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给定一个包含不同变量的两个重复测量的数据框(即A1, A2, B1, B2

library(purrr)
library(tidyr)
library(broom)

set.seed(123)

my_df = data.frame(matrix(rnorm(80), nrow=10))
colnames(my_df) <- c("A1_BEFORE", "A1_AFTER", "A2_BEFORE", "A2_AFTER",
                     "B1_BEFORE", "B1_AFTER", "B2_BEFORE", "B2_AFTER")

如何使用函数式编程原则来迭代相同变量的对(之前,之后),并获得“整洁”的结果?这是我的尝试:

bef <- select(my_df, contains("BEFORE"))
aft <- select(my_df, contains("AFTER"))
result <- map2(bef, aft, t.test, paired = T)

以上结果产生了多个嵌套列表。我怎样才能获得“整洁”的结果?

result <- tidy(map2(bef, aft, t.test, paired = T))

结果 <- tidy(map2(bef, aft, t.test,paired = T))
tidy.list(map2(bef, aft, t.test,paired = T)) 中的错误:此列表中没有识别整理方法另外:警告消息:在 sort(names(x)) == c("d", "u", "v") 中:较长的对象长度不是较短对象长度的倍数

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2 回答 2

2

这是另一种方法,在进行 t 检验之前整理数据。显然得到了相同的结果,但是这种方法在最终输出中标记了正在测试的变量。

仅更改数据-添加了一个 id 变量来索引重复测量

需要broom并且tidyr除了dplyr

library(tidyr, dplyr, broom)

用于tidyr重组

my_tidy_df <- my_df %>% 
  mutate(id = row_number()) %>% # needs an id to group repeated measure
  gather(var, value, -id) %>% 
  extract(var, c("var", "timepoint"), "([[:alnum:]]+)_([[:alnum:]]+)") %>% 
  spread(timepoint, value) 

这给出了这个结构

   id var       AFTER     BEFORE
1   1  A1 -1.14854253 -0.9032172
2   1  A2  2.36114529 -0.6500869
3   1  B1  0.26204456 -0.5477532
4   1  B2 -1.34416890 -0.4696884
5   2  A1  0.53400345  1.2722203

然后,您可以为每个变量运行 t 检验,如下所示:

my_tidy_df %>% 
  group_by(var) %>% 
  do(broom::tidy(t.test(.$BEFORE, .$AFTER, data=., paired=T)))

结果:

# Groups:   var [4]
    var    estimate  statistic   p.value parameter   conf.low conf.high        method alternative
  <chr>       <dbl>      <dbl>     <dbl>     <dbl>      <dbl>     <dbl>        <fctr>      <fctr>
1    A1  0.16014628  0.3470400 0.7365381         9 -0.8837567 1.2040493 Paired t-test   two.sided
2    A2 -0.99798993 -1.6271640 0.1381451         9 -2.3854407 0.3894609 Paired t-test   two.sided
3    B1  0.04916586  0.1289803 0.9002097         9 -0.8131436 0.9114753 Paired t-test   two.sided
4    B2 -0.06919212 -0.1833619 0.8585784         9 -0.9228233 0.7844391 Paired t-test   two.sided
于 2017-09-27T14:06:27.053 回答
2

我们可以使用map_df,因为它是list

map2(bef, aft, t.test, paired = TRUE) %>%
            map_df(tidy)
#   estimate  statistic    p.value parameter   conf.low conf.high        method
#1 -0.1339963 -0.4613684 0.65548187         9 -0.7909999 0.5230073 Paired t-test
#2 -0.7466034 -1.8820475 0.09250351         9 -1.6439954 0.1507885 Paired t-test
#3 -0.2304015 -0.5740849 0.57997286         9 -1.1382891 0.6774860 Paired t-test
#4  0.4860015  1.3468795 0.21095133         9 -0.3302644 1.3022674 Paired t-test
#   alternative
#1   two.sided
#2   two.sided
#3   two.sided
#4   two.sided

或者更紧凑

map2_df(bef, aft, ~tidy(t.test(.x, .y, paired = TRUE)))
于 2017-09-27T07:02:09.890 回答