@Tom @Quentin @ballardw @Astounding  You guys are really great. Thanks a lot for the great job. It worked, after a little changes ... View more

@ballardw  Thanks for your remarks: Q: Editor values of 1,2,3, 4 sure don't look like two people entering data which was part of the prior description. A: I did not want to use 1, 2 for that this doesn't confuse with time and control   Q: Please restrict use of "data set" to discussion about SAS data sets. You are apparently talking about rows of values in the text file as a data set and that doesn't make much sense (at least not to me) and confuses where a data set may actually be needed. A: Sorry I have editted this above   Q:  You may need to provide a bit more detail about this. b) Check all data records to see whether they already have a control entry or not; these two groups should be outputted separately - a group with data records that do not yet have a 2nd input - a group with data records for which there is a 2nd input As in which records in your example meet each requirement and it would be nice for consistency sake to provide something like the name of an output data set to hold which "separate" outputs   A:  Identify the matching pairs of the 1st and 2nd input         - there should always be 2 observations that match in the variables with match in the id, pnr, pid and time         - The values of id, pnr, pid  should always be the same to become a matching pair. In variable "control" there should be a first and second entry to see that two different people entered the data              - Like in observation one and two  id=1 , since id, pnr, pid match they are pairs and there exist control 1 and 2 for the values of var1 to var5 could be checked to if if they are the same or not. If they are the same I output them in a dataset  "all_values_match". If they do not match I output them in a data set "values_doesnt_match"        -Then I check for the next pair with the same id,  pnr, pid but I have only "control"=2 so  I know control one is definitely missing.  I then check if these are duplicates if not I output them in dataset "missing_entry" etc   I hope I could answer your questions ... View more

 @Tom  @Quentin   @Astounding  @ballardw  Thankyou all for your contribution to find a solution to this problem After consultation with the customer. Here is a detailed explanation of his intentions with the data. I have adjusted (added the variable that contains the person who entered the data) the data a little.   data have; input id $ pnr pid time cn editor var1 var2 var3 var4 var5; datalines; 1 12 2 1 1 1 5 6 7 3 4 1 12 2 1 2 2 5 7 9 3 5 1 12 2 2 2 3 3 8 5 1 5 1 12 2 2 2 4 3 8 7 1 6 15 6 3 1 1 1 3 5 8 3 4 15 6 3 2 2 2 3 6 8 3 5 15 6 3 2 2 3 1 7 5 1 5 25 7 3 1 1 1 3 5 8 3 4 25 7 3 2 2 2 3 6 8 3 5 25 7 3 1 2 3 1 7 5 1 5 25 7 3 1 2 4 3 8 7 1 6 ; run; a) Identify the matching pairs of the 1st and 2nd input         - there should always be 2 observations that match in the variables with match in the id, pnr and pid          - these observations differ by the number of the control input in the variable 'control' (1 or 2)   b) Check all data records to see whether they already have a control entry or not; these two groups should be outputted                   separately       - a group with observations that do not yet have a 2nd input        - a group with observations for which there is a 2nd input   c) Compare the observations that exist twice with each other, prerequisite as described above with match in id, pnr, pid      - then compare whether the var1-var5 match (by rows) or not,       - observations that do not match should be output so that we know what needs to be corrected   So these are the notes I made. I hope someone can give me a helping hand. Even if it's just the start because am little bit confused of where to start from. Thanks ... View more

@Quentin Is there any short way to list the variables. I have more than 300 variables to compare. To list each single variable will not be funny ... View more

@Tom This is the error message I get when using this code on the original data ERROR: Array subscript out of range at row 411 column 5. ... View more

@Tom I don't want to split the data. The data  shouldn't be splitted. Am only suppose to compare what person A and B entered if they are the same or not ... View more

Sorry, this did not work even using the test data ... View more

Sorry, var1 to var5 are not numeric  data have; input id $3 pnr pid time cn var1 $ var2 $ var3 $ var4 $ var5 $ ; datalines; 1 12 2 1 1 5 6 7 3 4 1 12 2 1 2 5 7 9 3 5 1 12 2 2 2 3 8 5 1 5 1 12 2 2 2 3 8 7 1 6 ; run; and the results should look like this data want; input id $3 pnr pid time cn var1 $ flag1 var2 $ flag2 var3 $ flag3 var4 $ flag4 var5 $ flag5 ; datalines; 1 12 2 1 1 5 1 6 0 7 0 3 1 4 0 1 12 2 1 2 5 1 7 0 9 0 3 1 5 0 1 12 2 2 2 3 1 8 1 5 0 1 1 5 0 1 12 2 2 2 3 1 8 1 7 0 1 1 6 0 ; run; It's just that two different people entered this data at different times. And the values they entered should be same. So, the idea behind it is to check if the values entered by the two different people are the same, if not then the data should be corrected. If values have not been entered at time point(time) then this should be done. Time is either 1 or 2 , cn is the control which is also 1 or 2. So at the end I will like to print all id where error occured during entry. And also all ids with missing entries. That is why I wanted to use the flags (1=same, 0=not the same) ... View more

Thankyou all for your suggestions, I will talk to my boss and contact the SAS key account of my company for more details. I  need to make a plan on how to go about this before contacting a SAS key account and also make preparation towards possible questions we will have to ask (tring not to forget any important aspect) that was my reason for posting this issue here.    I appreciate all your answers, thanks ... View more

Okay, thanks for the hints ... View more

@s_lassen @Patrick @ErikLund_Jensen  Thanks for those examples. Combining your ideas, I think am almost getting what I want ... View more

@sbxkoenk @SASKiwi @ErikLund_Jensen @s_lassen @Patrick I have tried all the samples you posted they are all very good ways to solved this problem depending on ones needs. Thankyou very much. All the same I have a question, assuming I have an ID=12345 in the  main table and the ID in the sub table is assigned 12345_c because the patient had a chemotherapy and this id belongs to the chemotherapy table.  The other ID is 12345_r because of the radiotherapy treatment (This id's a have this extensions because the data comes from different centers).   My problem is if I use the hash function sha256 or md5 and use the format $hex64. I get different pseudonyms as the main ID's. The idea behinde all this is to assign both 12345_c and 12345_r to the main ID 12345   Is there any solution to this using the hash function or do I will need to define my own  values for the ID's ... View more

Thankyou all for all the suggestions and examples. I will try them to find out which will work for me. I will then give a feedback ... View more