R 神经网络的预测功能给出了奇怪的结果

我对R中的机器学习技术和编程都比较陌生，目前我正在尝试将神经网络与我掌握的一些数据相匹配。然而，由此产生的神经网络预测对我来说没有意义。我已经查看了StackOverflow，但找不到解决此问题的方法

我的数据（这是测试集的一部分，训练集的格式相同）

我的代码

resultsnn <- neuralnet(target~monday+tuesday+wednesday+thursday+friday+saturday+independent,data=training,hidden=3,threshold=0.01,linear.output = TRUE)
compute(resultsnn,test[,2:8])$net.result

我还试过什么？

我尝试过没有假人的版本（只包括自变量，这不会改变结果的类型）

我已经创建了一些合成数据并将其用作输入，对于相同的代码，这确实可以正常工作：

#building training set
input_train <- as.data.frame(c(1:100))
output_train <- as.data.frame(c(sqrt((1:100)+1)))

train <- cbind.data.frame(output_train,input_train)
colnames(train) <- c("output","input")

#building test set
input_test <- as.data.frame(c(101:150))
output_test <- as.data.frame(c(sqrt((101:150)+1)))

test <- cbind.data.frame(output_test,input_test)
colnames(test) <- c("output","input")

#NEURALNET PACKAGE
#neural network 3 neurons
res.train <- neuralnet(output~input,data=train,hidden=3,threshold=0.01) #train nn
compute(res.train,test[,2])$net.result #predict using nn on test set

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如果您需要任何其他信息，请告诉我！谢谢大家的帮助（：

看起来目标和独立之间没有信号。现在忽略工作日，如果你适合线性模型，有梯度和没有梯度：

# a linear model looking at response with indepedent (with intercept)
lm1 <- lm(target ~ indepedent, data = training)
lm1
#
# Call:
# lm(formula = target ~ indepedent, data = training)
#
# Coefficients:
#  (Intercept)    indepedent  
# 206.37312594    0.04853823 
# intercept only
lm0 <- lm(target ~ 1, data = training)
lm0
#
# Call:
# lm(formula = target ~ 1, data = training)
#
# Coefficients:
# (Intercept)  
#   310.0769  
# two models of the data equivalent to possible outcomes
plot(target ~ indepedent, data = training)
lines(fitted(lm1) ~ indepedent, data = training, lty = 2)
lines(fitted(lm0) ~ indepedent, data = training, col = 2)

这也是机器学习方法告诉你的。简单模型预测每个独立值的目标值。添加工作日变量显然不能改善这一点

---

您看到了玩具示例的预测，因为响应中有一个强信号。

当我添加更多的协变量时，神经网络是否可能工作（换句话说：我唯一的问题是当前的协变量集对目标没有任何解释力）？是的，有可能存在解释目标和独立性之间关系的协变量，除非你没有包括它们。如果你对这些变量之间的关系有任何机械上的理解，那可能建议尝试一个变量。如果你只是希望存在一种关系，那么就尝试很多变量；一旦你有了candidate解释变量，查看测试数据中是否存在该变量。在查看测试之前，尝试完成对训练数据的探索！

#building training set
input_train <- as.data.frame(c(1:100))
output_train <- as.data.frame(c(sqrt((1:100)+1)))

train <- cbind.data.frame(output_train,input_train)
colnames(train) <- c("output","input")

#building test set
input_test <- as.data.frame(c(101:150))
output_test <- as.data.frame(c(sqrt((101:150)+1)))

test <- cbind.data.frame(output_test,input_test)
colnames(test) <- c("output","input")

#NEURALNET PACKAGE
#neural network 3 neurons
res.train <- neuralnet(output~input,data=train,hidden=3,threshold=0.01) #train nn
compute(res.train,test[,2])$net.result #predict using nn on test set

str(test)
 'data.frame':  82 obs. of  8 variables:
  $ target     : int  277 204 309 487 289 411 182 296 212 308 ...
  $ monday     : int  1 0 0 0 0 0 0 1 0 0 ...
  $ tuesday    : int  0 1 0 0 0 0 0 0 1 0 ...
  $ wednesday  : int  0 0 1 0 0 0 0 0 0 1 ...
  $ thursday   : int  0 0 0 1 0 0 0 0 0 0 ...
  $ friday     : int  0 0 0 0 1 0 0 0 0 0 ...
  $ saturday   : int  0 0 0 0 0 1 0 0 0 0 ...
  $ independent: int  3317 1942 2346 2394 2023 1886 1750 1749 1810 2021 ...

str(training)
 'data.frame':  397 obs. of  8 variables:
  $ target     : int  1079 1164 1069 1038 629 412 873 790 904 898 ...
  $ monday     : int  0 0 0 0 0 0 1 0 0 0 ...
  $ tuesday    : int  1 0 0 0 0 0 0 1 0 0 ...
  $ wednesday  : int  0 1 0 0 0 0 0 0 1 0 ...
  $ thursday   : int  0 0 1 0 0 0 0 0 0 1 ...
  $ friday     : int  0 0 0 1 0 0 0 0 0 0 ...
  $ saturday   : int  0 0 0 0 1 0 0 0 0 0 ...
  $ independent: int  2249 2381 4185 2899 2387 2145 2933 2617 2378 3569 ...

# a linear model looking at response with indepedent (with intercept)
lm1 <- lm(target ~ indepedent, data = training)
lm1
#
# Call:
# lm(formula = target ~ indepedent, data = training)
#
# Coefficients:
#  (Intercept)    indepedent  
# 206.37312594    0.04853823 
# intercept only
lm0 <- lm(target ~ 1, data = training)
lm0
#
# Call:
# lm(formula = target ~ 1, data = training)
#
# Coefficients:
# (Intercept)  
#   310.0769  
# two models of the data equivalent to possible outcomes
plot(target ~ indepedent, data = training)
lines(fitted(lm1) ~ indepedent, data = training, lty = 2)
lines(fitted(lm0) ~ indepedent, data = training, col = 2)

# test which model is better
# large p-value suggests we're happy accepting the simple model
anova(lm0, lm1)
# Analysis of Variance Table
#
# Model 1: target ~ 1
# Model 2: target ~ indepedent
#   Res.Df       RSS Df  Sum of Sq       F  Pr(>F)
# 1     12 86990.923                             
# 2     11 81792.165  1 5198.7582 0.69917 0.42086
head(fitted(lm0))
#         428         429         430         431         432         433 
# 310.0769231 310.0769231 310.0769231 310.0769231 310.0769231 310.0769231