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[ML L4] Decision Tress

时间:2020-07-03 23:28:51      阅读:73      评论:0      收藏:0      [点我收藏+]

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Decision trees can handle none linear speratable dataset, in the picture, there is none separable dataset

技术图片

 

When we use dscision tree, we ask multi linear separable questions:

技术图片

For example, we can ask, 

1. Is windy?

2. Is Sunny?

Then we can reach our correct classified dataset.

 

Another Example:

技术图片

First, on the left hand side: ask ‘x<3‘ then we will got better answer.

Then ask ‘y < 2‘ to separae data.

 

Second, on the right hand side: 

技术图片

 

Code:

>>> from sklearn import tree
>>> X = [[0, 0], [1, 1]]
>>> Y = [0, 1]
>>> clf = tree.DecisionTreeClassifier()
>>> clf = clf.fit(X, Y)

Example of output for DT:

技术图片

 

 

Overfitting:

https://scikit-learn.org/stable/modules/generated/sklearn.tree.DecisionTreeClassifier.html#sklearn.tree.DecisionTreeClassifier

Sometimes, results migth be overfitting, one params we can modify to solve the problme is ‘min_sample_split‘:

技术图片

In the picture, right most only contains one dataset, so not able to split further.

 

Entropy:

Controls how a DT decides where to split the data

definition: measure of impurity in a bunch example:

技术图片

Try to get all dataset classified as correct as possible. In the example, right one is better.

 

 

Calculation Entropy:

技术图片

So Pi(slow) should be 2 / 4 = 0.5

Pi(fast) = 1 - 0.5 = 0.5

 

Entropy:

import math

e = -0.5 * math.log(0.5, 2) - 0.5 * math.log(0.5, 2) // 1

 E(parent) = 1

 

Information Gain:

Information gain calculates the reduction in entropy or surprise from transforming a dataset in some way.

It is commonly used in the construction of decision trees from a training dataset, by evaluating the information gain for each variable, and selecting the variable that maximizes the information gain, which in turn minimizes the entropy and best splits the dataset into groups for effective classification.

Number is the bigger the better.

技术图片

For example, we use ‘grade‘ as chid entropy, as we can see, there are two categories:

  • steep
  • flat

Based on partent: ‘ssff‘

     for ‘steep: we got 2 slow + 1 fast

     for ‘flat‘: we got 1 fast.

 

For the Node one the right (F): 

  entropy for this should be 0. because it only contains one category

 

For the Node one the left (SSF):

  P(slow) = 2/3

  P(fast) = 1/3

  E(left) = -2/3 * Log(2/3) - 1/3 * Log(1/3) = 0.9184

import math

p_slow = 2 / float(3)
p_fast = 1/float(3)

e = -1 * p_slow * math.log(p_slow, 2) - p_fast * math.log(p_fast ,2)

 

To calculate entropy(children) = 3/4*0.9184 + 1/4*0 = 0.6889

entropy(parent) = 1

Information Gain = E(parnet) - E(Children) = 1 - 0.6888 = 0.3112

 

The best separation would be ‘speed limit‘:

技术图片

 

[ML L4] Decision Tress

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原文地址:https://www.cnblogs.com/Answer1215/p/13222050.html

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