2022 Hanukkah of Data • Puzzle 1

Visiting #2 and doing some $WORK-work, but intrigued with Hanukkah of Data since Puzzle 0 was solvable with a ZIP password cracker (the calendar date math seemed too trivial to bother with).

Decided to fall back to R for this (vs Observable for the Advent of Code which I’ll dedicate time to finishing next week).

R has a {phonenumber} package, so we’ll cheat and use that despite it being very brutish in how it does the letterToNumber() conversion.

No spoilers besides the code.

library(phonenumber)
library(tidyverse)

cust <- read_csv("~/Downloads/noahs-csv/noahs-customers.csv")

cust |> 
  filter(!grepl("[01]", phone)) |> # only care abt letters
  mutate(
    last_name = stri_replace_all_regex(name, "(II|III|IV|Jr\\.)", ""), # get rid of suffix if any
  ) |> 
  separate( # get only the last name
    col = last_name,
    into = c("x1", "x2", "last_name"),
    sep = " ",
    fill = "left"
  ) |> 
  filter(
    nchar(last_name) == 10 # only complete last names
  ) |> 
  mutate(
    last_name = toupper(last_name),
    phone = gsub("-", "", phone) # we're going to compare so remove the '-'
  ) |> 
  select(last_name, phone) |> 
  mutate(
    trans = strsplit(xx$last_name, "") |> 
      map_chr(~map(.x, letterToNumber) |> paste0(collapse="")) # feels like I cld optimize this
  ) |> 
  filter(trans == phone)