Can you give me a poste for Science Writing
TOPIC: DECISION TREE
Decision Tree Algorithm Pseudocode:-
1) Place the best attribute of the dataset at the root node of the
tree.
2) Split the training set into subsets. Subsets should be make in
such a way that each subset contains data with the same value for
an attribute.
3) Repeat steps 1 and 2 on each subset until you find leaf nodes in
all the branches of the tree.
Two features for using the selection of attribute:-
1) Information gain
2) gini index
In case of information gain, the more the gain value,it is suitable
for selecting the attribute as the root node/internal node of the
tree.
In case of the gini index. the less the gini value,it is suitable
for selecting the attribute as the root node/internal node of the
decision tree.
Examples:-
Solution:-
similarly for all the sub trees we calculate the gini index for each feature, 2) Example for ID3:-
similarly do calculations for all the features for the internal nodes,then we get final decision tree for this dataset
Formulas:-
From what I have understood you want a good example to explain decision tree, impurity algorithms. Here is an elaborative example.
Let us say we want to create a tree that uses chest pain, good
blood circulation, and blocked artery status to predict whether or
not a patient has heart disease. (Data is as shown in the table
attached)
We have to decide which node will be at the top; in other words, we
need to decide which node will become the root node. To do so, we
have to calculate 'impurity.' Impurity is the state of actual
results with false positives. To find contamination, we use the
Gini index or Information Gain. Extending this example, we do it
something like this.
Assumptions:
number of people with heart disease = x
number of people with no heart disease = y
Let us say that from our data we got the following results:
a) Making 'Chest Pain' as root:
if yes: x=105 and y=39
if no: x=34 and y=125
This means, out of all the people having chest pain, 105 have heart
disease, whereas 39 do not. Also, out of all the people not having
chest pain, 34 have heart disease, whereas 125 do not.
b) Similarly for making 'Good blood Circulation' as root:
if yes: x=37 and y=127
if no: x=100 and y=33
c) Making 'Blocked Arteries' as root:
if yes: x=92 and y=31
if no: x= 45 and y = 129
1) Gini Impurity:
Algorithm:
1) Calculate all of the Gini impurity scores.
2) If the node itself has the lowest score, then there is no point
in separating the patients anymore, and it becomes a leave
node.
3) If separating the data results in an improvement, then pick the
separation with the lowest impurity value.
Formula:
GI = 1 - (probability of yes)2 - (probability of
no)2
a) For chest pain:
For yes:
GI = 1 - (105/(105+39))2 -
(39/(105+39))2
GI = 0.395
For no:
GI = 1 - (34/(34+125))2 -
(125/(34+125))2
GI = 0.336
Total GI:
Note: In both side (yes and no) the number of patients is not
equal. Thus, we take a weighted average.
TGI = ((Total of yes)/Total patients * GI of yes) + ((Total of no)/Total patients * GI of no)
TGI = (144/144+159)*0.395 + (159/144+159)*0.336
TGI = 0.364
b) Similarly, we calculate for good blood circulation:
TGI = 0.360
c) And for blocked arteries:
TGI = 0.381
Thus we find that for good blood circulation total Gini impurity is the least and therefore, we use it as the root node.
Note: Now the number of patients in each separated node is different, so the Gini impurity has to be calculated again for remaining features.
2) Information Gain.
Algorithm:
1) Calculate all of the gain scores.
2) If the node itself has the highest score, then there is no point
in separating the patients anymore, and it becomes a leave
node.
3) If separating the data results in an improvement, then pick the
separation with the highest score value.
Formula:
(Base of the log is 2)
Entropy of class(Ce) = -(p/p+n) (log(p/p+n)) - (n/p+n)
(log(n/p+n))
Information Gain of each attribute (IG) = -(p/p+n) (log(p/p+n)) -
(n/p+n) (log(n/p+n))
Entropy of attribute (Ea) = Sum(Pi + Ni)/p+n (IG)
Gain = Ce - Ea
Ce = -139/(139+164) (log(139/139+164)) - 164/(139+164)
(log(164/139+164))
Ce = 0.995 or Ce=1
a) For chest pain:
IG for yes:
IG = -105/(105+39) (log(105/105+39)) - 39/(105+39)
(log(39/105+39))
IG = 0.842
IG for no:
IG = -34/(34+125) (log(34/(34+125))) - 125/(34+125)
(log(125/34+125))
IG = 0.749
Ea = (105+39)/(303) * (0.842) + (125+34)/303 * 0.749
Ea = 0.794
Gain = Ce - Ea = 1 - 0.794
Gain = 0.206
Similarly, calculate for the other attributes and find the
highest score.
In this case, the score of Blocked Arteries comes out to be
highest, and thus, we make it the root node.
Note: Now the number of patients in each separated node is different, so the Gain has to be calculated again for remaining features.
Can you give me a poste for Science Writing TOPIC: DECISION TREE Decision Tree Algorithm Pseudocode:-...
1. Decision trees As part of this question you will implement and compare the Information Gain, Gini Index and CART evaluation measures for splits in decision tree construction.Let D= (x,y), D = n be a dataset with n samples. The entropy of the dataset is defined as H(D)= P(c|D)log2P(c|D), where P(CD) is the fraction of samples in class i. A split on an attribute of the form X, <c partitions the dataset into two subsets Dy and Dn based on...
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Summary
You will write an application to build a tree structure called
Trie for a dictionary of English words, and use the Trie to
generate completion lists for string searches.
Trie Structure
A Trie is a general tree, in that each node can have
any number of children. It is used to store a dictionary
(list) of words that can be searched on,
in a manner that allows for efficient generation of completion
lists.
The word list is originally stored...