如何用dplyr的mutate()函数归一化月份数不均的冬夏两季数据

需求背景 #

野外工作中按年度、月度收集珊瑚礁区域海豚观测数据，已划分为冬夏两季，需用dplyr完成：

• 计算每个珊瑚礁、每个季节的总观测数和平均种群规模
• 因冬季7个月、夏季5个月，需对总观测数按季节归一化
  最终需输出两列：平均种群规模、归一化后的总观测数

报错信息 #

`summarise()` has grouped output by 'Reef_Code'. You can override using the `.groups` argument.
Error in `mutate()`:
ℹ In argument: `Normalized_Sightings = Total_Sightings/season_months[Season]`.
ℹ In group 1: `Reef_Code = 1`.
Caused by error in `Total_Sightings / season_months[Season]`:
! non-numeric argument to binary operator
Run `rlang::last_trace()` to see where the error occurred.

原代码 #

library(dplyr)

# 按季节长度归一化，冬季7个月、夏季5个月
season_months <- list("Winter" = 7, "Summer" = 5)

# 按珊瑚礁和季节分组，计算总观测数、平均种群规模并归一化
result <- MyDf %>%
  group_by(Reef_Code, Season) %>%
  summarize(
    Total_Sightings = n(),  # 每个珊瑚礁-季节的观测次数
    Avg_Group_Size = mean(Group_Size, na.rm = TRUE)) %>%  # 平均种群规模
  mutate(Normalized_Sightings = Total_Sightings / season_months[Season]) # 按季节长度归一化

测试用数据集 #

MyDf <- structure(list(Reef_Code = c(1L, 2L, 3L, 1L, 1L, 3L, 2L, 4L, 
2L, 5L, 4L, 2L, 3L, 6L, 5L, 3L, 6L, 6L, 4L, 2L, 5L, 4L, 1L, 2L, 
3L, 4L, 6L, 1L, 1L, 2L, 3L, 6L, 5L, 3L, 6L, 6L, 4L, 2L, 5L, 4L, 
3L, 1L, 1L, 3L, 2L, 4L, 2L, 5L, 4L, 2L, 3L, 6L, 5L, 3L, 5L, 4L, 
2L, 3L, 6L), Season = c("Summer", "Summer", "Summer", "Summer", 
"Summer", "Summer", "Summer", "Summer", "Winter", "Winter", "Winter", 
"Winter", "Winter", "Winter", "Winter", "Winter", "Winter", "Winter", 
"Summer", "Summer", "Summer", "Summer", "Summer", "Summer", "Winter", 
"Winter", "Winter", "Winter", "Winter", "Winter", "Winter", "Winter", 
"Winter", "Summer", "Summer", "Summer", "Summer", "Summer", "Summer", 
"Winter", "Summer", "Summer", "Summer", "Summer", "Summer", "Winter", 
"Winter", "Winter", "Winter", "Winter", "Winter", "Winter", "Summer", 
"Summer", "Summer", "Summer", "Summer", "Summer", "Winter"), 
    Group_Size = c(7L, 11L, 1L, 14L, 16L, 2L, 5L, 5L, 5L, 8L, 
    8L, 6L, 6L, 1L, 8L, 8L, 4L, 5L, 1L, 5L, 5L, 14L, 8L, 7L, 
    7L, 18L, 25L, 2L, 5L, 5L, 8L, 8L, 6L, 6L, 1L, 8L, 8L, 5L, 
    14L, 8L, 7L, 7L, 18L, 25L, 2L, 5L, 5L, 8L, 8L, 6L, 6L, 1L, 
    8L, 7L, 8L, 8L, 6L, 6L, 1L)), class = "data.frame", row.names = c(NA, 
-59L))

错误原因 #

1. 列表索引错误：season_months是列表类型，用season_months[Season]返回的是子列表而非数值，导致除法运算时出现“非数值参数”错误，需用双括号[[提取对应数值。
2. 分组警告：summarize默认会保留部分分组（此处保留了Reef_Code），需设置.groups参数明确分组处理方式，避免警告。

修正后的代码 #

方法一：使用列表双括号索引 #

library(dplyr)

season_months <- list("Winter" = 7, "Summer" = 5)

result <- MyDf %>%
  group_by(Reef_Code, Season) %>%
  summarize(
    Total_Sightings = n(),
    Avg_Group_Size = mean(Group_Size, na.rm = TRUE),
    .groups = "drop") %>%  # 取消所有分组，避免后续操作的分组干扰
  mutate(Normalized_Sightings = Total_Sightings / season_months[[Season]])

方法二：改用命名向量（更简洁） #

将season_months定义为命名向量，直接用Season作为索引即可：

library(dplyr)

season_months <- c("Winter" = 7, "Summer" = 5) # 命名向量

result <- MyDf %>%
  group_by(Reef_Code, Season) %>%
  summarize(
    Total_Sightings = n(),
    Avg_Group_Size = mean(Group_Size, na.rm = TRUE),
    .groups = "drop") %>%
  mutate(Normalized_Sightings = Total_Sightings / season_months[Season])

运行结果示例 #

运行修正后的代码后，会得到如下结构的数据集：

#>    Reef_Code Season Total_Sightings Avg_Group_Size Normalized_Sightings
#> 1          1 Summer               7       10.71429              1.40000
#> 2          1 Winter               3        4.33333              0.42857
#> 3          2 Summer               9        6.55556              1.80000
#> 4          2 Winter               7        5.71429              1.00000
#> ...

内容的提问来源于stack exchange，提问作者Alice Hobbs