Multiplying and Adding Values across Rows

Here is a way.
The main trick is to strsplit into single characters and match these vectors with the legend. Then add or multiply the matching numbers.

set.seed(2022)
color <- c("AKZ", "ZZA", "KAK")    
color_1 <- sample(color, 100, replace=TRUE, prob=c(0.4, 0.3, 0.3))
id <- 1:100

sample_data = data.frame(id, color_1)

legend <- setNames(c(3, 4, 6), c("K", "A", "Z"))

add_mul <- function(x, l){
  add <- function(y, l){
    i <- match(y, names(l))
    sum(l[i])
  }
  mul <- function(y, l){
    i <- match(y, names(l))
    prod(l[i])
  }
  
  s <- strsplit(x, "")
  add_score <- sapply(s, add, l = l)
  mul_score <- sapply(s, mul, l = l)
  data.frame(add_score, mul_score)
}

sample_data <- cbind(sample_data, add_mul(sample_data$color_1, legend))
head(sample_data)
#>   id color_1 add_score mul_score
#> 1  1     ZZA        16       144
#> 2  2     KAK        10        36
#> 3  3     AKZ        13        72
#> 4  4     KAK        10        36
#> 5  5     AKZ        13        72
#> 6  6     KAK        10        36

^{Created on 2022-03-10 by the reprex package (v2.0.1)}

We can use stri_replace_all_regex to replace your color_1 into integers together with the arithmetic operator.

Here I've stored your values into a vector color_1_convert. We can use this as the input in stri_replace_all_regex for better management of the values.

library(dplyr)
library(stringi)

color_1_convert <- c("K" = "3", "A" = "4", "Z" = "6")

sample_data %>%
  group_by(id) %>%
  mutate(add_score = eval(parse(text = gsub("\\+$", "", stri_replace_all_regex(color_1, names(color_1_convert), paste0(color_1_convert, "+"), vectorize_all = F)))),
         multiply_score = eval(parse(text = gsub("\\*$", "", stri_replace_all_regex(color_1, names(color_1_convert), paste0(color_1_convert, "*"), vectorize_all = F)))))

# A tibble: 100 × 4
# Groups:   id [100]
      id color_1 add_score multiply_score
   <int> <chr>       <dbl>          <dbl>
 1     1 KAK            10             36
 2     2 ZZA            16            144
 3     3 AKZ            13             72
 4     4 ZZA            16            144
 5     5 AKZ            13             72
 6     6 AKZ            13             72
 7     7 AKZ            13             72
 8     8 KAK            10             36
 9     9 ZZA            16            144
10    10 AKZ            13             72
# … with 90 more rows

Here is another option using tidyverse. I use recode from dplyr to change the letters to numbers according to the legend.

library(tidyverse)

legend <- c(K = 3, A = 4, Z = 6)

sample_data %>%
  rowwise %>%
  mutate(code = list(recode(str_split(color_1, "", simplify = T), !!!legend)),
         add_score = sum(code),
         multiply_score = prod(code)) %>%
  select(-code)

Output

      id color_1 add_score multiply_score
   <int> <chr>       <dbl>          <dbl>
 1     1 AKZ            13             72
 2     2 AKZ            13             72
 3     3 KAK            10             36
 4     4 KAK            10             36
 5     5 AKZ            13             72
 6     6 AKZ            13             72
 7     7 KAK            10             36
 8     8 AKZ            13             72
 9     9 AKZ            13             72
10    10 AKZ            13             72
# … with 90 more rows

Data

set.seed(103)
color <- c("AKZ", "ZZA", "KAK")    
color_1 <- sample(color, 100, replace=TRUE, prob=c(0.4, 0.3, 0.3))
id = 1:100

sample_data = data.frame(id, color_1)

Benchmark

It looks like Rui Barradas' solution is the fastest of the answers so far.

Here’s an approach based on tidyr::separate_rows() followed by a grouped dplyr::summarize():

library(tidyverse)
set.seed(1)

legend <- c(K = 3, A = 4, Z = 6)

sample_data %>%
  mutate(decoded = color_1) %>%
  separate_rows(decoded, sep = "(?!^)") %>%
  mutate(decoded = legend[decoded]) %>%
  group_by(id, color_1) %>%
  summarize(
    add_score = sum(decoded),
    multiply_score = prod(decoded), 
    .groups = "drop"
  )

Output:

# A tibble: 100 x 4
      id color_1 add_score multiply_score
   <int> <chr>       <dbl>          <dbl>
 1     1 AKZ            13             72
 2     2 AKZ            13             72
 3     3 KAK            10             36
 4     4 ZZA            16            144
 5     5 AKZ            13             72
 6     6 ZZA            16            144
 7     7 ZZA            16            144
 8     8 KAK            10             36
 9     9 KAK            10             36
10    10 AKZ            13             72
# ... with 90 more rows

Yet another alternative, using some libraries optimized for speed, stringi for string manipulation and Rfast for matrix operations. Note that when any NA values are present in your data matrixStats is safer to use than Rfast.

set.seed(2022)
color <- c("AKZ", "ZZA", "KAK")    
color_1 <- sample(color, 100, replace=TRUE, prob=c(0.4, 0.3, 0.3))
id = 1:100
sample_data = data.frame(id, color_1)

m <- strsplit(sample_data[["color_1"]], "") |>
  unlist(use.names = F) |>
  stringi::stri_replace_all_regex(
    c("K", "A", "Z"),
    c("3", "4", "6"), vectorize_all = F) |>
  as.integer() |>
  matrix(ncol = 3, byrow = T)
sample_data$add_score <- Rfast::rowsums(m)
sample_data$mul_score <- Rfast::rowprods(m)

head(sample_data)
  id color_1 add_score mul_score
1  1     ZZA        16       144
2  2     KAK        10        36
3  3     AKZ        13        72
4  4     KAK        10        36
5  5     AKZ        13        72
6  6     KAK        10        36