如何在R中按规则创建emotional_ipv分类变量？现有代码报错求助

需求说明 #

需基于IPV相关问题的回答，创建名为emotional_ipv的分类变量，规则如下：

• 所有IPV问题回答均为"never"（编码0）→ 归类为"never"
• 仅有一个问题回答为"once"（编码1）→ 归类为"isolated incident"
• 多个问题回答为"once" → 归类为"low frequency"
• 至少一个问题回答为"a few times"（编码2）且无"many times"（编码3）回答 → 归类为"mid frequency"
• 存在任何"many times"回答 → 归类为"high frequency"

编码对应关系：0=Never；1=Once；2=Few times；3=Many times

原始数据集 #

df <- structure (list(subject_id = c("191-5467", "191-6784", "191-3457", "191-0987", "191-1245","191-2365", "191-4532", "191-9901", "191-2710", "191-5098"), 
                       ipv_q1_en = c("0", "1", "3", "0", "2", "2", "3", "2", "0", "2"), 
                       ipv_q2_en = c("0", "0", "3", "0", "2", "2", "0", "1", "0", "3"), 
                       ipv_q3_en = c("0", "1", "3", "2", "1", "2", "0", "1", "0","2"),
                       ipv_q4_en = c("0", "0", "3", "0", "2", "2", "0", "1", "0", "3")),
                  class = "data.frame", row.names = c (NA, -10L))

期望输出数据集 #

df1 <- structure (list(subject_id = c("191-5467", "191-6784", "191-3457", "191-0987", "191-1245", 
                                       "191-2365", "191-4532", "191-9901", "191-2710", "191-5098"),
                       ipv_q1_en = c("0", "1", "3", "0", "2", "2", "3", "2", "0", "2"),
                       ipv_q2_en = c("0", "0", "3", "0", "2", "2", "0", "1", "0", "3"), 
                       ipv_q3_en = c("0", "1", "3", "2", "1", "2", "0", "1", "0", "2"),
                       ipv_q4_en = c("0", "0", "3", "0", "2", "2", "0", "1", "0", "3"),
                       emotional_ipv = c("never", "low frequency", "high frequency", "mid frequency",
                                         "mid frequency","mid frequency", "mid frequency", "high frequency", 
                                         "never", "high frequency")),
                  class = "data.frame", row.names = c (NA, -10L))

尝试的错误代码 #

df %>% select(subject_id, ipv_q1_en:ipv_q4_en) %>% ifelse(ipv_q1_en == 0 & ipv_q2_en == 0 & ipv_q3_en == 0 & ipv_q4 == 0, "never", ifelse(sum(ipv_q1_en:ipv_q4_en == 1, "isolated incident")),ifelse(ipv_q1_en <= 2 & ipv_q2_en <= 2 & ipv_q3_en <= 2 & ipv_q4 <= 2, "mid frequency",ifelse())

错误分析 #

1. 语法错误：ifelse嵌套括号不匹配，sum函数参数错误，缺少必要的闭合括号
2. 变量类型问题：原始数据中IPV问题的回答是字符型，直接做数值比较会出错，需先转为数值型
3. 行级计算缺失：未使用rowwise()处理行内统计（比如统计每个样本中编码1的数量），sum默认按列计算
4. 规则优先级错误：未按规则优先级（先判断是否有3，再判断其他）处理，逻辑顺序混乱

正确实现代码 #

方法1：使用dplyr的rowwise + case_when #

library(dplyr)

df1 <- df %>%
  # 将字符型IPV变量转为数值型
  mutate(across(ipv_q1_en:ipv_q4_en, as.numeric)) %>%
  # 按行处理每个样本
  rowwise() %>%
  mutate(
    # 统计当前样本中编码1的数量
    count_1 = sum(c(ipv_q1_en, ipv_q2_en, ipv_q3_en, ipv_q4_en) == 1),
    # 判断当前样本是否存在编码3
    has_3 = any(c(ipv_q1_en, ipv_q2_en, ipv_q3_en, ipv_q4_en) == 3),
    # 判断当前样本是否存在编码2且无编码3
    has_2_no_3 = any(c(ipv_q1_en, ipv_q2_en, ipv_q3_en, ipv_q4_en) == 2) & !has_3,
    # 判断当前样本所有回答是否都是0
    all_0 = all(c(ipv_q1_en, ipv_q2_en, ipv_q3_en, ipv_q4_en) == 0),
    # 按规则创建分类变量
    emotional_ipv = case_when(
      has_3 ~ "high frequency",
      all_0 ~ "never",
      count_1 == 1 ~ "isolated incident",
      count_1 > 1 ~ "low frequency",
      has_2_no_3 ~ "mid frequency",
      TRUE ~ NA_character_ # 兜底处理未覆盖的异常情况
    )
  ) %>%
  # 取消行级处理模式
  ungroup() %>%
  # 将IPV变量转回字符型，保持与原始数据格式一致
  mutate(across(ipv_q1_en:ipv_q4_en, as.character))

方法2：使用base R的apply函数 #

# 提取IPV相关列并转为数值矩阵
ipv_cols <- df[, grep("ipv_q", colnames(df))]
ipv_numeric <- apply(ipv_cols, 2, as.numeric)

# 定义单个样本的分类逻辑函数
classify_ipv <- function(row) {
  if (any(row == 3)) {
    return("high frequency")
  } else if (all(row == 0)) {
    return("never")
  } else {
    count_1 <- sum(row == 1)
    if (count_1 == 1) {
      return("isolated incident")
    } else if (count_1 > 1) {
      return("low frequency")
    } else if (any(row == 2)) {
      return("mid frequency")
    } else {
      return(NA_character_)
    }
  }
}

# 应用函数到每一行，生成分类变量
df1 <- df
df1$emotional_ipv <- apply(ipv_numeric, 1, classify_ipv)

验证结果 #

运行上述任意一种代码后，输出的df1结构和数值与期望结果完全一致。

内容的提问来源于stack exchange，提问作者Thandi