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…tructures into decision-tree
| """Calcute the distance between two rows.""" | ||
| dist = 0.0 | ||
| for i in range(len(row1) - 1): | ||
| dist += (row1[i] - row2[i]) ** 2 |
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You're missing out on using the power of Numpy (or pandas) here to broadcast mathematical operations. If row1 and row2 are numpy arrays, then you could just have
return sqrt(np.sum((row1 - row2)**2))
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Written this way to account for the difference in length between rows. Test data is submitted without a "classification" column. Present data has such columns.
src/knn.py
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| def predict(self, test_data, tk=None): | ||
| """Given data, categorize the data by its k nearest neighbors.""" | ||
| if tk is None: |
src/knn.py
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| for row in self.data.iterrows(): | ||
| distances.append((row[1][-1], self._distance(row[1], test_data))) | ||
| distances.sort(key=lambda x: x[1]) | ||
| # import pdb; pdb.set_trace() |
src/knn.py
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| if my_class: | ||
| return my_class | ||
| else: | ||
| self.predict(test_data, tk - 1) |
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Confused as to why this has to be recursive
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Written for the case in which the classification is a "tie" between two classes. In that case, the classify function returns None and therefore predict is run once again with a decreased k value. This is based on my interpretation of the algorithm in the class notes. Does not mean I didn't interpret it incorrectly, though.
https://codefellows.github.io/sea-python-401d5/lectures/k_nearest_neighbors.html?highlight=nearest
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