I'm looking to find the day of year for a POSIXct class object with lubridate. For example, 12-9-2015 is day 343.
It's easy to find the day of the week or month with lubridate:
> lubridate::wday("2015-12-09 04:27:56 EST", labels = T)
Wed
> lubridate::day("2015-12-09 04:27:56 EST")
9
Is there an easy way to do so for the day of the year? I've searched the documentation and other questions but have not (yet) found an answer.
The correct function is yday, as in
lubridate::yday(Sys.time())
Figured out a more complicated way to do this before stumbling on this answer from u/blindjesse:
# compute the time interval from the first of the year until now
YTD = interval(floor_date(now(), unit='year'), now())
# compute the length of the interval in days, and discard the fractional part
as.integer(time_length(YTD, "day"))
By the way, an even more compact version of u/blindjesse's answer would be:
lubridate::yday(now())
Related
For example:
Year A
Year B
1990
2021
1980
2021
Thanks in advance.
It depends of what you expect.
If you only want to use the difftime() function with date objects composed only of years (as below), it will work (it will set the day and month to the ones of today for the calculation).
> a = as.Date("2021", format("%Y"))
> b = as.Date("2010", format("%Y"))
> difftime(a,b)
Time difference of 4018 days
But if you want to get the difference in year, it is not possible, as the function documentation clearly state that the return value unit must be: units = c("auto", "secs", "mins", "hours", "days", "weeks")
You might find better way to handle date/time data with the lubridate package.
difftime requests a date object to be used, I tried reproducing this using only years but was unable to.
Why not simply use absolute value (abs())If you're only interested in year difference?
as an example so you can see the difference added to a new column:
Year_A <- c(1990, 1980)
Year_B <- c(2021, 2021)
df <- data.frame(Year_A, Year_B)
df$diff <- abs(Year_A - Year_B)
P.S I noticed the answer above me was added while I was answering and I can't comment to it due to low rep, i see you can't use "years" as a unit value there, the biggest one being weeks, but you can manipulate that from days/weeks to years if that's what you're after.
How do I find the week number from an arbitrary start date in R. Let's say I want my start date to be august 1st.
Using lubridate, you can do:
interval(today(), dmy("21-08-2020"))/weeks(1)
[1] 30.42857
Or from the date of interest to another date:
interval(dmy("21-08-2020"), dmy("21-09-2020"))/weeks(1)
[1] 4.428571
You can use difftime for this:
difftime("2020-08-21", Sys.Date(), units = "weeks")
# Time difference of 30.45238 weeks
I'm trying to get the next week day for a vector of dates in R. My approach was to create a vector of weekdays and then find the date to the weekend date I have. The problem is that for Saturday and some holidays (which are a lot in my country) i end up getting the previous week day which doesn't work.
This is an example of my problem:
vecDates = as.Date(c("2011-01-11","2011-01-12","2011-01-13","2011-01-14","2011-01-17","2011-01-18",
"2011-01-19","2011-01-20","2011-01-21","2011-01-24"))
testDates = as.Date(c("2011-01-22","2011-01-23"))
findInterval(testDates,vecDates)
for both dates the correct answer should be 10 which is "2011-01-24" but I get 9.
I though of a solution where I remove all the previous dates to the date i'm analyzing, and then use findInterval. It works but it is not vectorized and therefore kind of slow which does not work for my actual purpose.
Does this do what you want?
vecDates = as.Date(c("2011-01-11","2011-01-12",
"2011-01-13","2011-01-14",
"2011-01-17","2011-01-18",
"2011-01-19","2011-01-20",
"2011-01-21","2011-01-24"))
testDates = as.Date(c("2011-01-20","2011-01-22","2011-01-23"))
get_next_biz_day <- function(testdays, bizdays){
o <- findInterval(testdays, bizdays) + 1
bizdays[o]
}
get_next_biz_day(testDates, vecDates)
#[1] "2011-01-21" "2011-01-24" "2011-01-24"
I want to create a single column with a sequence of date/time increasing every hour for one year or one month (for example). I was using a code like this to generate this sequence:
start.date<-"2012-01-15"
start.time<-"00:00:00"
interval<-60 # 60 minutes
increment.mins<-interval*60
x<-paste(start.date,start.time)
for(i in 1:365){
print(strptime(x, "%Y-%m-%d %H:%M:%S")+i*increment.mins)
}
However, I am not sure how to specify the range of the sequence of dates and hours. Also, I have been having problems dealing with the first hour "00:00:00"? Not sure what is the best way to specify the length of the date/time sequence for a month, year, etc? Any suggestion will be appreciated.
I would strongly recommend you to use the POSIXct datatype. This way you can use seq without any problems and use those data however you want.
start <- as.POSIXct("2012-01-15")
interval <- 60
end <- start + as.difftime(1, units="days")
seq(from=start, by=interval*60, to=end)
Now you can do whatever you want with your vector of timestamps.
Try this. mondate is very clever about advancing by a month. For example, it will advance the last day of Jan to last day of Feb whereas other date/time classes tend to overshoot into Mar. chron does not use time zones so you can't get the time zone bugs that code as you can using POSIXct. Here x is from the question.
library(chron)
library(mondate)
start.time.num <- as.numeric(as.chron(x))
# +1 means one month. Use +12 if you want one year.
end.time.num <- as.numeric(as.chron(paste(mondate(x)+1, start.time)))
# 1/24 means one hour. Change as needed.
hours <- as.chron(seq(start.time.num, end.time.num, 1/24))
How can I accurately convert the products (units is in days) of the difftime below to years, months and days?
difftime(Sys.time(),"1931-04-10")
difftime(Sys.time(),"2012-04-10")
This does years and days but how could I include months?
yd.conv<-function(days, print=TRUE){
x<-days*0.00273790700698851
x2<-floor(x)
x3<-x-x2
x4<-floor(x3*365.25)
if (print) cat(x2,"years &",x4,"days\n")
invisible(c(x2, x4))
}
yd.conv(difftime(Sys.time(),"1931-04-10"))
yd.conv(difftime(Sys.time(),"2012-04-10"))
I'm not sure how to even define months either. Would 4 weeks be considered a month or the passing of the same month day. So for the later definition of a month if the initial date was 2012-01-10 and the current 2012-05-31 then we'd have 0 years, 5 months and 21 days. This works well but what if the original date was on the 31st of the month and the end date was on feb 28 would this be considered a month?
As I wrote this question the question itself evolved so I'd better clarify:
What would be the best (most logical approach) to defining months and then how to find diff time in years, months and days?
If you're doing something like
difftime(Sys.time(), someDate)
It comes as implied that you must know what someDate is. In that case, you can convert this to a POSIXct class object that gives you the ability to extract temporal information directly (package chron offers more methods, too). For instance
as.POSIXct(c(difftime(Sys.time(), someDate, units = "sec")), origin = someDate)
This will return your desired date object. If you have a timezone tz to feed into difftime, you can also pass that directly to the tz parameter in as.POSIXct.
Now that you have your date object, you can run things like months(.) and if you have chron you can do years(.) and days(.) (returns ordered factor).
From here, you could do more simple math on the difference of years, months, and days separately (converting to appropriate numeric representations). Of course, convert someDate to POSIXct will be required.
EDIT: On second thought, months(.) returns a character representation of the month, so that may not be efficient. At least, it'll require a little processing (not too difficult) to give a numeric representation.
R has not implemented these features out of ignorance. difftime objects are transitive. A 700 day difference on any arbitrary start-date can yield a differing number of years depending on whether there was a leap year or not. Similarly for months, they take between 28-31 days.
For research purposes, we use these units a lot (months and years) and pragmatically, we define a year as 365.25 days and a month as 365.25/12 = 30.4375 days.
To do arithmetic on a given difftime, you must convert this value to numeric using as.numeric(difftime.obj) which is, in default, days so R stops spouting off the units.
You can not simply convert a difftime to month, since the definition of months depends on the absolute time at which the difftime has started.
You'll need to know the start date or the end date to accurately tell the number of months.
You could then, e.g., calculate the number of months in the first year of your timespan, the number of month in the last your of the timespan, and add the number of years between times 12.
Hmm. I think the most sensible would be to look at the various units themselves. So compare the day of the month first, then compare the month of the year, then compare the year. At each point, you can introduce a carry to avoid negative values.
In other words, don't work with the product of difftime, but recode your own difftime.