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Update decision_tree.py

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thor-harsh 2023-08-17 13:26:09 +05:30 committed by GitHub
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machine_learning/decision_tree.py
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@ -155,16 +155,19 @@ class TestDecisionTree:
def main():
"""
In this demonstration first we are generating x which is a numpy array containing values starting
from -1 to 1 with an interval of 0.005 i.e [-1,-0.995,....,0.995,1] this is what we are
getting by applying arange function of numpy.Then the we are generating y by applying sin function
on x which is an array containing values from -1 to 1 with difference of 0.005 i.e we are getting
an array y which contains sin of each value of x. We then train a decision tree on the data set
and use the decision tree to predict the label of 10 different test values. Here we should prefer
calculating Root Mean Squared Error over Mean Sqaured error beacause RMSE should be used
when you need to communicate your results in an understandable way to end users or when
penalising outliers is less of a priority.Interpretation will be easy in this case.
You can check out these https://stephenallwright.com/rmse-vs-mse/ to know the reason for this.
In this demonstration first we are generating x which is a numpy array
containing values starting from -1 to 1 with an interval of 0.005
i.e [-1,-0.995,....,0.995,1] this is what we are getting by applying arange
function of numpy.Then the we are generating y by applying sin function
on x which is an array containing values from -1 to 1 with difference
of 0.005 i.e we are getting an array y which contains sin of each value
of x. We then train a decision tree on the data set and use the decision tree
to predict the label of 10 different test values. Here we should prefer
calculating Root Mean Squared Error over Mean Sqaured error because RMSE
should be used when you need to communicate your results in an understandable
way to end users or when penalising outliers is less of a priority.Interpretation
will be easy in this case. You can check out https://stephenallwright.com/rmse-vs-mse/ to
know more.
"""
x = np.arange(-1.0, 1.0, 0.005)
y = np.sin(x)