假设我创建了一个分类决策树,如下所示:
HP(1:size(HP), end) = 0; LP(1:size(LP), end) = 1;
% the dt's input & target pop
x = [HP(:,1:end-1); LP(:,1:end-1)]; t = [HP(:,end); LP(:,end)];
dt = fitctree(x,t);
view(dt)
view(dt, 'mode', 'graph');
输出如下:
1 if x2<-21.4866 then node 2 elseif x2>=-21.4866 then node 3 else 1
2 class = 1
3 if x2<20.093 then node 4 elseif x2>=20.093 then node 5 else 0
4 if x1<27.8438 then node 6 elseif x1>=27.8438 then node 7 else 0
5 if x2<39.6866 then node 8 elseif x2>=39.6866 then node 9 else 1
6 if x1<-33.7504 then node 10 elseif x1>=-33.7504 then node 11 else 0
7 class = 1
8 class = 1
9 class = 0
10 class = 1
11 if x2<1.53772 then node 12 elseif x2>=1.53772 then node 13 else 0
12 if x2<-2.50063 then node 14 elseif x2>=-2.50063 then node 15 else 0
13 class = 0
14 if x1<14.2153 then node 16 elseif x1>=14.2153 then node 17 else 0
15 class = 1
16 class = 0
17 class = 1
并且
1)如何获得导致值为“0”的叶子的所有路径?
2)是否存在基于决策树创建新实例的精细方式(除了随机生成实例和循环直到所需输出)?例如,我想创建一个随机实例,上面的树将其分类为'0'
答案 0 :(得分:0)
如果您知道决策的结束节点,则可以找到路径。您可以从[lable, score, node, cusum] = predict(mdl,x);
获取结束节点号。如果您想从Treebagger获取每棵树的节点,则需要为每棵树循环运行相同的节点。 [lable, score, node, cusum] = mdl.Trees{i}.predict(x);