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#151–152

### Puzzles

Author: ExcelBI

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

### Puzzle #151

Today we simplified log of people work. They just written down date and time of start and end. Not really helpful, but fortunatelly their manager gave us certain rules: there are no overtime or weekend work, and prepare timetable when exactly each day they were suppose to work.
Lets utilize hms package and sequences to look for how many hours should be paid.

library(tidyverse)
library(hms)

test  = read_excel("Power Query/PQ_Challenge_151.xlsx", range = "G1:H6") %>%
janitor::clean_names()

read_excel_range <- function(file, range) {
read_excel(file, range = range) %>%
mutate(across(c(starts_with("Start Time"), starts_with("End Time")), as_hms),
across(c(starts_with("Start Date"), starts_with("End Date")), as_date)) %>%
janitor::clean_names()
}

input1 <- read_excel_range("Power Query/PQ_Challenge_151.xlsx", "A1:E6")
input2 <- read_excel_range("Power Query/PQ_Challenge_151.xlsx", "A9:D14")

#### Transformation

result <- input1 %>%
mutate(
start = as_datetime(start_date) + start_time,
end = as_datetime(end_date) + end_time,
datetime = map2(start, end, seq, by = "hour")
) %>%
unnest(datetime) %>%
mutate(
weekday = wday(datetime, week_start = 1),
time = as_hms(datetime)
) %>%
left_join(input2, by = "weekday") %>%
filter(datetime >= start & datetime <= end,
time >= start_time.y & time < end_time.y) %>%
group_by(employee) %>%
summarise(total_hours = n() %>% as.numeric())

#### Validation

identical(result, test)
#> [1] TRUE

### Puzzle #152

Another they, another HR issue. Now we have to calculate how many different types of leave and how many days certain workers take. Some conditional expressions and we will have it covered. Let’s go.

library(tidyverse)

input = read_excel("Power Query/PQ_Challenge_152.xlsx", range = "A1:D17") %>%
janitor::clean_names()
test  = read_excel("Power Query/PQ_Challenge_152.xlsx", range = "F1:I5") %>%
janitor::clean_names()

#### Transformation

result = input %>%
mutate(seq = map2(from_date, to_date, seq, by = "day")) %>%
unnest_longer(seq) %>%
select(-c(from_date, to_date)) %>%
mutate(value = 1) %>%
pivot_wider(names_from = type_of_leave, values_from = value, values_fill = 0) %>%
select(name, seq, ML, PL, CL) %>%
mutate(sum = ML + PL + CL,
concat = paste0(ML, PL, CL) %>% as.numeric(),
main_leave = case_when(sum == 1 & ML == 1 ~ "ML",
sum == 1 & PL == 1 ~ "PL",
sum == 1 & CL == 1 ~ "CL",
sum == 2 & concat >= 100 ~ "ML",
sum == 2 & concat < 100 ~ "PL",
sum == 3 ~ "ML",
TRUE ~ "NA"),
wday = wday(seq, week_start = 1)) %>%
filter(!wday %in% c(6, 7)) %>%
select(name, seq, main_leave) %>%
mutate(main_leave = str_to_lower(main_leave)) %>%
group_by(name, main_leave) %>%
summarise(days = n() %>% as.numeric()) %>%
ungroup() %>%
pivot_wider(names_from = main_leave, values_from = days, values_fill = 0)

#### Validation

identical(result, test)
#> [1] TRUE

Feel free to comment, share and contact me with advices, questions and your ideas how to improve anything. Contact me on Linkedin if you wish as well.

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.

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