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Hello.
In the lesson you say:
" Now, let’s compare that with a different feature we could have split on first, persons_2. In this case, the left branch will have a Gini impurity of 1 - (505/917)^2 - (412/917)^2 = 0.4949 "
Where 505,412 and 917 comes from ?
And why its not 1 - ( (505/917)^2 + (412/917)^2) as per Gini impurity formula 1 - (P1^2 + P2^2)
Further in the exercises, after placing everything as written in hints into preloaded functions gini and info_gain I get different all three answers.
1. Calculate gini and info gain for a root node split at safety_low<=0.5
y_train_sub = y_train[x_train[‘safety_low’]==0]
x_train_sub = x_train[x_train[‘safety_low’]==0]
gi = gini(y_train_sub)
print(f’Gini impurity at root: {gi}')
2. Information gain when using feature persons_2
left = y_train[x_train[‘persons_2’]==0]
right = y_train[x_train[‘persons_2’]==1]
print(f’Information gain for persons_2: {info_gain(left, right, gi)}')
3. Which feature split maximizes information gain?
info_gain_list =
for i in x_train.columns:
left = y_train_sub[x_train_sub[i]==0]
right = y_train_sub[x_train_sub[i]==1]
info_gain_list.append([i, info_gain(left, right, gi)])
info_gain_table = pd.DataFrame(info_gain_list).sort_values(1,ascending=False)
print(f’Greatest impurity gain at:{info_gain_table.iloc[0,:]}')
print(info_gain_table)
Output:
Gini impurity at root: 0.49534472145275465
Information gain for persons_2: 0.16699155320608106
Greatest impurity gain at:0 persons_2
1 0.208137
0.495 Doesn’t match 0.418
safety_low doesn’t give the largest information gain but person_2 gives
) below.
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