基于条件删除R数据框重复观测并新建richness列的实现方法

解决代码 #

以下实现完全匹配你的需求，1.27亿行大体积数据可以优先使用下方的优化方案，运算效率更高。

标准dplyr实现 #

# 加载依赖包
library(dplyr)

result <- df %>%
  # 按地点+物种分组，处理同一地点下同一物种的多记录问题
  group_by(location, species) %>%
  summarise(
    # 判断该物种在当前地点是否同时存在两种定植类型
    has_both_type = n_distinct(establishment) == 2,
    # stress字段优先取natural行的非NA值，单种类型时直接取现有非NA值
    stress = dplyr::if_else(
      has_both_type,
      stress[establishment == "natural" & !is.na(stress)][1],
      stress[!is.na(stress)][1]
    ),
    # maturity字段逻辑同stress
    maturity = dplyr::if_else(
      has_both_type,
      maturity[establishment == "natural" & !is.na(maturity)][1],
      maturity[!is.na(maturity)][1]
    ),
    # 存在两种类型时保留anthropogenic标识，否则保留原有标识
    establishment = dplyr::if_else(
      has_both_type,
      "anthropogenic",
      first(establishment)
    ),
    .groups = "drop"
  ) %>%
  # 按地点分组统计物种总数，填充到组内所有行
  group_by(location) %>%
  mutate(richness = n_distinct(species)) %>%
  ungroup()

大体积数据优化方案（适配1.27亿行规模） #

用dtplyr实现dplyr语法转data.table执行，运算效率比标准dplyr高数倍：

# 加载依赖包
library(dplyr)
library(dtplyr)
library(data.table)

result <- lazy_dt(df) %>%
  group_by(location, species) %>%
  summarise(
    has_both_type = n_distinct(establishment) == 2,
    stress = dplyr::if_else(
      has_both_type,
      stress[establishment == "natural" & !is.na(stress)][1],
      stress[!is.na(stress)][1]
    ),
    maturity = dplyr::if_else(
      has_both_type,
      maturity[establishment == "natural" & !is.na(maturity)][1],
      maturity[!is.na(maturity)][1]
    ),
    establishment = dplyr::if_else(
      has_both_type,
      "anthropogenic",
      first(establishment)
    ),
    .groups = "drop"
  ) %>%
  group_by(location) %>%
  mutate(richness = n_distinct(species)) %>%
  ungroup() %>%
  as.data.frame() # 运算结束后转为常规data.frame格式

说明 #

如果单个location+species分组下natural行存在多个非NA的stress/maturity值，代码默认取第一个出现的值，你可以根据需求调整取数逻辑（比如替换为取均值、最大值等）。所有因子类字段的层级会自动保留，输出结构和你给出的示例完全一致。

内容的提问来源于stack exchange，提问作者Jose