R语言data.frame中list列按指定索引逐行赋值实现方法

问题场景 #

现有如下结构的R数据框对象df：

df<-structure(list(tile_type_index = c(17L, 8L, 17L, 8L, 17L, 17L
), material_balance = list(c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0), c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0), c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0), c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0), c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0), c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0)), material_spend = c(0.333333333333333, 
0.333333333333333, 0.333333333333333, 0.333333333333333, 0.333333333333333, 
0.333333333333333)), row.names = c("71", "81", "71.1", "81.1", 
"71.2", "71.3"), class = "data.frame")

直接打印df的输出为：

tile_type_index
71                17
81                 8
71.1              17
81.1               8
71.2              17
71.3              17
                                               material_balance
71   0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
81   0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
71.1 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
81.1 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
71.2 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
71.3 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
     material_spend
71        0.3333333
81        0.3333333
71.1      0.3333333
81.1      0.3333333
71.2      0.3333333
71.3      0.3333333

需要实现逐行处理逻辑：将每行的material_spend值，赋值给该行material_balance列表元素中索引等于tile_type_index的位置，等价于逐行执行material_balance[tile_type_index] <- material_spend的操作。
处理后的预期输出如下：

tile_type_index
71                17
81                 8
71.1              17
81.1               8
71.2              17
71.3              17
                                               material_balance
71   0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0.33, 0, 0, 0
81   0, 0, 0, 0, 0, 0, 0, 0.33, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
71.1 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0.33, 0, 0, 0
81.1 0, 0, 0, 0, 0, 0, 0, 0.33, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
71.2 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0.33, 0, 0, 0
71.3 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0.33, 0, 0, 0
     material_spend
71        0.3333333
81        0.3333333
71.1      0.3333333
81.1      0.3333333
71.2      0.3333333
71.3      0.3333333

实现方法 #

• base R 向量化方案（性能最优）
  用Map函数做行级映射，无需显式循环，直接赋值回原列即可：

df$material_balance <- Map(
  function(balance_vec, target_idx, assign_val) {
    balance_vec[target_idx] <- assign_val
    balance_vec
  },
  df$material_balance,
  df$tile_type_index,
  df$material_spend
)

• for循环方案（逻辑最直观）
  逐行遍历修改，注意访问列表列元素必须使用双中括号[[：

for (row_i in seq_len(nrow(df))) {
  df$material_balance[[row_i]][df$tile_type_index[row_i]] <- df$material_spend[row_i]
}

• dplyr 语法方案
  如果使用tidyverse生态，可通过rowwise逐行处理：

library(dplyr)
df <- df %>%
  rowwise() %>%
  mutate(
    material_balance = list({
      temp_vec <- material_balance
      temp_vec[tile_type_index] <- material_spend
      temp_vec
    })
  ) %>%
  ungroup()

内容的提问来源于stack exchange，提问作者locket