Question

我有一个非常快速的问题，一个简单的可重复的例子，与我的bnlearn预测工作有关

    library(bnlearn)
    Learning.set4=cbind(c("Yes","Yes","Yes","No","No","No"),c(9,10,8,3,2,1))
    Learning.set4=as.data.frame(Learning.set4)
    Learning.set4[,c(2)]=as.numeric(as.character(Learning.set4[,c(2)]))
    colnames(Learning.set4)=c("Cause","Cons")
    b.network=empty.graph(colnames(Learning.set4))
    struct.mat=matrix(0,2,2)
    colnames(struct.mat)=colnames(Learning.set4)
    rownames(struct.mat)=colnames(struct.mat)
    struct.mat[1,2]=1
    bnlearn::amat(b.network)=struct.mat
    haha=bn.fit(b.network,Learning.set4)


    #Some predictions with "lw" method

    #Here is the approach I know with a SET particular modality. 
    #(So it's happening with certainty, here for example I know Cause is "Yes")
    classic_prediction=cpdist(haha,nodes="Cons",evidence=list("Cause"="Yes"),method="lw")
    print(mean(classic_prediction[,c(1)]))


    #What if I wanted to predict the value of Cons, when Cause has a 60% chance of being Yes and 40% of being no?
    #I decided to do this, according the help
    #I could also make a function that generates "Yes" or "No" with proper probabilities.
    prediction_idea=cpdist(haha,nodes="Cons",evidence=list("Cause"=c("Yes","Yes","Yes","No","No")),method="lw")
    print(mean(prediction_idea[,c(1)]))

以下是帮助说明的内容：

＆＃34;在离散或序数节点的情况下，也可以提供两个或更多个值。在这种情况下，将从指定值集合中以均匀概率对该节点的值进行采样＆＃34;

当我使用分类变量预测变量的值时，我现在只使用了所述变量的某种形态，如示例中的第一个预测。（将证据设置为＆＃34;是＆＃34;让Cons获得高价值）

但是如果我想在不知道变量的确切模态的情况下预测Cons，可以肯定地说，我可以使用我在第二次预测中所做的事情（只知道概率）吗？这是一种优雅的方式还是有更好的实施方式我不知道？

Answer 1

我与该软件包的创建者联系，我将在此处粘贴与该问题相关的答案：

对cpquery（）的调用是错误的：

Prediction_idea=cpdist(haha,nodes="Cons",evidence=list("Cause"=c("Yes","Yes","Yes","No","No")),method="lw")
print(mean(prediction_idea[,c(1)]))

具有40％-60％软证据的查询要求您首先将这些新概率放在网络中

haha$Cause = c(0.40, 0.60)

然后在没有证据参数的情况下运行查询。（因为你没有任何确凿的证据，实际上，只是原因的概率分布不同。）

我将发布代码，让我从示例中做出我想要的拟合网络。

change=haha$Cause$prob
change[1]=0.4
change[2]=0.6
haha$Cause=change
new_prediction=cpdist(haha,nodes="Cons",evidence=TRUE,method="lw")
print(mean(new_prediction[,c(1)]))

用cpdist预测＆＃34;概率＆＃34;作为证据

1 个答案: