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使用宽格式和长格式的总结

wide-format-data summarize tidyr r

Derf · 技术社区 · 3 周前

我有一个数据帧,格式如下。状态有两个级别(PRE、POST)。

SI_mean	TU_平均	ED_mean	平均值(_M)	DT_mean	SK_mean	ATT_mean	地位
2.6	2.75	2.6	2.8	3.4	2.5	3.8	PRE
3.	3.	2.4	2.4	3.	3.	4.	PRE
2.4	2.75	2.4	2.2	2.6	2.25	2.8	PRE

我想用wilcox.test比较每列状态级别的值。所以我立即尝试,

df |>
  summarise(across(contains("mean"),~wilcox.test(.x~Status)$p.value))

但迎接我的是

Error in `summarise()`:
â¹ In argument: `across(contains("mean"), ~wilcox.test(.x ~ Status)$p.value)`.
â¹ In row 1.
Caused by error in `across()`:
! Can't compute column `SI_mean`.
Caused by error in `wilcox.test.formula()`:
! grouping factor must have exactly 2 levels

所以我改为使用长格式,它如预期的那样工作,

df |> pivot_longer(contains("mean"),names_to = "Variable",values_to = "Mean") |>
  group_by(Variable) |>
  summarise(
    wilcox_p_value = wilcox.test(Mean~Status)$p.value
    )

但为什么 summarise 在宽格式中失败?

我只是对我误解的内容感兴趣 总结 功能,以及我将如何使它在宽格式上工作。

数据

df=structure(list(SI_mean = c(2.6, 3, 2.4, 3, 3, 3.2, 2.2, 4, 3.8, 
2.8, 3.6, 2, 3.6, 3.6, 3.8, 3.2, 3, 4, 4, 3, 3.2, 4, 4, 3.2, 
3.2, 3, 3.2, 3.8, 4, 4, 4, 3), TU_mean = c(2.75, 3, 2.75, 3, 
3, 2.75, 3, 3.5, 3.75, 2.5, 3.25, 2, 3.5, 4, 3, 3.25, 3, 4, 4, 
3, 4, 4, 4, 3.25, 3.25, 3, 3, 3.25, 4, 4, 4, 3), ED_mean = c(2.6, 
2.4, 2.4, 3, 2.8, 4, 2, 3.8, 2.6, 2, 2.8, 2, 3, 3.4, 3, 1, 3, 
4, 3.8, 3, 3, 4, 4, 3.2, 4, 2.6, 4, 4, 3.8, 3.6, 4, 3), MT_mean = c(2.8, 
2.4, 2.2, 3, 2.8, 3.4, 2.2, 3.6, 3.4, 3, 2.6, 1.8, 3.4, 3, 4, 
2, 3, 3.4, 3.4, 3, 4, 4, 4, 3.2, 4, 2.8, 4, 4, 3.8, 3.6, 4, 3
), DT_mean = c(3.4, 3, 2.6, 3, 3, 3.8, 2.4, 3.6, 3, 3, 2.8, 2.4, 
3.6, 3.6, 3, 2.2, 3, 4, 4, 4, 3.6, 4, 4, 3.6, 3.8, 2.8, 4, 4, 
4, 3.8, 4, 3), SK_mean = c(2.5, 3, 2.25, 3, 3, 3.5, 2.25, 4, 
3.25, 2.25, 2.5, 2.5, 3.75, 3.75, 4, 1, 2, 4, 3.25, 3, 3.75, 
4, 4, 2.75, 4, 3, 4, 4, 4, 4, 4, 3), ATT_mean = c(3.8, 4, 2.8, 
3, 3, 3.8, 3, 3.6, 3, 4, 4, 3, 3.8, 3.6, 4, 3.8, 4, 4, 4, 4, 
4, 4, 4, 3.6, 3.8, 3, 4, 4, 4, 4, 4, 4), Status = c("PRE", "PRE", 
"PRE", "PRE", "PRE", "PRE", "PRE", "PRE", "PRE", "PRE", "PRE", 
"PRE", "PRE", "PRE", "PRE", "PRE", "PRE", "POST", "POST", "POST", 
"POST", "POST", "POST", "POST", "POST", "POST", "POST", "POST", 
"POST", "POST", "POST", "POST")), class = c("rowwise_df", "tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -32L), groups = structure(list(
    .rows = structure(list(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 
        10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 20L, 
        21L, 22L, 23L, 24L, 25L, 26L, 27L, 28L, 29L, 30L, 31L, 
        32L), ptype = integer(0), class = c("vctrs_list_of", 
    "vctrs_vctr", "list"))), row.names = c(NA, -32L), class = c("tbl_df", 
"tbl", "data.frame")))

1 回复 | 直到 3 周前

Rui Barradas 3 周前

问题出在数据上。
您的tibble按行分组,可以在的第二行中看到

df %>% print(n = 1L)
#> # A tibble: 32 Ã 8
#> # Rowwise:

并且因此被逐行处理。然后,每个 Status 值就是该行中的值。但是 wilcox.test 需要一个具有两个级别的分组变量,并给出错误。

解决方案是先取消数据分组,然后运行测试。

suppressPackageStartupMessages(
  library(dplyr)
)

df %>% print(n = 1L)
#> # A tibble: 32 Ã 8
#> # Rowwise: 
#>   SI_mean TU_mean ED_mean MT_mean DT_mean SK_mean ATT_mean Status
#>     <dbl>   <dbl>   <dbl>   <dbl>   <dbl>   <dbl>    <dbl> <chr> 
#> 1     2.6    2.75     2.6     2.8     3.4     2.5      3.8 PRE   
#> # â¹ 31 more rows

attributes(df)
#> $names
#> [1] "SI_mean"  "TU_mean"  "ED_mean"  "MT_mean"  "DT_mean"  "SK_mean"  "ATT_mean"
#> [8] "Status"  
#> 
#> $class
#> [1] "rowwise_df" "tbl_df"     "tbl"        "data.frame"
#> 
#> $row.names
#>  [1]  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
#> [26] 26 27 28 29 30 31 32
#> 
#> $groups
#> # A tibble: 32 Ã 1
#>          .rows
#>    <list<int>>
#>  1         [1]
#>  2         [1]
#>  3         [1]
#>  4         [1]
#>  5         [1]
#>  6         [1]
#>  7         [1]
#>  8         [1]
#>  9         [1]
#> 10         [1]
#> # â¹ 22 more rows

df %>%
  ungroup() %>%
  summarise(across(contains("mean"), ~wilcox.test(.x ~ Status)$p.value))
#> Warning: There were 7 warnings in `summarise()`.
#> The first warning was:
#> â¹ In argument: `across(contains("mean"), ~wilcox.test(.x ~ Status)$p.value)`.
#> Caused by warning in `wilcox.test.default()`:
#> ! cannot compute exact p-value with ties
#> â¹ Run `dplyr::last_dplyr_warnings()` to see the 6 remaining warnings.
#> # A tibble: 1 Ã 7
#>   SI_mean TU_mean  ED_mean MT_mean  DT_mean SK_mean ATT_mean
#>     <dbl>   <dbl>    <dbl>   <dbl>    <dbl>   <dbl>    <dbl>
#> 1  0.0147 0.00512 0.000527 0.00114 0.000114 0.00252  0.00613

^{创建于2023-11-09

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