Combining Matches Within Subgroups
Ben B. Hansen, Mark Fredrickson, Josh Errickson, Josh Buckner
2024-11-08
Source:vignettes/matching-within-subgroups.Rmd
matching-within-subgroups.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 medium-to-large 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 0.9-11 and above, optmatch
objects can be
easily combined to facilitate breaking a problem into smaller
sub-problems 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 0-5yrs 26 6 1 1 2 1 3
## 2 0-5yrs 42 1 1 1 0 2 1
## 3 0-5yrs 39 6 2 1 0 3 4
## 4 0-5yrs 34 4 2 1 0 4 2
## 5 6-11yrs 35 3 1 1 1 5 32
## 6 6-11yrs 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$case)
##
## 0 1
## 0-5yrs 8 4
## 6-11yrs 80 40
## 12+ yrs 77 39
Due to the sample size, if we were to compute matches on the entire
data set, the fullmatch
call would generate a distance
matrix of size
.
However, if we were instead to compute a match within each level of the
“education” variable, we’d compute three different distance matrices, of
total size
,
a reduction of 55%.
We’ll do this by splitting the data within each match.
> f1 <- fullmatch(case ~ age, data = infert[infert$education == "0-5yrs", ])
> f2 <- fullmatch(case ~ age, data = infert[infert$education == "6-11yrs", ])
> f3 <- fullmatch(case ~ age, data = infert[infert$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$education == "6-11yrs", ],
+ max.controls = 4)
> f3 <- fullmatch(case ~ age, data = infert[infert$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.