Xgboost

XGBoost
eXtremeGradientBoosting
Tong He

Overview
Introduction
Basic Walkthrough
Real World Application
Model Specification
Parameter Introduction
Advanced Features
Kaggle Winning Solution
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Introduction
Nowadays we have plenty of machine learning models. Those most well-knowns are
Linear/Logistic Regression
k-Nearest Neighbours
Support Vector Machines
Tree-based Model
Neural Networks
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Decision Tree
Random Forest
Gradient Boosting Machine
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Introduction
XGBoost is short for eXtreme Gradient Boosting. It is
An open-sourced tool
A variant of the gradient boosting machine
The winning model for several kaggle competitions
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Computation in C++
R/python/Julia interface provided
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Tree-based model-
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Introduction
XGBoost is currently host on github.
The primary author of the model and the c++ implementation is Tianqi Chen.
The author for the R-package is Tong He.
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Introduction
XGBoost is widely used for kaggle competitions. The reason to choose XGBoost includes
Easy to use
Efficiency
Accuracy
Feasibility
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Easy to install.
Highly developed R/python interface for users.
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Automatic parallel computation on a single machine.
Can be run on a cluster.
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Good result for most data sets.-
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Customized objective and evaluation
Tunable parameters
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Basic Walkthrough
We introduce the R package for XGBoost. To install, please run
This command downloads the package from github and compile it automatically on your
machine. Therefore we need RTools installed on Windows.
devtools::install_github('dmlc/xgboost',subdir='R-package')
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Basic Walkthrough
XGBoost provides a data set to demonstrate its usages.
This data set includes the information for some kinds of mushrooms. The features are binary,
indicate whether the mushroom has this characteristic. The target variable is whether they are
poisonous.
require(xgboost)
##Loadingrequiredpackage:xgboost
data(agaricus.train,package='xgboost')
data(agaricus.test,package='xgboost')
train=agaricus.train
test=agaricus.test
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Basic Walkthrough
Let's investigate the data first.
We can see that the data is a dgCMatrixclass object. This is a sparse matrix class from the
package Matrix. Sparse matrix is more memory efficient for some specific data.
str(train$data)
##Formalclass'dgCMatrix'[package"Matrix"]with6slots
## ..@i :int[1:143286]26811182021242832...
## ..@p :int[1:127]036937233065845648965138380838410991...
## ..@Dim :int[1:2]6513126
## ..@Dimnames:Listof2
## ....$:NULL
## ....$:chr[1:126]"cap-shape=bell""cap-shape=conical""cap-shape=convex""cap-shape=flat
## ..@x :num[1:143286]1111111111...
## ..@factors:list()
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Basic Walkthrough
To use XGBoost to classify poisonous mushrooms, the minimum information we need to
provide is:
1. Input features
2. Target variable
3. Objective
4. Number of iteration
XGBoost allows dense and sparse matrix as the input.·
A numeric vector. Use integers starting from 0 for classification, or real values for
regression
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For regression use 'reg:linear'
For binary classification use 'binary:logistic'
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The number of trees added to the model·
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Basic Walkthrough
To run XGBoost, we can use the following command:
The output is the classification error on the training data set.
bst=xgboost(data=train$data,label=train$label,
nround=2,objective="binary:logistic")
##[0] train-error:0.000614
##[1] train-error:0.001228
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Basic Walkthrough
Sometimes we might want to measure the classification by 'Area Under the Curve':
bst=xgboost(data=train$data,label=train$label,nround=2,
objective="binary:logistic",eval_metric="auc")
##[0] train-auc:0.999238
##[1] train-auc:0.999238
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Basic Walkthrough
To predict, you can simply write
pred=predict(bst,test$data)
head(pred)
##[1]0.25824980.74332210.25824980.25824980.25765090.2750908
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Basic Walkthrough
Cross validation is an important method to measure the model's predictive power, as well as
the degree of overfitting. XGBoost provides a convenient function to do cross validation in a line
of code.
Notice the difference of the arguments between xgb.cvand xgboostis the additional nfold
parameter. To perform cross validation on a certain set of parameters, we just need to copy
them to the xgb.cvfunction and add the number of folds.
cv.res=xgb.cv(data=train$data,nfold=5,label=train$label,nround=2,
objective="binary:logistic",eval_metric="auc")
##[0] train-auc:0.998780+0.000590test-auc:0.998547+0.000854
##[1] train-auc:0.999114+0.000728test-auc:0.998736+0.001072
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Basic Walkthrough
xgb.cvreturns a data.tableobject containing the cross validation results. This is helpful for
choosing the correct number of iterations.
cv.res
## train.auc.meantrain.auc.stdtest.auc.meantest.auc.std
##1: 0.998780 0.000590 0.998547 0.000854
##2: 0.999114 0.000728 0.998736 0.001072
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Higgs Boson Competition
The debut of XGBoost was in the higgs boson competition.
Tianqi introduced the tool along with a benchmark code which achieved the top 10% at the
beginning of the competition.
To the end of the competition, it was already the mostly used tool in that competition.
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XGBoost offers the script on github.
To run the script, prepare a datadirectory and download the competition data into this
directory.
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Firstly we prepare the environment
require(xgboost)
require(methods)
testsize=550000
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Then we can read in the data
dtrain=read.csv("data/training.csv",header=TRUE)
dtrain[33]=dtrain[33]=="s"
label=as.numeric(dtrain[[33]])
data=as.matrix(dtrain[2:31])
weight=as.numeric(dtrain[[32]])*testsize/length(label)
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The data contains missing values and they are marked as -999. We can construct an
xgb.DMatrixobject containing the information of weightand missing.
xgmat=xgb.DMatrix(data,label=label,weight=weight,missing=-999.0)
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Next step is to set the basic parameters
param=list("objective"="binary:logitraw",
"scale_pos_weight"=sumwneg/sumwpos,
"bst:eta"=0.1,
"bst:max_depth"=6,
"eval_metric"="auc",
"eval_metric"="ams@0.15",
"silent"=1,
"nthread"=16)
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We then start the training step
bst=xgboost(params=param,data=xgmat,nround=120)
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Then we read in the test data
dtest=read.csv("data/test.csv",header=TRUE)
data=as.matrix(dtest[2:31])
xgmat=xgb.DMatrix(data,missing=-999.0)
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We now can make prediction on the test data set.
ypred=predict(bst,xgmat)
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Finally we output the prediction according to the required format.
Please submit the result to see your performance :)
rorder=rank(ypred,ties.method="first")
threshold=0.15
ntop=length(rorder)-as.integer(threshold*length(rorder))
plabel=ifelse(rorder>ntop,"s","b")
outdata=list("EventId"=idx,
"RankOrder"=rorder,
"Class"=plabel)
write.csv(outdata,file="submission.csv",quote=FALSE,row.names=FALSE)
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Besides the good performace, the efficiency is also a highlight of XGBoost.
The following plot shows the running time result on the Higgs boson data set.
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After some feature engineering and parameter tuning, one can achieve around 25th with a
single model on the leaderboard. This is an article written by a former-physist introducing his
solution with a single XGboost model.
On our post-competition attempts, we achieved 11th on the leaderboard with a single XGBoost
model.
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Training Objective
To understand other parameters, one need to have a basic understanding of the model behind.
Suppose we have trees, the model is
where each is the prediction from a decision tree. The model is a collection of decision trees.
K
∑
k=1
K
f
k
f
k
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Training Objective
Having all the decision trees, we make prediction by
where is the feature vector for the -th data point.
Similarly, the prediction at the -th step can be defined as
= ( )y
i
ˆ
∑
k=1
K
f
k
xi
xi
i
t
= ( )yi
ˆ (t)
∑
k=1
t
f
k
xi
/

Training Objective
To train the model, we need to optimize a loss function.
Typically, we use
Rooted Mean Squared Error for regression
LogLoss for binary classification
mlogloss for multi-classification
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- L = ( −1
N
∑N
i=1
y
i
y
i
ˆ )
2
·
- L = − ( log( ) + (1 − ) log(1 − ))
1
N
∑N
i=1
y
i
p
i
y
i
p
i
·
- L = − log( )
1
N
∑N
i=1
∑M
j=1
y
i,j
p
i,j
/

Training Objective
Regularization is another important part of the model. A good regularization term controls the
complexity of the model which prevents overfitting.
Define
where is the number of leaves, and is the score on the -th leaf.
Ω = γT + λ
1
2 ∑
j=1
T
w
2
j
T w
2
j
j
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Training Objective
Put loss function and regularization together, we have the objective of the model:
where loss function controls the predictive power, and regularization controls the simplicity.
Obj = L + Ω
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Training Objective
In XGBoost, we use gradient descent to optimize the objective.
Given an objective to optimize, gradient descent is an iterative technique which
calculate
at each iteration. Then we improve along the direction of the gradient to minimize the
objective.
Obj(y, )yˆ
Obj(y, )∂yˆ yˆ
yˆ
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Training Objective
Recall the definition of objective . For a iterative algorithm we can re-define the
objective function as
To optimize it by gradient descent, we need to calculate the gradient. The performance can also
be improved by considering both the first and the second order gradient.
Obj = L + Ω
Ob = L( , ) + Ω( ) = L( , + ( )) + Ω( )j
(t)
∑
i=1
N
y
i
yˆ
(t)
i
∑
i=1
t
f
i
∑
i=1
N
y
i
yi
ˆ (t−1)
f
t
xi
∑
i=1
t
f
i
Ob∂y
i
ˆ (t) j
(t)
Ob∂2
y
i
ˆ
(t) j
(t)
/

Training Objective
Since we don't have derivative for every objective function, we calculate the second order taylor
approximation of it
where
Ob ≃ [L( , ) + ( ) + ( )] + Ω( )j
(t)
∑
i=1
N
y
i
yˆ(t−1)
g
i
f
t
xi
1
2
hi f
2
t
xi
∑
i=1
t
f
i
· = l( , )g
i
∂yˆ(t−1) y
i
yˆ(t−1)
· = l( , )hi ∂2
yˆ
(t−1) y
i
yˆ(t−1)
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Training Objective
Remove the constant terms, we get
This is the objective at the -th step. Our goal is to find a to optimize it.
Ob = [ ( ) + ( )] + Ω( )j
(t)
∑
i=1
n
g
i
f
t
xi
1
2
hi f
2
t
xi f
t
t f
t
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Tree Building Algorithm
The tree structures in XGBoost leads to the core problem:
how can we find a tree that improves the prediction along the gradient?
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Every decision tree looks like this
Each data point flows to one of the leaves following the direction on each node.
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The core concepts are:
Internal Nodes
Leaves
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Each internal node split the flow of data points by one of the features.
The condition on the edge specifies what data can flow through.
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Data points reach to a leaf will be assigned a weight.
The weight is the prediction.
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Two key questions for building a decision tree are
1. How to find a good structure?
2. How to assign prediction score?
We want to solve these two problems with the idea of gradient descent.
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Let us assume that we already have the solution to question 1.
We can mathematically define a tree as
where is a "directing" function which assign every data point to the -th leaf.
This definition describes the prediction process on a tree as
(x) =f
t
wq(x)
q(x) q(x)
Assign the data point to a leaf by
Assign the corresponding score on the -th leaf to the data point.
· x q
· wq(x)
q(x)
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Define the index set
This set contains the indices of data points that are assigned to the -th leaf.
= {i|q( ) = j}Ij xi
j
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Then we rewrite the objective as
Since all the data points on the same leaf share the same prediction, this form sums the
prediction by leaves.
Ob = [ ( ) + ( )] + γT + λj
(t)
∑
i=1
n
gi
ft
xi
1
2
hi f
2
t
xi
1
2 ∑
j=1
T
w
2
j
= [( ) + ( + λ) ] + γT
∑
j=1
T
∑
i∈Ij
g
i
wj
1
2 ∑
i∈Ij
hi w
2
j
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It is a quadratic problem of , so it is easy to find the best to optimize .
The corresponding value of is
wj
wj
Obj
= −w
∗
j
∑i∈Ij
g
i
+ λ∑i∈Ij
hi
Obj
Ob = − + γTj
(t)
1
2 ∑
j=1
T
(∑i∈Ij
g
i
)
2
+ λ∑i∈Ij
hi
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The leaf score
relates to
= −wj
∑i∈Ij
g
i
+ λ∑i∈Ij
hi
The first and second order of the loss function and
The regularization parameter
· g h
· λ
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Now we come back to the first question: How to find a good structure?
We can further split it into two sub-questions:
1. How to choose the feature to split?
2. When to stop the split?
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In each split, we want to greedily find the best splitting point that can optimize the objective.
For each feature
1. Sort the numbers
2. Scan the best splitting point.
3. Choose the best feature.
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Now we give a definition to "the best split" by the objective.
Everytime we do a split, we are changing a leaf into a internal node.
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Let
Recall the best value of objective on the -th leaf is
be the set of indices of data points assigned to this node
and be the sets of indices of data points assigned to two new leaves.
· I
· IL
IR
j
Ob = − + γj
(t)
1
2
(∑i∈Ij
g
i
)
2
+ λ∑i∈Ij
hi
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The gain of the split is
gain =
[
+ −
]
− γ
1
2
(∑i∈IL
gi
)
2
+ λ∑i∈IL
hi
(∑i∈IR
gi
)
2
+ λ∑i∈IR
hi
(∑i∈I
g
i
)
2
+ λ∑i∈I
hi
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To build a tree, we find the best splitting point recursively until we reach to the maximum
depth.
Then we prune out the nodes with a negative gain in a bottom-up order.
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XGBoost can handle missing values in the data.
For each node, we guide all the data points with a missing value
Finally every node has a "default direction" for missing values.
to the left subnode, and calculate the maximum gain
to the right subnode, and calculate the maximum gain
Choose the direction with a larger gain
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To sum up, the outline of the algorithm is
Iterate for nroundtimes·
Grow the tree to the maximun depth
Prune the tree to delete nodes with negative gain
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Find the best splitting point
Assign weight to the two new leaves
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XGBoost has plenty of parameters. We can group them into
1. General parameters
2. Booster parameters
3. Task parameters
Number of threads·
Stepsize
Regularization
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Objective
Evaluation metric
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After the introduction of the model, we can understand the parameters provided in XGBoost.
To check the parameter list, one can look into
The documentation of xgb.train.
The documentation in the repository.
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General parameters:
nthread
booster
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Number of parallel threads.-
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gbtree: tree-based model.
gblinear: linear function.
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Parameter for Tree Booster
eta
gamma
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Step size shrinkage used in update to prevents overfitting.
Range in [0,1], default 0.3
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Minimum loss reduction required to make a split.
Range [0, ], default 0
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max_depth
min_child_weight
max_delta_step
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Maximum depth of a tree.
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- ∞
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Minimum sum of instance weight needed in a child.
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Maximum delta step we allow each tree's weight estimation to be.
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- ∞
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subsample
colsample_bytree
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Subsample ratio of the training instance.
Range (0, 1], default 1
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Subsample ratio of columns when constructing each tree.
Range (0, 1], default 1
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Parameter for Linear Booster
lambda
alpha
lambda_bias
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L2 regularization term on weights
default 0
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L1 regularization term on weights
default 0
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L2 regularization term on bias
default 0
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objectives·
"reg:linear": linear regression, default option.
"binary:logistic": logistic regression for binary classification, output probability
"multi:softmax": multiclass classification using the softmax objective, need to specify
num_class
User specified objective
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eval_metric·
"rmse"
"logloss"
"error"
"auc"
"merror"
"mlogloss"
User specified evaluation metric
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Guide on Parameter Tuning
It is nearly impossible to give a set of universal optimal parameters, or a global algorithm
achieving it.
The key pointsof parameter tuning are
Control Overfitting
Deal with Imbalanced data
Trust the cross validation
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The "Bias-Variance Tradeoff", or the "Accuracy-Simplicity Tradeoff" is the main idea for
controlling overfitting.
For the booster specific parameters, we can group them as
Controlling the model complexity
Robust to noise
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max_depth, min_child_weightand gamma-
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subsample, colsample_bytree-
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Sometimes the data is imbalanced among classes.
Only care about the ranking order
Care about predicting the right probability
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Balance the positive and negative weights, by scale_pos_weight
Use "auc" as the evaluation metric
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Cannot re-balance the dataset
Set parameter max_delta_stepto a finite number (say 1) will help convergence
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To select ideal parameters, use the result from xgb.cv.
Trust the score for the test
Use early.stop.roundto detect continuously being worse on test set.
If overfitting observed, reduce stepsize etaand increase nroundat the same time.
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Advanced Features
There are plenty of highlights in XGBoost:
Customized objective and evaluation metric
Prediction from cross validation
Continue training on existing model
Calculate and plot the variable importance
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Customization
According to the algorithm, we can define our own loss function, as long as we can calculate the
first and second order gradient of the loss function.
Define and . We can optimize the loss function if we can calculate
these two values.
grad = l∂y
t−1 hess = l∂2
y
t−1
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Customization
We can rewrite logloss for the -th data point as
The is calculated by applying logistic transformation on our prediction .
Then the logloss is
i
L = log( ) + (1 − ) log(1 − )y
i
p
i
y
i
p
i
p
i
yî
L = log + (1 − ) logy
i
1
1 + e
−yî
y
i
e
−yî
1 + e
−yî
/

Customization
We can see that
Next we translate them into the code.
· grad = − = −1
1+e
−yˆ
i
y
i
p
i
y
i
· hess = = (1 − )
1+e
−yˆ
i
(1+e
−yˆ
i )
2
p
i
p
i
/

Customization
The complete code:
logregobj=function(preds,dtrain){
labels=getinfo(dtrain,"label")
preds=1/(1+exp(-preds))
grad=preds-labels
hess=preds*(1-preds)
return(list(grad=grad,hess=hess))
}
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Customization
The complete code:
#Extractthetruelabelfromthesecondargument
grad=preds-labels
}
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Customization
The complete code:
#applylogistictransformationtotheoutput
grad=preds-labels
}
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Customization
The complete code:
#Calculatethe1stgradient
grad=preds-labels
}
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Customization
The complete code:
grad=preds-labels
#Calculatethe2ndgradient
}
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Customization
The complete code:
grad=preds-labels
#Calculatethe2ndgradient
#Returntheresult
}
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Customization
We can also customize the evaluation metric.
evalerror=function(preds,dtrain){
err=as.numeric(sum(labels!=(preds>0)))/length(labels)
return(list(metric="error",value=err))
}
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Customization
}
/

Customization
#Calculatetheerror
}
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Customization
#Calculatetheerror
#Returnthenameofthismetricandthevalue
}
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Customization
To utilize the customized objective and evaluation, we simply pass them to the arguments:
param=list(max.depth=2,eta=1,nthread=2,silent=1,
objective=logregobj,eval_metric=evalerror)
bst=xgboost(params=param,data=train$data,label=train$label,nround=2)
##[0] train-error:0.0465223399355136
##[1] train-error:0.0222631659757408
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Prediction in Cross Validation
"Stacking" is an ensemble learning technique which takes the prediction from several models. It
is widely used in many scenarios.
One of the main concern is avoid overfitting. The common way is use the prediction value from
cross validation.
XGBoost provides a convenient argument to calculate the prediction during the cross
validation.
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Prediction in Cross Validation
res=xgb.cv(params=param,data=train$data,label=train$label,nround=2,
nfold=5,prediction=TRUE)
##[0] train-error:0.046522+0.001041 test-error:0.046523+0.004164
str(res)
##Listof2
## $dt :Classes'data.table'and'data.frame': 2obs.of 4variables:
## ..$train.error.mean:num[1:2]0.04650.0223
## ..$train.error.std:num[1:2]0.001040.00122
## ..$test.error.mean:num[1:2]0.04650.0223
## ..$test.error.std :num[1:2]0.004160.00488
## ..-attr(*,".internal.selfref")=<externalptr>
## $pred:num[1:6513]2.58-1.04-1.122.57-3.04...
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xgb.DMatrix
XGBoost has its own class of input data xgb.DMatrix. One can convert the usual data set into
it by
It is the data structure used by XGBoost algorithm. XGBoost preprocess the input dataand
labelinto an xgb.DMatrixobject before feed it to the training algorithm.
If one need to repeat training process on the same big data set, it is good to use the
xgb.DMatrixobject to save preprocessing time.
dtrain=xgb.DMatrix(data=train$data,label=train$label)
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xgb.DMatrix
An xgb.DMatrixobject contains
1. Preprocessed training data
2. Several features
Missing values
data weight
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·
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Continue Training
Train the model for 5000 rounds is sometimes useful, but we are also taking the risk of
overfitting.
A better strategy is to train the model with fewer rounds and repeat that for many times. This
enable us to observe the outcome after each step.
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Continue Training
bst=xgboost(params=param,data=dtrain,nround=1)
##[0] train-error:0.0465223399355136
ptrain=predict(bst,dtrain,outputmargin=TRUE)
setinfo(dtrain,"base_margin",ptrain)
##[1]TRUE
##[0] train-error:0.0222631659757408
/

Continue Training
#Trainwithonlyoneround
##[0] train-error:0.0222631659757408
##[1]TRUE
##[0] train-error:0.00706279748195916
/

Continue Training
##[0] train-error:0.00706279748195916
#marginmeansthebaselineoftheprediction
##[1]TRUE
##[0] train-error:0.0152003684937817
/

Continue Training
##[0] train-error:0.0152003684937817
#Setthemargininformationtothexgb.DMatrixobject
##[1]TRUE
##[0] train-error:0.00706279748195916
/

Continue Training
##[0] train-error:0.00706279748195916
#Setthemargininformationtothexgb.DMatrixobject
##[1]TRUE
#Trainbasedonthepreviousresult
##[0] train-error:0.00122831260555811
/

Importance and Tree plotting
The result of XGBoost contains many trees. We can count the number of appearance of each
variable in all the trees, and use this number as the importance score.
bst=xgboost(data=train$data,label=train$label,max.depth=2,verbose=FALSE,
eta=1,nthread=2,nround=2,objective="binary:logistic")
xgb.importance(train$dataDimnames[[2]],model=bst)
## Feature Gain CoverFrequence
##1: 280.676154840.4978746 0.4
##2: 550.171353520.1920543 0.2
##3: 590.123172410.1638750 0.2
##4: 1080.029319220.1461960 0.2
/

Importance and Tree plotting
We can also plot the trees in the model by xgb.plot.tree.
xgb.plot.tree(agaricus.train$data@Dimnames[[2]],model=bst)
/

Early Stopping
When doing cross validation, it is usual to encounter overfitting at a early stage of iteration.
Sometimes the prediction gets worse consistantly from round 300 while the total number of
iteration is 1000. To stop the cross validation process, one can use the early.stop.round
argumetn in xgb.cv.
bst=xgb.cv(params=param,data=train$data,label=train$label,
nround=20,nfold=5,
maximize=FALSE,early.stop.round=3)
/

Early Stopping
##[10]train-error:0.000422+0.000582 test-error:0.000768+0.001086
##Stopping.Bestiteration:12
/

To get a higher rank, one need to push the limit of
1. Feature Engineering
2. Parameter Tuning
3. Model Ensemble
The winning solution in the recent Otto Competition is an excellent example.
/

Then they used a 3-layer ensemble learning model, including
33 models on top of the original data
XGBoost, neural network and adaboost on 33 predictions from the models and 8 engineered
features
Weighted average of the 3 prediction from the second step
·
·
·
/

The data for this competition is special: the meanings of the featuers are hidden.
For feature engineering, they generated 8 new features:
Distances to nearest neighbours of each classes
Sum of distances of 2 nearest neighbours of each classes
Sum of distances of 4 nearest neighbours of each classes
Distances to nearest neighbours of each classes in TFIDF space
Distances to nearest neighbours of each classed in T-SNE space (3 dimensions)
Clustering features of original dataset
Number of non-zeros elements in each row
X (That feature was used only in NN 2nd level training)
·
·
·
·
·
·
·
·
/

This means a lot of work. However this also implies they need to try a lot of other models,
although some of them turned out to be not helpful in this competition. Their attempts include:
A lot of training algorithms in first level as
Some preprocessing like PCA, ICA and FFT
Feature Selection
Semi-supervised learning
·
Vowpal Wabbit(many configurations)
R glm, glmnet, scikit SVC, SVR, Ridge, SGD, etc...
-
-
·
·
·
/

Influencers in Social Networks
Let's learn to use a single XGBoost model to achieve a high rank in an old competition!
The competition we choose is the Influencers in Social Networks competition.
It was a hackathon in 2013, therefore the size of data is small enough so that we can train the
model in seconds.
/

First let's download the data, and load them into R
train=read.csv('train.csv',header=TRUE)
test=read.csv('test.csv',header=TRUE)
y=train[,1]
train=as.matrix(train[,-1])
test=as.matrix(test)
/

Observe the data:
colnames(train)
## [1]"A_follower_count" "A_following_count" "A_listed_count"
## [4]"A_mentions_received""A_retweets_received""A_mentions_sent"
## [7]"A_retweets_sent" "A_posts" "A_network_feature_1"
##[10]"A_network_feature_2""A_network_feature_3""B_follower_count"
##[13]"B_following_count" "B_listed_count" "B_mentions_received"
##[16]"B_retweets_received""B_mentions_sent" "B_retweets_sent"
##[19]"B_posts" "B_network_feature_1""B_network_feature_2"
##[22]"B_network_feature_3"
/

Observe the data:
train[1,]
## A_follower_count A_following_count A_listed_count
## 2.280000e+02 3.020000e+02 3.000000e+00
##A_mentions_receivedA_retweets_received A_mentions_sent
## 5.839794e-01 1.005034e-01 1.005034e-01
## A_retweets_sent A_postsA_network_feature_1
## 1.005034e-01 3.621501e-01 2.000000e+00
##A_network_feature_2A_network_feature_3 B_follower_count
## 1.665000e+02 1.135500e+04 3.446300e+04
## B_following_count B_listed_countB_mentions_received
## 2.980800e+04 1.689000e+03 1.543050e+01
##B_retweets_received B_mentions_sent B_retweets_sent
## 3.984029e+00 8.204331e+00 3.324230e-01
## B_postsB_network_feature_1B_network_feature_2
## 6.988815e+00 6.600000e+01 7.553030e+01
##B_network_feature_3
## 1.916894e+03
/

The data contains information from two users in a social network service. Our mission is to
determine who is more influencial than the other one.
This type of data gives us some room for feature engineering.
/

The first trick is to increase the information in the data.
Every data point can be expressed as . Actually it indicates <1-y, B, A> as well. We can simply
use extract this part of information from the training set.
new.train=cbind(train[,12:22],train[,1:11])
train=rbind(train,new.train)
y=c(y,1-y)
/

The following feature engineering steps are done on both training and test set. Therefore we
combine them together.
x=rbind(train,test)
/

The next step could be calculating the ratio between features of A and B seperately:
followers/following
mentions received/sent
retweets received/sent
followers/posts
retweets received/posts
mentions received/posts
·
·
·
·
·
·
/

Considering there might be zeroes, we need to smooth the ratio by a constant.
Next we can calculate the ratio with this helper function.
calcRatio=function(dat,i,j,lambda=1)(dat[,i]+lambda)/(dat[,j]+lambda)
/

A.follow.ratio=calcRatio(x,1,2)
A.mention.ratio=calcRatio(x,4,6)
A.retweet.ratio=calcRatio(x,5,7)
A.follow.post=calcRatio(x,1,8)
A.mention.post=calcRatio(x,4,8)
A.retweet.post=calcRatio(x,5,8)
B.follow.ratio=calcRatio(x,12,13)
B.mention.ratio=calcRatio(x,15,17)
B.retweet.ratio=calcRatio(x,16,18)
B.follow.post=calcRatio(x,12,19)
B.mention.post=calcRatio(x,15,19)
B.retweet.post=calcRatio(x,16,19)
/

Combine the features into the data set.
x=cbind(x[,1:11],
A.follow.ratio,A.mention.ratio,A.retweet.ratio,
A.follow.post,A.mention.post,A.retweet.post,
x[,12:22],
B.follow.ratio,B.mention.ratio,B.retweet.ratio,
B.follow.post,B.mention.post,B.retweet.post)
/

Then we can compare the difference between A and B. Because XGBoost is scale invariant,
therefore minus and division are the essentially same.
AB.diff=x[,1:17]-x[,18:34]
x=cbind(x,AB.diff)
train=x[1:nrow(train),]
test=x[-(1:nrow(train)),]
/

Now comes to the modeling part. We first investigate how far can we gowith a single model.
The parameter tuning step is very important in this step. We can see the performance from
cross validation.
/

Here's the xgb.cvwith default parameters.
set.seed(1024)
cv.res=xgb.cv(data=train,nfold=3,label=y,nrounds=100,verbose=FALSE,
objective='binary:logistic',eval_metric='auc')
/

We can see the trend of AUC on training and test sets.
/

It is obvious our model severly overfits. The direct reason is simple: the default value of etais
0.3, which is too large for this mission.
Recall the parameter tuning guide, we need to decrease etaand inccrease nroundsbased on
the result of cross validation.
/

After some trials, we get the following set of parameters:
set.seed(1024)
cv.res=xgb.cv(data=train,nfold=3,label=y,nrounds=3000,
objective='binary:logistic',eval_metric='auc',
eta=0.005,gamma=1,lambda=3,nthread=8,
max_depth=4,min_child_weight=1,verbose=F,
subsample=0.8,colsample_bytree=0.8)
/

Next we extract the best number of iterations.
We calculate the AUC minus the standard deviation, and choose the iteration with the largest
value.
bestRound=which.max(as.matrix(cv.res)[,3]-as.matrix(cv.res)[,4])
bestRound
##[1]2442
cv.res[bestRound,]
## train.auc.meantrain.auc.stdtest.auc.meantest.auc.std
##1: 0.934967 0.00125 0.876629 0.002073
/

Then we train the model with the same set of parameters:
set.seed(1024)
bst=xgboost(data=train,label=y,nrounds=3000,
objective='binary:logistic',eval_metric='auc',
eta=0.005,gamma=1,lambda=3,nthread=8,
max_depth=5,min_child_weight=1,
subsample=0.8,colsample_bytree=0.8)
preds=predict(bst,test,ntreelimit=bestRound)
/

Finally we submit our solution
This wins us into top 10 on the leaderboard!
result=data.frame(Id=1:nrow(test),
Choice=preds)
write.csv(result,'submission.csv',quote=FALSE,row.names=FALSE)
/

Xgboost

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