'R: Determine if each date interval overlaps with all other date intervals in a dataframe
For each date interval row in my dataframe, I would like to determine whether it overlaps with all other date intervals or not. Excluding itself.
A dataframe with start and end date, representing intervals:
`data <- read.table(header=TRUE,text="
start.date end.date
2019-09-01 2019-09-10
2019-09-05 2019-09-07
2019-08-25 2019-09-05
2019-10-10 2019-10-15
")`
This function lubridate::int_overlaps() checks if two date intervals overlap or not by returning logical TRUE or FALSE.
`int_overlaps(interval(ymd("2019-09-01"),ymd("2019-09-10")), interval(ymd("2019-09-05"), ymd("2019-09-07")))
[1] TRUE
int_overlaps(interval(ymd("2019-09-01"),ymd("2019-09-10")), interval(ymd("2019-10-10"), ymd("2019-10-15")))
[1] FALSE`
I would like to iterate each date interval with the all other date intervals excluding itself using int_overlap() to determine whether it overlaps with other date or not.
The output should look like this:
`data <- read.table(header=TRUE,text="
start.date end.date overlaps
2019-09-01 2019-09-10 TRUE
2019-09-05 2019-09-07 TRUE
2019-08-25 2019-09-05 TRUE
2019-10-10 2019-10-15 FALSE
")
`
Solution 1:[1]
Here is one option using dplyr and purrr, we loop through Int's indexes comparing the current interval with the other intervals.
library(dplyr)
library(purrr)
library(lubridate)
data %>% mutate(Int = interval(start.date, end.date),
overlaps = map(seq_along(Int), function(x){
#browser()
#Get all Int indexes other than the current one
y = setdiff(seq_along(Int), x)
#The interval overlaps with all other intervals
#return(all(int_overlaps(Int[x], Int[y])))
#The interval overlaps with any other intervals
return(any(int_overlaps(Int[x], Int[y])))
}))
start.date end.date Int overlaps
1 2019-09-01 2019-09-10 2019-09-01 UTC--2019-09-10 UTC TRUE
2 2019-09-05 2019-09-07 2019-09-05 UTC--2019-09-07 UTC TRUE
3 2019-08-25 2019-09-05 2019-08-25 UTC--2019-09-05 UTC TRUE
4 2019-10-10 2019-10-15 2019-10-10 UTC--2019-10-15 UTC FALSE
Solution 2:[2]
I think this can be accomplished nicely through a combination of dplyr and the ivs package, which is a package for working with interval vectors, which is what you have here.
library(ivs)
library(dplyr)
data <- tribble(
~start.date, ~end.date,
"2019-09-01", "2019-09-10",
"2019-09-05", "2019-09-07",
"2019-08-25", "2019-09-05",
"2019-10-10", "2019-10-15"
)
# Parse the dates and then convert them into an interval vector
data <- data %>%
mutate(
start = as.Date(start.date),
end = as.Date(end.date),
.keep = "unused"
) %>%
mutate(interval = iv(start, end), .keep = "unused")
# Note that interval vectors are half-open! You may need to adjust your end
# dates by 1 depending on how you interpret them.
data
#> # A tibble: 4 × 1
#> interval
#> <iv<date>>
#> 1 [2019-09-01, 2019-09-10)
#> 2 [2019-09-05, 2019-09-07)
#> 3 [2019-08-25, 2019-09-05)
#> 4 [2019-10-10, 2019-10-15)
# Use `iv_identify_group()` to identify the wider "overlap group" that rows 1-3
# fall in, noting that row 4 gets its own group. Then it is just a matter of
# grouping by `groups` and checking if there is more than one value in each group
data %>%
mutate(groups = iv_identify_group(interval)) %>%
group_by(groups) %>%
mutate(overlaps = n() > 1)
#> # A tibble: 4 × 3
#> # Groups: groups [2]
#> interval groups overlaps
#> <iv<date>> <iv<date>> <lgl>
#> 1 [2019-09-01, 2019-09-10) [2019-08-25, 2019-09-10) TRUE
#> 2 [2019-09-05, 2019-09-07) [2019-08-25, 2019-09-10) TRUE
#> 3 [2019-08-25, 2019-09-05) [2019-08-25, 2019-09-10) TRUE
#> 4 [2019-10-10, 2019-10-15) [2019-10-10, 2019-10-15) FALSE
Created on 2022-04-05 by the reprex package (v2.0.1)
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | |
| Solution 2 | Davis Vaughan |
