Question

早上好，

我目前正在尝试运行回归神经网络，使用数据集上的“rnn”包，名为BostonHousing的数值;具体来说，这是结构：

Classes ‘tbl_df’, ‘tbl’ and 'data.frame': 1038 obs. of 3 variables: 
$ date : Date, format: "2013-11-19" "2013-11-20" "2013-11-21" "2013-11-22" ... 
$ Quantità : num 0.85 0.85 -0.653 -0.653 -0.653 ... 
$ Giacenza_In: num 0.945 1.648 -0.694 -0.694 -0.694 ...

#Split into train and test
cutoff = round(0.8*nrow(BostonHousing))

train_x <- BostonHousing[1:cutoff,]
test_x <- BostonHousing[-(1:cutoff),]

str(train_x)
#I apply the model and remove the first column because it's made up of dates

require(rnn)
model <- trainr( Y = train_x[,2], 
                 X = train_x[,3],
                 learningrate = 0.05,
                 hidden_dim = 4,
                 numepochs = 100)

pred <- predictr( model, test_x[,3])

每当我尝试运行代码时，它都会给出标题中报告的错误。

基本上，我想预测“Quantità”（这意味着订购的数量），给定当前库存的产品数量（Giacenza_In）

最诚挚的问候，亚历山德罗

Answer 1

似乎包rnn中的trainr需要二进制格式的输入和输出值。因此，您必须使用“ int2bin ”

转换每列

因此，首先必须将数值转换为整数值（使用舍入乘以10 ^ n）

Classes ‘tbl_df’, ‘tbl’ and 'data.frame': 1038 obs. of 3 variables: 
$ date : Date, format: "2013-11-19" "2013-11-20" "2013-11-21" "2013-11-22" ... 
$ Quantità : num 0.85 0.85 -0.653 -0.653 -0.653 ... 
$ Giacenza_In: num 0.945 1.648 -0.694 -0.694 -0.694 ...

#Split into train and test
cutoff = round(0.8*nrow(BostonHousing))

train_x <- BostonHousing[1:cutoff,]
test_x <- BostonHousing[-(1:cutoff),]

str(train_x)
#I apply the model and remove the first column because it's made up of dates
n<-5 #Number of decimal values
require(rnn)
model <- trainr( Y = int2bin(round(train_x[,2])*10^n), 
                 X = int2bin(round(train_x[,3])*10^n),
                 learningrate = 0.05,
                 hidden_dim = 4,
                 numepochs = 100)

pred <- predictr( model, int2bin(round(test_x[,3])-10^n))
pred[pred>=0.5]<-1
pred[pred<0.5]<-0

然后你必须再次将二进制值转换为整数

Y * 0出错：二进制运算符的非数字参数 - RNN

1 个答案: