Combining Matches Within Subgroups
Ben B. Hansen, Mark Fredrickson, Josh Errickson, Josh Buckner
20220815
Source:vignettes/matchingwithinsubgroups.Rmd
matchingwithinsubgroups.Rmd
When utilizing full matching, a user may want to introduce restrictions to the potential match sets. There are two main reasons to do this.
 There may be certain variables which matches should either always or never agree upon. For example, if there is a binary gender variable, you may wish to only match treatment members to control members of the same gender.
 When matching on a mediumtolarge data set, speed concerns can become paramount. By splitting the matching problem into a series of smaller subproblems, we can realize substantial performance improvements.
Combining matches
In optmatch 0.911 and above, optmatch
objects can be easily combined to facilitate breaking a problem into smaller subproblems and reconstituting a matched structure on the entire data set. To demonstrate this, let’s consider the infert
data set.
> data(infert)
> head(infert)
## education age parity induced case spontaneous stratum pooled.stratum
## 1 05yrs 26 6 1 1 2 1 3
## 2 05yrs 42 1 1 1 0 2 1
## 3 05yrs 39 6 2 1 0 3 4
## 4 05yrs 34 4 2 1 0 4 2
## 5 611yrs 35 3 1 1 1 5 32
## 6 611yrs 36 4 2 1 1 6 36
The “case” variable indicates treatment (1) versus control (0) status. We’ll want to match upon “age”.
> table(infert$case)
##
## 0 1
## 165 83
> table(infert$education, infert$case)
##
## 0 1
## 05yrs 8 4
## 611yrs 80 40
## 12+ yrs 77 39
Due to the sample size, if we were to compute matches on the entire data set, the fullmatch
call would generate a distance matrix of size \(165\times 83 = 13,695\). However, if we were instead to compute a match within each level of the “education” variable, we’d compute three different distance matrices, of total size \(8\times 4 + 80\times 40 + 77\times 39 = 6,235\), a reduction of 55%.
We’ll do this by splitting the data within each match.
> f1 < fullmatch(case ~ age, data = infert[infert$education == "05yrs", ])
> f2 < fullmatch(case ~ age, data = infert[infert$education == "611yrs", ])
> f3 < fullmatch(case ~ age, data = infert[infert$education == "12+ yrs", ])
> summary(f1)
## Structure of matched sets:
## 1:2
## 4
## Effective Sample Size: 5.3
## (equivalent number of matched pairs).
> summary(f2)
## Structure of matched sets:
## 1:1 1:2 1:3 1:4 1:5+
## 20 8 6 4 2
## Effective Sample Size: 49.4
## (equivalent number of matched pairs).
> summary(f3)
## Structure of matched sets:
## 1:1 1:2 1:3 1:4 1:5+
## 23 5 6 2 3
## Effective Sample Size: 47
## (equivalent number of matched pairs).
Some of the matched sets are quite large (1:5+) so let’s put some restrictions.
> f2 < fullmatch(case ~ age, data = infert[infert$education == "611yrs", ],
+ max.controls = 4)
> f3 < fullmatch(case ~ age, data = infert[infert$education == "12+ yrs", ],
+ max.controls = 4)
> summary(f2)
## Structure of matched sets:
## 1:1 1:2 1:3 1:4
## 18 10 6 6
## Effective Sample Size: 49.9
## (equivalent number of matched pairs).
> summary(f3)
## Structure of matched sets:
## 1:1 1:2 1:3 1:4
## 20 6 7 6
## Effective Sample Size: 48.1
## (equivalent number of matched pairs).
Now we simply combine the three matches.
Using the within
argument
An alternative approach would be using the within
argument and the exactMatch
function to define subproblems.
> fwithin < fullmatch(case ~ age, data = infert, max.controls = 4,
+ within = exactMatch(case ~ education, data = infert))
> summary(fwithin)
## Structure of matched sets:
## 1:1 1:2 1:3 1:4
## 38 20 13 12
## Effective Sample Size: 103.4
## (equivalent number of matched pairs).
Observe that we obtain equivalent matched structure. A few notes comparing the two approaches:

When using the
within
argument, restrictions must be the same across subproblems. That is,max.controls
,min.controls
andomit.fraction
will be equivalent. By running the subproblems separately, you can set different restrictions per subproblem. E.g., While the matched structures will be equivalent between these two approaches (if the restrictions are the same across subproblems), the actual matched sets themselves may differ if you have observations of equal distance. In general this should not be considered a problem.