PowerQuery Puzzle solved with R

#159–160

Puzzles

Author: ExcelBI

All files (xlsx with puzzle and R with solution) for each and every puzzle are available on my Github. Enjoy.

Puzzle #159

Today we have data about some salespeople. But they have only data about the months when they exceeded “norm” which is 100. Nonetheless those 100’s are also cost of employee so… We need to fill missing months with 100. Of course only those months that we know that person works. So again some time magic again

Load libraries and data

library(tidyverse)
library(readxl)

input = read_excel("Power Query/PQ_Challenge_159.xlsx", range = "A1:D19")
test  = read_excel("Power Query/PQ_Challenge_159.xlsx", range = "F1:I73")

Transformation

calendar = input %>%
  select(-c(Sales,Month)) %>%
  group_by(Name) %>%
  expand_grid(Y = unique(Year), M = 1:12) %>%
  distinct() %>%
  filter(Y == Year) %>%
  select(Name, Year, Month = M) %>%
  ungroup()

result = calendar %>%
  left_join(input, by = c("Name", "Year", "Month")) %>%
  replace_na(list(Sales = 100))

Validation

identical(result, test)
# [1] TRUE

Puzzle #160

Geography now! (It is a name of interesting YT channel.) But I really like Geo so, lets check it out. We have coordinates of several cities and our output should be distance matrix from each city to each city. I didn’t really check if coordinates are correct but it is not crucial to this task. Way of solving is crucial, because we need to use Pythagorean Theorem to find distnaces. Good luck!

Loading libraries and data

library(tidyverse)
library(readxl)

input = read_excel("Power Query/PQ_Challenge_160.xlsx", range = "A1:C8")
test  = read_excel("Power Query/PQ_Challenge_160.xlsx", range = "F1:M8")

Transformation

grid = crossing(city1 = input$Cities, city2 = input$Cities) %>%
  mutate(x1 = input$x[match(city1, input$Cities)],
         y1 = input$y[match(city1, input$Cities)],
         x2 = input$x[match(city2, input$Cities)],
         y2 = input$y[match(city2, input$Cities)], 
         dist = round(sqrt((x2 - x1)^2 + (y2 - y1)^2),2)) %>%
  select(Cities = city1, city2, dist) %>%
  pivot_wider(names_from = city2, values_from = dist)

Validation

identical(test, grid)
# [1] TRUE

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