我正在尝试在matlab中创建一个用于创建决策树的树类。当我尝试运行这个脚本时,第一次运行代码时,我可以看到它达到了F和V的有利值。但是,在左右新建节点的初始化之后,我甚至是当前对象为空。如何正确地将引用嵌套到类中的同一个类中,以便它们不会相互干扰
classdef dtree
properties (Access = public)
MaxDepth;
CurrentDepth;
Features;
Left;
Right;
F;
V;
end
methods
function this=dtree(md, cd, nf, Xtrain, Ytrain)
% Now the real initialization begins
this.MaxDepth = md
this.CurrentDepth = cd;
this.Features = nf;
this.train(Xtrain, Ytrain);
end
function s = gini(dt, Labels)
s = 1;
s = s - (sum(Labels > 0.0) ./ numel(Labels)) ^ 2;
s = s - (sum(Labels < 0.0) ./ numel(Labels)) ^ 2;
end
function train(dt, Xtrain, Ytrain)
if (size(unique(Ytrain)) == 1 | dt.CurrentDepth > dt.MaxDepth)
return;
end
minGINI = Inf;
minF = 0;
minV = Inf;
for i = dt.Features
for n = 1:size(Xtrain, 1)
idx = Xtrain(:, i) > Xtrain(n, i);
GINI = dt.gini(Ytrain(idx)) + dt.gini(Ytrain(~idx));
if GINI < minGINI
minGINI = GINI;
minF = i;
minV = Xtrain(n, i);
end
end
end
dt.F = minF
dt.V = minV
lIdx = Xtrain(:, dt.F) > Xtrain(dt.V, dt.F);
dt.Left = dtree(dt.MaxDepth, dt.CurrentDepth + 1, dt.Features,Xtrain(lIdx, :), Ytrain(lIdx))
dt.Right = dtree(dt.MaxDepth, dt.CurrentDepth + 1, dt.Features, Xtrain(~lIdx, :), Ytrain(~lIdx));
end
end
端
ans = dtree(1,1,[2,5,6],XTrain,YTrain);
执行期间 具有属性的dtree:
MaxDepth: 1
CurrentDepth: 1
Features: [4 5 2]
Left: []
Right: []
F: 5
V: 7
执行后,当我输入ans时 ans =
带有属性的dtree:
MaxDepth: 1
CurrentDepth: 1
Features: [4 5 2]
Left: []
Right: []
F: []
V: []
在列车运行之前,这是一个空对象。
答案 0 :(得分:0)
处理类方法时,需要显式传递并返回对象
function this = dtree(...)
this = this.train(...)
end
function dt = train(dt, XTrain, YTrain)
...
end
如果您不这样做,则不会更新对您所拥有对象的引用。您总是需要在Matlab中更新引用!