如何在R中基于联系人属性计算自我子网络的平均亲密度？

我有一份自我中心网络（ego-network）数据，参与者把自己的3位朋友分为A、B、C三类，同时给朋友间的亲密度打了分。现在想新增一列，计算每位参与者的C类朋友之间的平均亲密度。
比如第一行里，只有朋友1和朋友2都是C类，所以只用friend1_closeness_friend2的值计算，另外两个亲密度评分忽略。

数据示例 #

library(dplyr)

data <- data.frame(
  ID = c("001", "002", "003"),
  friendshipType_1 = c("C", "C", "A"),
  friendshipType_2 = c("C", "C", "B"),
  friendshipType_3 = c("A", "C" , "A"),
  friend1_closeness_friend2 = c(1, 2, 3),
  friend1_closeness_friend3 = c(4, 3, 2),
  friend2_closeness_friend3 = c(1, 1, 2)
)

期望输出 #

data <- data.frame(
  ID = c("001", "002", "003"),
  friendshipType_1 = c("C", "C", "A"),
  friendshipType_2 = c("C", "C", "B"),
  friendshipType_3 = c("A", "C" , "A"),
  friend1_closeness_friend2 = c(1, 2, 3),
  friend1_closeness_friend3 = c(4, 3, 2),
  friend2_closeness_friend3 = c(1, 1, 2),
  mean_c = c(1, 2, NA)
)

解决方案 #

可以利用dplyr的行处理功能，逐行判断每对朋友是否都属于C类，筛选出符合条件的亲密度值后计算平均值：

方法一：简洁版 #

data <- data %>%
  rowwise() %>%
  mutate(
    mean_c = mean(
      c(
        # 判断朋友1和朋友2是否都是C类，是则取对应亲密度，否则设为NA
        ifelse(friendshipType_1 == "C" & friendshipType_2 == "C", friend1_closeness_friend2, NA),
        # 判断朋友1和朋友3是否都是C类
        ifelse(friendshipType_1 == "C" & friendshipType_3 == "C", friend1_closeness_friend3, NA),
        # 判断朋友2和朋友3是否都是C类
        ifelse(friendshipType_2 == "C" & friendshipType_3 == "C", friend2_closeness_friend3, NA)
      ),
      na.rm = TRUE
    )
  ) %>%
  ungroup() %>%
  # 把全NA情况产生的NaN转为NA，匹配期望输出
  mutate(mean_c = ifelse(is.nan(mean_c), NA, mean_c))

方法二：分步清晰版 #

如果希望逻辑更直观，可以先收集符合条件的亲密度值，再计算均值：

data <- data %>%
  rowwise() %>%
  mutate(
    # 收集所有C类朋友间的亲密度值，不符合条件的不加入列表
    c_closeness = list(
      c(
        if (friendshipType_1 == "C" & friendshipType_2 == "C") friend1_closeness_friend2,
        if (friendshipType_1 == "C" & friendshipType_3 == "C") friend1_closeness_friend3,
        if (friendshipType_2 == "C" & friendshipType_3 == "C") friend2_closeness_friend3
      )
    ),
    # 计算列表中数值的均值，空列表时返回NA
    mean_c = if (length(c_closeness) == 0) NA else mean(unlist(c_closeness))
  ) %>%
  # 移除临时中间列
  select(-c_closeness) %>%
  ungroup()

两种方法都能得到与期望输出完全一致的结果。

内容的提问来源于stack exchange，提问作者Marie