# INBO CODING CLUB

28 February 2023

Welcome!

## Share your code snippet

If you want to share your code snippet, copy paste your snippet within a section of three backticks (```):

As an **example**:

library(tidyverse)

(*you can copy paste this example and add your code further down*)

## Yellow sticky notes

No yellow sticky notes online. Put your name + " | " and add a "*" each time you solve a challenge (see below).

## Participants

Name | Challenges
--- | ---
Damiano Oldoni | ***
Pieter Huybrechts | ****
Dirk Maes | ***
Heleen Deroo | *
An Leyssen | **
Wouter Depaepe | **
Raïsa Carmen | ****
Nele Mullens |
Arc'hantael Labriere |
Zoé Hermans |
Wim Kuypers | *
Amber Mertens | ***
Joppe Massant | **
Ioannis Rallis | **
Adriaan Seynaeve | ***

## Challenge 1

### Pieter's solution

# Display first 8 rows; display the last 8 rows.
head(partridge,8)
tail(partridge,8)

## dplyr equivalents, I'm not sure when you'd use one over the other
slice_head(partridge,n = 8)
slice_tail(partridge,n = 8)

## or you can even do this with square brackets!
partridge[1:8,]
partridge[{nrow(partridge)-7}:nrow(partridge),]

# Select columns wbe and date.
select(partridge,wbe,date) #tidy evaluation
select(partridge,c(wbe,date)) # usin
select(partridge,c("wbe","date")) #standard evaluation

# Display the values of type. How many different values of type are?

unique(pull(partridge,type))
length(unique(pull(partridge,type)))
# data.table actually has a function for counting unique values
data.table::uniqueN(partridge,"type")

# you can also use distinct instead of unique, distinct is like unique but for
# whole data.frames or columns
partridge %>%
  select(type) %>%
  distinct %>%
  nrow

# How to remove observations with type Sound? How to remove observations with
# type Sound and wbe NA?

filter(partridge, type != "Sound")
## you can keep adding filter expresssions
filter(partridge, type != "Sound", !is.na(wbe))
## or combine them
filter(partridge, type != "Sound" & !is.na(wbe))

# Add a new column called seen which is TRUE if type is not "Sound". Add a
# column month with the month of the observation. Tip: use lubridate::month()
# function

mutate(partridge,
       month = lubridate::month(date),
       seen = type != "Sound")

## or with case_when, useful for more complex cases

## I like to keep my assignment (<-) on a new line, so it's very clear I'm
## creating a new object (overwriting one in this case)
partridge <-
  partridge %>%
  mutate(month = lubridate::month(date, label = TRUE)) %>%
  mutate(seen = case_when(type == "Sound" ~ FALSE,
                          .default = TRUE))
# Place the column seen after type and month after date

partridge %>%
  relocate(month, .after = date) %>%
  relocate(seen, .after = type)

## select also allows you to reorder columns, it'll return them in the same order
## as you request them, you can use this together with tidyselect
select(partridge, date, month,
       year, wbe, fieldworker, type, seen, xlambert, ylambert)

### An's solution

#   Display first 8 rows; display the last 8 rows.
partridge %>% 
  slice_head(n = 8)
partridge %>% 
  slice_tail(n = 8)

# Select columns wbe and date.
partridge %>% 
  select(wbe, date)

# Display the values of type. How many different values of type are?
unique(partridge$type)
partridge %>% 
  distinct(type)

#   How to remove observations with type Sound? How to remove observations with type Sound and wbe NA?
partridge %>% 
  filter(!(type == "Sound"))
partridge %>% 
  filter(!(type == "Sound"), !is.na(wbe)) 

#   Add a new column called seen which is TRUE if type is not "Sound". Add a column month with the month of the observation. Tip: use lubridate::month() function. Install lubridate package if not yet done.
partridge_2 <- partridge %>% 
  mutate(seen = ifelse(!(type == "Sound"), TRUE, FALSE),
         month = lubridate::month(date))
 
# Place the column seen after type and month after date
partridge_3 <- partridge_2 %>% 
  select(date, month, wbe, fieldworker, type, seen, xlambert:year)
partridge_3 <- partridge_2 %>% 
  relocate(seen, .after = type) %>% 
  relocate(month, .after = date)

### Amber's solution

## CHALLENGE 1

# Display first 8 rows; display the last 8 rows.
head(partridge, 8)
tail(partridge, 8)

# Select columns wbe and date.
select(partridge, wbe, date)

# Display the values of type. How many different values of type are?
unique(partridge$type)
# or:
distinct(partridge, type)

count(distinct(partridge, type))
   
#   How to remove observations with type Sound? How to remove observations with type Sound and wbe NA?
partridge %>% filter(type != "Sound") 

partridge %>% filter(type != "Sound", 
                      !is.na(wbe))

#   Add a new column called seen which is TRUE if type is not "Sound". Add a column month with the month of the observation. Tip: use lubridate::month() function. Install lubridate package if not yet done.
# Place the column seen after type and month after date

partridge %>% mutate(
  seen = case_when(
    type != "Sound" ~ TRUE,
    TRUE ~ FALSE),
  month = lubridate::month(date, label = TRUE, abbr = FALSE)
  ) %>%
  relocate(
    seen, .after = type
    ) %>%
  relocate(month, .after = date)

### Joppe's Solution

### 1. Display first 8 rows; display the last 8 rows.
head(partridge, 8)
tail(partridge, 8)

### 2. Select columns wbe and date.
partridge$wbe
partridge$date

### 3. Display the values of type. How many different values of type are?
distinct(partridge, type)
count(distinct(partridge, type))

### 4. How to remove observations with type Sound? How to remove observations with type Sound and wbe NA?
filter(partridge, !(type == 'Sound'))
filter(partridge, !(type == 'Sound') & !is.na(wbe))
partridge <- mutate(partridge, seen=!(type == 'Sound'))
partridge <- mutate(partridge, month=month(date))
partridge

### 5.
# Relocate was wellicht cleaner
col_order <- c("date", "month", "wbe", "fieldworker", "type", "seen", "xlambert",
               "ylambert", "year")
partridge <- partridge[, col_order]
partridge

### Arc'hantael's solution

# 1- Display the first 8 rows; display the last 8 rows
head(partridge, 8)
tail(partridge, 8)

# 2- Select columns wbe and date
partridge %>% select(wbe, date)

# 3- display the values of type. How many different values of type are there?
partridge %>% count(type)

# 4- How to remove observations with type Sound? How to remove observations with
# type Sound and wbe NA?
partridge %>% filter(type != "Sound", wbe != "NA")

# 5- Add a new column called seen which is TRUE if type is not "Sound". Add a 
# new column month with the month of the observation. Tip: use lubridate::month
# function.
partridge %>% mutate(seen = case_when(type != "Sound" ~ TRUE), 
                     month = month(date))

## Challenge 2

### Pieter's solution

# How many observations per WBE? Save it as overview_partridge as we will use it
# in challenge 3.
overview_partridge <-
  partridge %>%
  group_by(wbe) %>%
  tally

# How many observations per WBE and type?
partridge %>%
  group_by(wbe, type) %>%
  tally

# Order the previous output per n_obs (descending order), wbe (ascending order)
# and type (alphabetically).
partridge %>%
  group_by(wbe, type) %>%
  tally(name = "n_obs") %>% # or add_tally() or add_count()
  arrange(-n_obs,wbe,type)

# For each month, return the wbe with the highest number of observations and
# report the number of observations for those wbe that month.

## filter can actually group internally, summarise is probably easier to read?
partridge %>%
  group_by(wbe, month) %>%
  tally(name = "n_obs") %>%
  ungroup() %>%
  filter(n_obs == max(n_obs), .by = month)

### Nele's solution

## 2.1 How many observations per WBE? Save it as overview_partridge as we will use it in challenge 3.

overview_partridge <- count(partridge, wbe)

## 2.2 How many observations per WBE and type?
count(partridge, wbe, type)  %>%  View()

## 2.3 Order the previous output per n_obs (descending order), 
## wbe (ascending order) and type (alphabetically).

partridge %>% 
  group_by(wbe, type) %>% 
  count(name = "n_obs") %>% 
  arrange(desc(n_obs), wbe, type)

### Amber's solution

## CHALLENGE 2 - Summaries
# How many observations per WBE? Save it as overview_partridge as we will use it in challenge 3.
overview_partridge <- partridge %>% group_by(wbe) %>% summarise(n_obs = n())

# How many observations per WBE and type?
#   Order the previous output per n_obs (descending order), wbe (ascending order) and type (alphabetically).
partridge %>% group_by(wbe, type) %>% 
  summarise(n_obs = n()) %>%
  arrange(desc(n_obs),
          wbe,
          type,
          .by_group = FALSE)


# For each month, return the wbe with the highest number of observations and report the number of observations for those wbe that month.

partridge %>% 
  group_by(month, wbe) %>%
  summarise(n_obs = n()) %>%
  arrange(month, desc(n_obs)) %>% 
  slice(1) %>% 
  ungroup()

### Joppe's solution

### 1. How many observations per WBE? Save it as overview_partridge as we will use it in challenge 3.
overview_partridge <- count(partridge, wbe)
overview_partridge


### 2. How many observations per WBE and type?
count(partridge, wbe, type)


### 3. Order the previous output per n_obs (descending order), wbe (ascending order) and type (alphabetically).
partridge %>%
  count(wbe, type) %>%
  arrange(desc(n), wbe, type)


### 4. For each month, return the wbe with the highest number of observations and report the number of observations for those wbe that month.
partridge %>%
  group_by(month, wbe) %>%
  count() %>%
  group_by(wbe) %>%
  summarise(n = max(n))

## INTERMEZZO

## Challenge 3

### Raisa

WBE <- read_csv("./data/20230228/20230228_wbe_info.txt", na = "")
overview_partridge  %>%
  left_join(WBE, by = join_by(wbe))

WBE %>% right_join(overview_partridge, by = join_by(wbe))

WBE %>% anti_join(partridge, by = "wbe")

overview_partridge %>% inner_join(WBE, by = join_by(wbe))

overview_partridge %>% 
  full_join(WBE, by = join_by(wbe)) %>%
  replace_na(list(n_obs = 0))

### Pieter's solution

# read the game management units details
WBE <- read_csv("./data/20230228/20230228_wbe_info.txt", na = "")

# Add WBE details in WBE to overview_partridge (where available)

## I don't need to tell it to join by what column, as the data.frames only have
## one column with the same name
overview_partridge %>% 
  left_join(WBE, by = "wbe") 

# How are the columns ordered? Are the columns of overview_partridge on the
# left? Try to put them on the right.

inner_join(WBE, overview_partridge, keep = TRUE) %>% 
  filter(!is.na(n)) # keep only partridge wbe rows
## this approach will drop rows with no wbe (with wbe is NA)

## check first if you actually have a match in WBE for all overview_partridge
right_join(WBE, overview_partridge)

## I prefer reordering columns after a join in a seperate step
overview_partridge %>% 
  left_join(WBE, by = "wbe") %>% 
  select(colnames(WBE), colnames(overview_partridge))

# Which WBEs are not in partridge, i.e. are not linked to any observation?

anti_join(WBE, partridge) %>% 
  pull(wbe) %>% 
  unique

# You don't need to use joins to get to this answer
setdiff(pull(WBE,wbe), pull(partridge,wbe))

# Some observations have a missing value for column wbe and some WBEs have no
# observations at all. Get rid of them while joining.
inner_join(WBE, overview_partridge)

# Now, join both again but retain ALL EXISTING WBEs. Set the number of
# observations equal to 0 for all WBEs where no partridges were observed.
full_join(overview_partridge, WBE) %>% 
  mutate(n = case_when(is.na(n) ~ 0,
                           .default = n)) %>% 
  filter(!is.na(wbe))

### Nele's solution

# 3.1 Add WBE details in WBE to overview_partridge (where available)
overview_partridge <- left_join(overview_partridge, WBE, by = "wbe")

# 3.2 How are the columns ordered? Are the columns of overview_partridge on the left? Try to put them on the right.

#overview_partridge <- overview_partridge %>%
  relocate(wbe, .after = province) %>% relocate(n, .after = wbe)
  
WBE %>% right_join(overview_partridge, by = join_by(wbe))
#or
right_join(WBE, overview_partridge, by = "wbe") %>% view()

## Bonus challenge 1

### Pieter's solution

# if there are a lot of columns to pivot, you can use tidyselect
pivot_longer(
  ias,
  cols = c("english_name", "dutch_name", "french_name"),
  names_to = "language",
  values_to = "vernacular_name"
) %>% 
  # regex magic is also possible instead 
  mutate(language = stringr::str_remove(language, stringr::fixed("_name"))) %>% 
  mutate(language = dplyr::case_match(language,
                                      "english" ~ "en",
                                      "dutch" ~ "nl",
                                      "french" ~ "fr"))

### Raisa

ias <- ias %>% pivot_longer(cols = 2:4,
                     values_to = "vernacular_name",
                     names_to = "language",
                     names_pattern = "(.*)_name") %>%
  mutate(language = ifelse(language == "dutch", 
                           "nl",
                           str_sub(language, 1, 2)))

## Bonus challenge 2

### Example code Amber

Stukje code wat ik een tijd geleden geschreven heb toen ik pas net met R begon te werken, grote kans dat er dus ruimte is voor verbetering.

# Minimal reproducible example for coding club 20230228

# starting from a dataframe (result from other part of the process) with surface area per vegetation sub-categories for two different years and the difference in the area for these two years
# goal is to summarize these values per simplified/main categories (and removing NAs). 
# Names of these simplified categories are given in a separate vector
# and column names are given as variables (because it had to be executed for many different year combinations)

# data
df <- structure(list(Code = c("A", "B1", "B2", "I1", "I2", "I3", "I4", "U1", "U2", "W"), 
                     Habitat = c("Akker", "Wilg", "Berk", "Riet", "Lisdodde", "Mix Riet & Lisdodde", "Heen", "Ruigte type 1", "Ruigte type 2 ", "Water"), 
                     `2013` = c(NA, 0.049591440864666, NA, 0.042735889211246, 0.248358365784165, 0.0977590620561381, NA, NA, 0.559824776974386, 3.68812422563596), 
                     `2017` = c(NA, 0.049591440864666, NA, 0.042735889211246, 0.248358589051378, 0.0977590620561381, NA, NA, 0.352468819308082, 3.68812409798934), 
                     Verschil = c(NA, 0, NA, 0, 2.23267212828837e-07, 0, NA, NA, -0.207355957666304, -1.27646623937494e-07)), 
                row.names = c(NA, -10L), 
                class = "data.frame")

categories_simple = c(
  "A" = "Akker",
  "B" = "Bos & struweel",
  "I" = "Riet",
  "M" = "Moeras",
  "P" = "Pionier",
  "U" = "Ruigte",
  "W" = "Water",
  "Z" = "Open zand"
)


year1 <- "2013"
year2 <- "2017"

df %>% 
  # get rid of na rows
  filter(!is.na(year1) & !is.na(year2) & !is.na(Verschil)) %>%
  # take only the letter part of the code using regex (to go from sub-categories to main categories)
  mutate(code_simple = str_extract(Code, "[A-Za-z]+")) %>%
  # group by main category and summarize the surface area values
  group_by(code_simple) %>%
  summarise(!!as.name(year1) := sum(!!as.name(year1)),
            !!as.name(year2) := sum(!!as.name(year2)),
            Verschil = sum(Verschil)) %>%
  # change the code of main categories to the names found in the vector given above
  mutate(Hoofdeenheid = recode(code_simple, !!!categories_simple)) %>%
  # change the position of the new column
  relocate(Hoofdeenheid, .after = code_simple) %>%
  # remove the code column
  select(-code_simple)