R check if date is 2 years apart - r

Question

I have a dataset with two columns Id and Date as shown below using a toy dataset.
Id Date
5373283 2010-11-05
5373283 2014-11-05
5373283 2001-07-13
5373283 2007-12-01
5373283 2015-07-07
3475684 2015-05-19
3475684 2010-06-24
I want to check if any of the dates for each id are within 2 years range. If they are then a column will show yes, if not, No. The final output would look like this
Id Status
5373283 Yes
3475684 No
Yes for Id 5373283 because the two dates 2014-11-05 and 2015-07-07 are within two years of each other. No for Id 3475684 because the two dates are more than 2 years apart. Any help on accomplishing this much appreciated.

Answer 1

Hypothetical data.
DF <- data.frame(id = c(1, 1, 1, 2, 2),
date = c("2010-10-9", "2012-10-8", "2008-10-5",
"2007-7-5", "2009-7-5"), stringsAsFactors = FALSE)
The code below gets the minimal interval by ID in days.
What is happening is:
mutate redefines the date as Date class
arrange sort the data by date
group_by tells the following computation shall be done for each ID,
summarize computes the minimum difference.
library(dplyr)
DF %>% mutate(date = as.Date(date)) %>%
arrange(date) %>%
group_by(id) %>%
summarize(diffmin = as.numeric(min(diff(date)), units = "days"))
# id diffmin
# (dbl) (dbl)
#1 1 730
#2 2 731
If you can ignore leap years, this being smaller than or equal to 730 means within 2 years. Note that difference between 2007-7-5 and 2009-7-5 is 731 days, and thus judged as out of 2 years.
If this is not good to you, simple days-difference is not enough. I would need to define a custom checker function.
check2years <- function(a, b) {
# check if b - a <= 2 years
# assumes a and b are Date
yr_a <- format(a, "%Y") %>% as.integer()
yr_b <- format(b, "%Y") %>% as.integer()
dy_a <- format(a, "%m-%d")
dy_b <- format(b, "%m-%d")
(yr_b - yr_a < 2) | ((yr_b - yr_a == 2) & (dy_b >= dy_a))
}
Then, you can check if any combination is within 2 years by the following.
DF %>% mutate(date = as.Date(date)) %>%
arrange(date) %>%
group_by(id) %>%
summarize(within2yr = any(check2years(head(date, length(date)-1),
tail(date, length(date)-1))))
# id within2yr
# (dbl) (lgl)
#1 1 TRUE
#2 2 TRUE

Answer 2

You can also solve this without any library:
Using your example:
Id = c(5373283,5373283,5373283,5373283,5373283,3475684,3475684)
Date = as.Date(c("2010-11-05","2014-11-05","2001-07-13","2007-12-01","2015-07-07","2015-05-19","2010-06-24"))
df = data.frame(Id,Date)
> df
Id Date
7 3475684 2010-06-24
6 3475684 2015-05-19
3 5373283 2001-07-13
4 5373283 2007-12-01
1 5373283 2010-11-05
2 5373283 2014-11-05
5 5373283 2015-07-07
Do the following:
First order your data first by Id then by Date
df = df[order(df$Id,df$Date),]
Do an aggregate by Id using the function min(diff(x)), where x are the dates for each Id.
z = aggregate(df$Date,by = list(Id = df$Id),FUN = function(x){min(diff(x))})
What this function does is it returns the lowest difference between adjacent dates. This is why you need to order the data frame first.
This returns:
> z
Id x
1 3475684 1790
2 5373283 244
Where column x is the minimum difference in days.
Here, you only need to evaluate is if column x is less than or equal to 2*365
z$result = z$x<=2*365
Giving:
Id x result
1 3475684 1790 FALSE
2 5373283 244 TRUE
Final code
df = df[order(df$Id,df$Date),]
z = aggregate(df$Date,by = list(Id = df$Id),FUN = function(x){min(diff(x))})
z$result = z$x>=2*365

Answer 3

You can use something like this with library dplyr with the idea of taking top two dates in sorted order for each ID and see if they differ by two years:
library(dplyr)
df$Date <- as.Date(df$Date)
df %>%
group_by(Id) %>%
summarise(Status = as.numeric(difftime(max(Date), Date[order(Date, decreasing = TRUE)][2], units = 'days')) < 730)
Output will be as follows:
Source: local data frame [2 x 2]
Id Status
(int) (lgl)
1 3475684 FALSE
2 5373283 TRUE