables you will need to edit these commands-- VD> COMPUTE max_seq VD> MATCH (RENAME and DROP VD> VECTOR VD> LOOP VD> RECODE, RENAME VARIABLES, VARIABLE LABELS VD> VD> If your data file extends to more cases, you will need to edit the DO VD> REPEAT command. VD> VD> Note this solution requires an ID variable in your original file (and a VD> saved copy of that file). VD> VD> This is tested: VD> VD> ** SAMPLE data. VD> VD> DATA LIST LIST/id (f8.0) v1 TO v5 (5 F8.0). VD> BEGIN DATA VD> 1 1 0 1 1 0 VD> 2 1 1 1 1 1 VD> 3 1 0 1 1 1 VD> 4 1 1 0 1 1 VD> 5 1 1 1 1 0 VD> 6 1 0 0 1 0 VD> 7 0 1 1 1 0 VD> 8 0 0 0 0 0 VD> END DATA. VD> VD> SAVE OUTFILE = 'c:\spss-test\conseq.sav'. VD> VD> ** we use flip so we can sum the lags VD> ** and determine the largest consecutive sequence. VD> VD> FLIP. VD> VD> DO REPEAT a = var001 to var008. VD> VD> DO IF $casenum GT 2 AND a = 1. VD> COMPUTE a= a+ lag(a). VD> END IF. VD> VD> END REPEAT. VD> VD> FLIP. VD> VD> COMPUTE max_seq = MAX(v1 to v5). VD> VD> ** join the largest sequence variable back to the original file. VD> VD> MATCH FILES FILE = * VD> /RENAME (case_lbl v1 v2 v3 v4 v5 = d0 d1 d2 d3 d4 d5) VD> /FILE = 'c:\spss-test\conseq.sav' VD> /BY id VD> /DROP d0 d1 d2 d3 d4 d5. VD> VD> ** now calculate the "b" variables. VD> VECTOR b(5f8.0). VD> LOOP #i = 1 TO 5. VD> COMPUTE b(#i) = max_seq = (#i). VD> END LOOP. VD> IF max_seq = 0 b1 = 1. VD> VD> RECODE b1 to b5 (SYSMIS = 0). VD> VD> ** requested variable names and labels (for clarity). VD> RENAME VARIABLES (b1 b2 b4 b5 = b5 b4 b2 b1). VD> VARIABLE LABELS VD> b1 '5 1 values' VD> b2 '4 1 values' VD> b3 '3 1 values' VD> b4 '2 1 values' VD> b5 '1 1 value (or no 1 values)' VD> . VD> EXECUTE. VD> VD> --jim VD> -----Original Message----- VD> From: SPSSX(r) Discussion [mailto:SPSSX-L@LISTSERV.UGA.EDU] On Behalf Of VD> Marta García-Granero VD> Sent: Friday, December 02, 2005 1:55 AM VD> To: SPSSX-L@LISTSERV.UGA.EDU VD> Subject: Re: COUNTING CONSECUTIVES NUMBERS IN A SET OF VARIABLES VD> VD> Hi Victor VD> VD> First of all, your coding scheme would leave out (b1 to b5 missing) VD> those cases with only one '1' among the 5 variables, because you reserve VD> b5=1 only for the particular case where v1 to v5 are all set to 0. What VD> happens to those cases with only one '1' among the 5 variables?. VD> VD> I'm sure there is a more elegant and simple approach, but this one works VD> (not easily transformed to more than 5 variables, sorry). I have listed VD> the whole set of pattern (32: 2**5) and sorted them taking into account VD> wether they had 5 '1', 4 consecutive '1', and so on. I have assigned VD> b5=1 to cases with 5 '0' or only one '1'. Change it if you don't like it VD> (change the "ELSE." to "ELSE IF pattern EQ '00000'.", and a new "ELSE" VD> at then end to give another solution to all patterns with only one '1'. VD> VD> DATA LIST LIST/v1 TO v5 (5 F8.0). VD> BEGIN DATA VD> 1 0 1 1 0 VD> 1 1 1 1 1 VD> 1 0 1 1 1 VD> 1 1 0 1 1 VD> 1 1 1 1 0 VD> END DATA. VD> VD> STRING pattern(A5). VD> COMPUTE pattern = VD> STRING((10**4)*v1+(10**3)*v2+(10**2)*v3+10*v4+v5,'N5.0'). VD> VD> DO IF pattern EQ '11111'. VD> - COMPUTE b1=1. /* Five '1' *. VD> - COMPUTE b2=0. VD> - COMPUTE b3=0. VD> - COMPUTE b4=0. VD> - COMPUTE b5=0. VD> ELSE IF ANY(pattern,'11110','01111'). VD> - COMPUTE b1=0. VD> - COMPUTE b2=1. /* Four consecutive '1' *. VD> - COMPUTE b3=0. VD> - COMPUTE b4=0. VD> - COMPUTE b5=0. VD> ELSE IF ANY(pattern,'00111','01110','11100','11101','10111'). VD> - COMPUTE b1=0. VD> - COMPUTE b2=0. VD> - COMPUTE b3=1. /* Three consecutive '1' *. VD> - COMPUTE b4=0. VD> - COMPUTE b5=0. VD> ELSE IF VD> ANY(pattern,'00011','00110','01100','11000','01011','11010','10110','011 VD> 01','10011','11001'). VD> - COMPUTE b1=0. VD> - COMPUTE b2=0. VD> - COMPUTE b3=0. VD> - COMPUTE b4=1. /* Only two consecutive '1' *. VD> - COMPUTE b5=0. VD> ELSE IF pattern EQ '11011'. VD> - COMPUTE b1=0. VD> - COMPUTE b2=0. VD> - COMPUTE b3=0. VD> - COMPUTE b4=2. /* Particular case where b4=2 *. VD> - COMPUTE b5=0. VD> ELSE. /*All patterns not explicited above have no consecutive '1' *. VD> - COMPUTE b1=0. VD> - COMPUTE b2=0. VD> - COMPUTE b3=0. VD> - COMPUTE b4=0. VD> - COMPUTE b5=1. VD> END IF. VD> EXECUTE . VD> FORMAT b1 TO b5 (F8.0). VD> VD> Regards, VD> Marta mailto:biostatistics@terra.es