Question

有几个类似的问题可以抓取chi-square结果，但这解决了我的问题。我想为chi-square中所有列的data_frame测试计算p.values，并将其存储在原始data_frame中的列中。会有重复的值，我很好。最后，我希望select data_frame中所有列的p.value都低于x，并且我的选择变量。

require(dplyr)

my_df <- data_frame(
  one_f = sample(LETTERS[1:5],100,T),
  two_f = sample(LETTERS[4:5],100,T),
  three_f = sample(LETTERS[5],100,T)
)
my_df %>% 
  head()

my_df %>% 
  summarise_all(funs(chisq.test(.,my_df$two_f)$p.value))

给我这个错误：

Error in summarise_impl(.data, dots) : 
  Evaluation error: 'x' and 'y' must have at least 2 levels.


my_df %>% 
  mutate_if(n_distinct>1,fun(chisq.test(.,my_df$two_f)$p.value))

给我这个错误：

Error in n_distinct > 1 : 
  comparison (6) is possible only for atomic and list types

我正在寻找类似的东西。

my_df %>% 
      mutate(p.value = sample(c(0.043,0.87,0.00),nrow(.),T)) %>% 
      head()

然后我计划使用gather和filter然后spread根据我的chi-square测试获取显着关联的变量。

我想

my_df %>% filter(foo,bar >= 0.05)#function that finds p.values and filters by 
# alpha level

将是我的最终目标。

Answer 1

require(dplyr)
require(tidyr)

my_df <- data_frame(
  one_f = sample(LETTERS[1:5],100,T),
  two_f = sample(LETTERS[4:5],100,T),
  three_f = sample(LETTERS[5],100,T)
)

# select all column names where the column has more than 1 distinct values
my_df %>% 
  summarise_all(function(x) length(unique(x))) %>%
  gather() %>%
  filter(value > 1) %>%
  pull(key) -> list_cols

# apply function only to those columns
my_df %>% 
  select(list_cols) %>%
  summarise_all(funs(chisq.test(.,my_df$two_f)$p.value))

# # A tibble: 1 x 2
#     one_f                      two_f
#     <dbl>                      <dbl>
#   1 0.880 0.000000000000000000000120

使用dplyr在data_frame的所有列中运行卡方检验

1 个答案: