PowerQuery Puzzle solved with R

#161–162

Puzzles

Author: ExcelBI

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

Puzzle #161

This week comes with some of hardest challenges I faced on ExcelBI series. It was looking pretty straightforward, but later I realised what is going on.
One of the least used measures describing set of data is so called mode or dominant. It points most commonly used value, and it is barely similar to mean or median.

So the story… we have weeks with winning numbers, and for each week we need to find most commonly occuring number in period of 8 weeks (current and 7 priors). I must admit that in two rows I don’t have exactly the same output as in given solution, but… it can not finish any closer. Let’s look at it.

Load libraries and data

library(tidyverse)
library(readxl)

input = read_excel("Power Query/PQ_Challenge_161.xlsx", range = "A1:C30") %>%   janitor::clean_names()
test  = read_excel("Power Query/PQ_Challenge_161.xlsx", range = "E1:H30") %>%   janitor::clean_names()

Transformation

find_mode <- function(x) {
  x <- na.omit(x)
if (length(x) == 0) return(NA)
  freq <- table(x)
  modes <- as.numeric(names(freq)[freq == max(freq)])
  return(modes)
}

input2 = input %>%
  group_by(group) %>%
  mutate(group_min_week = min(week_no)) %>%
  ungroup() %>%
  mutate(week_start_date = as.Date(paste0(week_no, "1"), format = "%Y%U%u")) %>%
  mutate(WM0 = week_start_date, 
         WM1 = week_start_date - weeks(1),
         WM2 = week_start_date - weeks(2),
         WM3 = week_start_date - weeks(3),
         WM4 = week_start_date - weeks(4),
         WM5 = week_start_date - weeks(5),
         WM6 = week_start_date - weeks(6),
         WM7 = week_start_date - weeks(7)) %>%
  select(group, group_min_week,  week_no, WM0, WM1, WM2, WM3, WM4, WM5, WM6, WM7) %>%
  pivot_longer(cols = starts_with("WM"), names_to = "week_marker", values_to = "valid_week_start") %>%
  mutate(group_min_week = as.Date(paste0(group_min_week, "1"), format = "%Y%U%u")) %>%
  left_join(input %>%
              mutate(week_start_date = as.Date(paste0(week_no, "1"), format = "%Y%U%u")) , 
            by = c("group", "valid_week_start" = "week_start_date")) %>%
  filter(valid_week_start >= group_min_week) %>%
  group_by(group, week_no.x) %>%
  mutate(no_groups = n()) %>%
  ungroup() %>%
  group_by(group, week_no.x, no_groups) %>%
  summarise(winning_nos = list(winning_no), .groups = 'drop') %>%
  ungroup() %>%
  arrange(group, desc(week_no.x)) %>%
  mutate(winning_nos_ = winning_nos) %>%
  mutate(winning_nos = map(winning_nos, ~na.omit(.x))) 
  
input3 = input2 %>%
  mutate(mode = map(winning_nos, find_mode)) %>%
  mutate(mode = map_chr(mode, ~paste(., collapse = ", "))) %>%
  mutate(mode = if_else(no_groups < 8, NA, mode))

input4 = input %>%
  left_join(input3, by = c("group", "week_no" = "week_no.x")) %>%
  left_join(test, by = c("group", "week_no" = "week_no")) %>%
  select(1,2,7,9) %>%
  mutate(check = mode == max_occurred_no)

If you can find flaws in my code, and fix it, be brave and do it.

#Puzzle 162

Usually after storm we see a rainbow… After hard challenge, we get something that we can have fun solving without headache caused by level of dificulty. And today we have pattern extracting. REGEXP rules again. In random string we need to find Letter and two Digits that are divided by exactly one character that is not Letter or Digit. Can be space, dollar sign, hyphen… whatever you want. And as result we need to give extracted patterns without those special signs in, concatenated. Easy Peasy…

Load libraries and data

library(tidyverse)
library(readxl)

input = read_excel("Power Query/PQ_Challenge_162.xlsx", range = "A1:A10")
test  = read_excel("Power Query/PQ_Challenge_162.xlsx", range = "C1:D10")

Transformation

result = input %>%
  mutate(result = str_extract_all(String, "([[:alpha:]])[^[:alnum:]]([[:digit:]]{2})")) %>%
  unnest_longer(result, keep_empty = TRUE) %>%
  mutate(result = str_remove(result, "[^[:alnum:]]")) %>%
  group_by(String) %>%
  summarise(Result = paste(result, collapse = ", ")) %>%
  ungroup() %>%
  mutate(Result = if_else(Result == "NA", NA, Result))

test1 = test %>%
  left_join(result, by = c("String" = "String"), suffix = c(".test", ".result"))

Validation

all.equal(test1$Result.test, test1$Result.result)
# [1] TRUE

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PowerQuery Puzzle solved with R was originally published in Numbers around us on Medium, where people are continuing the conversation by highlighting and responding to this story.