Transcript Document

Decision trees,
cross-validation
Peter Fox
Data Analytics – ITWS-4600/ITWS-6600
Week 7a, March 8, 2016
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Contents
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Numeric v. non-numeric
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In R – data frame and types
• Almost always at input R sees categorical
data as “strings” or “character”
• You can test for membership (as a type)
using is.<x> (x=number, factor, etc.)
• You can “coerce” it (i.e. change the type)
using as.<x> (same x)
• To tell R you have categorical types (also
called enumerated types), R calls them
“factor”…. Thus – as.factor()
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In R
factor(x = character(), levels, labels = levels,
exclude = NA, ordered = is.ordered(x), nmax = NA),
ordered(x, ...), is.factor(x), is.ordered(x), as.factor(x),
as.ordered(x), addNA(x, ifany = FALSE)
levels - values that x might have taken.
labels - either an optional vector of labels for the levels (in the
same order as levels after removing those in exclude), or a
character string of length 1.
Exclude Ordered Nmax - upper bound on the number of levels;
Ifany 5
Relate to the datasets…
• Abalone - Sex = {F, I, M}
• Eye color?
• EPI - GeoRegions?
• Sample v. population – levels and names IN
the dataset versus all possible levels/names
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Weighted KNN
require(kknn)
data(iris)
m <- dim(iris)[1]
val <- sample(1:m, size = round(m/3), replace =
FALSE, prob = rep(1/m, m))
iris.learn <- iris[-val,] # train
iris.valid <- iris[val,]
# test
iris.kknn <- kknn(Species~., iris.learn, iris.valid,
distance = 1, kernel = "triangular") # Possible choices
are "rectangular" (which is standard unweighted knn),
"triangular", "epanechnikov" (or beta(2,2)), "biweight"
(or beta(3,3)), "triweight" (or beta(4,4)), "cos", "inv",
"gaussian", "rank" and "optimal".
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names(iris.kknn)
•
•
•
•
•
•
•
fitted.values Vector of predictions.
CL Matrix of classes of the k nearest neighbors.
W Matrix of weights of the k nearest neighbors.
D Matrix of distances of the k nearest neighbors.
C Matrix of indices of the k nearest neighbors.
prob
Matrix of predicted class probabilities.
responseType of response variable, one of
continuous, nominal or ordinal.
• distance Parameter of Minkowski distance.
• call
The matched call.
• terms
The 'terms' object used.
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Look at the output
> head(iris.kknn$W)
[,1]
[,2]
[,3]
[,4]
[,5]
[,6]
[,7]
[1,] 0.4493696 0.2306555 0.1261857 0.1230131 0.07914805 0.07610159
0.014184110
[2,] 0.7567298 0.7385966 0.5663245 0.3593925 0.35652546 0.24159191
0.004312408
[3,] 0.5958406 0.2700476 0.2594478 0.2558161 0.09317996 0.09317996
0.042096849
[4,] 0.6022069 0.5193145 0.4229427 0.1607861 0.10804205 0.09637177
0.055297983
[5,] 0.7011985 0.6224216 0.5183945 0.2937705 0.16230921 0.13964231
0.053888244
[6,] 0.5898731 0.5270226 0.3273701 0.1791715 0.15297478 0.08446215
0.010180454
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Look at the output
> head(iris.kknn$D)
[,1]
[,2]
[,3]
[,4]
[,5]
[,6]
[,7]
[1,] 0.7259100 1.0142464 1.1519716 1.1561541 1.2139825 1.2179988
1.2996261
[2,] 0.2508639 0.2695631 0.4472127 0.6606040 0.6635606 0.7820818
1.0267680
[3,] 0.6498131 1.1736274 1.1906700 1.1965092 1.4579977 1.4579977
1.5401298
[4,] 0.2695631 0.3257349 0.3910409 0.5686904 0.6044323 0.6123406
0.6401741
[5,] 0.7338183 0.9272845 1.1827617 1.7344095 2.0572618 2.1129288
2.3235298
[6,] 0.5674645 0.6544263 0.9306719 1.1357241 1.1719707 1.2667669
1.3695454
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Look at the output
> head(iris.kknn$C)
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,] 86 38 43 73 92 85 60
[2,] 31 20 16 21 24 15 7
[3,] 48 80 44 36 50 63 98
[4,] 4 21 25 6 20 26 1
[5,] 68 79 70 65 87 84 75
[6,] 91 97 100 96 83 93 81
> head(iris.kknn$prob)
setosa versicolor virginica
[1,]
0 0.3377079 0.6622921
[2,]
1 0.0000000 0.0000000
[3,]
0 0.8060743 0.1939257
[4,]
1 0.0000000 0.0000000
[5,]
0 0.0000000 1.0000000
[6,]
0 0.0000000 1.0000000
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Look at the output
> head(iris.kknn$fitted.values)
[1] virginica setosa versicolor setosa
Levels: setosa versicolor virginica
virginica virginica
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Contingency tables
fitiris <- fitted(iris.kknn)
table(iris.valid$Species, fitiris)
fitiris
setosa versicolor virginica
setosa
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0
0
versicolor
0
18
2
virginica
0
1
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# rectangular – no weight
fitiris2
setosa versicolor virginica
setosa
17
0
0
versicolor
0
18
2
virginica
0
2
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(Weighted) kNN
• Advantages
– Robust to noisy training data (especially if we use inverse
square of weighted distance as the “distance”)
– Effective if the training data is large
• Disadvantages
– Need to determine value of parameter K (number of
nearest neighbors)
– Distance based learning is not clear which type of
distance to use and which attribute to use to produce the
best results. Shall we use all attributes or certain
attributes only?
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Additional factors
• Dimensionality – with too many dimensions
the closest neighbors are too far away to be
considered close
• Overfitting – does closeness mean right
classification (e.g. noise or incorrect data, like
wrong street address -> wrong lat/lon) –
beware of k=1!
• Correlated features – double weighting
• Relative importance – including/ excluding
features
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More factors
• Sparseness – the standard distance measure
(Jaccard) loses meaning due to no overlap
• Errors – unintentional and intentional
• Computational complexity
• Sensitivity to distance metrics – especially
due to different scales (recall ages, versus
impressions, versus clicks and especially
binary values: gender, logged in/not)
• Does not account for changes over time
• Model updating as new data comes in
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Glass
library(e1071)
library(rpart)
data(Glass, package="mlbench")
index <- 1:nrow(Glass)
testindex <- sample(index, trunc(length(index)/3))
testset <- Glass[testindex,]
trainset <- Glass[-testindex,]
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Now what?
# now what happens?
> rpart.model <- rpart(Type ~ ., data = trainset)
> rpart.pred <- predict(rpart.model, testset[,-10],
type = "class”)
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General idea behind trees
• Although the basic philosophy of all the classifiers
based on decision trees is identical, there are many
possibilities for its construction.
• Among all the key points in the selection of an algorithm
to build decision trees some of them should be
highlighted for their importance:
– Criteria for the choice of feature to be used in each
node
– How to calculate the partition of the set of examples
– When you decide that a node is a leaf
– What is the criterion to select the class to assign to
each leaf
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• Some important advantages can be pointed to the
decision trees, including:
– Can be applied to any type of data
– The final structure of the classifier is quite simple and can
be stored and handled in a graceful manner
– Handles very efficiently conditional information,
subdividing the space into sub-spaces that are handled
individually
– Reveal normally robust and insensitive to
misclassification in the training set
– The resulting trees are usually quite understandable and
can be easily used to obtain a better understanding of the
phenomenon in question. This is perhaps the most
important of all the advantages listed
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Stopping – leaves on the tree
• A number of stopping conditions can be used
to stop the recursive process. The algorithm
stops when any one of the conditions is true:
– All the samples belong to the same class, i.e.
have the same label since the sample is already
"pure"
– Stop if most of the points are already of the same
class. This is a generalization of the first
approach, with some error threshold
– There are no remaining attributes on which the
samples may be further partitioned
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– There are no samples for the branch test attribute
Recursive partitioning
• Recursive partitioning is a fundamental tool in data mining. It
helps us explore the structure of a set of data, while
developing easy to visualize decision rules for predicting a
categorical (classification tree) or continuous (regression
tree) outcome.
• The rpart programs build classification or regression models
of a very general structure using a two stage procedure; the
resulting models can be represented as binary trees.
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Recursive partitioning
• The tree is built by the following process:
– first the single variable is found which best splits
the data into two groups ('best' will be defined
later). The data is separated, and then this
process is applied separately to each sub-group,
and so on recursively until the subgroups either
reach a minimum size or until no improvement
can be made.
– second stage of the procedure consists of using
cross-validation to trim back the full tree.
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Why are we careful doing this?
• Because we will USE these trees, i.e. apply
them to make decisions about what things
are and what to do with them!
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> printcp(rpart.model)
Classification tree:
rpart(formula = Type ~ ., data = trainset)
Variables actually used in tree construction:
[1] Al Ba Mg RI
Root node error: 92/143 = 0.64336
n= 143
CP
nsplit rel error xerror xstd
1 0.206522
0 1.00000 1.00000 0.062262
2 0.195652
1 0.79348 0.92391 0.063822
3 0.050725
2 0.59783 0.63043 0.063822
4 0.043478
5 0.44565 0.64130 0.063990
5 0.032609
6 0.40217 0.57609 0.062777
6 0.010000
7 0.36957 0.51087 0.061056
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plotcp(rpart.model)
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> rsq.rpart(rpart.model)
Classification tree:
rpart(formula = Type ~ ., data = trainset)
Variables actually used in tree construction:
[1] Al Ba Mg RI
Root node error: 92/143 = 0.64336
n= 143
CP
nsplit rel error xerror xstd
1 0.206522
0 1.00000 1.00000 0.062262
2 0.195652
1 0.79348 0.92391 0.063822
3 0.050725
2 0.59783 0.63043 0.063822
4 0.043478
5 0.44565 0.64130 0.063990
5 0.032609
6 0.40217 0.57609 0.062777
6 0.010000
7 0.36957 0.51087 0.061056
Warning message:
In rsq.rpart(rpart.model) : may not be applicable for this method
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rsq.rpart
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> print(rpart.model)
n= 143
node), split, n, loss, yval, (yprob)
* denotes terminal node
1) root 143 92 2 (0.3 0.36 0.091 0.056 0.049 0.15)
2) Ba< 0.335 120 70 2 (0.35 0.42 0.11 0.058 0.058 0.0083)
4) Al< 1.42 71 33 1 (0.54 0.28 0.15 0.014 0.014 0)
8) RI>=1.517075 58 22 1 (0.62 0.28 0.086 0.017 0 0)
16) RI< 1.518015 21 1 1 (0.95 0 0.048 0 0 0) *
17) RI>=1.518015 37 21 1 (0.43 0.43 0.11 0.027 0 0)
34) RI>=1.51895 25 10 1 (0.6 0.2 0.16 0.04 0 0)
68) Mg>=3.415 18 4 1 (0.78 0 0.22 0 0 0) *
69) Mg< 3.415 7 2 2 (0.14 0.71 0 0.14 0 0) *
35) RI< 1.51895 12 1 2 (0.083 0.92 0 0 0 0) *
9) RI< 1.517075 13 7 3 (0.15 0.31 0.46 0 0.077 0) *
5) Al>=1.42 49 19 2 (0.082 0.61 0.041 0.12 0.12 0.02)
10) Mg>=2.62 33 6 2 (0.12 0.82 0.061 0 0 0) *
11) Mg< 2.62 16 10 5 (0 0.19 0 0.37 0.37 0.062) *
3) Ba>=0.335 23 3 7 (0.043 0.043 0 0.043 0 0.87) *
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Tree plot
> plot(rpart.model,compress=TRUE)
> text(rpart.model, use.n=TRUE)
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plot(object, uniform=FALSE, branch=1, compress=FALSE, nspace, margin=0, minbranch=.3, args)
And if you are brave
summary(rpart.model)
… pages….
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Remember to LOOK at the data
> names(Glass)
[1] "RI" "Na" "Mg" "Al" "Si" "K" "Ca" "Ba" "Fe"
"Type"
> head(Glass)
RI
Na Mg Al Si K
Ca Ba Fe Type
1 1.52101 13.64 4.49 1.10 71.78 0.06 8.75 0 0.00 1
2 1.51761 13.89 3.60 1.36 72.73 0.48 7.83 0 0.00 1
3 1.51618 13.53 3.55 1.54 72.99 0.39 7.78 0 0.00 1
4 1.51766 13.21 3.69 1.29 72.61 0.57 8.22 0 0.00 1
5 1.51742 13.27 3.62 1.24 73.08 0.55 8.07 0 0.00 1
6 1.51596 12.79 3.61 1.62 72.97 0.64 8.07 0 0.26 1
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rpart.pred
> rpart.pred
91 163 138 135 172 20 1 148 169 206 126 157 107 39 150
203 151 110 73 104 85 93 144 160 145 89 204 7 92 51
1 1 2 1 5 2 1 1 5 7 2 1 7 1 1 7 2 2 2 1
2 2 2 2 1 2 7 1 2 1
186 14 190 56 184 82 125 12 168 175 159 36 117 114 154
62 139 5 18 98 27 183 42 66 155 180 71 83 123 11
7 1 7 2 2 2 1 1 5 5 2 1 1 1 1 7 2 1 1 1
1 5 1 1 1 5 2 1 2 2
195 101 136 45 130 6 72 87 173 121 3
7 2 1 1 5 2 1 2 5 1 2
Levels: 1 2 3 5 6 7
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plot(rpart.pred)
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Cross-validation
• Cross-validation is a model validation
technique for assessing how the results of a
statistical analysis will generalize to an
independent data set.
• It is mainly used in settings where the goal is
prediction, and one wants to estimate how
accurately a predictive model will perform in
practice.
• I.e. predictive and prescriptive analytics…
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Cross-validation
In a prediction problem, a model is usually
given a dataset of known data on which training
is run (training dataset), and a dataset of
unknown data (or first seen data) against which
the model is tested (testing dataset).
Sound familiar?
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Cross-validation
The goal of cross validation is to define a
dataset to "test" the model in the training phase
(i.e., the validation dataset), in order to limit
problems like overfitting
And, give an insight on how the model will
generalize to an independent data set (i.e., an
unknown dataset, for instance from a real
problem), etc.
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Common type of x-validation
•
•
•
•
K-fold
2-fold (do you know this one?)
Rep-random-subsample
Leave out-subsample
• Lab in a few weeks … to try these out
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K-fold
• Original sample is randomly partitioned into k
equal size subsamples.
• Of the k subsamples, a single subsample is
retained as the validation data for testing the
model, and the remaining k − 1 subsamples
are used as training data.
• Repeat cross-validation process k times
(folds), with each of the k subsamples used
exactly once as the validation data.
– The k results from the folds can then be
averaged (usually) to produce a single
estimation.
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Leave out subsample
• As the name suggests, leave-one-out crossvalidation (LOOCV) involves using a single
observation from the original sample as the
validation data, and the remaining
observations as the training data.
• i.e. K=n-fold cross-validation
• Leave out > 1 = bootstraping and jackknifing
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boot(strapping)
• Generate replicates of a statistic applied to
data (parametric and nonparametric).
– nonparametric bootstrap, possible methods:
ordinary bootstrap, the balanced bootstrap,
antithetic resampling, and permutation.
• For nonparametric multi-sample problems
stratified resampling is used:
– this is specified by including a vector of strata in
the call to boot.
– importance resampling weights may be specified.
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Jackknifing
• Systematically recompute the statistic
estimate, leaving out one or more
observations at a time from the sample set
• From this new set of replicates of the statistic,
an estimate for the bias and an estimate for
the variance of the statistic can be calculated.
• Often use log(variance) [instead of variance]
especially for non-normal distributions
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Repeat-random-subsample
• Random split of the dataset into training and
validation data.
– For each such split, the model is fit to the training
data, and predictive accuracy is assessed using
the validation data.
• Results are then averaged over the splits.
• Note: for this method can the results will vary
if the analysis is repeated with different
random splits.
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Advantage?
• The advantage of K-fold over repeated
random sub-sampling is that all observations
are used for both training and validation, and
each observation is used for validation
exactly once.
– 10-fold cross-validation is commonly used
• The advantage of rep-random over k-fold
cross validation is that the proportion of the
training/validation split is not dependent on
the number of iterations (folds).
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Disadvantage
• The disadvantage of rep-random is that some
observations may never be selected in the
validation subsample, whereas others may be
selected more than once.
– i.e., validation subsets may overlap.
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New dataset to work with trees
fitK <- rpart(Kyphosis ~ Age + Number + Start, method="class",
data=kyphosis)
printcp(fitK) # display the results
plotcp(fitK) # visualize cross-validation results
summary(fitK) # detailed summary of splits
# plot tree
plot(fitK, uniform=TRUE, main="Classification Tree for
Kyphosis")
text(fitK, use.n=TRUE, all=TRUE, cex=.8)
# create attractive postscript plot of tree
post(fitK, file = “kyphosistree.ps", title = "Classification Tree for
Kyphosis") # might need to convert to PDF (distill)
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> pfitK<- prune(fitK, cp= fitK$cptable[which.min(fitK$cptable[,"xerror"]),"CP"])
> plot(pfitK, uniform=TRUE, main="Pruned Classification Tree for Kyphosis")
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> text(pfitK, use.n=TRUE, all=TRUE, cex=.8)
> post(pfitK, file = “ptree.ps", title = "Pruned Classification Tree for Kyphosis”)
> fitK <- ctree(Kyphosis ~ Age + Number + Start, data=kyphosis)
> plot(fitK, main="Conditional Inference Tree for Kyphosis”)
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> plot(fitK, main="Conditional Inference Tree for Kyphosis",type="simple")
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randomForest
> require(randomForest)
> fitKF <- randomForest(Kyphosis ~ Age + Number + Start, data=kyphosis)
> print(fitKF)
# view results
Call:
randomForest(formula = Kyphosis ~ Age + Number + Start, data = kyphosis)
Type of random forest: classification
Number of trees: 500
No. of variables tried at each split: 1
OOB estimate of error rate: 20.99%
Confusion matrix:
absent present class.error
absent
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5 0.0781250
present 12
5 0.7058824
> importance(fitKF) # importance of each predictor
MeanDecreaseGini
Age
8.654112
Number
5.584019
Start
10.168591
Random forests improve
predictive accuracy by
generating a large number of
bootstrapped trees (based on
random samples of variables),
classifying a case using each
tree in this new "forest", and
deciding a final predicted
outcome by combining the
results across all of the trees 51
(an average in regression, a
majority vote in classification).
Trees for the Titanic
data(Titanic)
rpart, ctree, hclust, etc. for Survived ~ .
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More on another dataset.
# Regression Tree Example
library(rpart)
# build the tree
fitM <- rpart(Mileage~Price + Country + Reliability + Type,
method="anova", data=cu.summary)
printcp(fitM) # display the results
….
Root node error: 1354.6/60 = 22.576
n=60 (57 observations deleted due to missingness)
CP nsplit rel error xerror xstd
1 0.622885
0 1.00000 1.03165 0.176920
2 0.132061
1 0.37711 0.51693 0.102454
3 0.025441
2 0.24505 0.36063 0.079819
4 0.011604
3 0.21961 0.34878 0.080273
5 0.010000
4 0.20801 0.36392 0.075650
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Mileage…
plotcp(fitM) # visualize cross-validation results
summary(fitM) # detailed summary of splits
<we will look more at this in a future lab>
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par(mfrow=c(1,2))
rsq.rpart(fitM) # visualize cross-validation results
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# plot tree
plot(fitM, uniform=TRUE, main="Regression Tree for
Mileage ")
text(fitM, use.n=TRUE, all=TRUE, cex=.8)
# prune the tree
pfitM<- prune(fitM, cp=0.01160389) # from cptable
# plot the pruned tree
plot(pfitM, uniform=TRUE, main="Pruned Regression
Tree for Mileage")
text(pfitM, use.n=TRUE, all=TRUE, cex=.8)
post(pfitM, file = ”ptree2.ps", title = "Pruned
Regression Tree for Mileage”)
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# Conditional Inference Tree for Mileage
fit2M <- ctree(Mileage~Price + Country +
Reliability + Type, data=na.omit(cu.summary))
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Enough of trees!
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Assignment 6 on Website – later today
• Your term projects should fall within the scope of a data analytics
problem of the type you have worked with in class/ labs, or know of
yourself – the bigger the data the better. This means that the work must
go beyond just making lots of figures. You should develop the project to
indicate you are thinking of and exploring the relationships and
distributions within your data. Start with a hypothesis, think of a way to
model and use the hypothesis, find or collect the necessary data, and do
both preliminary analysis, detailed modeling and summary
(interpretation). Grad students must develop two types of models.
– Note: You do not have to come up with a positive result, i.e. disproving the hypothesis
is just as good.
• Introduction (2%) % may change…
•
•
•
•
•
Data Description (3%)
Analysis (5%)
Model Development (12%)
Conclusions and Discussion (3%)
Oral presentation (5%) (~5 mins)
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Coming weeks
• Spring break – March 14 – 18
• Your project proposals (Assignment 5) are on
March 22/25. Come prepared (5% grade).
• Support Vector Machines = the
“red pill”?
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