运行`map :: purrr()`函数后，用列名替换`.x` [英] replacing `.x` with column name after running `map::purrr()` function
问题描述
我使用purrr:map()函数对数据集中的每一列运行lm()，其中一列作为因变量.
I run lm() for every column of a dataset with one of the column as the dependent variable, using purrr:map() function.
除此以外，结果几乎是完美的-我想用运行lm()的变量替换结果中的.x.
The results are almost perfect except for this - I want to replace .x in the results with the variable that i run lm() for.
帖子 R purrr映射在输出中显示列名称是相关的，但我想避免创建一个函数.
The post R purrr map show column names in output is related but I want to avoid creating a function.
下面是使用mtcars数据集的代码.我知道，例如，第一个输出的.x指的是$mpg.我不确定setNames()是否可以工作.
Below, are the codes using the mtcars dataset. I know, for example that .x for the first output refers to $mpg. I am not sure if setNames() will work.
library(tidyverse)
library(broom)
mod3 <- map(mtcars, ~ lm(mpg ~ .x, data = mtcars)) %>%
map(~tidy(.x))
#> Warning in summary.lm(x): essentially perfect fit: summary may be
#> unreliable
mod3
#> $mpg
#> # A tibble: 2 x 5
#> term estimate std.error statistic p.value
#> <chr> <dbl> <dbl> <dbl> <dbl>
#> 1 (Intercept) -5.02e-15 9.94e-16 -5.06e 0 0.0000198
#> 2 .x 1.00e+ 0 4.74e-17 2.11e16 0
#>
#> $cyl
#> # A tibble: 2 x 5
#> term estimate std.error statistic p.value
#> <chr> <dbl> <dbl> <dbl> <dbl>
#> 1 (Intercept) 37.9 2.07 18.3 8.37e-18
#> 2 .x -2.88 0.322 -8.92 6.11e-10
#>
#> $disp
#> # A tibble: 2 x 5
#> term estimate std.error statistic p.value
#> <chr> <dbl> <dbl> <dbl> <dbl>
#> 1 (Intercept) 29.6 1.23 24.1 3.58e-21
#> 2 .x -0.0412 0.00471 -8.75 9.38e-10
#>
#> $hp
#> # A tibble: 2 x 5
#> term estimate std.error statistic p.value
#> <chr> <dbl> <dbl> <dbl> <dbl>
#> 1 (Intercept) 30.1 1.63 18.4 6.64e-18
#> 2 .x -0.0682 0.0101 -6.74 1.79e- 7
#>
#> $drat
#> # A tibble: 2 x 5
#> term estimate std.error statistic p.value
#> <chr> <dbl> <dbl> <dbl> <dbl>
#> 1 (Intercept) -7.52 5.48 -1.37 0.180
#> 2 .x 7.68 1.51 5.10 0.0000178
#>
#> $wt
#> # A tibble: 2 x 5
#> term estimate std.error statistic p.value
#> <chr> <dbl> <dbl> <dbl> <dbl>
#> 1 (Intercept) 37.3 1.88 19.9 8.24e-19
#> 2 .x -5.34 0.559 -9.56 1.29e-10
#>
#> $qsec
#> # A tibble: 2 x 5
#> term estimate std.error statistic p.value
#> <chr> <dbl> <dbl> <dbl> <dbl>
#> 1 (Intercept) -5.11 10.0 -0.510 0.614
#> 2 .x 1.41 0.559 2.53 0.0171
#>
#> $vs
#> # A tibble: 2 x 5
#> term estimate std.error statistic p.value
#> <chr> <dbl> <dbl> <dbl> <dbl>
#> 1 (Intercept) 16.6 1.08 15.4 8.85e-16
#> 2 .x 7.94 1.63 4.86 3.42e- 5
推荐答案
这是一种实现方法
library(tidyverse)
library(broom)
names(mtcars)[-1] %>%
set_names() %>%
map(~ lm(as.formula(paste0('mpg ~ ', .x)), data = mtcars)) %>%
map_dfr(., broom::tidy, .id = "variable")
#> # A tibble: 20 x 6
#> variable term estimate std.error statistic p.value
#> <chr> <chr> <dbl> <dbl> <dbl> <dbl>
#> 1 cyl (Intercept) 37.9 2.07 18.3 8.37e-18
#> 2 cyl cyl -2.88 0.322 -8.92 6.11e-10
#> 3 disp (Intercept) 29.6 1.23 24.1 3.58e-21
#> 4 disp disp -0.0412 0.00471 -8.75 9.38e-10
#> 5 hp (Intercept) 30.1 1.63 18.4 6.64e-18
#> 6 hp hp -0.0682 0.0101 -6.74 1.79e- 7
#> 7 drat (Intercept) -7.52 5.48 -1.37 1.80e- 1
#> 8 drat drat 7.68 1.51 5.10 1.78e- 5
#> 9 wt (Intercept) 37.3 1.88 19.9 8.24e-19
#> 10 wt wt -5.34 0.559 -9.56 1.29e-10
#> 11 qsec (Intercept) -5.11 10.0 -0.510 6.14e- 1
#> 12 qsec qsec 1.41 0.559 2.53 1.71e- 2
#> 13 vs (Intercept) 16.6 1.08 15.4 8.85e-16
#> 14 vs vs 7.94 1.63 4.86 3.42e- 5
#> 15 am (Intercept) 17.1 1.12 15.2 1.13e-15
#> 16 am am 7.24 1.76 4.11 2.85e- 4
#> 17 gear (Intercept) 5.62 4.92 1.14 2.62e- 1
#> 18 gear gear 3.92 1.31 3.00 5.40e- 3
#> 19 carb (Intercept) 25.9 1.84 14.1 9.22e-15
#> 20 carb carb -2.06 0.569 -3.62 1.08e- 3
由 reprex软件包(v0.2.1.9000)创建于2019-02-10
Created on 2019-02-10 by the reprex package (v0.2.1.9000)
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