Propensity Match in 1:1:1 fashion

[GS2]

Hello,

Using SAS 9.4. I am trying to propensity match in a 1:1:1 fashion. I have 3 procedure type open, lap and robo and I would like to propensity match evenly across the 3 groups based on age, sex, side and primary. I understand that proc psmatch is not able to accomplish this, is there another SAS procedure that can handle this type of match or is there a workaround others have discovered? Any help is appreciated. Thank you 

[OsoGris]

You are correct that PROC PSMATCH does not match with more than two groups (treatment and control).  I am not aware of a procedure that does matching of 3 groups. 

[SteveDenham]

I have absolutely no idea whether this would work or not, but couldn't you match for the first two levels, ignoring the third, and then do a 2:1 match from the first two levels to the third as a separate procedure. That might give relatively equal group sizes.

An alternative might be to consider a blocked design. Block by age, sex, side and primary. using the three binary categories (sex, side, primary) to generate 8 levels, and then sort by age within each of those. Assign each combination a random number drawn from a uniform distribution, then identify blocks of 3 subjects within these. Sort again by all of the design factors you want to match on and the random number. Assign the low, middle and high random number values in each block to open, lap or robo. That should give you a triple of the three methods with the most similar subjects in each triple. Where this may get much harder is if the variable 'primary' has more than two values, which might be the case if the variable refers to a primary physician, and a given primary physician has not referred at least 3 (or a multiple of 3) subjects to your trial. 

SteveDenham

[GS2]

Steve,

Thank you for the comment, it was helpful. I attempted the first method and unfortunately performing a 1:1 and then a 2:1 led to 1 of 3 groups being overrepresented in the 2:1. However, this thought did lead me to perform a 1:1 as suggested and then to take 1 of the groups from the 1:1 and match it to the remaining group in a 1:1 fasion. I then combined the matched from the second 1:1 with the previous 1:1 matches. This ended up leading to equal groups across the 3 procedure types. This method requires 2 different match keys and to create a new match key to combine the previous two but it appears to have worked for my purposes. Thank you again for the comment.