2022 Hanukkah of Data • Puzzle 1
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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)
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