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正在从mutate(across())中减去{.col}

tidyr dplyr r

C.Robin · 技术社区 · 10 月前

假设我有以下数据:

df <- structure(list(treat = structure(1:4, levels = c("Control", "Treatment 1", 
"Treatment 2", "Treatment 3"), class = "factor"), 
    female_n = c(314709L, 10456L, 10481L, 10455L), female_mean = c(0.506, 
    0.506, 0.504, 0.5), female_sd = c(0.5, 0.5, 0.5, 0.5), birth_year_n = c(314709L, 
    10456L, 10481L, 10455L), birth_year_mean = c(1973.74, 1973.654, 
    1973.486, 1973.766), birth_year_sd = c(16.867, 16.997, 16.869, 
    16.89), provided_phone_no_n = c(314709L, 10456L, 10481L, 
    10455L), provided_phone_no_mean = c(0.656, 0.666, 0.663, 
    0.647), provided_phone_no_sd = c(0.475, 0.472, 0.473, 0.478
    ), dem_n = c(314709L, 10456L, 10481L, 10455L), dem_mean = c(0.48, 
    0.474, 0.482, 0.478), dem_sd = c(0.5, 0.499, 0.5, 0.5), rep_n = c(314709L, 
    10456L, 10481L, 10455L), rep_mean = c(0.136, 0.141, 0.142, 
    0.138), rep_sd = c(0.343, 0.348, 0.349, 0.345), uaf_n = c(314709L, 
    10456L, 10481L, 10455L), uaf_mean = c(0.363, 0.365, 0.357, 
    0.363), uaf_sd = c(0.481, 0.481, 0.479, 0.481)), class = c("tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -4L))

我想添加一个新的 *_se 列,该列将现有的 *_n 和 *_sd 我的数据中每个变量组的列。即每个一个 female_* , birth_year_* , provided_phone_no_* , dem_* , rep_* 和 uaf_* .

我想,试图做到这一点 mutate(across()) 可能是正确的helper函数,但我在substring中遇到了一些问题 {.col} 并让R将其识别为列名。这是我迄今为止的尝试:

df %>%
    mutate(
    across(ends_with("_sd"),
           list(
            se = ~.x / sqrt(!!ensym("{str_replace(.col, '_sd', '_n')}"))
           )
    )

以上返回错误:

Error in `ensym()`:
! `arg` must be a symbol
Backtrace:
  1. ... %>% ...
 10. rlang::abort(message = message)

有人能看到我哪里错了吗?

1 回复 | 直到 10 月前

LMc 10 月前

这确实是将数据转换为长格式,然后再转换回宽格式的理想情况:

library(tidyr)
library(dplyr)

df |>
  pivot_longer(cols = -treat, 
               names_pattern = "(.*)_(.*)", 
               names_to = c("grp", ".value")) |>
  mutate(se = sd / sqrt(n)) |>
  pivot_wider(names_from = grp, 
              values_from = n:se, 
              names_glue = "{grp}_{.value}", 
              names_vary = "slowest")
#> # A tibble: 4 Ã 25
#>   treat    female_n female_mean female_sd female_se birth_year_n birth_year_mean
#>   <fct>       <int>       <dbl>     <dbl>     <dbl>        <int>           <dbl>
#> 1 Control    314709       0.506       0.5  0.000891       314709           1974.
#> 2 Alexandâ¦    10456       0.506       0.5  0.00489         10456           1974.
#> 3 Politicâ¦    10481       0.504       0.5  0.00488         10481           1973.
#> 4 Mark yoâ¦    10455       0.5         0.5  0.00489         10455           1974.
#> # â¹ 18 more variables: birth_year_sd <dbl>, birth_year_se <dbl>,
#> #   provided_phone_no_n <int>, provided_phone_no_mean <dbl>,
#> #   provided_phone_no_sd <dbl>, provided_phone_no_se <dbl>, dem_n <int>,
#> #   dem_mean <dbl>, dem_sd <dbl>, dem_se <dbl>, rep_n <int>, rep_mean <dbl>,
#> #   rep_sd <dbl>, rep_se <dbl>, uaf_n <int>, uaf_mean <dbl>, uaf_sd <dbl>,
#> #   uaf_se <dbl>

如果你真的想以宽格式进行,这里有一个可能的解决方案:

library(dplyr)
library(stringr)

df |> 
  mutate(across(ends_with("_sd"), \(x) {
    x / sqrt(pick(all_of(str_replace(cur_column(), "_sd", "_n")))[[1]])
  }, 
  .names = "{str_remove(.col, '_sd')}_se"
  ))