8d040 basis beeldverwerking feature extraction
DESCRIPTION
8D040 Basis beeldverwerking Feature Extraction. Anna Vilanova i Bartrol í Biomedical Image Analysis Group bmia.bmt.tue.nl. N=M=30. What is an image?. Image is a 2D rectilinear array of pixels (picture element). N=M=256. Binary Image L=1 (1 bit). L=3 (2 bits). L=15(4 bits). - PowerPoint PPT PresentationTRANSCRIPT
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8D040 Basis beeldverwerking
Feature Extraction
Anna Vilanova i BartrolíBiomedical Image Analysis Groupbmia.bmt.tue.nl
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N=M=30
What is an image?
Image is a 2D rectilinear array of pixels (picture element)
N=M=256
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L=15(4 bits)L=255 (8 bits)
What is an image?No continuous values - Quantization
255
170
15
8
Binary Image L=1 (1 bit)L=3 (2 bits)
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An image is just 2D?
No! – It can be in any dimensionExample 3D:
Voxel-Volume Element
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Segmentation
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Reduction of dimensionality
Why feature extraction ?
Pixel levelImage of 256x256 and
8 bits
256 65536 ~ 10 157826
possible images
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• Incorporation of cues from human perception• Transcendence of the limits of human perception
• The need for invariance
Why feature extraction ?
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Apple detection …
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Transformation (Rotation)
cos( ) sin( ) 0sin( ) cos( ) 0
0 0 1
2a
1a
1b2b
P( )T P
1 2( , ,1)P x x
1
1 2 2
cos( ) sin( ) 0( ) ( , ,1) sin( ) cos( ) 0
0 0 1 A
aT P x x a
O
1 1 2cos( ) sin( )b a a
2 1 2sin( ) cos( ) b a a
cos( )
sin( )
sin( )
cos( )
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How do we transform an image?
• We transform a point P
• How do we transform an image f(P)?
1
1 2 2( ) ( , ,1)
A
aT P x x a
OT ( ) .T P P T
( ) ( )newf Q f Pnewff
QPHow do we know
which Q belongs to P?
1b
2a
1a
2
b
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.Q P T
How do we transform an image?
• How do we transform an image f(P)?
1b
1. Q PT
We know T which is the transformation we want to achieve.
1( ) ( . )newf Q f Q T
( ) ( )newf Q f P newff
QP2
a
1a
2
bHow do we know
which Q belongs to P?
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Apple detection …
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Feature Characteristics
• Invariance (e.g., Rotation, Translation)• Robust (minimum dependence on)• Noise, artifacts, intrinsic variations• User parameter settings
• Quantitative measures
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We extract features from…
Region of InterestSegmented Objects
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Classification
Features
Texture Based(Image & ROI)
Shape(Segmented objects)
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Shape Based Features
• Object based• Topology based (Euler Number)• Effective Diameter
(similarity to a circle to a box)• Circularity• Compactness• Projections• Moments (derived by Hu 1962)
• …
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4-neighbourhood of
8-neighbourhood of
Adjacency and Connectivity – 2D
ppp
*( ) ( ) { }k kN p N p p
Notation: k-Neighbourhood of is p ( )kN p
[ , ]p i j
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Adjacency and Connectivity – 3D
6-neighbourhood 18-neighbourhood 26-neighbourhood
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Objects or Components (Jordan Theorem)
• In 2D – (8,4) or (4,8)-connectivity• In 3D – (6,26)-,(26,6)-,(18,6)- or
(6,18)-connectivity
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Connected Components Labeling
Each object gets a different label
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Connected Components Labeling
A
B
CRaster Scan
Note: We want to label A. Assuming objects are 4-connected B, C are already labeled.
Cortesy of S. Narasimhan
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Connected Components Labeling
1
0
0 label(A) = new label
0
X
X label(A) = “background”
1
0
C label(A) = label(C)
1
B
0 label(A) = label(B)
1
B
C If
label(B) = label(C)then, label(A) = label(B)
Cortesy of S. Narasimhan
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24 2 2 2 34 4 4 4 ?
22 2 2 2 32 2 2 2 2
22 2 2 2 32 2 2 2 2 2 2 32 2 2 2 2 2 2
2 2 2
What if label(B) not equal to label(C)?
Connected Components Labeling
1
B
C
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Connected Components Labeling
Each object gets a different label
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Classification
Features
Texture Based(Image & ROI)
Shape(Segmented objects)
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Topology based – Euler Number
E C H
Euler Number E describes topology. C is # connected components H is # of holes.
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Euler Number 3D
E C Cav G
Euler Number E describes topology. C is # connected components Cav is # of cavitiesG is # of genus
E=1+0-1=0 E=1+0-1=0 E=1+1-0=2
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Euler Number 3D
E=2+0-0=2 E=1+1-0=2
Euler Number E describes topology. C is # connected components Cav is # of cavitiesG is # of genus E C Cav G
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3D Euler Number
• The Euler Number in 3D can be computed with local operations• Counting number of vertices, edges and faces of the
surfaces of the objects
1
# # #
C Cav
i i ii
E vertices edges faces
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Simple Shape Measurements
• 2D area - 3D volume • Summing elements
• 2D perimeter - 3D surface area• Selection of border elements • Sum of elemets with weights
• Error of precision
,
( , )x y
A f x y
1 22 where #ele. with 4c background ele.i
P N NN i
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Similarity to other Shape
• Effective Diameter
2
4 ACP
2PCompA
4
• Circularity (Circle C=1)
• Compactness – (Actually non-compactness)(Circle Comp= )
Ar
2A r
2P r
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Moments• Definition
• Order of a moment is• Moments identify an object uniquely
• ? is the Area• Centroid
pqr
, ,
[ , , ]p q rpqr
x y z
x y z f x y z p q r
000
100 010 001
000 000 000
( , , ) ( , , )
x y zc c c
[ ] : [1, ] {0,1}nf x N
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Central Moments
• Moments invariant to position
• Invariant to scaling
, ,
( ) ( ) ( ) [ , , ]p q rpqr x y z
x y z
c x c y c z c f x y z
, ,
13
000
( ) ( ) ( ) [ , , ]p q rx y z
x y zpqr p q r
x c y c z c f x y z
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Moments to Define Shape and Orientation
( , , )x y zc c c200
020
002
110
101
011
xx xy xz
yx yy yz
zx zy zz
xx
yy
zz
xy yx
xz zx
yz zy
I I II I I I
I I I
II
II I
I II I
Inertia Tensor
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Eigenanalysis of a Matrix
• Given a matrix S , we solve the following equation
( ) =S I x 0 j j jSv = λ v
we find the eigenvectors and eigenvalues
• Eigenvectors and eigenvalues go in couples an usually are ordered as follows:
jjv
det( ) =S I 0
1 2 3
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Eigenanalysis of the Inertia Matrix
Eigenanalysis
Sphere
Flatness
Elongated
( , , )x y zc c c
1 1v
2 2v
3 3v3
2
1
3
1
2
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Eigenanalysis of the Inertia Matrix
Eigenanalysis
Sphere
Flatness
Elongated
( , , )x y zc c c
1 1v
2 2v
3 3v2
3
1
3
1
2
1 2 3
1 2 3
1 2 3
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Orientation in 2D
• Using similar concepts than 3D
• Covariance or Inertia Matrix
• Eigenanalysis we obtain 2 eigenvalues and 2 eigenvectors of the ellipse
20 11
11 02IC
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Moments Invariance
• Translation• Central moments are invariant
• Rotation• Eigenvalues of Inertia Matrix are invariant
• Scaling• If moment scaled by (3D) (2D) 1
3000
p q r
12
00
p q
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Moments invariant rotation-translation-scaling
• For 3D three moments (Sadjadi 1980)
For 2D seven moments
1 200 020 002
2 2 22 200 020 200 002 020 002 101 110 011
23 200 020 002 002 110 110 101 011
2 2020 101 200 011
2
J
J
J
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Classification
Features
Texture Based(Image & ROI)
Shape(Segmented objects)
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Image Based Features
• Using all pixels individually• Histogram based features
− Statistical Moments (Mean, variance, smoothness)− Energy− Entropy− Max-Min of the histogram− Median
• Co-occurrance Matrix
Gonzalez & Woods – Digital Image ProcessingChapter 11 – 11.3.3 Texture
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Histogram
0 1 2 3 4 5 6 7 8 9
Quantized , 0,...,
( ) # of voxels [ ] total # of voxels
i
i i
b i L
N b f x bN
L=9
( ) ( ) /i iP b N b N
bi
P(bi)
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How do the histograms of this images look like?
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Bimodal Histogram
60 80 100 120
0.1
0.2
0.3
0.4
0.5
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Trimodal Features
50 100 150 200 250
0.005
0.01
0.015
0.02
0.025
0.03
0.035
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Histogram Features
• Mean
• Central Moments0
( )
L
i ii
m b P b
0
( ) ( )
L
nn i i
i
c b m P b
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Histogram Features
• Mean
• Variance
• Relative Smoothness
• Skewness
0
( )
L
i ii
m b P b
2 22
0
( ) ( )
L
i ii
b m P b c
2
111
R
33
0
( ) ( )
L
i ii
c b m P b
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Histogram Features
• Energy (Uniformity)
• Entropy
2
0
[ ( )]
L
ii
E P b
20
2
( ) log ( ( ))
if ( ) 0 then ( ) log ( ( )) 0
L
i ii
i i i
H P b P b
P b P b P b
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Examples of Energy and Entropy
0
0,1
0,2
0,3
0 2 4 6 8 10 12 14 bi
P(bi)
0
0,02
0,04
0,06
0,08
0 2 4 6 8 10 12 14 bi
P(bi)
0
0,1
0,2
0,3
0,4
0 2 4 6 8 10 12 14 bi
P(bi)
0 2 4 6 8 10 12 140
0.10.20.30.40.50.60.70.80.9
11.1
bi
P(bi) Energy=1Entropy=0
Energy=0,111Entropy=3,327
Energy=0,255Entropy=2,018
Energy=0,0625Entropy= 4
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Examples
Texture Mean std R 3rd moment Energy Entropy1 82.64 11.79 0.002 -0.105 0.026 5.434
2 143.56 74.63 0.079 -0.151 0.005 7.783
3 99.72 33.73 0.017 0.750 0.013 6.374
1 2 3
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The next slides were not given, during the lecture and will not be asked in the exam
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Intensity Co-occurrance Matrix
• Operator Q defines the position between two pixels (e.g, pixel to the right)
• Co-occurance matrix G is (L+1) x (L+1) (6x6). Counts how often Q occurs
0 1 4
0 4 5
2 3 1
2 3 4
4
4
1
1
6
3
5
1
1 6 55 1
0 1 2 3 4 5 6
0 0 1 0 0 1 0 0
1 0 0 2 0 1 0 1
2 0 0 0 2 0 0 0
3 0 1 0 0 1 0 0
4 2 1 0 0 0 1 1
5 0 2 0 1 0 0 0
6 0 0 0 0 0 1 0
Image G
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Example
• L=256• Q “one pixel immediately to the right”
Image
G - Matrix
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Features based on the co-ocurrence Matrix
• The elements of G (gij) is converted to probability (pij) by dividing by the amount of pairs in G
• Based on the probability density function we can use• Maximum• Energy (uniformity)• Entropy
0 0
1
L L
iji j
p
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Features based on the co-ocurrence Matrix
• Homogenity – closeness to a diagonal matrix
• Contrast
0 0 1 | | L L
ij
i j
pi j
2
0 0
( )
L L
iji j
i j p
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Features based on the co-ocurrence Matrix
• Correlation – measure of correlation with neighbours
0 0
( )( )L Lr c ij
i j r c
i m j m p
2 2
0 0 0
( ) ( ) ( ) ( )
L L L
i ij r i r r ii i i
P i p m iP i i m P i
2 2
0 0 0
( ) ( ) ( ) ( )L L L
j ij c j c c ij j i
P j p m jP j j m P i
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Example
• L=256• Q “one pixel immediately to the right”
Image
G - Matrix
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Example
Image
G - Matrix
Correlation Contrast Homogeneity1 - 0.0005 10838 0.0366
2 0.9650 570 0.0824
3 0.8798 1356 0.2048
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Moments • Definition
• Order of a moment is• Moments identify an object uniquely
• Centroid
, ,
[ , , ]p q rpqr
x y z
x y z f x y z pqr p q r
100 010 001
000 000 000
( , , ) ( , , )x y zc c c
[ ] :[1, ] [1, ]nf x N L
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Central Moments
• Moments invariant to position
• Normalized central moments, ,
( ) ( ) ( ) ( , , )p q rpqr x y z
x y z
c x c y c z c f x y z
, ,
13
000
( ) ( ) ( ) ( , , )p q rx y z
x y zpqr p q r
x c y c z c f x y z
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Moments invariant rotation-translation-scaling• For 3D three moments (Sadjadi 1980)
• For 2D seven moments (Hu’s 1962)
1 200 020 002
2 2 22 200 020 200 002 020 002 101 110 011
23 200 020 002 002 110 110 101 011
2 2020 101 200 011
2
J
J
J
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Moments invariant rotation-translation-scaling-mirroring (within minus sign)
1 2 3 4
5 6
, , , ,,
are all
equal Mirroring7 7
7 7
or
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Projections
1
( ) ( , )
yN
xy
proj x f x y
1
( ) ( , )xN
yx
proj y f x y
x
y