map_dfr 输出一行而不是一列

这与purrr::map_dfr binds by columns, not row as expected相似，但那里的解决方案对我不起作用.我有一个数据框

beta_df <- structure(list(intercept = c(-2.75747056032685, -2.90831892599742, 
-2.92478082251453, -2.99701559041538, -2.88885796048347, -3.09564193631675
), B1 = c(0.0898235360814854, 0.0291839369781567, 0.0881023522236231, 
0.231703026085554, 0.0441573699433149, 0.258219673780526), B2 = c(-0.222367437619057, 
0.770536384299238, 0.199648657850609, 0.0529038155448773, 0.00310458335580774, 
0.132604387458483), B3 = c(1.26339268033385, 1.29883641278223, 
0.949504940387809, 1.26904511447941, 0.863882674439083, 0.823907268679309
), B4 = c(2.13662994525526, 1.02340744740827, 0.959079691725652, 
1.60672779812489, 1.19095838867883, -0.0693120654049908)), row.names = c(NA, 
-6L), class = c("tbl_df", "tbl", "data.frame"))

#> # A tibble: 6 × 5
#>   intercept     B1       B2    B3      B4
#>       <dbl>  <dbl>    <dbl> <dbl>   <dbl>
#> 1     -2.76 0.0898 -0.222   1.26   2.14  
#> 2     -2.91 0.0292  0.771   1.30   1.02  
#> 3     -2.92 0.0881  0.200   0.950  0.959 
#> 4     -3.00 0.232   0.0529  1.27   1.61  
#> 5     -2.89 0.0442  0.00310 0.864  1.19  
#> 6     -3.10 0.258   0.133   0.824 -0.0693

我想把它转换成一个tibble，用列表示平均值，0.025和0.975分位数.对于quantile功能，此功能起作用:

beta_df %>%
    map_dfr(quantile,0.025)

#> # A tibble: 5 × 1
#>    `2.5%`
#>     <dbl>
#> 1 -3.08  
#> 2  0.0311
#> 3 -0.194 
#> 4  0.829 
#> 5  0.0592

这就得到了两个分位数

bind_cols(beta_df %>%
    map_dfr(quantile, 0.025),
    beta_df %>%
    map_dfr(quantile, 0.975))

#> # A tibble: 5 × 2
#>    `2.5%` `97.5%`
#>     <dbl>   <dbl>
#> 1 -3.08    -2.77 
#> 2  0.0311   0.255
#> 3 -0.194    0.699
#> 4  0.829    1.30 
#> 5  0.0592   2.07

但是对于mean，

beta_df %>%
    map_dfr(mean)

#> # A tibble: 1 × 5
#>   intercept    B1    B2    B3    B4
#>       <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1     -2.93 0.124 0.156  1.08  1.14

给我一个长的行而不是列.如何将原始数据框中每列的平均值转换为标记为平均值的单列数据框中的一行？