Information Entropy and Information Gain
TIL about Information Entropy and Information Gain, which are key pieces towards determining the relevance of a decision when constructing a decision tree.
Wait… Information has Entropy?
Before now I always thought about entropy in the context of the physics domain, as in the degree of randomness in a physical system.
After studying decision trees as part of my AI and Machine Learning course through Georgia Tech, I learned that entropy could be used to describe the amount of unpredictability in a random variable.
The Entropy
of a random variable can be calculated using the following formula:
d
B(k) = Σ P(k_d) * log2(P(k_d))
1
Where k
is an attribute of your decision tree.
A few examples of entropy values:

A fair coin has an entropy value of 1 bit, because the H/T values are equally weighted and can be encoded by a single bit of information.

A weighted coin with 99% probability of heads an 1% of tails has an entropy value of 0.08 bits. There is clearly less randomness in this weighted coin, so it makes sense that the required information to store the result of this coin is much lower.
Say I have data describing the random variable Will I go Running?
. Each row of the table represents a day of the week Day1  Day7
where I may or may not have gone running.
Each row has 3 attributes: Weather
, Just Ate
, and Late at Work
which influence the Yes/No decision of whether I go running.
Day  Weather  Just Ate  Late at Work  Will I go Running?
        
1  'Sunny'  'yes'  'no'  'yes'
2  'Rainy'  'yes'  'yes'  'no'
3  'Sunny'  'no'  'yes'  'yes'
4  'Rainy'  'no'  'no'  'no'
5  'Rainy'  'no'  'no'  'yes'
6  'Sunny'  'yes'  'no'  'yes'
7  'Rainy'  'no'  'yes'  'no'
Calculating the total entropy
of the system, or the total amount of variability in the random variable, is possible by summing the entropy of a goal (a yes
response that I will go running) and the entropy of a negative result (a no
response that I will go running).
The probabilities can be read out of the data table itself, by counting.
B(Will I go Running) = p('yes')log2(p('yes'))  p('no')log2(p('no'))
B(Will I go Running) = (4/7)log2(4/7)  p(3/7)log2(3/7)
B(Will I go Running) = 0.985228136034
So the total entropy for the variable Will I go running is a small amount less than 1, indicating that there is slightly less than a 50% chance that the decision to go running will be yes/no.
Information Gain
How is this useful when constructing decision trees?
Entropy is used when determining how much information is encoded in a particular decision. This information gain
is useful when, upon being presented with a set of attributes about your random variable, you need to decide on which attribute tells you the most info about the variable.
When building decision trees, placing attributes with the highest information gain at the top of the tree will lead to the highest quality decisions being made first. This will result in more succinct and compact decision trees.
Calculating information gain
A quick plug for an information gain calculator that I wrote recently. Check it out, and continue reading to understand how it works.
We’ve already calculated the total entropy for the system above. The next step is to calculate the entropy remainders from that total entropy after each attribute in the data set is processed and data is classified.
Information Gain
is calculated as following:
Gain(attribute) = total_entropy  remainder(attribute)
branch_n
remainder(attribute) = Σ P(attribute_branch_n)*B(branch)
branch
When calculating the remainder for an attribute, you iterate over the count of goal values classified by each branch of your attribute. In the data table above, when looking at the Weather attribute, each branch would contain counts of “yes” and “no” Will I Go Running
values classified by the ‘sunny’ and ‘rainy’ branches of the Weather attribute.
P(attribute_branch_n)
is the number of goal values classified by the branch of the attribute, divided by the total number of entries in the data table. B(branch)
is the entropy of the branch of the attribute itself.
In the example above, the remainder and information gain for the Weather
attribute can be calculated as follows:
remainder(Weather) = P('Sunny and yes running')*B('Sunny') + P('Sunny and no running')*B('Sunny') + P('Rainy and yes running')*B('Rainy') + P('Rainy and no running)*B('Rainy')
remainder(Weather) = (2/7)*B('Sunny') + (1/7)*B('Sunny') + (2/7)*B('Rainy') + (2/7)*B('Rainy')
remainder(Weather) = 0.463587499691
Gain(Weather) = 0.985228136034  0.463587499691 = 0.5216406363433186
The information gain for the Weather attribute is 0.5216 bits, which is over half of the information stored in the random variable. For comparison, here are the gains for all 3 attributes together:
Gain('Just Ate') = 0.02024420715375619
Gain('Weather') = 0.5216406363433186
Gain('Late at Work') = 0.12808527889139454
The Weather attribute tells us the most about our Will I Go Running
random variable, since its information gain is the highest of the 3 available attributes.
If we were to construct a decision tree for classifying new instances of this random variable, we would want to place the Weather attribute at the top of the tree since it contains the most information about the inevitable result of the variable.
Comments