Hi Ursula, As others have pointed out, this is for more likely an Excel problem than a SAS problem. I suggest you look at the csv file in a plain text editor such as Notepad, not Excel, to see exactly what SAS is putting out. In Excel (at least on my computer), if I enter 11/2007 in a cell, Excel automatically converts it to the date value for November 1, 2007, and displays Nov-07. As Eric@Stats suggests, you can format this with the custom format mm/yyyy so that it displays as 11/2007, but the internal value is still the date value for 11/1/2007, so if you later transfer the data to another program or another format, you may not get what you want. Essentially, Excel corrupts the data when it imports it. Eric ... View more

It might be helpful to provide the example data as it looks in a plain text file or in a SAS printout, as Excel may apply formatting that makes it look different. In your program fragment, you have at least two errors that will make it not work, even if the data are as they appear in the Excel screenshot. 1. the value returned from the input function needs to be assigned to a variable, like salary2 = input(salary, dollar7.); 2. when you compare the salary to a particular number in the if statement, that number cannot have commas in it. For example (leaving out the other parts of your if statement for clarity), you want something like: if salary2 > 10000 then .... ; ... View more

From Brenda's example, there does not have to be a week 53 in every year, since up to 3 days at the beginning of the year may be counted in week 53 of the previous year. ... View more

This is what I get when I run code that I cut-and-pasted from my first post. The missing week 1 suggests that something went missing when you edited the code. I am not sure of the week 52/53 discrepancy, since the rule I used may be different from yours.                           Obs    cdcweek      dth_date ...                     1084 201151 12/23/2011                      1085 201151 12/24/2011                      1086 201152 12/25/2011                      1087 201152 12/26/2011                      1088 201152 12/27/2011                      1089 201152 12/28/2011                      1090 201152 12/29/2011                      1091 201152 12/30/2011                      1092 201152 12/31/2011                      1093 201201 01/01/2012                      1094 201201 01/02/2012                      1095 201201 01/03/2012                      1096 201201 01/04/2012                      1097 201201 01/05/2012                      1098 201201 01/06/2012                     1099 201201 01/07/2012                      1100 201202 01/08/2012                      1101 201202 01/09/2012                      1102 201202 01/10/2012                      1103 201202 01/11/2012                      1104 201202 01/12/2012                      1105 201202 01/13/2012                      1106 201202 01/14/2012 ... ... View more

This may do what you want, although I am not sure it follows exactly the same rule that you described. /* compute year/week according to this rule: If Jan 1 falls on a Sunday through Wednesday, then the week that includes Jan 1 is week 1. Otherwise the following week is week 1. Then weeks are numbered consecutively up to the last week of the year, which may be a 52 or a 53. */ data date1(keep=cdcweek dth_date);    length cdcweek $ 6;    do dth_date = '01jan2009'd to '31dec2012'd;       year = year(dth_date);       yyyy = put(year,4.);       pyyy = put(year-1,4.);       first_sat = mdy(1,1,year);       jan01     = mdy(1,1,year);       prevjan01 = mdy(1,1,year-1);       do while(weekday(first_sat) ne 7);          first_sat = first_sat + 1;       end;       if dth_date = . then cdcweek = '';       else do;          if weekday(jan01) le 4 then do;             if dth_date le first_sat then cdcweek = yyyy||"01";             else cdcweek = put(((year*100) + int((dth_date-first_sat-1)/7) + 2),6.);             end;          else do;             if dth_date le first_sat then do;                if (weekday(prevjan01) = 4) or ((weekday(prevjan01) = 3) and                   (jan01-prevjan01 = 366)) then  cdcweek = pyyy||"53";                else cdcweek = pyyy||"52";                end;             else cdcweek = put(((year*100)+int((dth_date-first_sat-1)/7) + 1),6.);             end;    /* weekday(jan01) le 4 */          end;  /* date missing */          if cdcweek ge 200901 then output;       end; /* dth_date from 1/1/2009 to 12/31/2012 */ run; ... View more