library(tidyverse)
library(readxl)
path = "files/CH-227 Data Normalization Min-Max.xlsx"
input = read_excel(path, range = "B2:G8")
test = read_excel(path, range = "J2:O8")
result = input %>%
mutate(across(where(is.numeric), ~ (. - min(.)) / (max(.) - min(.))))
all.equal(result, test)
#> [1] TRUEOmid - Challenge 227
data-challenges
advanced-exercises
🔰 For this challenge, calculate the minimum and maximum values for each column.

Challenge Description
🔰 For this challenge, calculate the minimum and maximum values for each column.
Solutions
Logic:
Reads the workbook ranges needed for the challenge
Builds the intermediate columns that drive the final result
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
path = "CH-227 Data Normalization Min-Max.xlsx"
input = pd.read_excel(path, usecols="B:G", skiprows=1, nrows=7)
test = pd.read_excel(path, usecols="J:O", skiprows=1, nrows=7).rename(columns=lambda col: col.split('.1')[0] if '.1' in col else col)
result = (input - input.min()) / (input.max() - input.min())
print(result.equals(test)) # TrueLogic:
- Reads the workbook ranges needed for the challenge
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 business rule is readable, but the workbook still requires careful implementation to reach the expected layout.