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我正在尝试用递归实现决策树：到目前为止，我已经写了以下内容：

  1. 从给定的数据集中，找到最佳拆分并返回分支，以提供更多详细信息，假设我的数据具有矩阵列的特征，最后一列表示数据 1、-1 的类别。
  2. 基于 1。我有一个最佳特征与该拆分下的分支一起拆分，假设基于信息增益，我得到特征 9 是最佳拆分，特征 9 {1,3,5} 中的唯一值是9
  3. 我已经想出了如何获取与每个分支相关的数据，然后我需要遍历每个分支的数据以获得下一组拆分。我在计算这种递归时遇到了麻烦。

这是我到目前为止的代码，我现在正在做的递归看起来不正确：我该如何解决这个问题？

function [indeces_of_node, best_split] = split_node(X_train, Y_train)

    %cell to save split information
    feature_to_split_cell = cell(size(X_train,2)-1,4);

    %iterate over features
    for feature_idx=1:(size(X_train,2) - 1)
        %get current feature
        curr_X_feature = X_train(:,feature_idx);

        %identify the unique values
        unique_values_in_feature = unique(curr_X_feature);

        H = get_entropy(Y_train); %This is actually H(X) in slides
        %temp entropy holder

        %Storage for feature element's class
        element_class = zeros(size(unique_values_in_feature,1),2);

        %conditional probability H(X|y)
        H_cond = zeros(size(unique_values_in_feature,1),1); 

        for aUnique=1:size(unique_values_in_feature,1)
            match = curr_X_feature(:,1)==unique_values_in_feature(aUnique);
            mat = Y_train(match);
            majority_class = mode(mat);
            element_class(aUnique,1) = unique_values_in_feature(aUnique);
            element_class(aUnique,2) = majority_class;
            H_cond(aUnique,1) = (length(mat)/size((curr_X_feature),1)) * get_entropy(mat);
        end

        %Getting the information gain
        IG = H - sum(H_cond);

        %Storing the IG of features
        feature_to_split_cell{feature_idx, 1} = feature_idx;
        feature_to_split_cell{feature_idx, 2} = max(IG);
        feature_to_split_cell{feature_idx, 3} = unique_values_in_feature;
        feature_to_split_cell{feature_idx, 4} = element_class;
    end
    %set feature to split zero for every fold
    feature_to_split = 0;

    %getting the max IG of the fold
    max_IG_of_fold = max([feature_to_split_cell{:,2:2}]);

    %vector to store values in the best feature
    values_of_best_feature = zeros(size(15,1));

    %Iterating over cell to get get the index and the values under best
    %splited feature.
    for i=1:length(feature_to_split_cell)
        if (max_IG_of_fold == feature_to_split_cell{i,2});
            feature_to_split = i;
            values_of_best_feature = feature_to_split_cell{i,4};
        end
    end
    display(feature_to_split)
    display(values_of_best_feature(:,1)')

    curr_X_feature = X_train(:,feature_to_split);

    best_split = feature_to_split
    indeces_of_node = unique(curr_X_feature)

    %testing
    for k = 1 : length(values_of_best_feature)
        % Condition to stop the recursion, if clases are pure then we are
        % done splitting, if both classes have save number of attributes
        % then we are done splitting.
        if (sum(values_of_best_feature(:,2) == -1) ~= sum(values_of_best_feature(:,2) == 1))
            if((sum(values_of_best_feature(:,2) == -1) ~= 0) || (sum(values_of_best_feature(:,2) == 1) ~= 0))
                mat1 = X_train(X_train(:,5)== values_of_best_feature(k),:);
                [indeces_of_node, best_split] = split_node(mat1, Y_train);
            end
        end
    end
end

这是我的代码：在我的递归中看起来像一些我只深入一个分支，之后我再也不会回到其余的分支

feature_to_split =

     5


ans =

     1     2     3     4     5     6     7     8     9


feature_to_split =

     9


ans =

     3     5     7     8    11


feature_to_split =

    21


feature_to_split =

    21


feature_to_split =

    21


feature_to_split =

    21

如果您有兴趣运行此代码：git

4

1 回答 1

0

经过多轮调试，我想出了答案，希望有人能从中受益：

for k = 1 : length(values_of_best_feature)
    % Condition to stop the recursion, if clases are pure then we are
    % done splitting, if both classes have save number of attributes
    % then we are done splitting.
    if((sum(values_of_best_feature(:,2) == -1) ~= 0) || (sum(values_of_best_feature(:,2) == 1) ~= 0))
        X_train(:,feature_to_split) = [];
        mat1 = X_train(X_train(:,5)== values_of_best_feature(k),:);
        %if(level >= curr_level)
        split_node(mat1, Y_train, 1, 2, level-1);
        %end
    end

end
return;
于 2012-10-30T05:40:38.270 回答