数据如下 No. x1 x2 x3 x4 y1 y2 1 -17.8 1.2 17.5 4.2 24.7 -56.4 2 -11.2 -3.7 0 14.4 -53.8 34.4 3 3.9 -5.7 20.9 7.8 18.3 22 4 4.2 -5.9 -10 -6.8 -47.5 43.3 5 -1.8 -13.8 -4.6 -10.4 -185.1 164.8 6 0 -3.4 10.6 6.1 8.2 8.5 ...共200笔有y1,y2 资料80笔未知y1,y2 参考网路使用mxnet 但目前都跑不出来 =.= 小弟目前找不到可以解的方法 也爬过文 希望有大大可以帮忙一下 Error in symbol$infer.shape(list(...)) : Error in operator dnn: Shape inconsistent, Provided=[200], inferred shape=[200,10] library(neuralnet) # for neuralnet(), nn model library(nnet) # for class.ind() library(xlsx) library(mxnet) #Load data Data_Path<-"D:/2019/W21/R/DM_Assignment_4_NN.xlsx" Data<-read.xlsx(file=Data_Path,sheetIndex=1 , header=T) Data<-na.omit(Data) #删除没有y1,y2的资料 str(Data) head(Data) Data<-as.matrix(Data) # 输入层 data <- mx.symbol.Variable("data") # 第一隐藏层: 500节点，状态是Full-Connected fc1 <- mx.symbol.FullyConnected(data, name="1-fc", num_hidden=500) # 第一隐藏层的激发函数: Relu act1 <- mx.symbol.Activation(fc1, name="relu1", act_type="relu") # 这里引入dropout的概念 drop1 <- mx.symbol.Dropout(data=act1, p=0.5) # 第二隐藏层: 400节点，状态是Full-Connected fc2 <- mx.symbol.FullyConnected(drop1, name="2-fc", num_hidden=500) # 第二隐藏层的激发函数: Relu act2 <- mx.symbol.Activation(fc2, name="relu2", act_type="relu") # 这里引入dropout的概念 drop2 <- mx.symbol.Dropout(data=act2, p=0.5) # 输出层：因为预测数字为0~9共十个，节点为10 output <- mx.symbol.FullyConnected(drop2, name="output", num_hidden=10) # Transfer the Evidence to Probability by Softmax-function dnn <- mx.symbol.LinearRegressionOutput(output, name="dnn") #分割资料 train.ind <- c(1:200) train.x = data.matrix(Data[train.ind, 2:5]) train.y1 = c(Data[train.ind, 6]) train.y2 = data.matrix(Data[train.ind, 7]) test.x = data.matrix(Data[-train.ind, 2:5]) test.y1 = data.matrix(Data[-train.ind, 6]) test.y2 = data.matrix(Data[-train.ind, 7]) mx.set.seed(0) # 训练刚刚创造/设计的模型 dnn.model <- mx.model.FeedForward.create( dnn, # 刚刚设计的DNN模型 X=train.x, # train.x y=train.y1, # train.y ctx=mx.cpu(), # 可以决定使用cpu或gpu num.round=4, # iteration round array.batch.size=200, # batch size learning.rate=0.05, # learn rate momentum=0.9, # momentum array.layout = "rowmajor", eval.metric=mx.metric.accuracy, # 评估预测结果的基准函式* initializer=mx.init.uniform(0.07), # 初始化参数 epoch.end.callback=mx.callback.log.train.metric(100) ) --

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