我在R波士顿数据集中尝试了神经网络。
data("Boston",package="MASS")
data <- Boston
仅保留我们想要使用的变量：
keeps <- c("crim", "indus", "nox", "rm" , "age", "dis", "tax" ,"ptratio", "lstat" ,"medv" )
data <- data[keeps]
在这种情况下，公式存储在名为f的R对象中。 响应变量medv将对其余九个属性进行“回归”。我做了如下：
f <- medv ~ crim + indus + nox + rm + age + dis + tax + ptratio + lstat
使用样本方法收集506行数据中没有替换的列车样本400：
set.seed(2016)
n = nrow(data)
train <- sample(1:n, 400, FALSE)
适合R的神经网络功能。
fit<- neuralnet(f, data = data[train ,], hidden=c(10 ,12 ,20),
algorithm = "rprop+", err.fct = "sse", act.fct = "logistic",
threshold =0.1, linear.output=TRUE)
但警告信息显示为算法未收敛。
警告讯息： 算法没有收敛于stepmax中的1次重复中的1次
尝试使用compute进行预测，
pred <- compute(fit,data[-train, 1:9])
显示以下错误信息
Error in nrow[w] * ncol[w] : non-numeric argument to binary operator
In addition: Warning message:
In is.na(weights) : is.na() applied to non-(list or vector) of type 'NULL'
为什么会出现错误以及如何从中进行预测以进行预测。我想在该数据集上使用神经网络功能。
答案 0 :(得分：3)
当neuralnet不收敛时，生成的神经网络不完整。你可以通过致电attributes(fit)$names来判断。当训练收敛时，它将如下所示：
[1] "call" "response" "covariate" "model.list" "err.fct"
[6] "act.fct" "linear.output" "data" "net.result" "weights"
[11] "startweights" "generalized.weights" "result.matrix"
如果没有，则不会定义某些属性：
[1] "call" "response" "covariate" "model.list" "err.fct" "act.fct" "linear.output"
[8] "data"
这解释了为什么compute不起作用。
当训练没有收敛时，第一个可能的解决方案可能是增加stepmax（默认为100000）。您还可以添加lifesign = "full"，以便更好地了解培训流程。
另外，看看你的代码，我会说有10层，12层和20层神经元的三层太多了。我将从一个具有与输入数量相同数量的神经元的层开始，在您的情况下为9。
编辑：
通过缩放（记住缩放训练和测试数据，以及'缩小'compute结果），它收敛得更快。另请注意，我减少了层数和神经元数量，仍然降低了误差阈值。
data("Boston",package="MASS")
data <- Boston
keeps <- c("crim", "indus", "nox", "rm" , "age", "dis", "tax" ,"ptratio", "lstat" ,"medv" )
data <- data[keeps]
f <- medv ~ crim + indus + nox + rm + age + dis + tax + ptratio + lstat
set.seed(2016)
n = nrow(data)
train <- sample(1:n, 400, FALSE)
# Scale data. Scaling parameters are stored in this matrix for later.
scaledData <- scale(data)
fit<- neuralnet::neuralnet(f, data = scaledData[train ,], hidden=9,
algorithm = "rprop+", err.fct = "sse", act.fct = "logistic",
threshold = 0.01, linear.output=TRUE, lifesign = "full")
pred <- neuralnet::compute(fit,scaledData[-train, 1:9])
scaledResults <- pred$net.result * attr(scaledData, "scaled:scale")["medv"]
+ attr(scaledData, "scaled:center")["medv"]
cleanOutput <- data.frame(Actual = data$medv[-train],
Prediction = scaledResults,
diff = abs(scaledResults - data$medv[-train]))
# Show some results
summary(cleanOutput)
答案 1 :(得分：0)
问题似乎出现在你的论证中linear.output = TRUE。
使用您的数据，但稍微改变代码（不定义公式并添加一些解释性注释）：
library(neuralnet)
fit <- neuralnet(formula = medv ~ crim + indus + nox + rm + age + dis + tax + ptratio + lstat,
data = data[train,],
hidden=c(10, 12, 20), # number of vertices (neurons) in each hidden layer
algorithm = "rprop+", # resilient backprop with weight backtracking,
err.fct = "sse", # calculates error based on the sum of squared errors
act.fct = "logistic", # smoothing the cross product of neurons and weights with logistic function
threshold = 0.1, # of the partial derivatives for error function, stopping
linear.output=FALSE) # act.fct applied to output neurons
print(net)
Call: neuralnet(formula = medv ~ crim + indus + nox + rm + age + dis + tax + ptratio + lstat, data = data[train, ], hidden = c(10, 12, 20), threshold = 0.1, rep = 10, algorithm = "rprop+", err.fct = "sse", act.fct = "logistic", linear.output = FALSE)
10 repetitions were calculated.
Error Reached Threshold Steps
1 108955.0318 0.03436116236 4
5 108955.0339 0.01391790099 8
3 108955.0341 0.02193379592 3
9 108955.0371 0.01705056758 6
8 108955.0398 0.01983134293 8
4 108955.0450 0.02500006437 5
6 108955.0569 0.03689097762 5
7 108955.0677 0.04765829189 5
2 108955.0705 0.05052776877 5
10 108955.1103 0.09031966778 7
10 108955.1103 0.09031966778 7
# now compute will work
pred <- compute(fit, data[-train, 1:9])