合并排除哑变量：重新归类种族变量的R代码问题求助

我有一个包含8个种族变量的数据集，需要按特定条件调整变量：

• 受访者可多选种族，其中ethnicity_2代表白人
• 要创建新的white变量，仅标记那些只选了ethnicity_2、未选择其他任何种族的受访者
• 尝试的代码出现三个问题：
  1. dnresp变量（标记未选任何种族的样本）被错误赋值给几乎所有样本
  2. 部分同时选了ethnicity_2和其他种族的受访者被错误标记为white
  3. 用{{ethnicities.19}}的写法时，无论判断==1还是==0，都得到所有受访者未选任何种族的错误结果

原尝试代码：

ethnicities.19 <- c("ethnicity_1", "ethnicity_2", "ethnicity_3", "ethnicity_4",
                    "ethnicity_5", "ethnicity_6", "ethnicity_7", "ethnicity_8")
bar <- foo %>%
    select(ID, ethnicity_1:ethnicity_8) %>%
    mutate(across(.cols=ethnicity_1:ethnicity_8, .fns=function(x) { ifelse(is.na(x), 0, x)} )) %>%
    rowwise() %>%
     mutate(dnresp=ifelse(sum(eval(as.name(ethnicities.19)))==0, 1, 0),
##         dnresp=ifelse(!any(eval(as.name(ethnicities.19))==1), 1, 0),
            white=ifelse(eval(as.name(ethnicities.19[2]))==1 & sum(eval(as.name(ethnicities.19[-c(2)])))==0, 1, 0))

错误尝试的代码片段：

dnresp=ifelse(!any({{ethnicities.19}}==1), 1, 0))

dnresp=ifelse(!any({{ethnicities.19}}==0), 1, 0))

数据样本：

structure(list(ID = c("ATL_01", "ATL_02", "ATL_03", "ATL_04", 
"ATL_05", "ATL_06", "ATL_07", "ATL_08", "ATL_09", "ATL_10", "ATL_11", 
"ATL_12", "ATL_13", "ATL_14", "ATL_15", "ATL_16", "ATL_17", "ATL_18", 
"ATL_19", "ATL_20"), ethnicity_1 = c(NA_real_, NA_real_, NA_real_, 
NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, 
NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, 
NA_real_, NA_real_, NA_real_), ethnicity_2 = c(1, 1, 1, NA, 1, 
NA, 1, NA, NA, 1, NA, NA, 1, NA, NA, NA, NA, 1, NA, NA), ethnicity_3 = c(NA, 
NA, NA, 1, NA, 1, NA, 1, 1, NA, 1, 1, NA, 1, 1, 1, 1, NA, 1, 
1), ethnicity_4 = c(NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, 
NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, 
NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, 
NA_real_), ethnicity_5 = c(NA_real_, NA_real_, NA_real_, NA_real_, 
NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, 
NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, 
NA_real_, NA_real_), ethnicity_6 = c(NA_real_, NA_real_, NA_real_, 
NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, 
NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, 
NA_real_, NA_real_, NA_real_), ethnicity_7 = c(NA_real_, NA_real_, 
NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, 
NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, 
NA_real_, NA_real_, NA_real_, NA_real_), ethnicity_8 = c(NA_real_, 
NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, 
NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, 
NA_real_, NA_real_, NA_real_, NA_real_, NA_real_)), row.names = c(NA, 
-20L), class = c("tbl_df", "tbl", "data.frame"))

原代码的核心问题是**eval(as.name())无法处理字符串向量**：

• as.name(ethnicities.19)只会把向量的第一个元素转为符号，后续元素被忽略，导致sum()和any()的计算完全错误
• {{ethnicities.19}}是tidy eval语法，用于传递变量名而非字符串向量，直接使用逻辑不通

不用rowwise()，改用rowSums()更高效准确，直接针对列进行计算：

# 定义种族变量名向量
ethnicities.19 <- c("ethnicity_1", "ethnicity_2", "ethnicity_3", "ethnicity_4",
                    "ethnicity_5", "ethnicity_6", "ethnicity_7", "ethnicity_8")

bar <- foo %>%
  select(ID, all_of(ethnicities.19)) %>%
  # 将NA替换为0
  mutate(across(all_of(ethnicities.19), ~ifelse(is.na(.), 0, .))) %>%
  # 计算总选择数
  mutate(total_selected = rowSums(across(all_of(ethnicities.19)))) %>%
  # 生成dnresp：未选任何种族的样本标记为1
  mutate(dnresp = ifelse(total_selected == 0, 1, 0)) %>%
  # 生成white：仅选了ethnicity_2的样本标记为1
  mutate(white = ifelse(ethnicity_2 == 1 & total_selected == 1, 1, 0)) %>%
  # 可选：不需要total_selected可删除
  select(-total_selected)

验证结果 #

用提供的样本数据运行后：

• white变量会标记ATL_01、ATL_02、ATL_03、ATL_05、ATL_07、ATL_10、ATL_13、ATL_18为1（这些样本仅选了ethnicity_2）
• dnresp变量所有样本都是0（样本中没有未选任何种族的记录）

如果一定要用rowwise()，可以用c_across()获取行内的列值：

bar <- foo %>%
  select(ID, all_of(ethnicities.19)) %>%
  mutate(across(all_of(ethnicities.19), ~ifelse(is.na(.), 0, .))) %>%
  rowwise() %>%
  mutate(
    dnresp = ifelse(sum(c_across(all_of(ethnicities.19))) == 0, 1, 0),
    white = ifelse(ethnicity_2 == 1 & sum(c_across(all_of(ethnicities.19[-2]))) == 0, 1, 0)
  ) %>%
  ungroup()

内容的提问来源于stack exchange，提问作者Stuart