I have data of patient prescription of oral DM drugs, i.e. DPP4 and SU, and would like to find out if patients had taken the drugs concurrently (i.e. whether there are overlapping intervals for DPP4 and SU within the same patient ID
).
Sample data:
ID DRUG START END
1 1 DPP4 2020-01-01 2020-01-20
2 1 DPP4 2020-03-01 2020-04-01
3 1 SU 2020-03-15 2020-04-30
4 2 SU 2020-10-01 2020-10-31
5 2 DPP4 2020-12-01 2020-12-31
In the sample data above,
ID == 1
, patient had DPP4 and SU concurrently from 2020-03-15
to 2020-04-01
.ID == 2
, patient had consumed both medications at separate intervals.I thought of splitting the data into 2, one for DPP4 and another for SU. Then, do a full join, and compare each DPP4 interval with each SU interval. This may be okay for small data, but if a patient has like 5 rows for DPP4 and another 5 for SU, we will have 25 comparisons, which may not be efficient. Add that with 10000+ patients.
I am not sure how to do it.
New data:
Hope to have a new df that looks like this. Or anything that is tidy.
ID DRUG START END
1 1 DPP4-SU 2020-03-15 2020-04-01
2 2 <NA> <NA> <NA>
Data Code:
df <- structure(list(ID = c(1L, 1L, 1L, 2L, 2L), DRUG = c("DPP4", "DPP4",
"SU", "SU", "DPP4"), START = structure(c(18262, 18322, 18336,
18536, 18597), class = "Date"), END = structure(c(18281, 18353,
18382, 18566, 18627), class = "Date")), class = "data.frame", row.names = c(NA,
-5L))
df_new <- structure(list(ID = 1:2, DRUG = c("DPP4-SU", NA), START = structure(c(18336,
NA), class = "Date"), END = structure(c(18353, NA), class = "Date")), class = "data.frame", row.names = c(NA,
-2L))
Edit: I think from the sample data I gave, it may seem that there can only be 1 intersecting interval. But there may be more. So, I think this would be better data to illustrate.
structure(list(ID = c(3, 3, 3, 3, 3, 3, 3), DRUG = c("DPP4",
"DPP4", "SU", "SU", "DPP4", "DPP4", "DPP4"), START = structure(c(17004,
17383, 17383, 17418, 17437, 17649, 17676), class = c("IDate",
"Date")), END = structure(c(17039, 17405, 17405, 17521, 17625,
17669, 17711), class = c("IDate", "Date")), duration = c(35L,
22L, 22L, 103L, 188L, 20L, 35L), INDEX = c(1L, 0L, 0L, 0L, 0L,
0L, 0L)), row.names = c(NA, -7L), class = c("tbl_df", "tbl",
"data.frame"))
1) Sort all intervals in increasing order of start time. This step takes O(nLogn) time. 2) In the sorted array, if start time of an interval is less than end of previous interval, then there is an overlap.
Let's take the following overlapping intervals example to explain the idea: If both ranges have at least one common point, then we say that they're overlapping. In other words, we say that two ranges and are overlapping if: On the other hand, non-overlapping ranges don't have any points in common.
Java Program to find overlapping intervals among a given set of intervals. In this approach, we are going to take a count array and for each given interval start time we will do count[startTime]++ and for end time, do count[endTime]--. sum > 2 it means there is overlapping intervals.
A Simple Solution is to consider every pair of intervals and check if the pair overlaps or not. A better solution is to Use Sorting. Following is complete algorithm. 1) Sort all intervals in increasing order of start time.
Check if any two intervals intersects among a given set of intervals 1 Sort all intervals in increasing order of start time. This step takes O (nLogn) time. 2 In the sorted array, if start time of an interval is less than end of previous interval, then there is an overlap. More ...
A Simple Solution is to consider every pair of intervals and check if the pair overlaps or not. The time complexity of this solution is O (n 2 ). A better solution is to use Sorting. Following is the complete algorithm. 1) Sort all intervals in increasing order of start time.
The problem statement asks to check if any of the intervals overlap each other if overlaps then print ‘YES’ else print ‘NO’. Because, the time interval (2, 3) and (1, 4) overlaps with each other. The image clearly depicts that intervals (2,3) and (1,4) overlap. Because none of the intervals overlaps each other.
It's way more complicated than dear @AnoushiravanR's but as an alternative you could try
library(dplyr)
library(tidyr)
library(lubridate)
df %>%
full_join(x = ., y = ., by = "ID") %>%
# filter(DRUG.x != DRUG.y | START.x != START.y | END.x != END.y) %>%
filter(DRUG.x != DRUG.y) %>%
group_by(ID, intersection = intersect(interval(START.x, END.x), interval(START.y, END.y))) %>%
drop_na(intersection) %>%
filter(START.x == first(START.x)) %>%
summarise(DRUG = paste(DRUG.x, DRUG.y, sep = "-"),
START = as_date(int_start(intersection)),
END = as_date(int_end(intersection)),
.groups = "drop") %>%
select(-intersection)
returning
# A tibble: 1 x 4
ID DRUG START END
<int> <chr> <date> <date>
1 1 DPP4-SU 2020-03-15 2020-04-01
Edit: Changed the filter condition. The former one was flawed.
Updated Solution
I have made considerable modifications based on the newly provided data set. This time I first created interval for each START
and END
pair and extract the intersecting period between them. As dear Martin nicely made use of them we could use lubridate::int_start
and lubridate::int_end
to extract the START
and END
date of each interval:
library(dplyr)
library(lubridate)
library(purrr)
library(tidyr)
df %>%
group_by(ID) %>%
arrange(START, END) %>%
mutate(int = interval(START, END),
is_over = c(NA, map2(int[-n()], int[-1],
~ intersect(.x, .y)))) %>%
unnest(cols = c(is_over)) %>%
select(-int) %>%
filter(!is.na(is_over) | !is.na(lead(is_over))) %>%
select(!c(START, END)) %>%
mutate(grp = cumsum(is.na(is_over))) %>%
group_by(grp) %>%
summarise(ID = first(ID),
DRUG = paste0(DRUG, collapse = "-"),
is_over = na.omit(is_over)) %>%
mutate(START = int_start(is_over),
END = int_end(is_over)) %>%
select(!is_over)
# A tibble: 1 x 5
grp ID DRUG START END
<int> <int> <chr> <dttm> <dttm>
1 1 1 DPP4-SU 2020-03-15 00:00:00 2020-04-01 00:00:00
Second data set:
# A tibble: 2 x 5
grp ID DRUG START END
<int> <dbl> <chr> <dttm> <dttm>
1 1 3 DPP4-SU 2017-08-05 00:00:00 2017-08-27 00:00:00
2 2 3 SU-DPP4 2017-09-28 00:00:00 2017-12-21 00:00:00
As per updated df
df <- structure(list(ID = c(3, 3, 3, 3, 3, 3, 3), DRUG = c(
"DPP4",
"DPP4", "SU", "SU", "DPP4", "DPP4", "DPP4"
), START = structure(c(
17004,
17383, 17383, 17418, 17437, 17649, 17676
), class = c(
"IDate",
"Date"
)), END = structure(c(
17039, 17405, 17405, 17521, 17625,
17669, 17711
), class = c("IDate", "Date")), duration = c(
35L,
22L, 22L, 103L, 188L, 20L, 35L
), INDEX = c(
1L, 0L, 0L, 0L, 0L,
0L, 0L
)), row.names = c(NA, -7L), class = c(
"tbl_df", "tbl",
"data.frame"
))
we obtain
> dfnew
ID DRUG start end
3.3 3 DPP4-SU 2017-08-05 2017-08-27
3.7 3 SU-DPP4 2017-09-28 2017-12-21
A base R option (not as fancy as the answers by @Anoushiravan R or @Martin Gal)
f <- function(d) {
d <- d[with(d, order(START, END)), ]
idx <- subset(
data.frame(which((u <- with(d, outer(START, END, `<`))) & t(u), arr.ind = TRUE)),
row > col
)
if (nrow(idx) == 0) {
return(data.frame(ID = unique(d$ID), DRUG = NA, start = NA, end = NA))
}
with(
d,
do.call(rbind,
apply(
idx,
1,
FUN = function(v) {
data.frame(
ID = ID[v["row"]],
DRUG = paste0(DRUG[sort(unlist(v))], collapse = "-"),
start = START[v["row"]],
end = END[v["col"]]
)
}
))
)
}
dfnew <- do.call(rbind, Map(f, split(df, ~ID)))
gives
> dfnew
ID DRUG start end
1 1 DPP4-SU 2020-03-15 2020-04-01
2 2 <NA> <NA> <NA>
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With