3

有没有办法通过包中的map()功能进行 Levene 测试purrr？还是有另一种简单的方法来计算各种变量的 Levene 测试？

我的数据框包含各种因子和数字列，所以我尝试使用map_if()，它工作正常，例如用于夏皮罗测试。但是，我不知道如何指定公式。我想根据“治疗”因素测试我所有的数字变量。

library("tidyverse")
library("rstatix")

data <- data.frame(site = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L), 
                                    .Label = c("S1 ", "S2 ", "S3 "), class = "factor"), 
                   plot = structure(c(1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L), 
                                    .Label = c(" Tree 1 ", " Tree 2 ", " Tree 3 "), class = "factor"), 
                   Treatment = structure(c(2L, 2L, 2L, 1L, 1L, 1L, 2L, 2L, 2L, 1L), .Label = c("T1", "T2"), class = "factor"), 
                   flux1 = c(11.52188065, 8.43156699, 4.495312274, -1.866676811, 3.861102035, -0.814742373, 6.51039536, 4.767950345, 10.36544542, 1.065963875), 
                   flux2 = c(0.142259208, 0.04060245, 0.807631744, 0.060127596, -0.157762562, 0.062464942, 0.043147603, 0.495001652, 0.34363348, 0.134183704), 
                   flux3 = c(0.147506197, 1.131009714, 0.038860728, 0.0176834, 0.053191593, 0.047591306, 0.00573377, -0.034926075, 0.123379247, 0.018882469))

map_if(data, is.numeric, levene_test(. ~ Treatment))

有什么建议么？谢谢你的帮助！

现在还有一个可重现的例子；）

4

3 回答 3

4

这是另一种选择：首先转向长数据，

然后group_by应用公式（这里通量应该是因素！）

library(tidyr)
library(dplyr)

data %>% 
  pivot_longer(
    cols = starts_with("flux"),
    names_to = "flux",
    values_to = "value"
  ) %>%
  mutate(flux = as.factor(flux)) %>% 
  group_by(flux) %>% 
  levene_test(value ~ Treatment)

  flux    df1   df2 statistic     p
  <fct> <int> <int>     <dbl> <dbl>
1 flux1     1     8     0.410 0.540
2 flux2     1     8     2.85  0.130
3 flux3     1     8     1.11  0.323
于 2021-12-22T20:15:15.203 回答
3

您也可以更直接地使用汇总。然后旋转并取消嵌套结果。

library(dplyr)
library(tidyr)

data %>% 
  summarize(across(where(is.numeric),
                   ~ list(levene_test(cur_data(), . ~ Treatment)))) %>% 
  pivot_longer(everything(), names_to = "flux", values_to = "levene_test") %>% 
  unnest(levene_test)

另一种选择是将变量名称输入映射并创建公式。

library(purrr)

names(data)[map_lgl(data, is.numeric)] %>% 
  set_names() %>% 
  map_dfr(~ levene_test(data, as.formula(paste(.x, "~ Treatment"))), .id = "flux")

结果（两者）：

# A tibble: 3 x 5
  flux    df1   df2 statistic     p
  <chr> <int> <int>     <dbl> <dbl>
1 flux1     1     8     0.410 0.540
2 flux2     1     8     2.85  0.130
3 flux3     1     8     1.11  0.323
于 2021-12-22T20:39:20.383 回答
2

问题是map在列上循环并且它不再是 data.frame 而levene_test需要一个data.frame/tibble. 根据?levene_test

data - 用于评估公式或模型的数据框

因此，不要map_if直接使用select数字select(where(is.numeric))列（namesmapselectreformulatelevene_test

library(rstatix)
library(dplyr)
library(purrr)
data %>% 
   select(where(is.numeric)) %>%
   names %>%
   map_dfr(~ data %>%
             select(Treatment, all_of(.x)) %>% 
       {levene_test(reformulate("Treatment", response = names(.)[2]), data = .)
         })

-输出

# A tibble: 3 × 4
    df1   df2 statistic     p
  <int> <int>     <dbl> <dbl>
1     1     8     0.410 0.540
2     1     8     2.85  0.130
3     1     8     1.11  0.323

它也可以使用across虽然 - 即循环中的across列，使用as ，使用创建公式，应用，在 a 中返回输出，然后使用（因为将从中删除 data.frame 属性）numericsummarisedatacur_data()reformulatelevene_testlistunclassbind_rowsunclasslist

data %>%
   summarise(across(where(is.numeric),
    ~  list(cur_data() %>%
     levene_test(reformulate("Treatment", response = cur_column()))))) %>% 
   unclass %>%
   unname %>%
   bind_rows
# A tibble: 3 × 4
    df1   df2 statistic     p
  <int> <int>     <dbl> <dbl>
1     1     8     0.410 0.540
2     1     8     2.85  0.130
3     1     8     1.11  0.323

如果我们需要“通量”列标识符，请使用该summarise步骤而不将输出包装在 a 中list，然后使用bind_rowswith.id

data %>%
    summarise(across(where(is.numeric),
     ~  cur_data() %>%
      levene_test(reformulate("Treatment", response = cur_column())))) %>%
    unclass %>% 
    bind_rows(.id = 'flux')
# A tibble: 3 × 5
  flux    df1   df2 statistic     p
  <chr> <int> <int>     <dbl> <dbl>
1 flux1     1     8     0.410 0.540
2 flux2     1     8     2.85  0.130
3 flux3     1     8     1.11  0.323

或者另一种选择是OPmap_if本身

map_if(data, is.numeric, 
    ~ levene_test(. ~ Treatment, 
  data = tibble(.x, Treatment = data$Treatment) ), .else = ~ NULL) %>% 
   bind_rows(.id = 'flux')
# A tibble: 3 × 5
  flux    df1   df2 statistic     p
  <chr> <int> <int>     <dbl> <dbl>
1 flux1     1     8     0.410 0.540
2 flux2     1     8     2.85  0.130
3 flux3     1     8     1.11  0.323
于 2021-12-22T20:11:21.177 回答