dplyr group_by %>% summarize_if()中的错误

Question

我正在处理一个相当小的数据集，试图按平均值汇总列，同时按第一列进行分组。目前，我有一个如下所示的df (LitterMean)：

      date3 TotalBorn LiveBorn StillBorn Mummies
1   7/6        12       12         0       0
2   7/6        20       15         2       3
3  6/29        14       14         0       0
4   7/6        11       10         1       0
5   7/6        16       15         1       0
6   7/6        11       11         0       0

我试着跑

LitterMean %>%
  group_by(date3) %>%
  summarize_if(LitterMean, is.numeric, mean, na.rm=TRUE)

回传

Error: `.predicate` must have length 1, not 5.
Run `rlang::last_error()` to see where the error occurred.

所以我运行rlang::last_error()并接收

<error/rlang_error>
`.predicate` must have length 1, not 5.
Backtrace:
  1. `%>%`(...)
  2. dplyr::summarize_if(., LitterMean, is.numeric, mean, na.rm = TRUE)
  3. dplyr:::manip_if(...)
  4. dplyr:::tbl_if_syms(.tbl, .predicate, .env, .include_group_vars = .include_group_vars)
  8. dplyr:::tbl_if_vars(.tbl, .p, .env, ..., .include_group_vars = .include_group_vars)
  9. dplyr:::bad_args(".predicate", "must have length 1, not {length(.p)}.")
 10. dplyr:::glubort(fmt_args(args), ..., .envir = .envir)
Run `rlang::last_trace()` to see the full context.

以下显示我确实有NA文章。

sum(is.na(LitterMean))
[1] 5

是否有人知道我在代码中遗漏了什么可以防止上述错误的东西？

Answer 1

您只需要正确调用summarize_if，如下所示：

LitterMean %>%
  group_by(date3) %>%
  summarize_if(is.numeric, mean, na.rm=TRUE)

预期结果：

> LitterMean %>%
+   group_by(date3) %>%
+   summarize_if(is.numeric, mean, na.rm=TRUE)
# A tibble: 2 × 5
  date3 TotalBorn LiveBorn StillBorn Mummies
  <chr>     <dbl>    <dbl>     <dbl>   <dbl>
1 6/29         14     14         0       0  
2 7/6          14     12.6       0.8     0.6

Answer 2

你应该使用across

作用域动词(_if，_at，_all)已被现有动词中的across()取代。详见vignette("colwise")。

https://dplyr.tidyverse.org/reference/summarise_all.html

library(dplyr)
df %>%
  group_by(date3) %>%
  summarise(across(where(is.numeric), mean))

  date3 TotalBorn LiveBorn StillBorn Mummies
  <chr>     <dbl>    <dbl>     <dbl>   <dbl>
1 6/29         14     14         0       0  
2 7/6          14     12.6       0.8     0.6