R list within a dataframe calculation-CodePudding

I have the following dataframe df:

     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

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"

For each row of df, I want to add material_spend to the element of material_balance that has the index given tile_type_index. So I want to do something like material_balance[tile_type_index]<-material_spend but I'm not sure how to do this.

The result should look like the following:

     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

CodePudding user response：

You can use dplyr for this. Normally dplyr assumes that you are operating on entire columns at once, but this is a special row-wise operation so we use the rowwise() verb. Then we can also use the replace function to replace part of a vector and return the updated value. Because you have a list column, we need to wrap the result in a list so the resulting value has a length of 1. So this should work

df %>% 
  rowwise() %>% 
  mutate(material_balance = list(replace(material_balance, tile_type_index, material_spend)))

CodePudding user response：

in base R you could do:

df[[2]] <- do.call(Map, c(`[<-`, df[c(2,1,3)]))

df[[2]] <-  Map(`[[<-`, df$material_balance, df$tile_type_index,df$material_spend)