Geography Reference
In-Depth Information
3.2.4. Example
Let us consider for example the image window below, centered on a pixel having the level
of gray 2. Let us estimate the textural parameter «asymmetry» on this window, with the rule
of connexion
2,45°
.
0 1 2 4 3
4 0 0 2 3
4 4 2 0 1
4 3 2 1 2
4 2 4 4 4
1.
Computation by the generic tree approach
The maximal level of gray of this image window is 4. A vector
of size 5 is
then created and initialized to the value zero.
The five following values are calculated:
,
1
,
2
,
3
and
4
according to the equation (8), by respecting the rule of connexion
2,45°
and the
neighbouring conditions
⋯
, with
=1
(order 2). The following
values are then obtained :
0
=
|
0−4
|
+
|
0−2
|
=6
1=|1−4|=3
2=|2−4|+|2−3|+|2−4|+|2−3|+|2−3|+|2−0|=9
3=|3−2|+|3−2|+|3−2|=3
4
=
|
4−4
|
+
|
4−2
|
+
|
4−4
|
+
|
4−0
|
+
|
4−2
|
+
|
4−1
|
=11
The textural parameter
2
is equal to the result of the summation of the vector
elements, according to the equation 9. One obtains:
2=0+1+2+3+4=32
2.
Computation with classical approach
Let us make the same calculation by the classical approach of the co-occurrence matrix of
levels of gray. In that case the asymmetry is calculated in the considered window by the
classical formula:
2=∑∑|−|×
,
being the number of time when the pair
of levels of gray
,
appears in the window, by respecting the rule of connexion
2,45°
.
The following result is obtained:
2=|0−1|×
+|0−2|×
+|0−3|×
+|0−4|×
+|1−0|×
+|1−2|×
+|1−3|×
+|1−4|×
+
|
2−0
|
×
+
|
2−1
|
×
+
|
2−3
|
×
+
|
2−4
|
×
+|3−0|×
+|3−1|×
+|3−2|×
+|3−4|×
+
|
4−0
|
×
+
|
4−1
|
×
+
|
4−2
|
×
+
|
4−3
|
×
=1×0+2×1+3×0+4×1
