Omid - Challenge 22

data-challenges
advanced-exercises
🔰 Convert Number To Convert
Published

March 24, 2026

Illustration for Omid - Challenge 22

Challenge Description

🔰 Convert Number To Convert

Solutions

library(tidyverse)
library(readxl)
library(english)

input = read_excel("files/CH-022 Convert Number To Text.xlsx" , range = "B2:B10")
test  = read_excel("files/CH-022 Convert Number To Text.xlsx" , range = "G2:H10")

result = input %>%
  mutate(
    number = as.character(Number),
    integer_part = map(number, ~ strsplit(.x, "\\.")[[1]][1]) %>% as.integer(),
    decimal_part = map(number, ~ strsplit(.x, "\\.")[[1]][2]) %>% as.integer()
  ) %>%
  mutate(
    integer_part_text = map_chr(integer_part, ~ as.character(as.english(.x))),
    decimal_part_text = map_chr(decimal_part, ~ as.character(as.english(.x))),
    Text = ifelse(is.na(decimal_part), 
                  integer_part_text, 
                  paste0(integer_part_text, " point ", decimal_part_text)) %>%
      str_to_sentence() %>%
      str_remove_all(" and")
  ) %>%
  select(Number, Text)
    
identical(result, test)
# [1] TRUE

  • Logic:

    • Reads the workbook ranges needed for the challenge

    • Builds the intermediate columns that drive the final result

    • Parses the text patterns directly instead of relying on manual cleanup

  • Strengths:

    • The R solution stays close to the workbook rule and keeps the transformation compact.

  • Areas for Improvement:

    • The code assumes the sheet structure and source ranges remain stable.

  • Gem:

    • The strongest part of the solution is choosing the right intermediate representation before shaping the final output.
import pandas as pd

input_data = pd.read_excel("CH-022 Convert Number To Text.xlsx", usecols="B", skiprows=1, nrows=9)
test = pd.read_excel("CH-022 Convert Number To Text.xlsx", usecols="G:H", skiprows=1, nrows=9)

ones = {
    0: "zero", 1: "one", 2: "two", 3: "three", 4: "four", 5: "five", 6: "six", 7: "seven", 8: "eight", 9: "nine",
    10: "ten", 11: "eleven", 12: "twelve", 13: "thirteen", 14: "fourteen", 15: "fifteen", 16: "sixteen",
    17: "seventeen", 18: "eighteen", 19: "nineteen"
}
tens = {2: "twenty", 3: "thirty", 4: "forty", 5: "fifty", 6: "sixty", 7: "seventy", 8: "eighty", 9: "ninety"}

def textify_whole(number: str) -> str:
    if not number:
        return ""
    n = int(number)
    if n < 20:
        return ones[n]
    if n < 100:
        t, o = divmod(n, 10)
        return tens[t] if o == 0 else f"{tens[t]}-{ones[o]}"
    if n < 1000:
        h, rem = divmod(n, 100)
        tail = textify_whole(str(rem)).strip()
        return f"{ones[h]} hundred" if rem == 0 else f"{ones[h]} hundred {tail}"
    th, rem = divmod(n, 1000)
    tail = textify_whole(str(rem)).strip()
    return f"{ones[th]} thousand" if rem == 0 else f"{ones[th]} thousand {tail}"

def textify_number(number):
    text = str(number)
    if "." in text:
        whole, dec = text.split(".", 1)
        out = f"{textify_whole(whole)} point {textify_whole(str(int(dec)))}"
    else:
        out = textify_whole(text)
    return out.capitalize()

result = input_data.assign(Text=input_data["Number"].map(textify_number))
print(result.equals(test))

  • Logic:

    • Reads the workbook ranges needed for the challenge

    • Builds the intermediate columns that drive the final result

  • Strengths:

    • The Python version follows the same rule in a direct dataframe-oriented implementation.

  • Areas for Improvement:

    • The code assumes the workbook layout remains stable, so any sheet redesign would require small adjustments.

  • Gem:

    • The implementation stays close to the original workbook rule instead of adding unnecessary abstraction.

Difficulty Level

This task is moderate:

  • The core logic is clear, but the correct transformation pattern is not obvious from the raw input.

  • The challenge combines multiple reshaping, grouping, or parsing steps.