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ID3 ALGORITHM
Abstract
• ID3 builds a decision tree from a fixed set of examples.
• Using this decision tree, future samples are classified.
• The example has several attributes and belongs to a class.
• The leaf nodes of the decision tree contain the class name whereas a non-leaf
node is a decision node.
• The decision node is an attribute test with each branch being a possible value of
the attribute.
• ID3 uses information gain to help it decide which attribute goes into a decision
node.
Algorithm
• Calculate the entropy of every attribute using the data set.
• Split the set into subsets using the attribute for which entropy is minimum (or
equivalently, information gain is maximum).
• Make a decision tree node containing that attribute.
• Recurse on subsets using remaining attributes.
Entropy and Information gain
• The entropy is a measure of the randomness in the information being processed.
• If the sample is completely homogeneous the entropy is zero and if the sample is
equally divided then it has entropy of one.
• Entropy can be calculated as:
Entropy(S) = ∑ – p(I) . log2p(I)
• The information gain is based on the decrease in entropy after a data-set is split
on an attribute.
• Information gain can be calculated as:
Gain(S, A) = Entropy(S) – ∑ [ p(S|A) . Entropy(S|A) ]
Decision tree for deciding if tennis is playable, using
data from past 14 days
Entropy(Decision) = – p(Yes) . log2p(Yes) – p(No) . log2p(No)
Entropy(Decision) = – (9/14) . log2(9/14) – (5/14) . log2(5/14)
= 0.940
Entropy
Wind factor on decision
• Wind attribute has two labels: weak and strong.
Gain(Decision, Wind) = Entropy(Decision) – [ p(Decision|Wind=Weak) .
Entropy(Decision|Wind=Weak) ] – [ p(Decision|Wind=Strong) . Entropy(Decision|Wind=Strong) ]
• We need to calculate (Decision|Wind=Weak) and (Decision|Wind=Strong) respectively.
Weak wind factor
• There are 8 instances for weak wind. Decision of 2 items
are no and 6 items are yes.
• Entropy(Decision|Wind=Weak) = – p(No) . log2p(No) –
p(Yes) . log2p(Yes)
Entropy(Decision|Wind=Weak) = – (2/8) . log2(2/8) – (6/8) . log2(6/8)
=0.811
Strong wind factor
• Here, there are 6 instances for strong wind. Decision is
divided into two equal parts.
Entropy(Decision|Wind=Strong) = – (3/6) . log2(3/6) – (3/6).log2(3/6)
= 1
Wind factor on decision
• Information Gain can be calculated as:
Gain(Decision, Wind) = Entropy(Decision) – [ p(Decision|Wind=Weak) . Entropy(Decision|Wind=Weak) ] –
[p(Decision|Wind=Strong) . Entropy(Decision|Wind=Strong) ]
= 0.940 – [ (8/14) . 0.811 ] – [ (6/14). 1]
= 0.048
Other factor on decision
On applying similar calculation on the other columns, we get:
• Gain(Decision, Outlook) = 0.246
• Gain(Decision,Temperature) = 0.029
• Gain(Decision, Humidity) = 0.151
Outlook
Sunny Overcast Rainy
Overcast outlook on decision
• Decision will always be yes if outlook
were overcast.
Outlook
Sunny Overcast Rainy
Yes
Sunny outlook on decision
We have 5 instances for sunny outlook.
Decision would be probably 3/5 percent no,
2/5 percent yes
• Gain(Outlook=Sunny|Temperature) = 0.570
• Gain(Outlook=Sunny|Humidity) = 0.970
• Gain(Outlook=Sunny|Wind) = 0.019
Outlook
Sunny
Overcast Rainy
Yes
Humidity
NormalHigh
• Decision will always be no when humidity is
high.
• Decision will always be yes when humidity is
normal.
Outlook
Sunny
Overcast Rainy
Yes
Humidity
NormalHigh
YesNo
Rain outlook on decision
Information gain for Rain outlook are:
• Gain(Outlook=Rain |Temperature) = 0.02
• Gain(Outlook=Rain | Humidity) = 0.02
• Gain(Outlook=Rain | Wind) = 0.971
Outlook
Sunny
Overcast
Rainy
Yes
Humidity
NormalHigh
YesNo
Wind
Strong Weak
• Decision will always be yes if wind
were weak and outlook were rain.
• Decision will always be no if wind
were strong and outlook were
rain.
Outlook
Sunny
Overcast
Rainy
Yes
Humidity
NormalHigh
YesNo
Wind
Strong Weak
No Yes
THANK YOU

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Id3 algorithm

  • 2. Abstract • ID3 builds a decision tree from a fixed set of examples. • Using this decision tree, future samples are classified. • The example has several attributes and belongs to a class. • The leaf nodes of the decision tree contain the class name whereas a non-leaf node is a decision node. • The decision node is an attribute test with each branch being a possible value of the attribute. • ID3 uses information gain to help it decide which attribute goes into a decision node.
  • 3. Algorithm • Calculate the entropy of every attribute using the data set. • Split the set into subsets using the attribute for which entropy is minimum (or equivalently, information gain is maximum). • Make a decision tree node containing that attribute. • Recurse on subsets using remaining attributes.
  • 4. Entropy and Information gain • The entropy is a measure of the randomness in the information being processed. • If the sample is completely homogeneous the entropy is zero and if the sample is equally divided then it has entropy of one. • Entropy can be calculated as: Entropy(S) = ∑ – p(I) . log2p(I) • The information gain is based on the decrease in entropy after a data-set is split on an attribute. • Information gain can be calculated as: Gain(S, A) = Entropy(S) – ∑ [ p(S|A) . Entropy(S|A) ]
  • 5. Decision tree for deciding if tennis is playable, using data from past 14 days
  • 6. Entropy(Decision) = – p(Yes) . log2p(Yes) – p(No) . log2p(No) Entropy(Decision) = – (9/14) . log2(9/14) – (5/14) . log2(5/14) = 0.940 Entropy
  • 7. Wind factor on decision • Wind attribute has two labels: weak and strong. Gain(Decision, Wind) = Entropy(Decision) – [ p(Decision|Wind=Weak) . Entropy(Decision|Wind=Weak) ] – [ p(Decision|Wind=Strong) . Entropy(Decision|Wind=Strong) ] • We need to calculate (Decision|Wind=Weak) and (Decision|Wind=Strong) respectively.
  • 8. Weak wind factor • There are 8 instances for weak wind. Decision of 2 items are no and 6 items are yes. • Entropy(Decision|Wind=Weak) = – p(No) . log2p(No) – p(Yes) . log2p(Yes) Entropy(Decision|Wind=Weak) = – (2/8) . log2(2/8) – (6/8) . log2(6/8) =0.811
  • 9. Strong wind factor • Here, there are 6 instances for strong wind. Decision is divided into two equal parts. Entropy(Decision|Wind=Strong) = – (3/6) . log2(3/6) – (3/6).log2(3/6) = 1
  • 10. Wind factor on decision • Information Gain can be calculated as: Gain(Decision, Wind) = Entropy(Decision) – [ p(Decision|Wind=Weak) . Entropy(Decision|Wind=Weak) ] – [p(Decision|Wind=Strong) . Entropy(Decision|Wind=Strong) ] = 0.940 – [ (8/14) . 0.811 ] – [ (6/14). 1] = 0.048
  • 11. Other factor on decision On applying similar calculation on the other columns, we get: • Gain(Decision, Outlook) = 0.246 • Gain(Decision,Temperature) = 0.029 • Gain(Decision, Humidity) = 0.151 Outlook Sunny Overcast Rainy
  • 12. Overcast outlook on decision • Decision will always be yes if outlook were overcast.
  • 14. Sunny outlook on decision We have 5 instances for sunny outlook. Decision would be probably 3/5 percent no, 2/5 percent yes • Gain(Outlook=Sunny|Temperature) = 0.570 • Gain(Outlook=Sunny|Humidity) = 0.970 • Gain(Outlook=Sunny|Wind) = 0.019
  • 16. • Decision will always be no when humidity is high. • Decision will always be yes when humidity is normal.
  • 18. Rain outlook on decision Information gain for Rain outlook are: • Gain(Outlook=Rain |Temperature) = 0.02 • Gain(Outlook=Rain | Humidity) = 0.02 • Gain(Outlook=Rain | Wind) = 0.971
  • 20. • Decision will always be yes if wind were weak and outlook were rain. • Decision will always be no if wind were strong and outlook were rain.

Editor's Notes

  1. Are your classroom colors different than what you see in this template? That’s OK! Click on Design -> Variants (the down arrow) -> Pick the color scheme that works for you! Feel free to change any “You will…” and “I will…” statements to ensure they align with your classroom procedures and rules!
  2. Are your classroom colors different than what you see in this template? That’s OK! Click on Design -> Variants (the down arrow) -> Pick the color scheme that works for you! Feel free to change any “You will…” and “I will…” statements to ensure they align with your classroom procedures and rules!