![]() v WHEN NOT MATCHED THEN INSERT ( k, v ) VALUES ( b. ![]() v - Use GROUP BY in the source clause to ensure that each target row joins against one row - in the source: CREATE OR REPLACE TABLE target CLONE target_orig MERGE INTO target USING ( select k, max ( v ) as v from src group by k ) AS b ON target. The Linux kernel is a free and open-source, : 4 monolithic, modular, multitasking, Unix-like operating system kernel. CREATE OR REPLACE TABLE target CLONE target_orig MERGE INTO target USING src ON target. Merge succeeds and the target row is set to target.v = 11. All logically deleted records are available for query or reactivation, because they are. v <= 12 THEN DELETE - Joined values that do not match any clause do not prevent an update (src.v = 12, 13). better performance than the DATA step merge lookup method. Merge succeeds and the target row is deleted. v - Multiple deletes do not conflict with each other - joined values that do not match any clause do not prevent the delete (src.v = 13). v = 11 THEN DELETE WHEN MATCHED THEN UPDATE SET target. ![]() 12 or 13) from one of the duplicate rows (row not defined). If ERROR_ON_NONDETERMINISTIC_MERGE=true, returns an error - otherwise either deletes the row or updates target.v with a value (e.g. v - Updates and deletes conflict with each other. 11, 12, or 13) from one of the duplicate rows (row not defined). If ERROR_ON_NONDETERMINISTIC_MERGE=true, returns an error - otherwise updates target.v with a value (e.g. CREATE TABLE target_orig ( k NUMBER, v NUMBER ) INSERT INTO target_orig VALUES ( 0, 10 ) CREATE TABLE src ( k NUMBER, v NUMBER ) INSERT INTO src VALUES ( 0, 11 ), ( 0, 12 ), ( 0, 13 ) - Multiple updates conflict with each other. But if there are some 1's, then it is the first value=1 date.- Setup for example. If there are no 1's, then the _cutoff_date is the first value=0 date. If first.category then _cutoff_date=date The first 2 steps are used to apply a numeric format to the ID as it imports as character sometimes. Each case should have an identical match of the ID. I am attempting to merge 2 datasets with the common identifier investigationid. Set have (where=(value=1) in=firstpass) have (in=secondpass) SAS Match-Merge statement creating duplicate cases. ![]() If you want to keep them all, then the following program will do: If you want to delete them all, then above program works. what do you want to do if a given sub/category has only zeroes? You haven't answer question (or my comment above). In such a case, the program above will drop all obs for that sub/category. Here is how the above two input files are match-merged in SAS. The information in the matched records is combined to form one output record. What you haven't explained is what you want to do if a given sub/category has no observations with value^=0. A match-merge in SAS means that records from the one file will be matched up with the records of the second file that have the same ID. So assume data sorted by sub/category/date. This assumes that data are already sorted by sub/category (and presumably you require the data to be sorted by date within each sub/category). If first.category=1 then _n_of_ones=value You apparently want to delete "leading zeroes" for each sub/category combination.
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