Date:         Fri, 26 Aug 2011 05:53:35 -0400
Reply-To:     Dave Brewer <david.brewer@UC.EDU>
Sender:       "SAS(r) Discussion" <SAS-L@LISTSERV.UGA.EDU>
From:         Dave Brewer <david.brewer@UC.EDU>
Subject:      Re: Gap Question, Part 3
Comments: To: Søren Lassen <s.lassen@POST.TELE.DK>
Content-Type: text/plain; charset=ISO-8859-1

Hi Søren,

Thanks much for your help...works like a charm!

Dave

On Fri, 26 Aug 2011 03:38:08 -0400, S=?ISO-8859-1?Q?=C3=B8ren?= Lassen
<s.lassen@POST.TELE.DK> wrote:

>Dave,
>Here is a simple solution (no PRX, no loop, just character functions):
>
>data dave2 (keep=Ngaps number_zeros Avg_Gap_Length);
>input jan1 feb1 mar1 apr1 may1 jun1 jul1 aug1 sep1 oct1 nov1 dec1
>jan2 feb2 mar2 apr2 may2 jun2 jul2 aug2 sep2 oct2 nov2 dec2;
>seq=catt(of jan1--dec2);
>/* convert zeros to blanks */
>seq=translate(seq,' ','0');
>/* delete single ones (neither coverage nor gaps) and preceding zeros */
>seq=left(tranwrd(' '!!seq,' 1 ','  '));
>if length(seq) then do; /* if any coverage at all */
>  /* ngaps=number of "words" -1  - in SAS 9.2 you can also use COUNTW */
>  ngaps=count(seq,'1 ')-1;
>  /* total length of gaps */
>  number_zeros=length(seq)-length(compress(seq));
>  /* if there are no gaps, the average length stays missing */
>  if ngaps then
>    avg_gap_length=number_zeros/ngaps;
>  end;
>cards;
>1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 0 0 0 0 1 1 1 1 1
>1 1 0 0 1 1 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0
>1 1 0 1 0 1 1 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0
>;
>run;
>
>Regards,
>Søren
>
>
>
>On Thu, 25 Aug 2011 08:42:11 -0400, Dave <David.brewer@UC.EDU> wrote:
>
>>Hi SAS-Lers,
>>
>>I am still struggling trying to get a solution that will solve my
problem.
>>
>>I am trying to capture the following:
>>
>>The pattern we are looking for is anyone with at least two consecutive
>>months of insurance coverage (the one's), at least one month without
>>coverage (the zeroes), and then regains coverage for at least two
>>consecutive months. WE need to calculate the number of patterns like
this,
>>the average time without insurance during the pattern, and the total
>>number of cumulative months without insurance.
>>
>>So:
>>1 1 0 0 1 1 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0
>>
>>Average gap length should be 2.5 (2+3/2). Number of months without
>>insurance within the pattern should be 5. Number of gaps should be 2.
>>
>>Another example:
>>1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 0 0 0 0 1 1 1 1 1
>>
>>The average gap length should be 3 (not 2.667); number of gaps should be
>>2, cumulative months within gap should be 6.
>>
>>Yu Zhang provided the following code, and now I have new criteria that I
>>need to modify his code, but I just can't quite get it to work.
>>
>>data dave2 (keep=Ngaps number_zeros Avg_Gap_Length);
>>call missing(Ngaps,number_zeros);
>>input jan1 feb1 mar1 apr1 may1 jun1 jul1 aug1 sep1 oct1 nov1 dec1
>>jan2 feb2 mar2 apr2 may2 jun2 jul2 aug2 sep2 oct2 nov2 dec2;
>>seq=catt(of jan1--dec2);
>>
>>ExpressionID=prxparse('/1{2}0+(?=1{2})/');
>>
>>call prxsubstr(ExpressionID, seq, position, length);
>>number_zeros=0;
>>do while(position~=0);
>>  Ngaps+1;
>>  match = substr(seq, position, length);
>>  seq=substr(seq,length+1);
>>  call prxsubstr(ExpressionID, seq, position, length);
>>  x=count(match,'0');
>>  number_zeros+count(match,'0');
>>end;
>>Avg_Gap_Length=round(number_zeros/Ngaps);
>>
>>cards;
>>1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 0 0 0 0 1 1 1 1 1
>>1 1 0 0 1 1 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0
>>;
>>run;
>>
>>proc print noobs;
>>run;
>>
>>Can anyone help me find the number of gaps, gap length, and average gap
>>length (number of months w/o insurance)?
>>
>>Thanks everybody for your help. Hopefully I made myself clear.
>>Dave