About skyvalley81

skyvalley81 · ‎12-15-2017

This is one good paper that I found useful and explains quite well the use of TYPES and WAYS statements http://www2.sas.com/proceedings/forum2008/087-2008.pdf

skyvalley81 · ‎12-14-2017

Just create 2 do loops. Something like this: data new (drop = i); set old; array DX{*} DX1-DX25; array DXCC{*} DXCC1-DXCC14; do i = 1 to dim(DX); if DX[i] in (4100, 4101) then dx1 = 1; if DX[i] in (4200, 4201) then dx2 = 1; end; do i = 1 to dim(DXCC); if DXCC[i] in (100,101,105) then dxcc1=1; if DXCC[i] in (200,201,205) then dxcc2=1; end; if sum(dx1,dxcc1) = 2 or sum(dx2,dxcc2) = 2; run; I would suggest to test your code first on a small subset 🙂

skyvalley81 · ‎12-14-2017

I think this code will work, with the assumption you have equal number of variables both for DX and DXCC. data new (drop = i); set old; array DX{*} DX1-DX25; array DXCC{*} DXCC1-DXCC25; do i = 1 to dim(DX); if DX[i] in (4100, 4101) then dx1 = 1; if DXCC[i] in (100,101,105) then dxcc1=1; if DX[i] in (4200, 4201) then dx2 = 1; if DXCC[i] in (200,201,205) then dxcc2=1; end; if sum(dx1,dxcc1) = 2 or sum(dx2,dxcc2) = 2; run;

skyvalley81 · ‎12-14-2017

It would be good if you provide it us your data in data step format in order to understand what is going wrong. The reasons that you get incorrect results might be different character cases in the schedule and today variables. If this is the case you can use the UPCASE function in order to ensure that the variables have the same character cases or the i modifier in the FIND functions. Check the codes below: output = ifc(index(strip(upcase(schedule)),strip(upcase(today)))>0, 'T', 'F'); output = ifc(find(strip(schedule),strip(today), "i")>0, 'T', 'F'); Also make sure that the variables have the same levels. For example if the schedule variable is MON and the today Monday this code wont work. Hope it helps.

skyvalley81 · ‎12-13-2017

Hi Fridge_wpg, Could you please provide a sample dataset and the desired output?

skyvalley81 · ‎12-13-2017

HI Reeza, i totally agree these are two alternative options too.

skyvalley81 · ‎12-13-2017

Hi Gil_ Check if this is what you want. data have; input id:$2. Schedule:$20. Today:$20.; cards; A1 MON-TUE WED A2 WED-SAT WED A3 MON-TUE-WED WED ; run; data want; set have; outputs = ifc(index(strip(upcase(Schedule)), strip(upcase(today)))>0, "TRUE", "FALSE"); run;

skyvalley81 · ‎12-13-2017

Thank you very much all guys @Ksharp, @PGStats, @Reeza for looking into this problem. I will try to digest it and I will come back if I have any questions. Again thank you very much @Ksharp for your elegant solution.

skyvalley81 · ‎12-11-2017

Hi Reeza, Thank you very much for getting back. That is right we reorder the drugs based on the drug on the last prescription date. I am involved in a research project and i would like to calculate persistence/adherence on some medications. Do you have any idea how i can approach it?

skyvalley81 · ‎12-11-2017

Hi Reeza, Thank you very much for looking into this problem. The logic here is the following: 1) We examine when there are 2 or more drugs on the same day dupDrugFn = 1. The first date that we have two different drugs is on line 7-8 (09JUN2009). 2) If any of those products is the same with the drug in the previous date (13AUG2008) then we bring this drug first. In this example, drug2 is already first so we leave it as it is. 3) Then we move to the next date 28-Oct-2009 (line 9-10). Since the last drug on 9-Jun-2009 is DRUG1 we reverse the order of drugs in lines 9-10 and bring first DRUG1 and then we put DRUG2. The same logic applies to other rows. I hope it makes more sense now.

skyvalley81 · ‎12-11-2017

Hi everyone, I have a dataset containing patients, date of treatment and the drug that was prescribed. When there are more than two different drugs on the same prescription date I need to check if one of the drugs was also the last drug on the previous prescription date and place this product first and order the other same date by drugcode. Could you please suggest me how I can approach this problem? Thanks in advance for your help. Below you can find the datasets: data have; input id:$3. dxdate:date9. treatment:$6. dupDrugFn:8.; format dxdate date9. ; cards; 100 09MAY2006 drug1 0 100 24OCT2006 drug1 0 100 19APR2007 drug1 0 100 17DEC2007 drug1 0 100 07JUL2008 drug1 0 100 13AUG2008 drug2 0 100 09JUN2009 drug2 1 100 09JUN2009 drug1 1 100 28OCT2009 drug2 1 100 28OCT2009 drug1 1 100 19MAR2010 drug3 1 100 19MAR2010 drug4 1 100 26MAR2010 drug3 1 100 26MAR2010 drug4 1 100 15DEC2010 drug3 1 100 15DEC2010 drug4 1 100 15NOV2011 drug3 1 100 15NOV2011 drug4 1 100 21MAY2012 drug3 1 100 21MAY2012 drug5 1 100 05FEB2013 drug6 0 100 14JUN2013 drug5 0 100 26JUL2013 drug7 0 100 02JAN2014 drug6 1 100 02JAN2014 drug5 1 100 02JAN2014 drug7 1 100 02FEB2016 drug8 1 100 02FEB2016 drug6 1 100 02FEB2016 drug5 1 ; run; data have; input id:$3. dxdate:date9. treatment:$6. dupDrugFn:8.; format dxdate date9. ; cards; 100 09MAY2006 drug1 0 100 24OCT2006 drug1 0 100 19APR2007 drug1 0 100 17DEC2007 drug1 0 100 07JUL2008 drug1 0 100 13AUG2008 drug2 0 100 09JUN2009 drug2 1 100 09JUN2009 drug1 1 100 28OCT2009 drug1 1 100 28OCT2009 drug2 1 100 19MAR2010 drug3 1 100 19MAR2010 drug4 1 100 26MAR2010 drug4 1 100 26MAR2010 drug3 1 100 15DEC2010 drug3 1 100 15DEC2010 drug4 1 100 15NOV2011 drug4 1 100 15NOV2011 drug3 1 100 21MAY2012 drug3 1 100 21MAY2012 drug5 1 100 05FEB2013 drug6 0 100 14JUN2013 drug5 0 100 26JUL2013 drug7 0 100 02JAN2014 drug7 1 100 02JAN2014 drug5 1 100 02JAN2014 drug6 1 100 02FEB2016 drug6 1 100 02FEB2016 drug5 1 100 02FEB2016 drug8 1 ; run;

skyvalley81 · ‎01-25-2017

Hi Patrick, Thank you so much, your approach it is also very helpful

skyvalley81 · ‎01-25-2017

Hi PGStats, Thank you very much for your solution the code is exactly what i was looking for 🙂 Could I ask you to explain me the second regex prxparse if it is possible prxparse("/[^',)]+$/")

skyvalley81 · ‎01-25-2017

Hi ballardw, Thank you very much for your prompt reply. Each line is a separate observation. Your idea is really good but because there is no consistent pattern the scan function might not work. For example, there are cases that there is no parenthesis at the end of the string. With this regex I can find the position of first match of the pattern. However this not suffice to extract the whole list. I was wondering if there is any way to find the last occurrence of the pattern then I could easily extract the substring. Thanks again

skyvalley81 · ‎01-25-2017

Hi Everyone, I would like to extract a list of codes from a particular dataset. I tried to use regular expressions but unfortunatly didnt manage to exract the whole list of codes but only first occurence of the pattern. The codes are quoted, consist from 3 to 5 digits, start with V or E or numbers {0-9} and continue with numbers. What I have: data test; length codelist $120; input codelist $; datalines; DX(i)('410','412')CC_GRP_1 DX(i)('39891','40201','40211','40291','40401','40403','40411','40413','40491','40493', '4254','4255','4257','4258','4259','428')CC_GRP_2 DX(i)('0930','4373','440','441','4431','4432','4438','4439','4471','5571','5579','V434') DX(i)('36234','430','431','432','433','434','435','436','437','438')CC_GRP_4 ; run; data test2; set test; retain pattern_num ; if _n_ = 1 then pattern_num = prxparse("/('E|V|\d\d*')/"); call prxsubstr(pattern_num,strip(codelist),start, lenght); run; Desired output: '410','412' '39891','40201','40211','40291','40401','40403','40411','40413','40491','40493' '4254','4255','4257','4258','4259','428' '0930','4373','440','441','4431','4432','4438','4439','4471','5571','5579','V434' '36234','430','431','432','433','434','435','436','437','438' Thanks in advance

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