英文:
Automatically select best columns for best subset
问题
The Leaps package允许识别子集，并且可以返回最佳子集的结果。例如：
library(leaps)mtcars.regsubsets <- regsubsets(mpg ~ ., data = mtcars)summary(mtcars.regsubsets, nvmax = 10)
选择算法：穷举cyl disp hp drat wt qsec vs am gear carb1 ( 1 ) " " " " " " " " " " " " "*" " " " " " " " " " " " " " " " "2 ( 1 ) "*" " " " " " " " " "*" " " " " " " " " " " " " " " " "3 ( 1 ) " " " " " " " " " " " " "*" "*" " " " " "*" " " " " "4 ( 1 ) " " " " " " "*" " " "*" "*" " " " " "*" " " " " "5 ( 1 ) " " " "*" "*" " " "*" "*" " " " " "*" " " " " "6 ( 1 ) " " " "*" "*" "*" "*" "*" " " " " "*" " " " " "7 ( 1 ) " " " "*" "*" "*" "*" "*" " " " " "*" "*" " " "8 ( 1 ) " " " "*" "*" "*" "*" "*" " " " " "*" "*" "*"
总结部分对于列数的最佳结果给出了 "*"。还可以使用特定的度量标准，例如Mallow's Cp来找到最佳结果：
which.min(summary(mtcars.regsubsets)$cp)
3
这返回结论，即3变量集具有Mallow's Cp的最低分数。
是否有一种方法可以自动选择数据集中的这些列（在本例中为wt、qsec和am），以便返回一个只包含这三列的新数据集？
mtcars[,c(5,6,8)]
显然，Leaps包的帮助文件已经查看过了，查看了StackOverflow和其他来源，但没有找到解决方案。
英文:
The Leaps package allows identification of subsets, and can return results for the best subsets. For example:
library(leaps)mtcars.regsubsets <- regsubsets(mpg ~ ., data = mtcars)summary(mtcars.regsubsets, nvmax = 10)
Selection Algorithm: exhaustivecyl disp hp drat wt qsec vs am gear carb1 ( 1 ) " " " " " " " " "*" " " " " " " " " " "2 ( 1 ) "*" " " " " " " "*" " " " " " " " " " "3 ( 1 ) " " " " " " " " "*" "*" " " "*" " " " "4 ( 1 ) " " " " "*" " " "*" "*" " " "*" " " " "5 ( 1 ) " " "*" "*" " " "*" "*" " " "*" " " " "6 ( 1 ) " " "*" "*" "*" "*" "*" " " "*" " " " "7 ( 1 ) " " "*" "*" "*" "*" "*" " " "*" "*" " "8 ( 1 ) " " "*" "*" "*" "*" "*" " " "*" "*" "*"
The summary gives an "*" for the best results for the number of columns. It's also possible to use a specific measure, such as Mallow's Cp to find the best result:
which.min(summary(mtcars.regsubsets)$cp)
3
This returns that conclusion that the 3-variable set has the lowest score for Mallow's Cp.
Is there a way to automatically select those columns (wt, sec and am in this case) in the data set, so it returns a new data set only with those three columns?
mtcars[,c(5,6,8)]
Obviously the help file for the Leaps package has been checked, looked on StackOverflow, other sources, but no solution was found.
答案1
得分: 2
以下是使用dplyr的一种方法：
library(leaps)library(dplyr)mtcars.regsubsets <- regsubsets(mpg ~ ., data = mtcars)cols_sel <- mtcars.regsubsets %>%broom::tidy() %>%filter(mallows_cp == min(mallows_cp)) %>%select(where(~.x == TRUE)) %>%colnames()select(mtcars, any_of(cols_sel))
英文:
Here is one way using dplyr:
library(leaps)library(dplyr)mtcars.regsubsets <- regsubsets(mpg ~ ., data = mtcars)cols_sel <- mtcars.regsubsets |>broom::tidy() |>filter(mallows_cp == min(mallows_cp)) |>select(where(~.x == TRUE)) |>colnames()select(mtcars, any_of(cols_sel))
答案2
得分: 0
你可以使用summary组件的"which"来扫描其列，逐行查找"TRUE"值，然后检索关联的列=变量名称并将其存储在列表中。然后，您可以从此列表中选择列名，使用所需数量的变量。
使用基本的R：
column_sets <-apply(regsubsets(mpg ~ ., data = mtcars) |summary() |(`[[`)('which'),MARGIN = 1,FUN = \(x) names(x)[x][-1] ## drop "(Intercept)")
mtcars[column_sets[['3']]]
+ mtcars[column_sets[['3']]]wt qsec amMazda RX4 2.620 16.46 1Mazda RX4 Wag 2.875 17.02 1Datsun 710 2.320 18.61 1Hornet 4 Drive 3.215 19.44 0Hornet Sportabout 3.440 17.02 0Valiant 3.460 20.22 0Duster 360 3.570 15.84 0## ...
英文:
You can use the summarys component "which" and scan its columns, row by row, for TRUE values, the retrieve the associated column = variable names and store it in a list. Then, you can pick the column names from this list, using the desired number of variables.
With base R:
column_sets <-apply(regsubsets(mpg ~ ., data = mtcars) |>summary() |>(`[[`)('which'),MARGIN = 1,FUN = \(x) names(x)[x][-1] ## drop "(Intercept)")
mtcars[column_sets[['3']]]
+ mtcars[column_sets[['3']]]wt qsec amMazda RX4 2.620 16.46 1Mazda RX4 Wag 2.875 17.02 1Datsun 710 2.320 18.61 1Hornet 4 Drive 3.215 19.44 0Hornet Sportabout 3.440 17.02 0Valiant 3.460 20.22 0Duster 360 3.570 15.84 0## ...
答案3
得分: 0
获取摘要，其中有一个 "which" 矩阵，获取不包括截距的列总和，并获取具有最多 TRUE 值的列，使用 tail：
ss <- summary(mtcars.regsubsets, nvmax = 10)names(tail(sort(colSums(ss$which[, -1 ])), which.min(ss$cp)))# [1] "qsec" "am" "wt"
(Note: The translation retains the code snippet as is, without translating the code itself.)
英文:
Get the summary, it has a "which" matrix, get column sums excluding intercept, and get the columns with most TRUE values, using tail:
ss <- summary(mtcars.regsubsets, nvmax = 10)names(tail(sort(colSums(ss$which[, -1 ])), which.min(ss$cp)))# [1] "qsec" "am" "wt"
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