R 插入符序列与给定度数的聚

Question

我在近似多项式的次数上滚动一个循环，以使用插入符进行训练，

ds = 1:20
for(i in 1:length(ds)){
  print(i)
  d=ds[i]
  fit = train(y~poly(x,degree=d),data=training,method="lm",trControl=fitCtrl)
  # other operations
}

运行代码给出的

Error in `[.data.frame`(data, 0, cols, drop = FALSE) : 
undefined columns selected

使用 d=4 不起作用，但修复调用中的次数（即 Degree=4）是有效的。

你猜猜这里发生了什么吗？

谢谢！

编辑：

library(caret)
set.seed(1)
omega = 0.5*pi
xi = 0.5
phi = 0.5*pi
f = function(t)1-exp(-xi*omega*t)*sin(sqrt(1-xi^2)*omega*t+phi)/sin(phi)
sigma = 0.03
train.n = 100
x = seq(0,2*pi,by=2*pi/(train.n-1))
y = f(x)+rnorm(train.n,mean=0,sd=sigma)
training = data.frame(x=x,y=y)
fitCtrl <- trainControl(method = "LOOCV",verboseIter = FALSE)
ds = 1:20
for(i in 1:length(ds)){
  print(i)
  d=4
fit=train(y~poly(x,degree=4),data=training,method="lm",trControl=fitCtrl)
}

Answer 1

这里的问题实际上是 caret 在底层公式上使用 all.vars() 来创建建模所需的数据帧。正如您所看到的，d 被认为是这些变量之一。

all.vars(y~poly(x, degree = d))
#> [1] "y" "x" "d"

通常，可以通过在公式中使用 I() 或围绕公式使用 force() 来解决这些问题，但不能使用 all.vars()< /代码>。

解决此问题的唯一方法是不在公式中发送 d，而是预先将其设置为数字。

在循环中使用 as.formula(paste0("y ~ poly(x, Degree=", d, ")") 将实现这一点（@akrun 也建议）。

这是一个工作基于您的代码的示例：

library(caret)
set.seed(1)
omega = 0.5 * pi
xi = 0.5
phi = 0.5 * pi
f = function(t)
  1 - exp(-xi * omega * t) * sin(sqrt(1 - xi ^ 2) * omega * t + phi) / sin(phi)
sigma = 0.03
train.n = 100
x = seq(0, 2 * pi, by = 2 * pi / (train.n - 1))
y = f(x) + rnorm(train.n, mean = 0, sd = sigma)
training = data.frame(x = x, y = y)
fitCtrl <- trainControl(method = "LOOCV", verboseIter = FALSE)

# Shorterning this to 2 to illustrate that it's working
ds = 1:2

for (i in 1:length(ds)) {
  print(i)
  d = ds[i]
  
  fit = train(
    as.formula(paste0("y ~ poly(x, degree=", d, ")")),
    data = training,
    method = "lm",
    trControl = fitCtrl
  )
  
}
#> [1] 1
#> [1] 2

# Notice that we have two poly-terms, so it's working
summary(fit)
#> 
#> Call:
#> lm(formula = .outcome ~ ., data = dat)
#> 
#> Residuals:
#>      Min       1Q   Median       3Q      Max 
#> -0.42507 -0.11917  0.00392  0.13771  0.27425 
#> 
#> Coefficients:
#>                        Estimate Std. Error t value Pr(>|t|)    
#> (Intercept)             0.94738    0.01655  57.237  < 2e-16 ***
#> `poly(x, degree = 2)1`  1.10450    0.16552   6.673 1.56e-09 ***
#> `poly(x, degree = 2)2` -1.62076    0.16552  -9.792 3.77e-16 ***
#> ---
#> Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
#> 
#> Residual standard error: 0.1655 on 97 degrees of freedom
#> Multiple R-squared:  0.5914, Adjusted R-squared:  0.583 
#> F-statistic: 70.21 on 2 and 97 DF,  p-value: < 2.2e-16

^{由 reprex 包于 2022 年 3 月 20 日创建 (v2.0.1)}

Answer 2

我们可以使用 paste 在这里创建公式

d <- 4
train(as.formula(paste0('y ~ poly(x, degree =', d, ')')), 
      data = training, method = "lm", trControl = fitCtrl)

-output

Linear Regression 

100 samples
  1 predictor

No pre-processing
Resampling: Leave-One-Out Cross-Validation 
Summary of sample sizes: 99, 99, 99, 99, 99, 99, ... 
Resampling results:

  RMSE        Rsquared   MAE       
  0.03790195  0.9779768  0.02937452

---

通过循环，我们可能需要将输出存储在 list 中

ds <- 1:20
fitlst <- vector('list', length(ds))
for(i in seq_along(ds)){
  print(i)
  d <- ds[i]
  
  fitlst[[i]] <- train(as.formula(paste0('y ~ poly(x, degree =', d, ')')), 
      data = training, method = "lm", trControl = fitCtrl)
      }

-output

[1] 1
[1] 2
[1] 3
[1] 4
[1] 5
[1] 6
[1] 7
[1] 8
[1] 9
[1] 10
[1] 11
[1] 12
[1] 13
[1] 14
[1] 15
[1] 16
[1] 17
[1] 18
[1] 19
[1] 20
> fitlst[[4]]
Linear Regression 

100 samples
  1 predictor

No pre-processing
Resampling: Leave-One-Out Cross-Validation 
Summary of sample sizes: 99, 99, 99, 99, 99, 99, ... 
Resampling results:

  RMSE        Rsquared   MAE       
  0.03790195  0.9779768  0.02937452

Tuning parameter 'intercept' was held constant at a value of TRUE