R中如何高效有条件批量填充DataFrame的多列需求字段

现有数据结构 #

现有结构如下的DataFrame：

CustomerID  Department  Price  SportswearDemand  HomeDemand  KidswearDemand  WomenswearDemand
-------------------------------------------------------------------------------------------
    1050091     Sportswear  497.6  0                 0           0               0                          
    1555018     Womenswear  336.0  0                 0           0               0                       
    210239      Womenswear  698.0  0                 0           0               0                       
    507556      Sportswear  209.0  0                 0           0               0                        
    1708193     Sportswear  209.0  0                 0           0               0                        
    1295733     Menswear    209.0  0                 0           0               0                        
    1213373     Sportswear  298.0  0                 0           0               0                       
    753471      Sportswear  209.0  0                 0           0               0                        
    82739       Menswear    349.0  0                 0           0               0                        
    1660995     Kidswear    424.6  0                 0           0               0
      .
      .
      .                

将SportswearDemand及右侧的所有列统称为需求列，需要按照如下规则填充：若某行的Department值和需求列名称去掉Demand后缀的前缀匹配，则将该行的Price值填入对应需求列替换原有0值，否则保留0。

预期效果 #

填充完成后效果如下：

CustomerID  Department  Price  SportswearDemand  HomeDemand  KidswearDemand  WomenswearDemand
-------------------------------------------------------------------------------------------
   1050091    Sportswear   497.6  497.6             0           0               0
   1555018    Womenswear   336.0  0                 0           0               336.0
   210239     Womenswear   698.0  0                 0           0               698.0
   507556     Sportswear   209.0  209.0             0           0               0
   1708193    Sportswear   209.0  209.0             0           0               0
   1295733    Menswear     209.0  0                 0           0               0
   1213373    Sportswear   298.0  298.0             0           0               0
   753471     Sportswear   209.0  209.0             0           0               0
   82739      Menswear     349.0  0                 0           0               0
   1660995    Kidswear     424.6  0                 0           424.6           0
     .
     .
     .

现有实现的问题 #

硬编码逐列写入可以实现需求：

df$SportswearDemand <- with(df, ifelse(df$Department == "Sportswear", df$Price, 0))
df$HomeDemand <- with(df, ifelse(df$Department == "Home", df$Price, 0))
df$KidswearDemand <- with(df, ifelse(df$Department == "Kidswear", df$Price, 0))
df$WomenswearDemand <- with(df, ifelse(df$Department == "Womenswear", df$Price, 0))

但需求列有30个，逐行写重复代码效率过低。尝试用for循环批量处理时出现错误：

DemandColumns # 存储了全部30个需求列名称的字符串数组
for (i in DemandColumns){
 df$i <- with(df, ifelse(df$Department == substr(i,1,nchar(i)-6), df$Price, 0))
}

运行后仅新增了一列名为i的全0字段，没有正确修改目标需求列。

错误原因 #

R语言中$运算符后的标识符会被直接当做固定列名解析，不会识别为变量的实际值，因此df$i永远是操作名为i的列，而非变量i存储的目标需求列名，导致循环逻辑错误。

修正方案 #

方案1：调整for循环写法 #

使用双括号[[]]替代$运算符，双括号支持解析变量内容作为列名，修正后代码如下：

# DemandColumns为你提前定义的所有需求列名的字符串数组
for (col_name in DemandColumns){
  # 提取列名对应的部门名（去掉Demand后缀）
  dept_name <- substr(col_name, 1, nchar(col_name)-6)
  df[[col_name]] <- ifelse(df$Department == dept_name, df$Price, 0)
}

方案2：tidyverse向量化写法（无需提前定义列名、无需循环） #

如果使用tidyverse生态工具，可以自动匹配所有Demand后缀的列，代码更简洁高效：

library(dplyr)
library(stringr)

df <- df %>%
  # 遍历所有结尾为Demand的列，批量赋值
  mutate(across(ends_with("Demand"), 
                ~ifelse(Department == str_remove(cur_column(), "Demand$"), Price, 0)))

内容的提问来源于stack exchange，提问作者Parseval