# Next 3 Friday the 13th

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Excel BI’s Excel Challenge #310 — solved in R

### Defining the Puzzle:

It was early Monday morning when hunt for Fridays come up. Of course new riddle about Fridays 13th.

Task was really short yet challenging: Find the next 3 Friday the 13th dates for dates given in column A.

### Loading Data from Excel:

Let me get needed libraries, exercise dates and solution from our excercise file.

library(tidyverse) library(lubridate) library(readxl) library(data.table) input = read_excel(“Next 3 Friday the 13th.xlsx”, range = “A1:A10”) test = read_excel(“next 3 friday the 13th.xlsx”, range = “B2:D10”, col_names = c(“friday_13th_1”, “friday_13th_2”, “friday_13th_3”))

### Approach 1: Tidyverse with purrr

get_3_f13_tv<- function(date) { start_date <- as.Date(paste0(year(date), “-”, month(date), “-13”)) end_date <- start_date + years(3) date_seq_13th <- seq.Date(from=start_date, to=end_date, by=”month”) friday_13ths <- date_seq_13th[wday(date_seq_13th) == 6 & date_seq_13th > date] head(friday_13ths, 3) } next_3_tv = input %>% rowwise() %>% mutate(friday_13th_1 = get_3_f13_tv(Date)[1] %>% as.character(), friday_13th_2 = get_3_f13_tv(Date)[2] %>% as.character() , friday_13th_3 = get_3_f13_tv(Date)[3] %>% as.character()) %>% ungroup() %>% select(-Date)

### Approach 2: Base R

get_3_f13_base <- function(date) { start_date <- as.Date(paste0(format(as.Date(date), “%Y-%m”), “-13”)) end_date <- start_date + (365.25 * 3) date_seq_13th <- seq(start_date, end_date, by=”month”) weekday <- weekdays(date_seq_13th ) friday_13ths <- date_seq_13th[weekday == “Friday” & date_seq_13th > date] head(friday_13ths, 3) } next_3_base <- tibble( friday_13th_1 = character(nrow(input)), friday_13th_2 = character(nrow(input)), friday_13th_3 = character(nrow(input)) ) for (i in 1:nrow(input)) { f13_dates <- get_3_f13_base(input$Date[i]) next_3_base$friday_13th_1[i] <- as.character(f13_dates[1]) next_3_base$friday_13th_2[i] <- as.character(f13_dates[2]) next_3_base$friday_13th_3[i] <- as.character(f13_dates[3]) }

### Approach 3: Data.table

input_dt = setDT(input) get_3_f13_dt <- function(date) { start_date <- as.Date(paste0(year(date), “-”, month(date), “-13”)) end_date <- start_date + years(3) date_seq_13th <- seq.Date(from=start_date, to=end_date, by=”month”) friday_13ths <- date_seq_13th[wday(date_seq_13th) == 6 & date_seq_13th > date] friday_13ths <- head(friday_13ths, 3) c(friday_13ths, rep(NA, 3 — length(friday_13ths))) } input_dt[, c(“friday_13th_1”, “friday_13th_2”, “friday_13th_3”) := as.list(get_3_f13_dt(Date)), by = 1:nrow(input_dt)][ , `:=`(friday_13th_1 = as.character(friday_13th_1), friday_13th_2 = as.character(friday_13th_2), friday_13th_3 = as.character(friday_13th_3))][ , Date := NULL] next_3_dt = as_tibble(as.data.frame(input_dt))

### Validating Our Solutions:

identical(test, next_3_tv) #> [1] TRUE identical(test, next_3_base) #> [1] TRUE identical(test, next_3_dt) #> [1] TRUE

If you like my publications or have your own ways to solve those puzzles in R, Python or whatever tool you choose, let me know.

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