R（tidyverse）-卡方检验的2个分类变量的汇总数据的列总和？

Question

有人可以给我他们的建议吗？

1. 我想总结一下我的专栏总数。
2. 我需要框架进行卡方独立性测试，因此，如果有更快的方法，请赐教！

做这个的最好方式是什么？

我尝试使用ColSums，但它给了我一个错误（colSums（。，mpaa_rating，na.rm = FALSE，dims = 1中的错误）：未使用的参数（mpaa_rating）。我显然没有正确使用它或在正确的地方输入它我尝试过：colSums（mpaa_rating，na.rm = FALSE，dims = 1）％>％刚刚高于价差。

提前致谢，克里斯汀

rereprex::reprex_info() 
movie_help<- data.frame(tribble(
             ~mpaa_rating,                       ~genre,
                     "PG",         "Action & Adventure",
                      "R",         "Mystery & Suspense",
                      "R",                      "Drama",
                      "R",                      "Drama",
                      "R",                      "Drama",
                     "PG",         "Action & Adventure",
                  "PG-13",                     "Comedy",
                      "R",                     "Comedy",
                      "R",         "Action & Adventure",
                      "R",                      "Drama",
                      "R",                      "Drama",
                      "G",                      "Drama",
                      "R",                     "Comedy",
                      "R",                      "Drama",
                      "R",         "Mystery & Suspense",
                      "R",  "Musical & Performing Arts",
                "Unrated",                      "Drama",
                      "R",                      "Drama",
                  "PG-13",                      "Drama",
                  "PG-13",                      "Drama"
             )) 
movie_help %>% 
filter(!is.na(genre), !is.na(mpaa_rating)) %>% 
count(genre, mpaa_rating) %>%
group_by(genre) %>%
mutate(prop = n) %>%
mutate(Total= sum(n)) %>%
select(-n) %>%
spread(key = mpaa_rating, value = prop) 
#> # A tibble: 5 x 7
#> # Groups:   genre [5]
#>                       genre Total     G    PG `PG-13`     R Unrated
#> *                     <chr> <int> <int> <int>   <int> <int>   <int>
#> 1        Action & Adventure     3    NA     2      NA     1      NA
#> 2                    Comedy     3    NA    NA       1     2      NA
#> 3                     Drama    11     1    NA       2     7       1
#> 4 Musical & Performing Arts     1    NA    NA      NA     1      NA
#> 5        Mystery & Suspense     2    NA    NA      NA     2      NA

Answer 1

为了获得总和，我喜欢使用janitor包中的janitor::adorn_totals函数。 管理员程序包具有许多小的帮助程序功能，适用于需要以所需方式清理表的情况。 在这里查看更多信息。 我最喜欢的还是janitor::clean_names ，它可以帮助您统一清理列名。

现在，您可以轻松地：

 movie_help %>% 
    filter(!is.na(genre), !is.na(mpaa_rating)) %>% 
    count(genre, mpaa_rating) %>% 
    group_by(genre) %>%
    mutate(prop = n) %>%
    mutate(Total= sum(n)) %>%  
    select(-n) %>%
    spread(key = mpaa_rating, value = prop, fill = 0) %>% 
    janitor::adorn_totals('row') %>% 
    janitor::clean_names() 

Answer 2

我们可以使用table和chisq.test来执行您想要的测试：

chisq.test(table(movie_help))

我们还可以手动计算总数：

dat <- movie_help %>%
  filter(!is.na(genre),!is.na(mpaa_rating)) %>%
  count(genre, mpaa_rating) %>%
  group_by(genre) %>%
  mutate(prop = n) %>%
  mutate(Total = sum(n)) %>%
  select(-n) %>%
  spread(key = mpaa_rating, value = prop) 

bind_rows(dat, 
          cbind(data_frame('genre' = 'Total'), summarise_all(dat[,-1], sum, na.rm = T)))

  genre                     Total     G    PG `PG-13`     R Unrated
  <chr>                     <int> <int> <int>   <int> <int>   <int>
1 Action & Adventure            3    NA     2      NA     1      NA
2 Comedy                        3    NA    NA       1     2      NA
3 Drama                        11     1    NA       2     7       1
4 Musical & Performing Arts     1    NA    NA      NA     1      NA
5 Mystery & Suspense            2    NA    NA      NA     2      NA
6 Total                        20     1     2       3    13       1