使用tidyverse从列表到数据框，选择特定的列表项

一个简单的问题，但我一直在寻找解决方案，但到目前为止没有任何结果。

假设我有一个列表对象，并且想提取特定的列表项并将其作为数据帧列并排输出。如何以简单的方式使用tidyverse /管道实现此目的？尝试在下面解决它。

数据

some_data <-
structure(list(x = c(23.7, 23.41, 23.87, 24.18, 24.15, 24.31, 
23.14, 23.72, 24.12, 23.47, 23.59, 23.29, 23.24, 23.5, 23.56, 
23.16, 23.62, 23.67, 23.84, 23.69, 23.7, 23.68, 24.2, 23.77, 
23.74, 23.64, 24.39, 24.05, 24.51, 23.6, 24.29, 23.31, 23.96, 
24.07, 24.37, 23.77, 23.64, 24, 23.68, 24.02, 23.36, 23.54, 23.34, 
23.69, 23.79, 23.8, 23.7, 24.45, 23.27, 23.57, 23.02, 24.23, 
23.41, 23.6, 24.02, 23.94, 24.06, 23.97, 23.38, 23.46, 24, 23.89, 
23.51, 23.72, 23.83, 23.96, 23.84, 23.52, 24.36, 23.94, 23.82, 
24.04, 24.05, 23.6, 23.52, 24.13, 23.43, 23.33, 24.01, 23.99, 
24.46, 24.23, 24.19, 23.83, 23.8, 23.93, 23.79, 23.48, 23.26, 
24.04, 23.93, 23.98, 23.86, 23.49, 24.17, 23.7, 23.54, 23.55, 
23.67, 23.66)), class = c("spec_tbl_df", "tbl_df", "tbl", "data.frame"
), row.names = c(NA, -100L), spec = structure(list(cols = list(
    x = structure(list(), class = c("collector_double", "collector"
    ))), default = structure(list(), class = c("collector_guess", 
"collector")), skip = 1), class = "col_spec"))

我想要此数据的hist（）函数的值输出

library(tidyverse)

some_data$x %>% 
   as.numeric() %>% 
   hist(breaks = seq(from = 23, to = 24.6, by = 0.2),
        plot = FALSE)

## $breaks
## [1] 23.0 23.2 23.4 23.6 23.8 24.0 24.2 24.4 24.6

## $counts
## [1]  3  9 20 23 19 16  7  3

## $density
## [1] 0.15 0.45 1.00 1.15 0.95 0.80 0.35 0.15

## $mids
## [1] 23.1 23.3 23.5 23.7 23.9 24.1 24.3 24.5

## $xname
## [1] "."

## $equidist
## [1] TRUE

## attr(,"class")
## [1] "histogram"

假设我要同时使用$ breaks和$ counts作为数据框 我将补充原始管道，以便：

some_data$x %>% 
   as.numeric() %>% 
   hist(breaks = seq(from = 23, to = 24.6, by = 0.2),
        plot = FALSE) %>%
##
   map_df(~.[1:30]) %>%
   select(bins = breaks, 
          frequency = counts)
##

## # A tibble: 30 x 2
##     bins frequency
##    <dbl>     <int>
##  1  23           3
##  2  23.2         9
##  3  23.4        20
##  4  23.6        23
##  5  23.8        19
##  6  24          16
##  7  24.2         7
##  8  24.4         3
##  9  24.6        NA
## 10  NA          NA
## # ... with 20 more rows

是的，它确实有效，但是map_df()我必须放置一个相对较大的“魔术”数字（任意输入30），以确保包括所有数据。有没有一种更简单的获取方式$breaks和$counts作为数据框？也许甚至只用一个步骤而不是合并map_df()然后select()呢？

谢谢！