Cem September 2013

cem@lists.gking.harvard.edu

2 participants
3 discussions

by Merlin Schaeffer

Hi, I am trying to use the cem function in R followed by the k2k function, because I have a very large dataset with few treated (about 20,000) and roughly ten times as many not treated respondents (200,000). For this reason I would like to reduce the dataset so that each treated is matched to one non treated person who is similar with respect to certain control variables. However, the k2k function does not seem to work and trying out the vignette suggests that the problem is not specific to my application: When I tried to run the simple example suggested in the vignette, k2k does not work either and of course following estimation commands simply reproduce the normal cem results after the k2k function is used unsuccessfully: data(LL) # cem match: automatic bin choice mat <- cem(treatment="treated", data=LL, drop="re78") mat mat$k2k # ATT estimate att(mat, re78 ~ treated, data=LL) # transform the match into k2k mat2 <- k2k(mat, LL, "euclidean", 1) mat2 mat2$k2k # ATT estimate after k2k att(mat2, re78 ~ treated, data=LL) When I run it, I get an error message for the k2k() function: There were 50 or more warnings (use warnings() to see the first 50) the warnings() function then tells 50 times: 1: In min(x, na.rm = TRUE) : no non-missing arguments to min; returning Inf … 50: In min(x, na.rm = TRUE) : no non-missing arguments to min; returning Inf I also exchanged "euclidean" for method = NULL, and here the 50 warnings read similar: 1: In min(x, na.rm = TRUE) : no non-missing arguments to min; returning Inf ... The concrete output is: > data(LL) > > # cem match: automatic bin choice > mat <- cem(treatment="treated", data=LL, drop="re78") > mat G0 G1 All 425 297 Matched 222 163 Unmatched 203 134 > mat$k2k [1] FALSE > > # ATT estimate > att(mat, re78 ~ treated, data=LL) G0 G1 All 425 297 Matched 222 163 Unmatched 203 134 Linear regression model on CEM matched data: SATT point estimate: 550.962564 (p.value=0.368242) 95% conf. interval: [-647.777701, 1749.702830] > > > # transform the match into k2k > mat2 <- k2k(mat, LL, "euclidean", 1) There were 50 or more warnings (use warnings() to see the first 50) > mat2 G0 G1 All 425 297 Matched 222 163 Unmatched 203 134 > mat2$k2k [1] FALSE > > # ATT estimate after k2k > att(mat2, re78 ~ treated, data=LL) G0 G1 All 425 297 Matched 222 163 Unmatched 203 134 Linear regression model on CEM matched data: SATT point estimate: 550.962564 (p.value=0.368242) 95% conf. interval: [-647.777701, 1749.702830] I also had a friend trying out the vignette on his macbook and he had the same results. I have a macbook too, if that is of any importance. I would be very glad for help with this problem. Kind regards, Merlin -- Dr. Merlin Schaeffer Wissenschaftszentrum Berlin für Sozialforschung (WZB) Department "Migration, Integration, Transnationalization" Reichpietschufer 50 10785 Berlin, Germany Phone: + 49 30 25491-459 Fax: + 49 30 25491-452 http://www.wzb.eu/en/persons/merlin-schaeffer

10 years, 6 months

Cutpoint = c(x)

by Lavelle, Bridget J (DSHS/RDA)

Can anyone tell me what the cutpoint parameter c(x) means for a character variable under the CEM matching procedure? I am using Spacegraph to identify potential matching solutions and dialog box associated with one of the solutions I click on in the graph shows this command: cem(treatment="trt", data=kids, drop=c("readscore7"), cutpoints=list( aptitude=c(5), female=c(0,0.5,1), readscore6=c(38, 57.67, 77.33, 97), var4=c(1)), keep.all=TRUE,eval.imbalance=TRUE, L1.breaks=sp2$raw.profile$medianCP ) I know that for continuous variables, c(5) would cut the variable as closely as possible into 4 equally sized intervals. However aptitude is a character variable ("VeryWeak","Weak","Fair","Strong" or "VeryStrong"). I don't know what c(5) means in this context, but if I change the value in the parentheses it doesn't appear that the matching solution changes at all. What does c(5) mean in this context? Thanks to the developers for making this new matching tool available! Bridget Lavelle, Ph.D. Senior Research Manager Program Research and Evaluation Section Research and Data Analysis Division Department of Social and Health Services PO Box 45204 Olympia, WA 98504-5204 Fax: (360) 902-0705 Email: Bridget.Lavelle(a)dshs.wa.gov<mailto:Bridget.Lavelle@dshs.wa.gov> Telephone: (360) 902-0741

10 years, 7 months

Specifying Breaks for L1 Measure

by Lavelle, Bridget J (DSHS/RDA)

I am trying to understand why the imbalance command produces a univariate L1 measurement of 0 on a variable (readscore6) that has significant imbalance (ctl mean = 83.9, trt mean = 76.0, ttest p<.001, standardized mean difference from MatchIt = -1.5385). I figure this must be because of the automatic breakpoints, so I examine how the univariate L1 measure changes using the L1.profile command (dropping all other variables). The L1 reportedly varies from 0 to 0.59 in the sets of cutpoints examined. I replicate 3 of these L1 measurements precisely "by hand" and they seem correct. But when I add breaks to the imbalance command, I cannot replicate the L1 values produced in L1.profile. Any idea what is going on here? Thanks in advance for any insights. Additional details follow: Univariate L1 is 0, even though multivariate L1 is 0.563 and imbalance test is restricted to single variable "readscore6": > imbalance(group=kids$trt,data=kids,drop=c("trt","readscore7","female","aptitude")) Multivariate Imbalance Measure: L1=0.563 Percentage of local common support: LCS=57.1% Univariate Imbalance Measures: statistic type L1 min 25% 50% 75% max readscore6 -16.45865 (diff) 0 -21 -18 -19 -21 -11 L1 varies between 0 and 0.59 in a small sample of sets of cutpoints: > a<-L1.profile(group=kids$trt,data=kids,drop=c("trt","readscore7","female","aptitude"),M=5) > a$L1 rnd(4) rnd(3) rnd(5) rnd(1) rnd(2) 0.0000000 0.4601504 0.4969925 0.5631579 0.5872180 > a$CP $`rnd(4)` $`rnd(4)`$readscore6 [1] 38 97 $`rnd(3)` $`rnd(3)`$readscore6 [1] 38.00 52.75 67.50 82.25 97.00 $`rnd(5)` $`rnd(5)`$readscore6 [1] 38.00000 57.66667 77.33333 97.00000 $`rnd(1)` $`rnd(1)`$readscore6 [1] 38.00000 44.55556 51.11111 57.66667 64.22222 70.77778 77.33333 83.88889 90.44444 97.00000 $`rnd(2)` $`rnd(2)`$readscore6 [1] 38.000 45.375 52.750 60.125 67.500 74.875 82.250 89.625 97.000 Cannot seem to replicate these L1 values in the imbalance command, specifying the same cutpoints: E.g. specifying 2 cutpoints (1 bin) produces L1 of 0.563 (should be 0 according to L1.profile as well as my understanding of how L1 is calculated) > imbalance(group=kids$trt,data=kids,drop=c("trt","readscore7","female","aptitude"),breaks=list((readscore6=c(38,97)))) Multivariate Imbalance Measure: L1=0.563 Percentage of local common support: LCS=57.1% Univariate Imbalance Measures: statistic type L1 min 25% 50% 75% max readscore6 -16.45865 (diff) 0.5631579 -21 -18 -19 -21 -11 E.g. specifying 5 cutpoints (4 bins) produces L1 of 0.563 (should be 0.46 according to L1.profile) > imbalance(group=kids$trt,data=kids,drop=c("trt","readscore7","female","aptitude"),breaks=list((readscore6=c(38.00,52.75,67.50,82.25,97.00)))) Multivariate Imbalance Measure: L1=0.563 Percentage of local common support: LCS=57.1% Univariate Imbalance Measures: statistic type L1 min 25% 50% 75% max readscore6 -16.45865 (diff) 0.5631579 -21 -18 -19 -21 -11 Bridget Lavelle, Ph.D. Senior Research Manager Program Research and Evaluation Section Research and Data Analysis Division Department of Social and Health Services PO Box 45204 Olympia, WA 98504-5204 Fax: (360) 902-0705 Email: Bridget.Lavelle(a)dshs.wa.gov<mailto:Bridget.Lavelle@dshs.wa.gov> Telephone: (360) 902-0741

10 years, 7 months

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Cem September 2013