Ordinal Classification. Rob Potharst Erasmus University Rotterdam. What is ordinal classification?. Company: catering service Swift. total liabilities / total assets 1 net income / net worth 3 … … managers’work experience 5 market nicheposition 3 … .
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Rob Potharst
Erasmus University Rotterdam
SIKSAdvanced Course on Computational Intelligence, October 2001
SIKSAdvanced Course on Computational Intelligence, October 2001
bankruptcy risk + (acceptable)
SIKSAdvanced Course on Computational Intelligence, October 2001
2 2 2 2 1 3 5 3 5 4 2 4 +
4 5 2 3 3 3 5 4 5 5 4 5 +
3 5 1 1 2 2 5 3 5 5 3 5 +
2 3 2 1 2 4 5 2 5 4 3 4 +
3 4 3 2 2 2 5 3 5 5 3 5 +
3 5 3 3 3 2 5 3 4 4 3 4 +
3 5 2 3 4 4 5 4 4 5 3 5 +
1 1 4 1 2 3 5 2 4 4 1 4 +
3 4 3 3 2 4 4 2 4 3 1 3 +
3 4 2 1 2 2 4 2 4 4 1 4 +
2 5 1 1 3 4 4 3 4 4 3 4 +
3 3 4 4 3 4 4 2 4 4 1 3 +
1 1 2 1 1 3 4 2 4 4 1 4 +
2 1 1 1 4 3 4 2 4 4 3 3 +
2 3 2 1 1 2 4 4 4 4 2 5 +
2 3 4 3 1 5 4 2 4 3 2 3 +
2 2 2 1 1 4 4 4 4 4 2 4 +
2 1 3 1 1 3 5 2 4 2 1 3 +
2 1 2 1 1 3 4 2 4 4 2 4 +
2 1 2 1 1 5 4 2 4 4 2 4 +
2 1 1 1 1 3 2 2 4 4 2 3 ?
1 1 3 1 2 1 3 4 4 4 3 4 ?
2 1 2 1 1 2 4 3 3 2 1 2 ?
1 1 1 1 1 1 3 2 4 4 2 3 ?
2 2 2 1 1 3 3 2 4 4 2 3 ?
2 2 1 1 1 3 2 2 4 4 2 3 ?
2 1 2 1 1 3 2 2 4 4 2 4 ?
1 1 4 1 3 1 2 2 3 3 1 2 ?
3 4 4 3 2 3 3 4 4 4 3 4 ?
3 1 3 3 1 2 2 3 4 4 2 3 ?
1 1 2 1 1 1 3 3 4 4 2 3 
3 5 2 1 1 1 3 2 3 4 1 3 
2 2 1 1 1 1 3 3 3 4 3 4 
2 1 1 1 1 1 2 2 3 4 3 4 
1 1 2 1 1 1 3 1 4 3 1 2 
1 1 3 1 2 1 2 1 3 3 2 3 
1 1 1 1 1 1 2 2 4 4 2 3 
1 1 3 1 1 1 1 1 4 3 1 3 
2 1 1 1 1 1 1 1 2 1 1 2 
20: + (acceptable)
9:  (unacceptable)
10: ? (uncertain)
from: Greco, Matarazzo, Slowinski (1996)
SIKSAdvanced Course on Computational Intelligence, October 2001
if man.exp. > 4, then class = ‘+’
if man.exp. < 4 and net.inc/net.worth = 1, then class = ‘’
all other cases: class = ‘?’
SIKSAdvanced Course on Computational Intelligence, October 2001
The act of assigning objects to classes, using the values of relevant features of those objects
SIKSAdvanced Course on Computational Intelligence, October 2001
SIKSAdvanced Course on Computational Intelligence, October 2001
SIKSAdvanced Course on Computational Intelligence, October 2001
man.exp. < 3
y
n
gen.exp./sales = 1
+
y
n
tot.liab/cashfl = 1
?
y
n
classifies 37 out of 39 ex’s correctly

?
SIKSAdvanced Course on Computational Intelligence, October 2001
SIKSAdvanced Course on Computational Intelligence, October 2001
A classifier is monotone iff:
if A < B, then also class(A) <class(B)
SIKSAdvanced Course on Computational Intelligence, October 2001
SIKSAdvanced Course on Computational Intelligence, October 2001
SIKSAdvanced Course on Computational Intelligence, October 2001
try to split subset T:
1) update D for subset T
2) ifD T is homogeneous then
assign class label to T and make T a leaf definitively
else
split T into two nonempty subsets TL and TR using entropy
try to split subset TL
try to split subset TR
SIKSAdvanced Course on Computational Intelligence, October 2001
update D for T:
1) if min(T) isnot in Dthen
 add min(T) to D
 class ( min(T) ) = the maximal value allowed, given D
2) if max(T) is not in Dthen
 add max(T) to D
 class ( max(T) ) = the minimal value allowed, given D
SIKSAdvanced Course on Computational Intelligence, October 2001
SIKSAdvanced Course on Computational Intelligence, October 2001
SIKSAdvanced Course on Computational Intelligence, October 2001
X:
0 0 0
1 0 0
0 1 0
….
2 2 2
D:
0 0 1 0
0 0 2 1
1 1 2 2
2 0 2 2
2 1 2 3
attr. 1: values 0,1,2
attr. 2: values 0,1,2
attr. 3: values 0,1,2
classes: 0, 1, 2, 3
Let us calculate the min and max poss value forx = 022:
minvalue: y* = 002, so the minvalue = 1
maxvalue: there is no y*, so the maxvalue = 3
SIKSAdvanced Course on Computational Intelligence, October 2001
Try to split subset T = X:
update D for X:
min(X) = 000 is not in D; maxvalue of 000 is 0
add 000 with class 0 to D
max(X) = 222 is not in D; minvalue of 222 is 3
add 222 with class 3 to D
D X is not homogeneous
so consider all the possible splits:
A1 0; A1 1; A2 0; A2 1; A3 0; A3 1
0 0 0 0
0 0 1 0
0 0 2 1
1 1 2 2
2 0 2 2
2 1 2 3
2 2 2 3
SIKSAdvanced Course on Computational Intelligence, October 2001
The split A1 0splits X into TL = [000,022] and TR= [100,222]
D TR
1 1 2 2
2 0 2 2
2 1 2 3
2 2 2 3
Entropy = 1
D TL
0 0 0 0
0 0 1 0
0 0 2 1
Entropy = 0.92
Average entropy of this split = 3/7 x 0.92 + 4/7 x 1 = 0.97
SIKSAdvanced Course on Computational Intelligence, October 2001
The split with lowest entropy is A1 0, so we go on with T = TL = [000,022]:
Try to split subset T= [000,022]:
update D for T:
min(T) = 000 is already in D
max(T) = 022 has minimum value 1, so it is added to D
0 0 0 0
0 0 1 0
0 0 2 1
0 2 2 1
1 1 2 2
2 0 2 2
2 1 2 3
2 2 2 3
D T is not homogeneous, so we go on to consider
the following splits: A2 0; A2 1; A3 0; A3 1
Lowest entropy
SIKSAdvanced Course on Computational Intelligence, October 2001
A1 0
A3 1
?
?
?
SIKSAdvanced Course on Computational Intelligence, October 2001
The split A3 1splits T into TL = [000,021] and TR= [002,022]
We go on with T = TL = [000,021]
Try to split subset T= [000,021]:
min(T) = 000 is already in D
max(T) = 021 has minimum value 0, so it is added to D
D T is homogeneous, so we stop and make T into a leaf with class value 0
Next, we go on with T = TR = [002,022], etc.
SIKSAdvanced Course on Computational Intelligence, October 2001
A1 0
A1 1
A3 1
A2 0
2
0
1
2
3
SIKSAdvanced Course on Computational Intelligence, October 2001
SIKSAdvanced Course on Computational Intelligence, October 2001
SIKSAdvanced Course on Computational Intelligence, October 2001