left_join报错‘Join columns must be present’：position列未识别求助

我需要汇总处于中间position（即第一个和最后一个position之间）的f频率数据。处理思路是先筛选这类数据，执行summarise操作，再将新数据与原数据集重新合并。这个方法在训练数据上运行正常：

library(tidyverse)
df %>%
  group_by(rowid) %>%
  # 筛选中间位置的数据并汇总频率
  filter(position != first(position) & position != last(position)) %>%
  summarise(across(position),
            middle_position = mean(f, na.rm = TRUE),
            word = str_c(word, collapse=" ")
            ) %>%
  left_join(df, ., by = c("rowid", "position"))

但应用到实际数据时，出现以下错误：

Error in `left_join()`:
! Join columns must be present in data.
✖ Problem with `position`.
Run `rlang::last_error()` to see where the error occurred.
> rlang::last_error()
<error/rlang_error>
Error in `left_join()`:
! Join columns must be present in data.
✖ Problem with `position`.
---
Backtrace:
 1. ... %>% left_join(bnc_X, ., by = c("rowid", "position"))
 3. dplyr:::left_join.data.frame(bnc_X, ., by = c("rowid", "position"))
Run `rlang::last_trace()` to see the full context.
> rlang::last_trace()
<error/rlang_error>
Error in `left_join()`:
! Join columns must be present in data.
✖ Problem with `position`.
---
Backtrace:
    ▆
 1. ├─... %>% left_join(bnc_X, ., by = c("rowid", "position"))
 2. ├─dplyr::left_join(bnc_X, ., by = c("rowid", "position"))
 3. └─dplyr:::left_join.data.frame(bnc_X, ., by = c("rowid", "position"))
 4.   └─dplyr:::join_mutate(...)
 5.     └─dplyr:::join_cols(...)
 6.       └─dplyr:::standardise_join_by(...)
 7.         └─dplyr:::check_join_vars(by$x, x_names, error_call = error_call)
 8.           └─rlang::abort(bullets, call = error_call)

核心问题似乎是变量position未被识别，尝试多时仍无法解决，求帮助！

示例数据：

df <- data.frame(
  size = c(3,3,3,
              3,3,3,
              4,4,4,4,
              5,5,5,5,5,
              3,3,3),
  rowid = c(1,1,1,2,2,2,3,3,3,3,4,4,4,4,4,5,5,5),
  turn = c(rep("How are you?",3),
           rep("I'm fine.",3),
           rep("How's the weather?",4),
           rep("It's really very cold.",5),
           rep("I love you",3)),
  word = c("how","are","you",
           "i","'m","fine",
           "how","'s","the","weather",
           "it","'s","really", "very","cold",
           "i","love","you"),
  f = c(400,300,250,
        600,555,1,
        400,500,700,20,
        390,500,177,200,35,
        600,199,400),
  position = c(1,2,3,
               1,2,3,
               1,2,3,4,
               1,2,3,4,5,
               1,2,3)
)

报错根源是summarise里的across(position)写法错误：across()用于批量处理多列，单独写across(position)不会保留每个rowid下的所有position值，反而会导致汇总后的数据集丢失完整的position列，最终在left_join时无法匹配原表的对应列。

方法一：修正原代码逻辑 #

先标记中间位置，再分组计算均值，最后按rowid合并（无需按position匹配，因为中间位置的均值是整个组的属性）：

library(tidyverse)

# 生成汇总表：每个rowid对应中间位置的均值和拼接单词
summary_df <- df %>%
  group_by(rowid) %>%
  mutate(is_middle = position != first(position) & position != last(position)) %>%
  filter(is_middle) %>%
  summarise(
    middle_position = mean(f, na.rm = TRUE),
    middle_words = str_c(word, collapse = " ")
  )

# 合并到原表
df %>%
  left_join(summary_df, by = "rowid")

方法二：更高效的直接计算（推荐） #

用group_by + mutate一步完成，无需额外合并操作：

df %>%
  group_by(rowid) %>%
  mutate(
    # 标记当前行是否为中间位置
    is_middle = position != first(position) & position != last(position),
    # 计算当前组中间位置的f均值，无中间位置则返回NA
    middle_position = if_else(any(is_middle), mean(f[is_middle], na.rm = TRUE), NA_real_),
    # 拼接当前组中间位置的单词，无中间位置则返回NA
    middle_words = if_else(any(is_middle), str_c(word[is_middle], collapse = " "), NA_character_)
  ) %>%
  ungroup()

两种方法都能避免合并时的列缺失问题，第二种写法更简洁高效，还能保留原表所有行的完整信息。

内容的提问来源于stack exchange，提问作者Chris Ruehlemann