#175–176

Puzzles

Author: ExcelBI

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

Puzzle #175

Sometimes even common tables tell the story. Today we have table of people that are family, but we need to find their relationships. We only have first and last name and number of generation. So to find relations we need to make manouver called self-join. Let check it out.

(Note: there are some flaws in file so code has some cleaning fragments.)

Loading libraries and data

library(tidyverse)
library(readxl)

input = read_excel("Power Query/PQ_Challenge_175.xlsx", range = "A1:C16")
test  = read_excel("Power Query/PQ_Challenge_175.xlsx", range = "E1:H19") %>%
  mutate(Relantionship = str_remove_all(Relantionship, " ")) # cleaned for purpose of validation

Transformation

result = input %>%
  left_join(input, by = c("Family" = "Family")) %>%
  filter(`Generation No.x` == `Generation No.y` - 1) %>%
  # there is mispronunciation in the challenge, it should be "Relationship" not "Relantionship"
  unite("Relantionship", `Generation No.x`, `Generation No.y`, sep = "-") %>% 
  select(Name = `Name.x`,Family,`Next Generation` = `Name.y`, Relantionship  ) %>%
  arrange(Family, Relantionship  , Name, `Next Generation`)

Validation

identical(result, test)
# [1] TRUE

Puzzle #176

Probably somebody make a note by hand, not thinking of further users and as usually we need to straight issue up. From data concatenated in two columns of table we need to make long table with running sum per group. Go on, lets do it.

Loading libraries and data

library(tidyverse)
library(readxl)

input = read_excel("Power Query/PQ_Challenge_176.xlsx", range = "A1:C5")
test  = read_excel("Power Query/PQ_Challenge_176.xlsx", range = "E1:G9")

Transformation

result = input %>%
  mutate(Column1 = map(Column1, ~strsplit(.x, ", ")),
         Column2 = map(Column2, ~strsplit(.x, ", "))) %>%
  unnest(cols = c(Column1, Column2)) %>%
  mutate(Column1 = map(Column1, ~tibble(Column1 = .x)),
         Column2 = map(Column2, ~tibble(Column2 = .x))) %>%
  mutate(n1 = map_dbl(Column1, ~nrow(.x)),
         n2 = map_dbl(Column2, ~nrow(.x))) %>%
  mutate(Column = map2(Column1, Column2, ~{
    n1 = nrow(.x)
    n2 = nrow(.y)
    if (n1 > n2) {
      .y = bind_rows(.y, tibble(Column2 = rep("0", n1 - n2)))
    } else if (n1 < n2) {
      .x = bind_rows(.x, tibble(Column1 = rep(NA, n2 - n1)))
    }
    bind_cols(.x, .y)
  })) %>%
  select(Group, Column) %>%
  unnest(cols = c(Column)) %>%
  drop_na()  %>%
  mutate(Column2 = cumsum(as.numeric(Column2)), .by = Group)

Validation

identical(result, test)
# [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.