R将10添加到数据框中列表的特定元素答案

【问题标题】：R adding 10 to a specific element of a list in a dataframeR将10添加到数据框中列表的特定元素
【发布时间】：2022-07-06 23:04:25
【问题描述】：

我有这个数据框 df:

df<-structure(list(tile_type_index = c(9, 15, 20, 5, 20), tile_type = c("Flowers", 
"Leather", "Outpost", "Wood 2", "Outpost"), material_on_hex = list(
    c(0, 0, 0, 0, 0, 0, 0, 0, 1000, 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, 1000, 
    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, 1000, 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))), row.names = c(NA, 
5L), class = "data.frame")

  tile_type_index tile_type
1               9   Flowers
2              15   Leather
3              20   Outpost
4               5    Wood 2
5              20   Outpost
                                                material_on_hex
1 0, 0, 0, 0, 0, 0, 0, 0, 1000, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
2 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1000, 0, 0, 0, 0, 0
3    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
4 0, 0, 0, 0, 1000, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
5    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

我想通过以下方式操作它：如果 tile_type 是“Outpost”，那么 material_on_hex 的行应该保持不变，但是如果 tile_type!="Outpost" 那么我想在 material_on_hex[tile_type_index] 中添加 10。结果应该是：

  tile_type_index tile_type
1               9   Flowers
2              15   Leather
3              20   Outpost
4               5    Wood 2
5              20   Outpost
                                                material_on_hex
1 0, 0, 0, 0, 0, 0, 0, 0, 1010, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
2 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1010, 0, 0, 0, 0, 0
3    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
4 0, 0, 0, 0, 1010, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
5    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

【问题讨论】：

标签： r dplyr

【解决方案1】：

简单的循环：

library(magrittr)
for (i in seq_len(nrow(df))) {
  if (df$tile_type[i] == 'Outpost') next 
  tidx = df$tile_type_index[i]
  df$material_on_hex[[i]][tidx] %<>% add(10)
}


#   tile_type_index tile_type                                               material_on_hex
# 1               9   Flowers 0, 0, 0, 0, 0, 0, 0, 0, 1010, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
# 2              15   Leather 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1010, 0, 0, 0, 0, 0
# 3              20   Outpost    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
# 4               5    Wood 2 0, 0, 0, 0, 1010, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
# 5              20   Outpost    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

也可以使用mapply()：

df$material_on_hex = mapply(
  \(x, y, z) {
    if (x!='Outpost') z[y] %<>% add(10)
    return(z)
  }, 
  x = df$tile_type, y = df$tile_type_index, z = df$material_on_hex, 
  SIMPLIFY = FALSE
)

【讨论】：