Example on how to use Quint in R
In this example we use data from a three-arm randomized controlled trial. The data contain information on women with early-stage breast cancer who were assigned randomly to different treatments. These treatments are a nutrition intervention, an education intervention, and standard care.
Quint is a method for two-arm randomized controlled trials. Therefore, in this example we restrict ourselves to the first two treatments, that is, the nutrition intervention and the education intervention.
data(bcrp)
head(bcrp)## physt1 cesdt1 physt3 cesdt3 negsoct1 uncomt1 disopt1 comorbid age
## 1 37.65374 14 52.62905 4 9 28 14 6 29.48392
## 2 53.64822 10 51.18797 14 7 36 10 2 44.66256
## 3 63.84140 8 66.45392 9 6 29 15 1 43.09925
## 4 38.72757 2 45.99656 4 5 30 17 13 46.93498
## 5 55.85700 0 60.66603 1 10 32 22 1 41.09514
## 6 53.06922 0 54.83928 0 8 24 21 3 44.44627
## wcht1 nationality marital trext cond
## 1 1 1 0 0.2589759 3
## 2 1 1 1 0.5557208 1
## 3 1 1 0 0.2589759 2
## 4 1 1 1 0.5557208 2
## 5 0 1 1 0.5557208 2
## 6 1 1 1 0.2589759 3bcrp2arm<-subset(bcrp,bcrp$cond<3)
head(bcrp2arm)## physt1 cesdt1 physt3 cesdt3 negsoct1 uncomt1 disopt1 comorbid age
## 2 53.64822 10 51.18797 14 7 36 10 2 44.66256
## 3 63.84140 8 66.45392 9 6 29 15 1 43.09925
## 4 38.72757 2 45.99656 4 5 30 17 13 46.93498
## 5 55.85700 0 60.66603 1 10 32 22 1 41.09514
## 7 47.87246 4 57.97079 3 9 22 20 1 45.76044
## 8 48.59693 4 58.66519 2 9 31 16 6 36.84052
## wcht1 nationality marital trext cond
## 2 1 1 1 0.5557208 1
## 3 1 1 0 0.2589759 2
## 4 1 1 1 0.5557208 2
## 5 0 1 1 0.5557208 2
## 7 0 1 0 -1.7742771 2
## 8 0 1 1 0.2589759 1As we have seen, the data contain the baseline measurements (t1) and the 9-month follow-up measurements (t3) and the pretreatment characteristics of the patients. For more information about the data, read the bcrp help file or ask for a summary.
summary(bcrp2arm)## physt1 cesdt1 physt3 cesdt3
## Min. :24.05 Min. : 0.000 Min. :24.64 Min. : 0.000
## 1st Qu.:44.45 1st Qu.: 2.000 1st Qu.:51.20 1st Qu.: 1.000
## Median :49.90 Median : 5.000 Median :55.37 Median : 3.000
## Mean :49.52 Mean : 6.417 Mean :54.47 Mean : 4.716
## 3rd Qu.:56.31 3rd Qu.:10.000 3rd Qu.:58.64 3rd Qu.: 7.000
## Max. :67.40 Max. :24.000 Max. :67.43 Max. :21.000
## NA's :20 NA's :20
## negsoct1 uncomt1 disopt1 comorbid
## Min. : 5.000 Min. :15.00 Min. : 5.00 Min. : 0.000
## 1st Qu.: 6.000 1st Qu.:26.00 1st Qu.:15.00 1st Qu.: 0.000
## Median : 7.500 Median :29.00 Median :17.00 Median : 2.000
## Mean : 7.821 Mean :29.45 Mean :16.74 Mean : 2.405
## 3rd Qu.: 9.000 3rd Qu.:33.00 3rd Qu.:19.00 3rd Qu.: 4.000
## Max. :16.000 Max. :42.00 Max. :24.00 Max. :13.000
##
## age wcht1 nationality marital
## Min. :29.29 Min. :0.0000 Min. :0.0000 Min. :0.000
## 1st Qu.:41.06 1st Qu.:0.0000 1st Qu.:1.0000 1st Qu.:0.000
## Median :44.81 Median :0.0000 Median :1.0000 Median :1.000
## Mean :43.96 Mean :0.4643 Mean :0.9286 Mean :0.744
## 3rd Qu.:48.17 3rd Qu.:1.0000 3rd Qu.:1.0000 3rd Qu.:1.000
## Max. :51.40 Max. :1.0000 Max. :1.0000 Max. :1.000
##
## trext cond
## Min. :-1.77428 Min. :1.000
## 1st Qu.: 0.25898 1st Qu.:1.000
## Median : 0.25898 Median :1.000
## Mean : 0.03413 Mean :1.494
## 3rd Qu.: 0.55572 3rd Qu.:2.000
## Max. : 2.58897 Max. :2.000
## Once we have the data, we need to set up the control parameters of the algorithm of quint. In this case, we use the default values for the following arguments: minimal sample size of patients assigned to treatment 1 and treatment 2 in each leaf (10% of the sample size of each treatment), the weights, and the minimum absolute standarized mean outcome difference (dmin = 0.3).
On the other hand, we set the type of treatment outcome difference used in the partitioning criterion to difference in treatment means, the maximum number of leaves to 5 and we decided to perform the bias-corrected bootstrap procedure using 10 bootstrap samples (the default number is 25).
control1 <- quint.control(crit="dm",maxl = 5,B = 10)Also, we need to define the formula to be used in quint. This formula should contain first the outcome variable (in this example the difference between the outcome measure at the baseline and at the 9-month follow-up). Next to it, the variable indicating the treatment, and finally the pretreatment characteristics.
formula1<- I(cesdt1-cesdt3) ~ cond | nationality+marital+wcht1+ age+trext+comorbid+disopt1+uncomt1+negsoct1Now, we are ready to perform quint.
set.seed(2)
quint1<-quint(formula1, data= bcrp2arm, control=control1 )## Treatment variable (T) equals 1 corresponds to cond = 1
## Treatment variable (T) equals 2 corresponds to cond = 2
## The sample size in the analysis is 148
## split 1
## #leaves is 2
## Bootstrap sample 1
## Bootstrap sample 2
## Bootstrap sample 3
## Bootstrap sample 4
## Bootstrap sample 5
## Bootstrap sample 6
## Bootstrap sample 7
## Bootstrap sample 8
## Bootstrap sample 9
## Bootstrap sample 10
## current value of C 3.233541
## split 2
## #leaves is 3
## Bootstrap sample 1
## Bootstrap sample 2
## Bootstrap sample 3
## Bootstrap sample 4
## Bootstrap sample 5
## Bootstrap sample 6
## Bootstrap sample 7
## Bootstrap sample 8
## Bootstrap sample 9
## Bootstrap sample 10
## current value of C 3.481599
## split 3
## #leaves is 4
## Bootstrap sample 1
## Bootstrap sample 2
## Bootstrap sample 3
## Bootstrap sample 4
## Bootstrap sample 5
## Bootstrap sample 6
## Bootstrap sample 7
## Bootstrap sample 8
## Bootstrap sample 9
## Bootstrap sample 10
## current value of C 3.636685
## split 4
## #leaves is 5
## Bootstrap sample 1
## Bootstrap sample 2
## Bootstrap sample 3
## Bootstrap sample 4
## Bootstrap sample 5
## Bootstrap sample 6
## Bootstrap sample 7
## Bootstrap sample 8
## Bootstrap sample 9
## Bootstrap sample 10After growing the binary tree using quint, we need to prune the tree. This is done using the prune function.
quint1pr <- prune(quint1)## The sample size in the analysis is 148
## split 1
## #leaves is 2
## current value of C 3.233541
## split 2
## #leaves is 3
## current value of C 3.481599
## split 3
## #leaves is 4Here, if you consider that the standard errors of the mean difference are very small (and therefore the CI), then we should use the function quint.bootstrapCI, which estimates the confidence intervals of the mean difference (up to version 2.1.0 it does not work with effects sizes) in each leaf using a bootstrap-based algorithm.
quint1pr_bootCI <- quint.bootstrapCI(quint1pr,n_boot = 5)## Treatment variable (T) equals 1 corresponds to cond = 1
## Treatment variable (T) equals 2 corresponds to cond = 2
## The sample size in the analysis is 148
## split 1
## #leaves is 2
## Bootstrap sample 1
## Bootstrap sample 2
## Bootstrap sample 3
## Bootstrap sample 4
## Bootstrap sample 5
## Bootstrap sample 6
## Bootstrap sample 7
## Bootstrap sample 8
## Bootstrap sample 9
## Bootstrap sample 10
## current value of C 3.643239
## split 2
## #leaves is 3
## Bootstrap sample 1
## Bootstrap sample 2
## Bootstrap sample 3
## Bootstrap sample 4
## Bootstrap sample 5
## Bootstrap sample 6
## Bootstrap sample 7
## Bootstrap sample 8
## Bootstrap sample 9
## Bootstrap sample 10
## current value of C 3.935028
## split 3
## #leaves is 4
## Bootstrap sample 1
## Bootstrap sample 2
## Bootstrap sample 3
## Bootstrap sample 4
## Bootstrap sample 5
## Bootstrap sample 6
## Bootstrap sample 7
## Bootstrap sample 8
## Bootstrap sample 9
## Bootstrap sample 10
## The sample size in the analysis is 148
## split 1
## #leaves is 2
## current value of C 3.643239
## split 2
## #leaves is 3
## current value of C 3.935028
## split 3
## #leaves is 4
## Treatment variable (T) equals 1 corresponds to cond = 1
## Treatment variable (T) equals 2 corresponds to cond = 2
## The sample size in the analysis is 148
## split 1
## #leaves is 2
## Bootstrap sample 1
## Bootstrap sample 2
## Bootstrap sample 3
## Bootstrap sample 4
## Bootstrap sample 5
## Bootstrap sample 6
## Bootstrap sample 7
## Bootstrap sample 8
## Bootstrap sample 9
## Bootstrap sample 10
## current value of C 3.425667
## split 2
## #leaves is 3
## Bootstrap sample 1
## Bootstrap sample 2
## Bootstrap sample 3
## Bootstrap sample 4
## Bootstrap sample 5
## Bootstrap sample 6
## Bootstrap sample 7
## Bootstrap sample 8
## Bootstrap sample 9
## Bootstrap sample 10
## current value of C 3.606798
## split 3
## #leaves is 4
## Bootstrap sample 1
## Bootstrap sample 2
## Bootstrap sample 3
## Bootstrap sample 4
## Bootstrap sample 5
## Bootstrap sample 6
## Bootstrap sample 7
## Bootstrap sample 8
## Bootstrap sample 9
## Bootstrap sample 10
## The sample size in the analysis is 148
## split 1
## #leaves is 2
## current value of C 3.425667
## split 2
## #leaves is 3
## current value of C 3.606798
## split 3
## #leaves is 4
## Treatment variable (T) equals 1 corresponds to cond = 1
## Treatment variable (T) equals 2 corresponds to cond = 2
## The sample size in the analysis is 148
## split 1
## #leaves is 2
## Bootstrap sample 1
## Bootstrap sample 2
## Bootstrap sample 3
## Bootstrap sample 4
## Bootstrap sample 5
## Bootstrap sample 6
## Bootstrap sample 7
## Bootstrap sample 8
## Bootstrap sample 9
## Bootstrap sample 10
## current value of C 3.591496
## split 2
## #leaves is 3
## Bootstrap sample 1
## Bootstrap sample 2
## Bootstrap sample 3
## Bootstrap sample 4
## Bootstrap sample 5
## Bootstrap sample 6
## Bootstrap sample 7
## Bootstrap sample 8
## Bootstrap sample 9
## Bootstrap sample 10
## current value of C 3.993374
## split 3
## #leaves is 4
## splitting process stopped after number of leaves equals 3 because new value of C was not higher than current value of C
## The sample size in the analysis is 148
## split 1
## #leaves is 2
## current value of C 3.591496
## split 2
## #leaves is 3
## Treatment variable (T) equals 1 corresponds to cond = 1
## Treatment variable (T) equals 2 corresponds to cond = 2
## The sample size in the analysis is 148
## split 1
## #leaves is 2
## Bootstrap sample 1
## Bootstrap sample 2
## Bootstrap sample 3
## Bootstrap sample 4
## Bootstrap sample 5
## Bootstrap sample 6
## Bootstrap sample 7
## Bootstrap sample 8
## Bootstrap sample 9
## Bootstrap sample 10
## current value of C 3.511183
## split 2
## #leaves is 3
## Bootstrap sample 1
## Bootstrap sample 2
## Bootstrap sample 3
## Bootstrap sample 4
## Bootstrap sample 5
## Bootstrap sample 6
## Bootstrap sample 7
## Bootstrap sample 8
## Bootstrap sample 9
## Bootstrap sample 10
## current value of C 3.80894
## split 3
## #leaves is 4
## Bootstrap sample 1
## Bootstrap sample 2
## Bootstrap sample 3
## Bootstrap sample 4
## Bootstrap sample 5
## Bootstrap sample 6
## Bootstrap sample 7
## Bootstrap sample 8
## Bootstrap sample 9
## Bootstrap sample 10
## The sample size in the analysis is 148
## split 1
## #leaves is 2
## current value of C 3.511183
## split 2
## #leaves is 3
## current value of C 3.80894
## split 3
## #leaves is 4
## Treatment variable (T) equals 1 corresponds to cond = 1
## Treatment variable (T) equals 2 corresponds to cond = 2
## The sample size in the analysis is 148
## split 1
## #leaves is 2
## Bootstrap sample 1
## Bootstrap sample 2
## Bootstrap sample 3
## Bootstrap sample 4
## Bootstrap sample 5
## Bootstrap sample 6
## Bootstrap sample 7
## Bootstrap sample 8
## Bootstrap sample 9
## Bootstrap sample 10
## current value of C 3.4231
## split 2
## #leaves is 3
## Bootstrap sample 1
## Bootstrap sample 2
## Bootstrap sample 3
## Bootstrap sample 4
## Bootstrap sample 5
## Bootstrap sample 6
## Bootstrap sample 7
## Bootstrap sample 8
## Bootstrap sample 9
## Bootstrap sample 10
## current value of C 3.819557
## split 3
## #leaves is 4
## Bootstrap sample 1
## Bootstrap sample 2
## Bootstrap sample 3
## Bootstrap sample 4
## Bootstrap sample 5
## Bootstrap sample 6
## Bootstrap sample 7
## Bootstrap sample 8
## Bootstrap sample 9
## Bootstrap sample 10
## The sample size in the analysis is 148
## split 1
## #leaves is 2
## current value of C 3.4231
## split 2
## #leaves is 3
## current value of C 3.819557
## split 3
## #leaves is 4To obtain the most interesting information about the results of the method we should use the summary function.
summary(quint1pr) #without bootstrap-based confidence intervals ## Partitioning criterion: Difference in treatment means criterion
##
## Fit information:
## Criterion
## - - - -
## split #leaves apparent biascorrected se
## 1 2 3.23 2.72 0.16
## 2 3 3.48 2.80 0.16
## 3 4 3.64 2.93 0.11
##
## Split information:
## parentnode childnodes splittingvar splitpoint
## Split 1 1 2,3 disopt1 18.5
## Split 2 2 4,5 negsoct1 5.5
## Split 3 5 10,11 uncomt1 34.5
##
## Leaf information:
## #(T=1) meanY|T=1 SD|T=1 #(T=2) meanY|T=2 SD|T=2 diff se class
## Leaf 1 11 1.00 3.10 7 3.71 4.75 -2.71 1.84 2
## Leaf 2 35 3.23 5.76 32 1.03 5.49 2.20 1.38 3
## Leaf 3 13 5.08 6.76 7 -6.00 4.04 11.08 2.81 1
## Leaf 4 19 -0.32 4.41 24 1.25 2.69 -1.57 1.09 2summary(quint1pr_bootCI$tree) #with bootstrap-based confidence intervals ## Partitioning criterion: Difference in treatment means criterion
##
## Fit information:
## Criterion
## - - - -
## split #leaves apparent biascorrected se
## 1 2 3.23 2.72 0.16
## 2 3 3.48 2.80 0.16
## 3 4 3.64 2.93 0.11
##
## Split information:
## parentnode childnodes splittingvar splitpoint
## Split 1 1 2,3 disopt1 18.5
## Split 2 2 4,5 negsoct1 5.5
## Split 3 5 10,11 uncomt1 34.5
##
## Leaf information:
## #(T=1) meanY|T=1 SD|T=1 #(T=2) meanY|T=2 SD|T=2 diff se class
## Leaf 1 11 1.00 3.10 7 3.71 4.75 -2.71 1.67 2
## Leaf 2 35 3.23 5.76 32 1.03 5.49 2.20 2.37 3
## Leaf 3 13 5.08 6.76 7 -6.00 4.04 11.08 3.09 1
## Leaf 4 19 -0.32 4.41 24 1.25 2.69 -1.57 1.21 2Another form of obtaining information about the solution is to plot the pruned tree.
plot(quint1pr) #without bootstrap-based confidence intervals 
plot(quint1pr_bootCI$tree) #with bootstrap-based confidence intervals 
With these results we can conclude that the first treatment (nutrition) works better than the second treatment (education) for patients with dispositional optimism at baseline (disopt1) below or equal to 18.5, negative social interaction at baseline (negsoct1) greater than 5.5 and unmitigated communion at baseline (uncomt1) greater than 34.5. Treatment 2 (education) works better for patients with disopt1 greater than 18.5, and patients with disopt1 less or equal than 18.5 and negsoct1 less or equal than 5.5. The rest of the patients could be assigned to any of the two treatments as the treatment effect does not differ for them according to quint. We can observe the different standard errors for the two methods, as the confidence intervals computed using bootstrap are larger.
The results might be slightly different for R versions below 3.6.0.