英文:
Smoothing splines, is there a way to automatically add s() to each column name?
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：```Rlibrary(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'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：``` rlibrary(MASS)library(gam)#> 加载所需的包: splines#> 加载所需的包: foreach#> 已加载 gam 1.22-2names_Boston <- names(Boston[,1:3])f1 <- as.formula(paste0('medv ~', paste0('s(', names_Boston, ')', collapse = '+')))gam(f1, data = Boston)#> 调用:#> gam(formula = f1, data = Boston)#>#> 自由度: 505 总数; 剩余 493.0002#> 剩余偏差: 26249.62
<sup>创建于2023-06-01，使用 reprex v2.0.2</sup>
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
英文:
This works for me:
library(MASS)library(gam)#> Loading required package: splines#> Loading required package: foreach#> Loaded gam 1.22-2names_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
<sup>Created on 2023-06-01 with reprex v2.0.2</sup>
But column 4 cannot be smoothed due to:
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 ResidualResidual Deviance: 26249.62gam(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 ResidualResidual Deviance: 26249.62
英文:
You may try
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 ResidualResidual Deviance: 26249.62gam(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 ResidualResidual Deviance: 26249.62
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