管道内的hist（）和dplyr中的group_by

I'm looking for a way to utilize dplyr 's group_by functionality to make counts, and a plot-less histogram for mpg after grouping_by gear and vs.

我正在寻找一种方法来利用dplyr的group_by功能进行计数,以及在grouping_by gear和vs.之后使用mpg的无绘图直方图

My code is:

我的代码是:

mtcars %>% 
group_by(gear,vs) %>% 
summarise(counts = count (n), hist(mpg, plot = FALSE, breaks = c(seq(10,40,1))))

The error is:

错误是:

Error in summarise_impl(.data, dots) : Column hist(mpg, plot = FALSE, breaks = c(seq(10, 40, 1)))` must be length 1 (a summary value), not 6

summarise_impl(.data,dots)出错:列hist(mpg,plot = FALSE,breaks = c(seq(10,40,1)))`必须是长度1(摘要值),而不是6

I'm not constrained to dplyr but that's all I'm familiar with in R at this point.

我并不局限于dplyr,但这就是我在R中所熟悉的。

Any help is appreciated.

任何帮助表示赞赏。

2 个解决方案

#1

I am not really sure what a plotless histogram is, but does this help?

我不确定无量纲直方图是什么,但这有帮助吗?

mtcars %>%
    mutate(mpgClasses = cut(mpg, 10:40)) %>%
    group_by(gear, vs, mpgClasses) %>%
    summarise(n())

You can also rearrange that a little like this

你也可以重新安排一下这样的事情

mtcars %>%
    mutate(mpgClasses = cut(mpg, 10:40)) %>%
    group_by(gear, vs, mpgClasses) %>%
    summarise(counts = n()) %>%
    spread(mpgClasses, counts)

If you can describe a little more, what you are heading for we can find a better solution.

如果你能描述一下,你前进的目标是什么,我们可以找到更好的解决方案。

#2

Here, I'm just extracting the counts from the hist. Since I have to make it one element for each group, I made it a list.

在这里,我只是从hist中提取计数。因为我必须为每个组制作一个元素,所以我把它作为一个列表。

library(dplyr)
x <- mtcars %>% group_by(gear,vs) %>%     
   summarise(counts = n(),
             hcounts =  list(hist(mpg, plot = FALSE, breaks = c(seq(10,40,1)))$counts))

x
# # A tibble: 6 x 4
# # Groups:   gear [?]
# gear    vs counts    hcounts
# <dbl> <dbl>  <int>     <list>
#   1     3     0     12 <int [30]>
#   2     3     1      3 <int [30]>
#   3     4     0      2 <int [30]>
#   4     4     1     10 <int [30]>
#   5     5     0      4 <int [30]>
#   6     5     1      1 <int [30]>

x$hcounts
# [[1]]
# [1] 2 0 0 1 2 3 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
# 
# [[2]]
# [1] 0 0 0 0 0 0 0 0 1 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
# 
# [[3]]
# [1] 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
# 
# [[4]]
# [1] 0 0 0 0 0 0 0 1 0 1 0 1 2 0 1 0 0 1 0 0 1 0 1 1 0 0 0 0 0 0
# 
# [[5]]
# [1] 0 0 0 0 1 1 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
# 
# [[6]]
# [1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0

#1