Excel BI - Excel Challenge 747

excel-challenges

excel-formulas

🔰 String Answer Expected 1 plus 2 18 minus 11 24 division 12 24 multiplication 11 36 modulo 8 3 plus 4 minus 2 24 division 3 plus 4 minus 3 34 plus 12 minus 2 division 2

Published

March 24, 2026

Illustration for Excel BI - Excel Challenge 747

Challenge Description

🔰 String Answer Expected 1 plus 2 18 minus 11 24 division 12 24 multiplication 11 36 modulo 8 3 plus 4 minus 2 24 division 3 plus 4 minus 3 34 plus 12 minus 2 division 2

Solutions

library(tidyverse)
library(readxl)
library(rlang)

path = "Excel/700-799/747/747 Mathematical Operations.xlsx"
input = read_excel(path, range = "A1:A10")
test  = read_excel(path, range = "B1:B10")

evaluate_text = function(text) {
  text %>% str_replace_all(c("minus" = "-",
                             "plus" = "+",
                             "multiplication" = "*",
                             "division" = "/",
                             "modulo" = "%%")) %>%
    parse_expr() %>%
    eval_bare()
}

result = input %>%
  mutate(`Answer Expected` = map_dbl(String, evaluate_text)) %>%
  select(-String)

all.equal(result, test)
# > [1] TRUE

Logic: Read the workbook ranges needed for the challenge; Derive the required intermediate columns.
Strengths: The code maps the workbook rule into a compact, reproducible pipeline.
Areas for Improvement: The solution assumes the workbook layout and selected ranges remain stable, so any structural change in the sheet would require small adjustments.
Gem: The elegant part is how little code is needed once the correct intermediate representation is chosen.

import pandas as pd
import re
import operator

path = "700-799/747/747 Mathematical Operations.xlsx"
input = pd.read_excel(path, usecols="A", nrows=10)
test = pd.read_excel(path, usecols="B", nrows=10)

replacements = {
    r"\bminus\b": "-",
    r"\bplus\b": "+",
    r"\bmultiplication\b": "*",
    r"\bdivision\b": "/",
    r"\bmodulo\b": "%"
}

exprs = input.iloc[:, 0]
for word, op in replacements.items():
    exprs = exprs.str.replace(word, op, regex=True)

input['Answer Expected'] = exprs.apply(lambda x: int(eval(x)) if pd.notnull(x) else float('nan'))
result = input[['Answer Expected']]

print(result.equals(test))

The Python version keeps the algorithm explicit, which helps when the challenge depends on a greedy or iterative rule.

Difficulty Level

Easy / Medium

The business rule is clear, though the workbook still needs a few transformation steps to reach the expected output.