Simulate data to work with.

library(tidyverse)

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# Generate sample data
# Sightings of Black Oystercatcher chicks at Santa Cruz beaches
beaches <- c("Cowell's", "Steamer Lane", "Natural Bridges", "Mitchell's", "Main")
# blue, green, black, white, yellow
band_colors <- c("B", "G", "K", "W", "Y") 
# Surveys took place weekly in the summer of 2023
surveys <- seq(as.Date("2023-06-01"), as.Date("2023-08-31"), by = 7)

# Setting the "seed" forces randomized functions (like sample()) to generate
# the same output
set.seed(1538)
# 3 band colors identify a bird. We want 12 birds.
birds <- paste0(
  sample(band_colors, 25, replace = TRUE),
  sample(band_colors, 25, replace = TRUE),
  sample(band_colors, 25, replace = TRUE)
) %>% 
  unique() %>%
  head(12)
bloy_chicks <- tibble(
  # Randomly generate survey data
  beach = sample(beaches, size = 100, replace = TRUE),
  bird = sample(birds, size = 100, replace = TRUE),
  survey = sample(surveys, size = 100, replace = TRUE)
) %>% 
  # Remove duplicates (see ?distinct)
  distinct() %>% 
  # Sort by survey date and location
  arrange(survey, beach)

Q1: We are all going to end up with the same data frames because we are using set seed, which starts the randomized functions all from the same point.

birds <- paste0(
  sample(band_colors, 25, replace = TRUE),
  sample(band_colors, 25, replace = TRUE),
  sample(band_colors, 25, replace = TRUE)
) %>% 
  unique() %>%
  head(12)

This code creates a fake dataset by taking 25 samples from band_colors with replacements. It then combines the first result in each list, so that it is a list of 25 sets of 3 band colors. It gets rid of any duplicates and then shows the first 12 results.

We generated 100 observations, but bloy_chicks has 94 observations. I think this is because it removes duplicates using distinct().

# Find most frequent beach per bird
beach_freq <- bloy_chicks %>% 
  group_by(bird) %>% 
  count(bird, beach) %>% 
  filter(n == max(n)) %>% 
  ungroup()

# Find first date for each bird+beach
beach_early <- bloy_chicks %>% 
  group_by(bird, beach) %>% 
  summarize(earliest = min(survey),
            .groups = "drop")

# Join the two conditions and retain most frequent beach, only earliest
hatch_beach <- beach_freq %>% 
  left_join(beach_early, by = c("bird", "beach")) %>% 
  group_by(bird) %>% 
  filter(earliest == min(earliest)) %>% 
  sample_n(1) %>% # Randomly choose 1 row. See ?sample_n
  ungroup()

Custom function!

The logic:

1. Put the logic for estimating the hatching beach in a single function.

2. Group the data by bird

3. Summarize each group using your custom function

Using this workflow: Most frequent site -> earliest day -> choose 1

find_hatching_beach <- function(site, date) {
  # Start with a data frame (or tibble) of site and date for *one* bird
  # Use pipes and dplyr functions to find the hatching beach
  bird_observations <- tibble(site, date)
  result <- bird_observations %>% 
    count(site) %>% 
    filter(n == max(n)) %>% 
    left_join(bird_observations, by = c("site")) %>% 
    filter(date == min(date)) %>% 
    sample_n(1)
    # use as many pipes and dplyr functions as necessary
  # result should end up as a data frame with one row for the hatching beach
  return(result$site) # return the hatching beach
}

# split-apply-combine
hatch <- bloy_chicks %>% 
  group_by(bird) %>% 
  summarize(hatch = find_hatching_beach(beach, survey))

The column I used for “date” was “survey”; the column I used for “site” was “beach”.

The hatching beach for both TWG and WYB was Mitchell’s.