## First steps

First, the necessary packages are loaded into memory.

```
library(dplyr) # data management
library(caret) # confusion matrix
library(party) # conditional inference random forests and trees
library(partykit) # conditional inference trees
library(pROC) # ROC curves
library(measures) # performance measures
library(varImp) # variable importance
library(pdp) # partial dependence
library(vip) # measure of interactions
library(moreparty) # surrogate trees, accumulated local effects, etc.
library(RColorBrewer) # color palettes
library(descriptio) # bivariate analysis
```

Now, we then import `titanic`

data set from `moreparty`

.

```
spc_tbl_ [1,309 × 5] (S3: spec_tbl_df/tbl_df/tbl/data.frame)
$ Survived: Factor w/ 2 levels "No","Yes": 2 2 1 1 1 2 2 1 2 1 ...
$ Sex : Factor w/ 2 levels "female","male": 1 2 1 2 1 2 1 2 1 2 ...
$ Pclass : Factor w/ 3 levels "1st","2nd","3rd": 1 1 1 1 1 1 1 1 1 1 ...
$ Age : num [1:1309] 29 0.917 2 30 25 ...
$ Embarked: Factor w/ 3 levels "Cherbourg","Queenstown",..: 3 3 3 3 3 3 3 3 3 1 ...
```

We have 1309 cases, one categorical explained variable,
`Survived`

, which codes whether or not an individual survived
the shipwreck, and four explanatory variables (three categorical and one
continuous): gender, age, passenger class, and port of embarkation. The
distribution of the variables is examined.

```
Survived Sex Pclass Age Embarked
No :809 female:466 1st:323 Min. : 0.1667 Cherbourg :270
Yes:500 male :843 2nd:277 1st Qu.:21.0000 Queenstown :123
3rd:709 Median :28.0000 Southampton:914
Mean :29.8811 NA's : 2
3rd Qu.:39.0000
Max. :80.0000
NA's :263
```

The distribution of the explained variable is not balanced, as
survival is largely in the minority. In addition, some explanatory
variables have missing values, in particular `Age`

.

We examine the bivariate statistical relationships between the variables.

```
$YX
variable measure assoc p.value criterion
1 Sex cramer 0.527 0.00000 0.000000000
2 Pclass cramer 0.313 0.00000 0.000000000
3 Embarked cramer 0.184 0.00000 0.000000001
4 Age eta2 0.002 0.26069 0.302040642
$XX
variable1 variable2 measure assoc p.value criterion
1 Pclass Age eta2 0.170 0.00000 0.0000000000
2 Pclass Embarked cramer 0.280 0.00000 0.0000000000
3 Sex Pclass cramer 0.125 0.00004 0.0000378611
4 Sex Embarked cramer 0.122 0.00006 0.0000563134
5 Age Embarked eta2 0.006 0.01789 0.0180491352
6 Sex Age eta2 0.003 0.03964 0.0404512887
```

Survival is primarily associated with gender, secondarily with the passenger class. The explanatory variables are weakly related to each other.

```
$variables
variable measure association
1 Sex Cramer V 0.529
2 Pclass Cramer V 0.313
3 Embarked Cramer V 0.190
4 Age Eta2 0.003
$bylevel
$bylevel$No
$bylevel$No$categories
categories freq pct.y.in.x pct.x.in.y overall.pct.x phi
1 Sex.male 682 80.9 84.3 64.4 0.529
2 Pclass.3rd 528 74.5 65.3 54.2 0.283
3 Embarked.Southampton 610 66.7 75.4 69.8 0.155
4 Embarked.Cherbourg 120 44.4 14.8 20.6 -0.182
5 Pclass.1st 123 38.1 15.2 24.7 -0.279
6 Sex.female 127 27.3 15.7 35.6 -0.529
$bylevel$No$continuous.var
variables mean.in.category overall.mean sd.in.category overall.sd correlation
1 Age 30.55 29.88 13.91 14.41 0.056
$bylevel$Yes
$bylevel$Yes$categories
categories freq pct.y.in.x pct.x.in.y overall.pct.x phi
1 Sex.female 339 72.7 67.8 35.6 0.529
2 Pclass.1st 200 61.9 40.0 24.7 0.279
3 Embarked.Cherbourg 150 55.6 30.0 20.6 0.182
4 Embarked.Southampton 304 33.3 60.8 69.8 -0.155
5 Pclass.3rd 181 25.5 36.2 54.2 -0.283
6 Sex.male 161 19.1 32.2 64.4 -0.529
$bylevel$Yes$continuous.var
variables mean.in.category overall.mean sd.in.category overall.sd correlation
1 Age 28.92 29.88 15.04 14.41 -0.056
```

Women, first class passengers and those who boarded at Cherbourg are over-represented among the survivors. Men, 3rd class passengers and those who boarded at Southampton are over-represented among the non-survivors.

Random forests imply a share of randomness (via resampling and
drawing of splitting variables), as well as some interpretation tools
(via variable permutations). From one program run to the next, the
results may therefore differ slightly. If you wish to obtain the same
results systematically and to ensure reproducibility, use the
`set.seed`

function.

## Classification tree

In order to build a classification tree with CTree conditional
inference algorithm, we use `partykit`

package, which allows more flexibility than `party`

package, in particular to deal with missing values.

The tree can be displayed in textual or graphical form.

```
arbre <- partykit::ctree(Survived~., data=titanic, control=partykit::ctree_control(minbucket=30, maxsurrogate=Inf, maxdepth=3))
print(arbre)
```

```
Model formula:
Survived ~ Sex + Pclass + Age + Embarked
Fitted party:
[1] root
| [2] Sex in female
| | [3] Pclass in 1st, 2nd
| | | [4] Pclass in 1st: Yes (n = 144, err = 3.5%)
| | | [5] Pclass in 2nd: Yes (n = 106, err = 11.3%)
| | [6] Pclass in 3rd
| | | [7] Embarked in Cherbourg, Queenstown: Yes (n = 87, err = 36.8%)
| | | [8] Embarked in Southampton: No (n = 129, err = 39.5%)
| [9] Sex in male
| | [10] Pclass in 1st
| | | [11] Age <= 53: No (n = 148, err = 38.5%)
| | | [12] Age > 53: No (n = 31, err = 12.9%)
| | [13] Pclass in 2nd, 3rd
| | | [14] Age <= 9: No (n = 77, err = 35.1%)
| | | [15] Age > 9: No (n = 587, err = 12.4%)
Number of inner nodes: 7
Number of terminal nodes: 8
```

`Sex`

is the first splitting variable, `Pclass`

is the second and all the explanatory variables are used in the
tree.

The proportion of survivors varies greatly from one terminal node to another.

`nodeapply(as.simpleparty(arbre), ids = nodeids(arbre, terminal = TRUE), FUN = function(x) round(prop.table(info_node(x)$distribution),3))`

```
$`4`
No Yes
0.035 0.965
$`5`
No Yes
0.113 0.887
$`7`
No Yes
0.368 0.632
$`8`
No Yes
0.605 0.395
$`11`
No Yes
0.615 0.385
$`12`
No Yes
0.871 0.129
$`14`
No Yes
0.649 0.351
$`15`
No Yes
0.876 0.124
```

Thus, 96.5% of women travelling in 1st class survive, compared with only 12.4% of men over 9 years of age travelling in 2nd or 3rd class.

The graphical representation can be parameterized to obtain a simpler and more readable tree.

`plot(arbre, inner_panel=node_inner(arbre,id=FALSE,pval=FALSE), terminal_panel=node_barplot(arbre,id=FALSE), gp=gpar(cex=0.6), ep_args=list(justmin=15))`

Note that the `ggparty`

package graphically represents Ctree trees with the `ggplot2`

grammar.

To measure the performance of the tree, the AUC is calculated, by comparing predicted and observed survival.

```
pred_arbre <- predict(arbre, type='prob')[,'Yes']
auc_arbre <- AUC(pred_arbre, titanic$Survived, positive='Yes')
auc_arbre %>% round(3)
```

`[1] 0.838`

AUC is 0.838, which is a relatively high performance.

To plot the ROC curve of the model :

```
pROC::roc(titanic$Survived, pred_arbre) %>%
ggroc(legacy.axes=TRUE) +
geom_segment(aes(x=0,xend=1,y=0,yend=1), color="darkgrey", linetype="dashed") +
theme_bw() +
xlab("TFP") +
ylab("TVP")
```

Other performance measures are based on the confusion matrix.

```
ifelse(pred_arbre > .5, "Yes", "No") %>%
factor %>%
caret::confusionMatrix(titanic$Survived, positive='Yes')
```

```
Confusion Matrix and Statistics
Reference
Prediction No Yes
No 760 212
Yes 49 288
Accuracy : 0.8006
95% CI : (0.7779, 0.8219)
No Information Rate : 0.618
P-Value [Acc > NIR] : < 2.2e-16
Kappa : 0.5497
Mcnemar's Test P-Value : < 2.2e-16
Sensitivity : 0.5760
Specificity : 0.9394
Pos Pred Value : 0.8546
Neg Pred Value : 0.7819
Prevalence : 0.3820
Detection Rate : 0.2200
Detection Prevalence : 0.2574
Balanced Accuracy : 0.7577
'Positive' Class : Yes
```

The `GetSplitStats`

function allows to examine the result
of the competition between covariates in the choice of splitting
variables. If, for a given node, the splitting variable is significantly
more associated with the explained variable than the other explanatory
variables, we can think that this split is stable.

```
$details
$details$`1`
statistic p.value criterion ratio
Sex 365.607431 6.771232e-81 -6.771232e-81 1.000000e+00
Pclass 127.761479 7.227818e-28 -7.227818e-28 1.067430e+53
Embarked 44.207893 1.005629e-09 -1.005629e-09 1.485150e+71
Age 3.220314 2.606920e-01 -3.020406e-01 4.460645e+79
$details$`2`
statistic p.value criterion ratio
Pclass 115.454414 2.549845e-25 -2.549845e-25 1.000000e+00
Embarked 24.336500 1.557812e-05 -1.557824e-05 6.109486e+19
Age 7.070397 2.332660e-02 -2.360298e-02 9.256632e+22
$details$`3`
statistic p.value criterion ratio
Pclass 5.9107118 0.04447125 -0.04549043 1.00000
Embarked 4.2482494 0.31745308 -0.38192401 8.39570
Age 0.2003411 0.95873811 -3.18781595 70.07663
$details$`6`
statistic p.value criterion ratio
Embarked 12.759466 0.003388277 -0.00339403 1.0000
Age 1.724815 0.342399641 -0.41915789 123.4986
$details$`9`
statistic p.value criterion ratio
Pclass 32.99374 2.054102e-07 -2.054103e-07 1.000
Embarked 17.71078 4.277725e-04 -4.278640e-04 2082.973
Age 10.83849 2.979392e-03 -2.983839e-03 14526.240
$details$`10`
statistic p.value criterion ratio
Age 9.079224 0.00516391 -0.005177289 1.0000
Embarked 2.193883 0.55629854 -0.812603321 156.9554
$details$`13`
statistic p.value criterion ratio
Age 25.40589207 1.393491e-06 -1.393492e-06 1.0
Embarked 5.27150955 1.999551e-01 -2.230874e-01 160092.3
Pclass 0.03486032 9.967509e-01 -5.729373e+00 4111521.7
$summary
node split_var best_var ratio
1 1 Sex Pclass 1.1e+53
2 2 Pclass Embarked 6.1e+19
3 3 Pclass Embarked 8.4
4 6 Embarked Age 123.5
5 9 Pclass Embarked 2.1e+03
6 10 Age Embarked 156.96
7 13 Age Embarked 1.6e+05
```

In this case, for each of the nodes, the result of the competition is
final (see `criterion`

). The tree therefore appears to be
stable.

## Random forest

Then we build a random forest with conditional inference algorithm,
`mtry`

=2 and `ntree`

=500.

`foret <- party::cforest(Survived~., data=titanic, controls=party::cforest_unbiased(mtry=2,ntree=500))`

To compare the performance of the forest to that of the tree, the predictions and then the AUC are calculated.

```
pred_foret <- predict(foret, type='prob') %>%
do.call('rbind.data.frame',.) %>%
select(2) %>%
unlist
auc_foret <- AUC(pred_foret, titanic$Survived, positive='Yes')
auc_foret %>% round(3)
```

`[1] 0.879`

The performance of the forest is 0.879, therefore slightly better than that of the tree.

The `OOB=TRUE`

option allows predictions to be made from
out-of-bag observations, thus avoiding optimism bias.

```
pred_oob <- predict(foret, type='prob', OOB=TRUE) %>%
do.call('rbind.data.frame',.) %>%
select(2) %>%
unlist
auc_oob <- AUC(pred_oob, titanic$Survived, positive='Yes')
auc_oob %>% round(3)
```

`[1] 0.847`

Calculated in this way, the performance is indeed slightly lower (0.847).

## Surrogate tree

The so-called “surrogate tree” can be a way to synthesize a complex model.

`[1] 0.96`

`plot(surro$tree, inner_panel=node_inner(surro$tree,id=FALSE,pval=FALSE), terminal_panel=node_boxplot(surro$tree,id=FALSE), gp=gpar(cex=0.6), ep_args=list(justmin=15))`

The surrogate tree reproduces here very faithfully the predictions of the random forest (R2 = 0.96). It is also very similar to the initial classification tree.

## Variable importance

To go further in the interpretation of the results, permutation
variable importance is calculated (using AUC as a performance measure).
Since the explanatory variables have little correlation with each other,
it is not necessary to use the “conditional permutation scheme”
(available with the `conditional=TRUE`

option in the
`varImpAUC`

function).

```
Sex Pclass Age Embarked
-0.213 -0.066 -0.026 -0.008
```

`Sex`

is the most important variable, ahead of
`Pclass`

, `Age`

and `Embarked`

(whose
importance is close to zero).

## First order effects

The calculation of partial dependences of all the covariates can then
be performed using the `GetPartialData`

function.

```
var cat value
1 Sex female 0.6705362
2 Sex male 0.2105944
3 Pclass 1st 0.5630163
4 Pclass 2nd 0.4123658
5 Pclass 3rd 0.2726452
6 Age 5 0.5535410
7 Age 14 0.4094217
8 Age 18 0.3835784
9 Age 19 0.3851776
10 Age 21 0.3849032
11 Age 22 0.3848739
12 Age 24 0.3860633
13 Age 25 0.3910394
14 Age 26 0.3897490
15 Age 28 0.3692686
16 Age 30 0.3742014
17 Age 31 0.3977995
18 Age 33 0.3602312
19 Age 36 0.3536723
20 Age 39 0.3411718
21 Age 42 0.3396432
22 Age 45 0.3421748
23 Age 50 0.3443513
24 Age 57 0.3187270
25 Embarked Cherbourg 0.4603041
[ reached 'max' / getOption("max.print") -- omitted 2 rows ]
```

They can be represented graphically.

The dashed line represents the average predicted probability of survival. We can see for example that the probability of survival is much higher for women or that it decreases when we go from 1st to 3rd class.

We’re zooming in on age to see its effect more closely. Partial dependencies are calculated for 40 quantiles of the age distribution, this high number allowing a more precise examination.

```
ggplot(pdep_age, aes(x=Age, y=yhat)) +
geom_line() +
geom_hline(aes(yintercept=mean(pred_foret)), size=0.2, linetype='dashed', color='black') +
ylim(c(0,1)) +
theme_bw() +
ylab("Probability of survival")
```

The probability of survival for young children is high, but drops rapidly with age. It is around the average for young adults and declines again after age 30.

We often limit ourselves to examining *average* survival
probabilities, but we can also look at their *distribution*, in
numerical form :

```
pdep_ind %>% group_by(var, cat) %>% summarise(prob = mean(value) %>% round(3),
Q1 = quantile(value, 0.25) %>% round(3),
Q3 = quantile(value, 0.75) %>% round(3))
```

```
# A tibble: 27 × 5
# Groups: var [4]
var cat prob Q1 Q3
<fct> <chr> <dbl> <dbl> <dbl>
1 Age 14 0.409 0.137 0.596
2 Age 18 0.384 0.12 0.568
3 Age 19 0.385 0.124 0.567
4 Age 21 0.385 0.131 0.562
5 Age 22 0.385 0.124 0.576
6 Age 24 0.386 0.139 0.572
7 Age 25 0.391 0.167 0.556
8 Age 26 0.39 0.16 0.556
9 Age 28 0.369 0.136 0.512
10 Age 30 0.374 0.155 0.509
# ℹ 17 more rows
```

Or in graphical form :

```
ggplot(pdep_ind, aes(x = value, y = cat, group = cat)) +
geom_boxplot(aes(fill=var), notch=TRUE) +
geom_vline(aes(xintercept=median(pred_foret)), size=0.2, linetype='dashed', color='black') +
facet_grid(var ~ ., scales = "free_y", space = "free_y") +
theme_bw() +
theme(panel.grid = element_blank(),
panel.grid.major.y = element_line(size=.1, color="grey70"),
legend.position = "none",
strip.text.y = element_text(angle = 0)) +
xlim(c(0,1)) +
xlab("Probability of survival") +
ylab("")
```