问题

I can provide the translated code for your request:

我正在使用MASS包中的波士顿房屋数据集。期望的目标是类似于这样的：

```R
library(MASS)
library(tidyverse)
library(gam)
Boston.splines <- gam(medv ~ s(crime) + s(zn) + s(indus), data = Boston)

我可以自动完成除了样条函数部分之外的所有内容：

names_Boston <- names(Boston[,1:4])

f1 <- paste("medv ~", paste(names_Boston, collapse = "+"))
f1 <- as.formula(f1)
Boston1.gam <- gam(f1, data = Boston)

但是，我无论如何都似乎无法在每个列名的前面添加s()函数。

我尝试过dplyr和基本的R，但都不起作用。例如，这个：

set_names(paste0('s(', paste0(names_Boston), paste0(')')))

返回了一个错误消息：

Error: unexpected string constant in "set_names(paste0('s(', paste0(names_Boston), paste0(')')))"

有什么方法可以自动将平滑样条函数添加到列名以得到类似于gam(medv ~ s(crime) + s(zn) + s(indus), data = Boston)的公式吗？

Please note that the code itself remains in English as you requested, and only the surrounding text is translated.

<details>
<summary>英文:</summary>

I&#39;m working with the Boston Housing data set in the MASS package. The desired goal is something like this:

library(MASS)
library(tidyverse)
library(gam)
Boston.splines <- gam(medv ~ s(crime) + s(zn) + s(indus), data = Boston)

I can get everything except the spline function to work automatically:

names_Boston <- names(Boston[,1:4])

f1 <- paste("medv ~", paste(names_Boston, collapse = "+"))
f1 <- as.formula(f1)
Boston1.gam <- gam(f1, data = Boston)

But for the life of me I can't seem to get the s() function to be added to the front of each of the column names.

I've tried dplyr and base R, nothing works. For example, this:

set_names(paste0('s(', paste0(names_Boston), paste0')))

returned an error message:

Error: unexpected string constant in "set_names(paste0('s(', paste0(names_Boston), paste0')'"

What is a way to automatically add the smoothing spline function to column names to result in a formula such as gam(medv ~ s(crime) + s(zn) + s(indus), data = Boston)?

</details>


# 答案1
**得分**: 2

这对我有用：

``` r
library(MASS)
library(gam)
#&gt; 加载所需的包: splines
#&gt; 加载所需的包: foreach
#&gt; 已加载 gam 1.22-2
names_Boston &lt;- names(Boston[,1:3])
f1 &lt;- as.formula(paste0(&#39;medv ~&#39;, paste0(&#39;s(&#39;, names_Boston, &#39;)&#39;, collapse = &#39;+&#39;)))
gam(f1, data = Boston)
#&gt; 调用:
#&gt; gam(formula = f1, data = Boston)
#&gt; 
#&gt; 自由度: 505 总数; 剩余 493.0002
#&gt; 剩余偏差: 26249.62

^{创建于2023-06-01，使用 reprex v2.0.2}

但第4列无法平滑处理，因为：

一个平滑变量遇到了3个或更少的唯一值；需要至少4个

library(MASS)
library(gam)
#> Loading required package: splines
#> Loading required package: foreach
#> Loaded gam 1.22-2
names_Boston <- names(Boston[,1:3])
f1 <- as.formula(paste0('medv ~', paste0('s(', names_Boston, ')', collapse = '+')))
gam(f1, data = Boston)
#> Call:
#> gam(formula = f1, data = Boston)
#> 
#> Degrees of Freedom: 505 total; 493.0002 Residual
#> Residual Deviance: 26249.62

A smoothing variable encountered with 3 or less unique values; at least 4 needed

答案2

得分: 1

你可以尝试：

names_Boston <- names(Boston[,1:3])
f1 <- paste("medv ~", paste0(sprintf("s(%s)", names_Boston), collapse = " + "))

f1
[1] "medv ~ s(crim) + s(zn) + s(indus)"

gam(as.formula(f1), data = Boston)

Call:
gam(formula = as.formula(f1), data = Boston)

Degrees of Freedom: 505 total; 493.0002 Residual
Residual Deviance: 26249.62

gam(medv ~ s(crim) + s(zn) + s(indus), data = Boston)

Call:
gam(formula = medv ~ s(crim) + s(zn) + s(indus), data = Boston)

Degrees of Freedom: 505 total; 493.0002 Residual
Residual Deviance: 26249.62

names_Boston <- names(Boston[,1:3])
f1 <- paste("medv ~", paste0(sprintf("s(%s)", names_Boston), collapse = " + "))

f1
[1] "medv ~ s(crim) + s(zn) + s(indus)"


gam(as.formula(f1), data = Boston)

Call:
gam(formula = as.formula(f1), data = Boston)

Degrees of Freedom: 505 total; 493.0002 Residual
Residual Deviance: 26249.62 

gam(medv ~ s(crim) + s(zn) + s(indus), data = Boston)

Call:
gam(formula = medv ~ s(crim) + s(zn) + s(indus), data = Boston)

Degrees of Freedom: 505 total; 493.0002 Residual
Residual Deviance: 26249.62