在 R 中每行应用函数

Question

我有一个数据集，我正在尝试应用shapiro.test。

test_data <- tibble(
  gene = rep(c(LETTERS[1:5]), times = 2, each = 5),
  treatment = c(rep('control', times = 25) , rep('treatment', times = 25)),
  day = rep(c(1:2), times = 5, each = 5),
  data = rnorm(50, mean = 25, sd = 5)
)

# A tibble: 50 x 4
   gene  treatment   day  data
   <chr> <chr>     <int> <dbl>
 1 A     control       1  28.8
 2 A     control       1  22.4
 3 A     control       1  24.8
 4 A     control       1  20.1
 5 A     control       1  15.6
 6 B     control       2  26.5
 7 B     control       2  26.2
 8 B     control       2  25.3
 9 B     control       2  21.4
10 B     control       2  35.0
# … with 40 more rows

我创建了一个函数来按基因、治疗和天运行测试：

normality_test <- function(x, y, z){
 with(test_data, shapiro.test(data[gene == x & treatment == y & day == z]))
} 

因此，如果我运行 normality_test('A', 'control', '1')，它将在 第 1 天在 control 中测试 gene A。

    Shapiro-Wilk normality test

data:  data[gene == x & treatment == y & day == z]
W = 0.99935, p-value = 0.9998

但是，我希望该函数循环遍历基因/治疗/天的所有组合并单独输出每个正态性测试，但无法弄清楚。
我已经能够创建一个循环，将每一行输出为单独的tibble，但未能成功分离行中的每个元素以添加到normality_test 函数中。
我还尝试了map 和lmap，但无济于事。
谢谢。

Answer 1

aggregate(test_data$data, 
          list(test_data$gene, test_data$treatment, test_data$day),
          FUN = shapiro.test, 
          simplify = FALSE)

导致：

   Group.1   Group.2 Group.3                                                                          x
1        A   control       1   0.88967912801297, 0.355492282929607, Shapiro-Wilk normality test, X[[i]]
2        C   control       1  0.872622601121686, 0.277206156703845, Shapiro-Wilk normality test, X[[i]]
3        E   control       1  0.886337216320072, 0.339020498506711, Shapiro-Wilk normality test, X[[i]]
4        B treatment       1  0.902918723585913, 0.426227078990388, Shapiro-Wilk normality test, X[[i]]
5        D treatment       1  0.850079117181635, 0.194768724952993, Shapiro-Wilk normality test, X[[i]]
6        B   control       2 0.810238329506977, 0.0979423168617567, Shapiro-Wilk normality test, X[[i]]
7        D   control       2  0.965126339172019, 0.843147936412715, Shapiro-Wilk normality test, X[[i]]
8        A treatment       2  0.933316692276928, 0.619155973369246, Shapiro-Wilk normality test, X[[i]]
9        C treatment       2 0.771672137756979, 0.0466697848359151, Shapiro-Wilk normality test, X[[i]]
10       E treatment       2   0.91332590966644, 0.487832813430558, Shapiro-Wilk normality test, X[[i]]

Answer 2

library(dplyr)

test_data %>% 
  nest_by(gene, treatment, day, .key = "nested") %>% 
  mutate(sw = list(shapiro.test(nested$data))) %>% 
  pull(sw)

输出

您的示例数据中有 10 个独特的 gene、treatment 和 day 组合。

显示前几个元素：

[[1]]

    Shapiro-Wilk normality test

data:  nested$data
W = 0.88041, p-value = 0.3112


[[2]]

    Shapiro-Wilk normality test

data:  nested$data
W = 0.96533, p-value = 0.8445

如果您不通过管道发送到pull，那么您的输出将是：

   gene  treatment   day             nested sw     
   <chr> <chr>     <int> <list<tibble[,1]>> <list> 
 1 A     control       1            [5 x 1] <htest>
 2 A     treatment     2            [5 x 1] <htest>
 3 B     control       2            [5 x 1] <htest>
 4 B     treatment     1            [5 x 1] <htest>
 5 C     control       1            [5 x 1] <htest>
 6 C     treatment     2            [5 x 1] <htest>
 7 D     control       2            [5 x 1] <htest>
 8 D     treatment     1            [5 x 1] <htest>
 9 E     control       1            [5 x 1] <htest>
10 E     treatment     2            [5 x 1] <htest>

sw 列包含测试结果。

---

或者只针对 p 值：

test_data %>% 
  nest_by(gene, treatment, day, .key = "nested") %>% 
  mutate(sw = shapiro.test(nested$data)$p.value)

输出

   gene  treatment   day             nested    sw
   <chr> <chr>     <int> <list<tibble[,1]>> <dbl>
 1 A     control       1            [5 x 1] 0.311
 2 A     treatment     2            [5 x 1] 0.845
 3 B     control       2            [5 x 1] 0.408
 4 B     treatment     1            [5 x 1] 0.204
 5 C     control       1            [5 x 1] 0.435
 6 C     treatment     2            [5 x 1] 0.316
 7 D     control       2            [5 x 1] 0.143
 8 D     treatment     1            [5 x 1] 0.236
 9 E     control       1            [5 x 1] 0.695
10 E     treatment     2            [5 x 1] 0.658