另外，尝试add_count()(以避开烦人的group_by .groups)。

mtcars %>% 
  count(am, gear) %>% 
  add_count(am, wt = n, name = "nn") %>% 
  mutate(proportion = n / nn)

为了这个常见问题的完整性，从dplyr的1.0.0版本开始，parameter .groups在group_by summary help之后控制了summary函数的组结构。

使用.groups = "drop_last"， summary删除分组的最后一层。这是版本1.0.0之前获得的唯一结果。

library(dplyr)
library(scales)

original <- mtcars %>%
  group_by (am, gear) %>%
  summarise (n=n()) %>%
  mutate(rel.freq =  scales::percent(n/sum(n), accuracy = 0.1))
#> `summarise()` regrouping output by 'am' (override with `.groups` argument)

original
#> # A tibble: 4 x 4
#> # Groups:   am [2]
#>      am  gear     n rel.freq
#>   <dbl> <dbl> <int> <chr>   
#> 1     0     3    15 78.9%   
#> 2     0     4     4 21.1%   
#> 3     1     4     8 61.5%   
#> 4     1     5     5 38.5%

new_drop_last <- mtcars %>%
  group_by (am, gear) %>%
  summarise (n=n(), .groups = "drop_last") %>%
  mutate(rel.freq =  scales::percent(n/sum(n), accuracy = 0.1))

dplyr::all_equal(original, new_drop_last)
#> [1] TRUE

使用.groups = "drop"，删除所有级别的分组。结果被转换为一个独立的tibble，没有前一个group_by的痕迹

# .groups = "drop"
new_drop <- mtcars %>%
  group_by (am, gear) %>%
  summarise (n=n(), .groups = "drop") %>%
  mutate(rel.freq =  scales::percent(n/sum(n), accuracy = 0.1))

new_drop
#> # A tibble: 4 x 4
#>      am  gear     n rel.freq
#>   <dbl> <dbl> <int> <chr>   
#> 1     0     3    15 46.9%   
#> 2     0     4     4 12.5%   
#> 3     1     4     8 25.0%   
#> 4     1     5     5 15.6%

如果.groups = "keep"，与.data(在本例中为mtcars)相同的分组结构。summary不会剥离group_by中使用的任何变量。

最后，使用.groups = "rowwise"，每一行都是它自己的组。在这种情况下相当于“keep”

# .groups = "keep"
new_keep <- mtcars %>%
  group_by (am, gear) %>%
  summarise (n=n(), .groups = "keep") %>%
  mutate(rel.freq =  scales::percent(n/sum(n), accuracy = 0.1))

new_keep
#> # A tibble: 4 x 4
#> # Groups:   am, gear [4]
#>      am  gear     n rel.freq
#>   <dbl> <dbl> <int> <chr>   
#> 1     0     3    15 100.0%  
#> 2     0     4     4 100.0%  
#> 3     1     4     8 100.0%  
#> 4     1     5     5 100.0%

# .groups = "rowwise"
new_rowwise <- mtcars %>%
  group_by (am, gear) %>%
  summarise (n=n(), .groups = "rowwise") %>%
  mutate(rel.freq =  scales::percent(n/sum(n), accuracy = 0.1))

dplyr::all_equal(new_keep, new_rowwise)
#> [1] TRUE

另一个值得注意的地方是，在应用group_by和summarise之后，有时可以使用总结行。

# create a subtotal line to help readability
subtotal_am <- mtcars %>%
  group_by (am) %>% 
  summarise (n=n()) %>%
  mutate(gear = NA, rel.freq = 1)
#> `summarise()` ungrouping output (override with `.groups` argument)

mtcars %>% group_by (am, gear) %>%
  summarise (n=n()) %>% 
  mutate(rel.freq = n/sum(n)) %>%
  bind_rows(subtotal_am) %>%
  arrange(am, gear) %>%
  mutate(rel.freq =  scales::percent(rel.freq, accuracy = 0.1))
#> `summarise()` regrouping output by 'am' (override with `.groups` argument)
#> # A tibble: 6 x 4
#> # Groups:   am [2]
#>      am  gear     n rel.freq
#>   <dbl> <dbl> <int> <chr>   
#> 1     0     3    15 78.9%   
#> 2     0     4     4 21.1%   
#> 3     0    NA    19 100.0%  
#> 4     1     4     8 61.5%   
#> 5     1     5     5 38.5%   
#> 6     1    NA    13 100.0%

由reprex包在2020-11-09创建(v0.3.0)

希望这个答案对你有用。

另外，尝试add_count()(以避开烦人的group_by .groups)。

mtcars %>% 
  count(am, gear) %>% 
  add_count(am, wt = n, name = "nn") %>% 
  mutate(proportion = n / nn)

@Henrik's的可用性更好，因为这将使列字符，不再是数字，但符合您的要求…

mtcars %>%
  group_by (am, gear) %>%
  summarise (n=n()) %>%
  mutate(rel.freq = paste0(round(100 * n/sum(n), 0), "%"))

##   am gear  n rel.freq
## 1  0    3 15      79%
## 2  0    4  4      21%
## 3  1    4  8      62%
## 4  1    5  5      38%

因为这是太空人要求的:-)

as.rel_freq <- function(x, rel_freq_col = "rel.freq", ...) {
    class(x) <- c("rel_freq", class(x))
    attributes(x)[["rel_freq_col"]] <- rel_freq_col
    x
}

print.rel_freq <- function(x, ...) {
    freq_col <- attributes(x)[["rel_freq_col"]]
    x[[freq_col]] <- paste0(round(100 * x[[freq_col]], 0), "%")   
    class(x) <- class(x)[!class(x)%in% "rel_freq"]
    print(x)
}

mtcars %>%
  group_by (am, gear) %>%
  summarise (n=n()) %>%
  mutate(rel.freq = n/sum(n)) %>%
  as.rel_freq()

## Source: local data frame [4 x 4]
## Groups: am
## 
##   am gear  n rel.freq
## 1  0    3 15      79%
## 2  0    4  4      21%
## 3  1    4  8      62%
## 4  1    5  5      38%

下面是一个基于R的答案，使用了aggregate和ave:

df1 <- with(mtcars, aggregate(list(n = mpg), list(am = am, gear = gear), length))
df1$prop <- with(df1, n/ave(n, am, FUN = sum))
#Also with prop.table
#df1$prop <- with(df1, ave(n, am, FUN = prop.table))
df1

#  am gear  n      prop
#1  0    3 15 0.7894737
#2  0    4  4 0.2105263
#3  1    4  8 0.6153846
#4  1    5  5 0.3846154 

我们也可以用道具。表，但输出显示不同。

prop.table(table(mtcars$am, mtcars$gear), 1)
   
#            3         4         5
#  0 0.7894737 0.2105263 0.0000000
#  1 0.0000000 0.6153846 0.3846154

这个答案是基于Matifou的回答。

首先，我修改了它，以确保我不会通过使用scipen选项将freq列作为科学符号列返回。

然后我将答案乘以100以得到一个百分比，而不是小数，以使频率列更容易以百分比的形式阅读。

getOption("scipen") 
options("scipen"=10) 
mtcars %>%
count(am, gear) %>% 
mutate(freq = (n / sum(n)) * 100)