问题 - 使用突变（）

Question

问题： 我想使用以下逻辑（示例）创建名为“ trig'的新变量（保留旧变量）（在末尾没有'.x'或.y'）：我希望新变量假设'。 y'值不丢失时，并且缺少它，新变量应该假设'.x'。当两者都缺少时，该变量应丢失。

代码：

> data_merged_ex
# A tibble: 5 × 8
  seqg    cyl piip.x piip.y dalo.x dalo.y tcct.x tcct.y
  <chr> <dbl>  <dbl>  <dbl>  <dbl>  <dbl>  <dbl>  <dbl>
1 7894      0    1.1    1.1    2.7   NA      242     NA
2 1597      2   NA     NA     NA      2.7     NA    196
3 2586      1    0.7    0.7    2.2    4.3    161     NA
4 4679      0    0.6   NA      1.4    2.2    239     NA
5 8371      0    0.5    1.3    1.6   NA      206     NA


data_merged_ex_t <- data_merged_ex %>%

  mutate(across(
    .cols = ends_with(".x"),
    .fns = ~ case_when(
      !is.na(.data[[cur_column()]]) & !is.na(.data[[str_replace(cur_column(), "\\.x$", ".y")]]) ~ .data[[str_replace(cur_column(), "\\.x$", ".y")]],
      is.na(.data[[cur_column()]]) & !is.na(.data[[str_replace(cur_column(), "\\.x$", ".y")]]) ~ .data[[str_replace(cur_column(), "\\.x$", ".y")]],
      !is.na(.data[[cur_column()]]) & is.na(.data[[str_replace(cur_column(), "\\.x$", ".y")]]) ~ .data[[cur_column()]],
      TRUE ~ NA_real_
    ),
    .names = "{str_remove(.col, '\\.x$')}"
  ))

Error in `cur_column()`:
! Must only be used inside `across()`.
Run `rlang::last_trace()` to see where the error occurred.
> rlang::last_trace()
<error/rlang_error>
Error in `cur_column()`:
! Must only be used inside `across()`.
---
Backtrace:
    ▆
 1. ├─mc_merged_sy %>% ...
 2. ├─dplyr::mutate(...)
 3. ├─dplyr:::mutate.data.frame(...)
 4. │ ├─dplyr:::mutate_cols(.data, dplyr_quosures(...), by)
 5. │ │ └─base::force(dots)
 6. │ └─dplyr:::dplyr_quosures(...)
 7. │   └─rlang::quos(..., .ignore_empty = "all")
 8. └─dplyr::cur_column()
Run rlang::last_trace(drop = FALSE) to see 4 hidden frames.
> rlang::last_trace(drop = FALSE)
<error/rlang_error>
Error in `cur_column()`:
! Must only be used inside `across()`.
---
Backtrace:
     ▆
  1. ├─mc_merged_sy %>% ...
  2. ├─dplyr::mutate(...)
  3. ├─dplyr:::mutate.data.frame(...)
  4. │ ├─dplyr:::mutate_cols(.data, dplyr_quosures(...), by)
  5. │ │ └─base::force(dots)
  6. │ └─dplyr:::dplyr_quosures(...)
  7. │   └─rlang::quos(..., .ignore_empty = "all")
  8. └─dplyr::cur_column()
  9.   └─dplyr:::peek_column()
 10.     └─dplyr:::context_peek("column", "`across()`", call)
 11.       ├─context_peek_bare(name) %||% ...
 12.       └─rlang::abort(glue("Must only be used inside {location}."), call = call)

您可以旋转到长时间，然后执行逻辑，然后向后旋转：

pivot_longer(data_merged_ex, -c(seqg, cyl),
             names_to=c("name", ".value"),
             names_pattern="(.*).(.)") %>%
  mutate(z=case_when(!is.na(y)~y,
                     !is.na(x)~x,
                     .default=NA), .by=c(seqg, name)) %>%
  pivot_wider(id_cols=c(seqg,cyl),
              values_from=c(x,y,z),
              names_vary = "slowest",
              names_glue = "{name}.{.value}") 

Answer 1