Filtering uneven data sets - r

Question

I am trying to filter out a data set into two months. I would like to filter out the ID and year that have data, and to remove the ID and year that do not have an associated pair.
For example if an ID and year has both the January and July month in the data set, I would like to include this ID and the year in my filtered data. If an ID has only the month of January and not July, I would like to remove this data and not include it in the filtered data set. Is there a good way to do this? Just a note that I wasn't sure how to simulate the uneven data set in the example.
After filtering for my desired output, I test by creating a list for each seasonal month where each ID and year has at least 15 rows associated with it.
library(lubridate)
library(dplyr)
set.seed(12345)
df <- tibble(
date = sample(seq(dmy("01-01-2010"), dmy("31-12-2013"), by = "days"),
1000, replace = TRUE),
x = runif(length(date), min = 60000, max = 80000),
y = runif(length(date), min = 800000, max = 900000),
ID = rep(1:5, 200),
month = month(date),
year =year(date)) %>%
arrange(ID, date)
df %>%
filter(month %in% c(1,7)) %>%
group_by(ID, year) %>%
mutate(complete = length(unique(month)) == 2) %>%
group_by(ID) %>%
filter(all(complete)) %>%
group_by(ID, year)
# Creates a list for each year and by ID
summer_list <- df %>%
filter(month %in% 7) %>%
filter(n() >= 15) %>%
group_split(year, ID)
# Renames the names in the list to AnimalID and year
names(summer_list) <- sapply(summer_list,
function(x) paste(x$ID[1],
x$year[1], sep = '_'))
# Creates a list for each year and by ID
winter_list <- df1 %>%
filter(month %in% 1) %>%
filter(n() >= 15) %>%
group_split(year, ID)
# Renames the names in the list to ID and year
names(winter_list) <- sapply(winter_list,
function(x) paste(x$ID[1],
x$year[1], sep = '_'))

Answer 1

You were really close. I think your filter can be simplified to the following. Just be sure to save it to df.
df <- df %>%
filter(month %in% c(1,7)) %>%
group_by(ID, year) %>%
mutate(complete = length(unique(month)) == 2) %>%
filter(complete)
# could add "%>% select(-c(complete))" to get rid of complete
On summer_list and winter_list, add a group_by between the filters. With the dataset you provided, there were no groups with 15 records, but I tested that this works by bumping up the size of df until I got some.
summer_list <- df %>%
filter(month == 7) %>% # used == since there's only one test value
group_by(ID, year) %>% # added this
filter(n() >= 15) %>%
group_split()
There's also a typo in your first use of winter_list -- the input data is df1, but I think you want df. Hope this works!
Here's the complete code including the larger df:
library(lubridate)
library(dplyr)
set.seed(12345)
df <- tibble(
date = sample(seq(dmy("01-01-2010"), dmy("31-12-2013"), by = "days"),
4000, replace = TRUE),
x = runif(length(date), min = 60000, max = 80000),
y = runif(length(date), min = 800000, max = 900000),
ID = rep(1:5, 800),
month = month(date),
year =year(date)) %>%
arrange(ID, date)
df <- df %>%
filter(month %in% c(1,7)) %>%
group_by(ID, year) %>%
mutate(complete = length(unique(month)) == 2) %>%
filter(complete)
# could add "%>% select(-c(complete))" to get rid of complete
# Creates a list for each year and by ID
summer_list <- df %>%
filter(month == 7) %>%
group_by(ID, year) %>%
filter(n() >= 15) %>%
group_split()
# Renames the names in the list to AnimalID and year
names(summer_list) <- sapply(summer_list,
function(x) paste(x$ID[1],
x$year[1], sep = '_'))
# Creates a list for each year and by ID
winter_list <- df %>%
filter(month == 1) %>%
group_by(ID, year) %>%
filter(n() >= 15) %>%
group_split()
# Renames the names in the list to ID and year
names(winter_list) <- sapply(winter_list,
function(x) paste(x$ID[1],
x$year[1], sep = '_'))