define an indicator for overlapping time interval [duplicate] - r

Question

This question already has an answer here:
finding overlapping time between start time and end time of individuals in a group
(1 answer)
Closed 3 years ago.
I have
household person start time end time
1 1 07:45:00 21:45:00
1 2 09:45:00 17:45:00
1 3 22:45:00 23:45:00
1 4 08:45:00 01:45:00
1 1 23:50:00 24:00:00
2 1 07:45:00 21:45:00
2 2 016:45:00 22:45:00
I want to find a column to find overlapping time between family members.
The indicator is: if a person's start and end time has intersection with another member's is 1 otherwise 0
In the above example first family, the time of first, second and forth persons have intersection so indicator is 1 and third and fifth rows doesn't have intersection with non of the other people in the household.
output:
household person start time end time overlap
1 1 07:45:00 21:45:00 1
1 2 09:45:00 17:45:00 1
1 3 22:45:00 23:45:00 0
1 4 08:45:00 01:45:00 1
1 1 23:50:00 24:00:00 0
2 1 07:45:00 21:45:00 1
2 2 016:45:00 22:45:00 1
data with dput format:
structure(list(SAMPN = c(1L, 1L, 1L, 2L, 2L, 2L), PERNO = c(1,
1, 1, 1, 1, 1), arr = structure(c(30300, 35280, 37200, 32400,
34200, 39600), class = c("hms", "difftime"), units = "secs"),
dep = structure(c(34200, 36300, 61800, 33600, 37800, 50400
), class = c("hms", "difftime"), units = "secs")), class = c("grouped_df",
"tbl_df", "tbl", "data.frame"), row.names = c(NA, -6L), groups = structure(list(
SAMPN = 1:2, PERNO = c(1, 1), .rows = list(1:3, 4:6)), row.names = c(NA,
-2L), class = c("tbl_df", "tbl", "data.frame"), .drop = TRUE))

Answer 1

I have tried a tidyverse solution:
library(tidyverse)
df = structure(list(SAMPN = c(1L, 1L, 1L, 2L, 2L, 2L),
PERNO = c(1:3, 1:3), arr = structure(c(30300, 35280, 37200, 32400,
34200, 39600), class = c("hms", "difftime"), units = "secs"),
dep = structure(c(34200, 36300, 61800, 33600, 37800, 50400), class = c("hms", "difftime"), units = "secs")), class = c("grouped_df","tbl_df", "tbl", "data.frame"), row.names = c(NA, -6L), groups = structure(list(SAMPN = 1:2, PERNO = c(1, 1), .rows = list(1:3, 4:6)), row.names = c(NA, -2L), class = c("tbl_df", "tbl", "data.frame"), .drop = TRUE))
Then, I added:
df %>% group_by(SAMPN) %>%
mutate(
arr_min = mapply(function(x) min(arr[-x]), 1:n()),
dep_max = mapply(function(x) max(dep[-x]), 1:n()),
overlap = ifelse(arr<arr_min | dep>dep_max, 0, 1)
)
You will get:
SAMPN PERNO arr dep arr_min dep_max overlap
<int> <int> <time> <time> <dbl> <dbl> <dbl>
1 1 1 08:25 09:30 35280 61800 0
2 1 2 09:48 10:05 30300 61800 1
3 1 3 10:20 17:10 30300 36300 0
4 2 1 09:00 09:20 34200 50400 0
5 2 2 09:30 10:30 32400 50400 1
6 2 3 11:00 14:00 32400 37800 0
You basically compare the current arr and dep with arr_min (min(arr) value excluding the current case) and dep_max (max(dep) excluding the current case).

Answer 2

Tidyverse solution
Here's a solution in tidyverse syntax. The basic idea is the same. We perform a many to many merge matching on household (sampn in your current example data) and remove the cases of comparing a person to themself (perno). We check for overlaps, then collapse to a single record per household and person. Note that this code will error if all records have the same perno.
compare <-
df %>%
left_join(df %>%
rename(compare_PERNO = PERNO,
compare_arr = arr,
compare_dep = dep), by = ("SAMPN")) %>%
filter(PERNO != compare_PERNO) %>%
mutate(overlap = arr <= compare_dep & dep >= compare_arr) %>%
group_by(SAMPN, PERNO) %>%
summarize(overlap = max(overlap))
SQL Solution with household grouping
Grouping the data by household actually makes this problem slightly easier. Again, I'm using SQL to solve it. In the inner SQL statement I do a many to many merge matching all members of a household to all other members, I remove any cases of matching a person to themself. Then, in the outer SQL statement we reduce to one record per household and person, which indicates if they ever overlapped.
df = data.frame(
household = c(rep(1,5), rep(2,2)),
person = c(1:5, 1:2),
start_time=as.Date(c("2017-05-31","2018-01-14", "2019-02-03", "2018-01-19", "2019-04-17",
"2018-02-03", "2018-03-03"),
format="%Y-%m-%d"),
end_time=as.Date(c("2018-01-17", "2018-01-20", "2019-04-15", "2018-02-20", "2019-05-17",
"2019-03-03", "2019-03-03"),
format="%Y-%m-%d"))
library(sqldf)
compare <- sqldf(
"
SELECT * FROM (
SELECT L.* ,
CASE WHEN L.start_time <= R.end_time AND L.end_time >= R.start_time THEN 1
ELSE 0 END AS overlap
FROM df as L
LEFT JOIN df as R ON L.household = R.household
WHERE L.person != R.person
)
GROUP BY household, person
HAVING overlap = MAX(overlap)
"
)
SQL Solution without household grouping
This is an SQL solution to your problem. I do a keyless many to many merge to compare each row to every other row (but dont compare a row to itself), then I parse the big data frame down to a single record per ID that records whether any matches were found. Your data isn't quite a reprex (use the dput function in R), so I used an example dataset I had lying around. If you have trouble adapting this to your exact data, post reproducible data and I can help you out.
df = data.frame(
id = 1:3,
start_time=as.Date(c("2017-05-31","2018-01-14", "2018-02-03"), format="%Y-%m-%d"),
end_time=as.Date(c("2018-01-17", "2018-01-20", "2018-04-17"), format="%Y-%m-%d"))
library(sqldf)
compare <- sqldf(
"
SELECT * FROM (
SELECT L.* ,
CASE WHEN L.start_time <= R.end_time AND L.end_time >= R.start_time THEN 1
ELSE 0 END AS overlap
FROM df as L
CROSS JOIN df as R
WHERE L.id != R.id
)
GROUP BY ID
HAVING overlap = MAX(overlap)
"
)