Excel BI - PowerQuery Challenge 286

excel-challenges
power-query
Split the amount on the basis of number of animals and / or food. If there are more than one food, then same number of entries would appear in Amount column also.
Published

March 24, 2026

Illustration for Excel BI - PowerQuery Challenge 286

Challenge Description

Split the amount on the basis of number of animals and / or food. If there are more than one food, then same number of entries would appear in Amount column also.

Solutions

library(tidyverse)
library(readxl)

path = "Power Query/200-299/286/PQ_Challenge_286.xlsx"
input = read_excel(path, range = "A1:C6")
test = read_excel(path, range = "F1:H13")

result = input %>%
  mutate(rn = row_number()) %>%
  separate_rows(1:3, sep = ", ") %>%
  group_by(rn, Food) %>%
  mutate(
    Amount = ifelse(
      n_distinct(Animals) > n_distinct(Food) & n_distinct(Amount) == 1,
      as.numeric(Amount) / n_distinct(Animals),
      as.numeric(Amount)
    )
  ) %>%
  ungroup() %>%
  select(Animal = Animals, Food, Amount)

all.equal(result, test, check.attributes = FALSE)
# TRUE

  • Logic:

    • Reads the workbook range needed for the challenge

    • Aggregates or ranks values at the relevant grouping level

    • Builds helper columns that drive the final output

  • Strengths:

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

  • Areas for Improvement:

    • The code assumes the workbook layout and selected ranges remain stable.

  • Gem:

    • The best part of the solution is choosing the right intermediate shape before formatting the final output.
import pandas as pd
import re

path = "200-299/286/PQ_Challenge_286.xlsx"
input = pd.read_excel(path, usecols="A:C", nrows=5)
test = pd.read_excel(path, usecols="F:H", nrows=13).rename(columns=lambda x: x.replace('.1', ''))

input['rn'] = input.index
cols = ['Animals', 'Food', 'Amount']
input[cols] = input[cols].applymap(lambda x: re.split(r',\s*', x) if isinstance(x, str) and ',' in x else [x])
input['max_len'] = input[cols].applymap(len).max(axis=1)
input['Food_set_len'] = input['Food'].apply(lambda x: len(set(x)))
input['Animals_set_len'] = input['Animals'].apply(lambda x: len(set(x)))
input['Amount'] = input['Amount'].apply(lambda x: [float(i) for i in x])

def expand_row(row):
    for col in cols:
        row[col] = (row[col] * row['max_len'])[:row['max_len']]
    return row

input = input.apply(expand_row, axis=1)
input = input.explode(cols, ignore_index=True)
input['Amount'] = input.apply(
    lambda row: int(row['Amount'] / row['Animals_set_len']) if row['Food_set_len'] < row['Animals_set_len'] else int(row['Amount']),
    axis=1
).astype('int64')
input = input[['Animals', 'Food', 'Amount']]
input = input.rename(columns={'Animals': 'Animal'})

print(input.equals(test)) # True

  • Logic:

    • Reads the workbook range needed for the challenge

    • Uses direct pattern parsing where the workbook encodes logic in text

    • Applies the rule iteratively until the output is complete

  • Strengths:

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

  • Areas for Improvement:

    • As with the R version, any workbook layout change would require small adjustments.

  • Gem:

    • The implementation stays close to the source challenge instead of adding unnecessary abstraction.

Difficulty Level

This task is moderate:

  • It combines reshaping, grouping, or parsing steps that are common in Power Query style problems.

  • The main challenge is reproducing the workbook output structure exactly.