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：
- 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。
- 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
- 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