The International Archives of the Photogrammetry, Remote Sensing and Spatial Information Sciences. Vol. XXXVII. Part B4. Beijing 2008 
1161 
SIj = — 
Where, SI,is the shape index; P, is the perimeter; A, is the area 
of image object i. 
The number of features used in this study is summarized in 
Table 1: 
Features 
The number of features 
Mean value 
3 
Mean difference to scene 
3 
Mean difference to 
neighbour 
3 
Standard Deviation 
3 
Area 
1 
Shape Index 
1 
Table 1. Features and the number of features used in one-class 
SVMs 
The values of features vary significantly. For example, the area 
of objects ranges from 1 to 10650 while the shape index ranges 
from 1 to 8.73. Therefore, before we implemented support 
vector machines to classify the objects, we firstly normalized 
the features by using a maximum and minim method, and the 
final values for each feature range from 0 - 1. 
2.2.3 Feature selection: The feature selection aims to select a 
subset of the features that could perform well in the 
classification algorithms. It could reduce the dimensionality of 
the feature space while still maintain high classification 
accuracy and avoid over-fitting. In this study, we used the 
brute-force search algorithm (also referred to as the exhaustive 
search), which searches for all of the possible combinations of 
features and finds the optimum solution for the classifier. 
Assuming the number of features is n, the number of possible 
combination of features will be: 
C 1 + C 2 + C 3 + ...+ C n = 2" -1 
n n n n 
Since there are 14 features in this study, which results in 16383 
possible combinations. The computational calculation is 
acceptable with current PC (the computer used in this study is 
Intel CPU 2.4, and takes approximately 2.6 hours to search all 
possible combinations as well as implemented one-class SVMs). 
Although the brute-force algorithm is very time-consuming, it 
solves the optimization problem directly and always finds the 
solution if it exists. However, it should be noted that when the 
number of features increase, the brutal search method could 
become too time-consuming as its cost is proportional to the 
number of candidate features. In these cases, other feature 
selection methods can be more appropriate (e.g. the stepwise 
method). The Kappa values and cross validation methods are 
used to evaluate model performance with difference feature 
combinations. 
2.2.4 Accuracy assessment: Ground truth data were acquired 
by manual photo interpretation. Totally, 100 house objects and 
20 non-house objects were selected. For one class support 
vector machines, we only used house data for training, and both 
house and non-housing data to evaluate the classification results. 
We used the Kappa value to measure the classification accuracy, 
Kappa values take into account an agreement that can occur by 
change (expected agreement). In General, Kappa values of 0.75 
and higher are considered good classification results (Eric et al 
2004). A five-fold cross-validation method is used to evaluate 
the model performance based on the training data. The cross 
validation method is implemented as follows (Guo et al. 2007): 
(1) Select 100 houses as set H and 20 non houses as set H n as 
training set for cross validation. 
(2) Randomly split set H into five subsets, i.e., every subset 
contains 20 houses. Every subset H in combination with H„ in 
turn composes of set T e . Other four subsets in set H compose of 
set T r . 
(3) The set T r is used as training samples; the set T e is used as 
testing samples to evaluate the model performance of one-class 
SVMs. 
(4) The Kappa value is calculated for each iteration. The 
average Kappa value is reported based on five implements. 
2.2.5. Implement SVMs for object-based classification 
Assuming we have / training points X ; (i=1, 2... /), we want 
to find a hypersphere as small as possible to contain the training 
points in multidimensional space. Meanwhile, we also allow a 
small portion of outliers to exist using a slack variable (): 
(1) 
Subject to: (x,. - c) T {x, - c) < R 2 + <£, > 0 for all i e [/] (2) 
Where c and R are the center and radius of the sphere, T is the 
transpose of a matrix, and V G (0, 1] is the trade-off between 
the volume of the sphere and the number of training points 
rejected. When V is large, the volume of the sphere is small 
thus more training points will be rejected than when V is small, 
where more training points will be contained within the sphere. 
V can be roughly explained as the percentage of outliers in the 
training dataset (Scholkopf et al., 2001). This optimization 
problem can be solved by the Lagrangian: 
L(R, £,c, a,, A ) = R 2 + ■- £ a, {/? 2 + £ - (x,. 2 - 2cx, +c 2 )} 
VI /=i ,=i 
-£m (3) 
/=1 
Where a ( . > 0 and > 0. Setting the partial derivative of L 
with respect to R, Cl i , c equal to 0, we get: 
i 
1^=1 
i=i 
(4) 
1 
0 < a, < — 
(5) 
vl 
i 
C = X a i X i 
(6) 
Substituting equation (4)-(6) to equation (3), we have the dual 
problem: 
minZ/j a > a j(*/ x j)~ L a i( x i • */) ( ? ) 
1 J-, 
Subject to: 0 < a, < — , = 1 
v/ ,-i 
To determine whether or not a test point ( X ) is within the 
sphere, we can calculate the distance between the test point and 
the center C. It can be expressed as: 
(x ■ x) - 2£ a, (x • x,.) + X afljO, •Xj)<R 2 (8) 
So far, we have assumed that the data are spherically distributed. 
In reality, the data are often not spherically distributed. To 
make the method more flexible and to capture the non-linearity 
such as multi-mode distribution, the kernel function K(x h Xj) can 
be introduced. We can express the inner product in equation 8 
as the kernel function: