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| def best_split_for_idx(tree, idx, X, y,num_parent,best_gini): | |
| """Find the best split for a node and a given index | |
| """ | |
| # Sort data along selected feature. | |
| thresholds, classes = zip(*sorted(zip(X[:, idx], y))) | |
| # print("Classes are: ", classes, " ", thresholds) | |
| # We could actually split the node according to each feature/threshold pair | |
| # and count the resulting population for each class in the children, but | |
| # instead we compute them in an iterative fashion, making this for loop | |
| # linear rather than quadratic. | |
| m = y.size | |
| num_left = [0] * tree.n_classes_ | |
| num_right = num_parent.copy() | |
| best_thr = float('NaN') | |
| for i in range(1, m): # possible split positions | |
| c = classes[i - 1] | |
| # print("c is ", c, "num left is", len(num_left)) | |
| num_left[c] += 1 | |
| num_right[c] -= 1 | |
| gini_left = 1.0 - sum( | |
| (num_left[x] / i) ** 2 for x in range(tree.n_classes_) | |
| ) | |
| gini_right = 1.0 - sum( | |
| (num_right[x] / (m - i)) ** 2 for x in range(tree.n_classes_) | |
| ) | |
| # The Gini impurity of a split is the weighted average of the Gini | |
| # impurity of the children. | |
| gini = (i * gini_left + (m - i) * gini_right) / m | |
| # The following condition is to make sure we don't try to split two | |
| # points with identical values for that feature, as it is impossible | |
| # (both have to end up on the same side of a split). | |
| if thresholds[i] == thresholds[i - 1]: | |
| continue | |
| if gini < best_gini: | |
| best_gini = gini | |
| best_thr = (thresholds[i] + thresholds[i - 1]) / 2 # midpoint | |
| return best_gini, best_thr | |
| @ray.remote | |
| def best_split_for_idx_remote(tree, idx, X, y,num_parent,best_gini): | |
| return best_split_for_idx(tree, idx, X, y,num_parent,best_gini) |
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